Samacheer Kalvi Class 10 Maths Solutions Chapter 4 Geometry More Ques

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions

I. Multiple choice questions

 

Question 1. If a straight line intersects the sides AB and AC of a \( \Delta ABC \) at D and E respectively and is parallel to BC, then \( \frac { AE }{ AC } \) = ...........
(1) \( \frac { AD }{ DB } \)
(2) \( \frac { AD }{ AB } \)
(3) \( \frac { DE }{ DC } \)
(4) \( \frac { AD }{ EC } \)
Answer: (2) \( \frac { AD }{ AB } \)
In simple words: When a line cuts two sides of a triangle and is parallel to the third side, it divides the two sides proportionally. This is called the Basic Proportionality Theorem. The ratio of AE to AC will be the same as AD to AB.

🎯 Exam Tip: Remember Thales' Theorem (Basic Proportionality Theorem) for questions involving parallel lines in a triangle, as it directly relates segment ratios.

 

Question 2. In \( \Delta ABC \), DE is \( || \) to BC, meeting AB and AC at D and E. If AD = 3 cm, DB = 2 cm and AE = 2.7 cm, then AC is equal to ...........
(1) 6.5 cm
(2) 4.5 cm
(3) 3.5 cm
(4) 3.5 cm
Answer: (2) 4.5 cm
Hint:
By BPT theorem,
\[ \frac { AD }{ DB } = \frac { AE }{ EC } \]
\[ \frac { 3 }{ 2 } = \frac { 2.7 }{ EC } \]
\[ EC = \frac{2.7 \times 2}{3} = 1.8 \text{ cm} \]
Now, calculate AC:
\[ AC = AE + EC \]
\[ = 2.7 + 1.8 = 4.5 \text{ cm} \]

In simple words: This problem uses the Basic Proportionality Theorem, which states that if a line is parallel to one side of a triangle and intersects the other two sides, it divides the two sides proportionally. We find the missing part EC first and then add AE and EC to get the total length AC.

🎯 Exam Tip: Always draw a clear diagram and label all given lengths for proportionality theorems to easily set up the correct ratios.

 

Question 3. In \( \Delta PQR \), RS is the bisector of \( \angle R \). If PQ = 6 cm, QR = 8 cm, RP = 4 cm then PS is equal to ...........
(1) 2 cm
(2) 4 cm
(3) 3 cm
(4) 6 cm
Answer: (2) 2 cm
Hint:
By ABT theorem (Angle Bisector Theorem),
\[ \frac { PS }{ SQ } = \frac { PR }{ QR } \]
Let \( PS = x \). Then \( SQ = PQ - PS = 6 - x \).
\[ \frac { x }{ 6-x } = \frac { 4 }{ 8 } \]
Now, cross-multiply to solve for x:
\[ 8x = 4(6-x) \]
\[ 8x = 24 - 4x \]
\[ 12x = 24 \]
\[ x = 2 \]
So, \( PS = 2 \text{ cm} \)

In simple words: When an angle bisector divides the opposite side of a triangle, it does so in a ratio equal to the ratio of the other two sides. By setting up this ratio and solving for the unknown length, we find PS is 2 cm. This is known as the Angle Bisector Theorem.

🎯 Exam Tip: The Angle Bisector Theorem is very useful for finding unknown lengths when an angle is bisected. Always write down the correct ratio of sides carefully.

 

Question 4. In figure, if \( \frac { AB }{ AC } = \frac { BD }{ DC } \), \( \angle B = 40^\circ \) and \( \angle C = 60^\circ \), then \( \angle BAD \) = ............
(1) 30°
(2) 50°<
(3) 80°
(4) 40°
Answer: (4) 40°
Hint:
Since \( \frac { AB }{ AC } = \frac { BD }{ DC } \), by the converse of the Angle Bisector Theorem, AD is the internal bisector of \( \angle BAC \).
First, find the angle A in \( \Delta ABC \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
\[ \angle A + 40^\circ + 60^\circ = 180^\circ \]
\[ \implies \angle A = 180^\circ - 100^\circ = 80^\circ \]
Since AD bisects \( \angle BAC \), then \( \angle BAD = \frac { \angle BAC }{ 2 } \).
\[ \angle BAD = \frac { 80^\circ }{ 2 } = 40^\circ \]

In simple words: The problem uses the Angle Bisector Theorem in reverse to show that line AD divides angle BAC into two equal parts. Once we find the total angle BAC in the triangle, we simply divide it by two to get angle BAD.

🎯 Exam Tip: If the sides are proportional, the line segment must be an angle bisector. Always find the total angle of the triangle first if other angles are given.

 

Question 5. In the figure, the value x is equal to ............
(1) 4.2
(2) 3.2
(3) 0.8
(4) 0.4
Answer: (2) 3.2
Hint:
By Thales theorem (Basic Proportionality Theorem), since DE is parallel to BC,
\[ \frac { AD }{ BD } = \frac { AE }{ EC } \]
Using the given values from the figure:
\[ \frac { x }{ 8 } = \frac { 4 }{ 10 } \]
To find x, cross-multiply:
\[ 10x = 8 \times 4 \]
\[ 10x = 32 \]
\[ x = \frac{32}{10} = 3.2 \]

In simple words: The figure shows two triangles that are similar because of a parallel line. This means their corresponding sides are in proportion. We set up a simple equation with the given lengths and solve for x.

🎯 Exam Tip: For problems with a line parallel to one side of a triangle, remember that it creates similar triangles and proportional segments. Ensure you match the correct sides when setting up the ratio.

 

Question 6. In triangles ABC and DEF, \( \angle B = \angle E \), \( \angle C = \angle F \), then ............
(1) \( \frac { AB }{ DE } = \frac { CA }{ EF } \)
(2) \( \frac { BC }{ EF } = \frac { AB }{ FD } \)
(3) \( \frac { AB }{ DE } = \frac { BC }{ EF } \)
(4) \( \frac { CA }{ FD } = \frac { AB }{ EF } \)
Answer: (3) \( \frac { AB }{ DE } = \frac { BC }{ EF } \)
Hint:
If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar (AA similarity criterion).
When \( \Delta ABC \sim \Delta DEF \), the ratios of their corresponding sides are equal.
The corresponding vertices are A to D, B to E, and C to F.
So, the correct proportion is \( \frac { AB }{ DE } = \frac { BC }{ EF } = \frac { AC }{ DF } \).

In simple words: When two triangles have two angles that are the same, they are similar triangles. This means their shapes are the same, but their sizes might be different. For similar triangles, the ratio of their matching sides is always equal. We just need to pick the option that correctly matches these sides.

🎯 Exam Tip: For similar triangles, always list the vertices in corresponding order (e.g., ABC to DEF) to easily identify the correct pairs of proportional sides.

 

Question 7. From the given figure, identify the wrong statement.
(1) \( \Delta ADB \sim \Delta ABC \)
(2) \( \Delta ABD \sim \Delta ABC \)
(3) \( \Delta BDC \sim \Delta ABC \)
(4) \( \Delta ADB \sim \Delta BDC \)
Answer: (2) \( \Delta ABD \sim \Delta ABC \)

In simple words: In this figure, we have a right-angled triangle where a perpendicular is drawn from the right angle to the hypotenuse. This creates three similar triangles. The question asks to find the incorrect similarity statement. We must compare each option with the actual similar triangles formed.

🎯 Exam Tip: When a perpendicular is drawn from the vertex of the right angle to the hypotenuse, three similar triangles are formed: the original triangle, and the two smaller triangles created by the altitude. Ensure you write the vertices in the correct order to represent similarity.

 

Question 8. If a vertical stick 12 m long casts a shadow 8 m long on the ground and at the same time a tower casts a shadow 40 m long on the ground, then the height of the tower is ............
(1) 40 m
(2) 50 m
(3) 75 m
(4) 60 m
Answer: (4) 60 m
Hint:
We can use the property of similar triangles. The stick and its shadow form one right-angled triangle, and the tower and its shadow form another. Since the sun's angle of elevation is the same at the same time, these two triangles are similar.
Let \( AB \) be the height of the stick and \( BC \) be its shadow.
Let \( DE \) be the height of the tower and \( EF \) be its shadow.
Given: \( AB = 12 \text{ m} \), \( BC = 8 \text{ m} \), \( EF = 40 \text{ m} \). We need to find \( DE = h \).
Since \( \Delta ABC \sim \Delta DEF \):
\[ \frac { AB }{ DE } = \frac { BC }{ EF } \]
Substitute the known values:
\[ \frac { 12 }{ h } = \frac { 8 }{ 40 } \]
Now, solve for \( h \):
\[ h = \frac{12 \times 40}{8} \]
\[ h = 12 \times 5 \]
\[ h = 60 \text{ m} \]

In simple words: At the same time of day, the sun creates shadows at the same angle for all objects. This means a stick and a tower standing upright form similar triangles with their shadows. We use the ratio of their heights to their shadows to find the unknown height of the tower.

🎯 Exam Tip: For problems involving shadows cast by objects at the same time, assume similar triangles. The ratio of height to shadow length will be constant for both objects.

 

Question 9. The sides of two similar triangles are in the ratio 2:3, then their areas are in the ratio ............
(1) 9:4
(2) 4:9
(3) 2:3
(4) 3:2
Answer: (2) 4:9
Hint:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Given ratio of sides \( = 2:3 \).
Ratio of areas \( = (\text{ratio of sides})^2 \)
\[ = 2^2 : 3^2 \]
\[ = 4:9 \]
In simple words: When triangles are similar, their areas don't follow the same simple ratio as their sides. Instead, the ratio of their areas is found by squaring the ratio of their matching sides. So, if sides are in a 2:3 ratio, areas are in a 4:9 ratio.

🎯 Exam Tip: Remember the critical relationship: ratio of areas of similar triangles = (ratio of corresponding sides)^2. Do not confuse it with the ratio of perimeters, which is simply the ratio of sides.

 

Question 10. Triangles ABC and DEF are similar. If their areas are 100 cm² and 49 cm² respectively and BC is 8.2 cm then EF = ............
(1) 5.47 cm
(2) 5.74 cm
(3) 6.47 cm
(4) 6.74 cm
Answer: (2) 5.74 cm
Hint:
For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Given: Area of \( \Delta ABC = 100 \text{ cm}^2 \), Area of \( \Delta DEF = 49 \text{ cm}^2 \), \( BC = 8.2 \text{ cm} \). We need to find \( EF \).
\[ \frac{\text{Area of } \Delta ABC}{\text{Area of } \Delta DEF} = \frac{BC^2}{EF^2} \]
Substitute the given values:
\[ \frac{100}{49} = \frac{(8.2)^2}{EF^2} \]
To solve for \( EF^2 \):
\[ EF^2 = \frac{(8.2)^2 \times 49}{100} \]
Now, take the square root to find \( EF \):
\[ EF = \sqrt{\frac{(8.2)^2 \times 49}{100}} \]
\[ EF = \frac{8.2 \times \sqrt{49}}{\sqrt{100}} \]
\[ EF = \frac{8.2 \times 7}{10} \]
\[ EF = \frac{57.4}{10} = 5.74 \text{ cm} \]
In simple words: We use the rule that for similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides. By plugging in the given areas and the length of one side, we can calculate the length of the matching side in the other triangle.

🎯 Exam Tip: Always remember to take the square root when converting from the ratio of areas back to the ratio of sides. Be careful with calculations involving decimals and squares.

 

Question 11. The perimeters of two similar triangles are 24 cm and 18 cm respectively. If one side of the first triangle is 8 cm, then the corresponding side of the other triangle is ............
(1) 4 cm
(2) 3 cm
(3) 15 cm
(4) 6 cm
Answer: (4) 6 cm
Hint:
For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.
Given: Perimeter of first triangle \( = 24 \text{ cm} \), Perimeter of second triangle \( = 18 \text{ cm} \).
Side of first triangle \( = 8 \text{ cm} \). We need to find the corresponding side of the second triangle.
Let the corresponding side of the second triangle be \( s_2 \).
\[ \frac{\text{Perimeter of the first triangle}}{\text{Perimeter of the second triangle}} = \frac{\text{Side of the first triangle}}{\text{Corresponding side of the second triangle}} \]
Substitute the given values:
\[ \frac{24}{18} = \frac{8}{s_2} \]
Now, solve for \( s_2 \):
\[ s_2 = \frac{18 \times 8}{24} \]
\[ s_2 = \frac{144}{24} \]
\[ s_2 = 6 \text{ cm} \]
In simple words: For similar triangles, their total boundary lengths (perimeters) are related in the same way as their individual matching sides. We use this direct relationship to find the length of the missing side.

🎯 Exam Tip: Distinguish between the ratios of perimeters (direct ratio of sides) and areas (square of the ratio of sides) for similar triangles. This is a common point of confusion.

 

Question 12. A point P is 26 cm away from the centre O of a circle and PT is the tangent drawn from P to the circle 10 cm, then OT is equal to ...........
(1) 36 cm
(2) 20 cm
(3) 18 cm
(4) 24 cm
Answer: (4) 24 cm
Hint:
We are given that PT is a tangent to the circle, and T is the point of tangency. We know that the radius drawn to the point of tangency is perpendicular to the tangent.
Therefore, \( \angle OTP = 90^\circ \). This forms a right-angled triangle \( \Delta OTP \).
Given: \( OP = 26 \text{ cm} \) (hypotenuse), \( PT = 10 \text{ cm} \) (one leg). We need to find \( OT \) (the other leg, which is the radius).
Using the Pythagorean theorem in \( \Delta OTP \):
\[ OT^2 + PT^2 = OP^2 \]
\[ OT^2 + 10^2 = 26^2 \]
\[ OT^2 + 100 = 676 \]
\[ OT^2 = 676 - 100 \]
\[ OT^2 = 576 \]
\[ OT = \sqrt{576} \]
\[ OT = 24 \text{ cm} \]

In simple words: When a line touches a circle at just one point (a tangent), the radius to that point always makes a right angle with the tangent. This creates a right-angled triangle. We can then use the Pythagorean theorem to find the missing side, which is the radius.

🎯 Exam Tip: Always remember that the radius is perpendicular to the tangent at the point of contact. This property creates right-angled triangles, making Pythagorean theorem applicable for finding lengths.

 

Question 13. In the figure, if \( \angle PAB = 120^\circ \) then \( \angle BPT \) = ............
(1) 120°
(2) 30°
(3) 40°
(4) 60°
Answer: (4) 60°
Hint:
We have a cyclic quadrilateral PABC, and PT is a tangent.
For a cyclic quadrilateral, the sum of opposite angles is 180°.
So, \( \angle BCP + \angle PAB = 180^\circ \).
Given \( \angle PAB = 120^\circ \).
\[ \angle BCP + 120^\circ = 180^\circ \]
\[ \angle BCP = 180^\circ - 120^\circ = 60^\circ \]
Now, using the tangent-chord theorem, the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.
Here, chord is BC and tangent is PT.
So, \( \angle BPT = \angle BCP \).
Therefore, \( \angle BPT = 60^\circ \).

In simple words: This problem involves two circle theorems. First, opposite angles of a shape inside a circle (cyclic quadrilateral) add up to 180 degrees. Second, the angle between a tangent and a chord is the same as the angle in the opposite part of the circle (alternate segment). We use these rules to find the unknown angle.

🎯 Exam Tip: Always look for cyclic quadrilaterals and tangent-chord relationships in circle geometry problems. Clearly identify the chord and the alternate segment for the tangent-chord theorem.

 

Question 14. If the tangents PA and PB from an external point P to circle with centre O are inclined to each other at an angle of 40°, then \( \angle POA \) = ............
(1) 70°
(2) 80°
(3) 50°
(4) 60°
Answer: (1) 70°
Hint:
We know that the line segment joining the center to the external point (OP) bisects the angle between the tangents ( \( \angle APB \) ).
Given: \( \angle APB = 40^\circ \).
So, \( \angle APO = \frac { \angle APB }{ 2 } = \frac { 40^\circ }{ 2 } = 20^\circ \).
Also, the radius OA is perpendicular to the tangent PA at the point of contact A. So, \( \angle OAP = 90^\circ \).
Now consider \( \Delta OAP \). The sum of angles in a triangle is 180°.
\[ \angle POA + \angle OAP + \angle APO = 180^\circ \]
\[ \angle POA + 90^\circ + 20^\circ = 180^\circ \]
\[ \angle POA + 110^\circ = 180^\circ \]
\[ \angle POA = 180^\circ - 110^\circ = 70^\circ \]

In simple words: When two tangents are drawn from an outside point to a circle, the line from the center to that point cuts the angle between the tangents exactly in half. Also, the radius always meets the tangent at a right angle. Using these facts and the sum of angles in a triangle, we can find the unknown angle.

🎯 Exam Tip: Remember the two key properties: the line from the center to the external point bisects the angle between tangents, and the radius is perpendicular to the tangent at the point of contact. These form right triangles crucial for calculations.

 

Question 15. In the figure, PA and PB are tangents to the circle drawn from an external point P. Also CD is a tangent to the circle at Q. If PA = 8 cm and CQ = 3 cm, then PC is equal to ............
(1) 11 cm
(2) 5 cm
(3) 24 cm
(4) 8 cm
Answer: (2) 5 cm
Hint:
We know that the lengths of tangents drawn from an external point to a circle are equal.
From external point P, tangents are PA and PB.
So, \( PA = PB \). Given \( PA = 8 \text{ cm} \), so \( PB = 8 \text{ cm} \).
From external point C, tangents are CQ and CB.
So, \( CQ = CB \). Given \( CQ = 3 \text{ cm} \), so \( CB = 3 \text{ cm} \).
From the solution's steps, PB is composed of PC and CB, implying P, C, B are collinear.
So, \( PB = PC + CB \).
Substitute the known values:
\[ 8 = PC + 3 \]
\[ PC = 8 - 3 \]
\[ PC = 5 \text{ cm} \]
In simple words: Tangent segments drawn from an external point to a circle are always equal in length. We use this rule for point P to get PB and for point C to get CB. Then, assuming C lies on segment PB, we subtract known lengths to find the unknown part PC.

🎯 Exam Tip: Carefully identify all external points and the tangents drawn from them. The key is to apply the "tangents from an external point are equal" theorem. When segments are collinear, their lengths can be added or subtracted.

 

Question 16. \( \Delta ABC \) is a right angled triangle where \( \angle B = 90^\circ \) and BD \( \perp \) AC. If BD = 8 cm, AD = 4 cm, then CD is ............
(1) 24 cm
(2) 16 cm
(3) 32 cm
(4) 8 cm
Answer: (2) 16 cm
Hint:
In a right-angled triangle, if an altitude is drawn from the vertex of the right angle to the hypotenuse, then the square of the altitude is equal to the product of the two segments it divides the hypotenuse into.
Here, BD is the altitude to the hypotenuse AC from the right angle B.
So, \( BD^2 = AD \times CD \).
Given: \( BD = 8 \text{ cm} \), \( AD = 4 \text{ cm} \). We need to find \( CD \).
\[ 8^2 = 4 \times CD \]
\[ 64 = 4 \times CD \]
\[ CD = \frac{64}{4} \]
\[ CD = 16 \text{ cm} \]

In simple words: When you drop a perpendicular line from the corner with the right angle down to the longest side (hypotenuse), that perpendicular line's length squared is equal to the product of the two parts it splits the longest side into. We use this special property to calculate the missing segment length.

🎯 Exam Tip: Remember the geometric mean theorem for right triangles: the altitude to the hypotenuse is the geometric mean of the two segments it divides the hypotenuse into (\( h^2 = xy \)). This is a direct formula for such problems.

 

Question 17. The areas of two similar triangles are 16 cm² and 36 cm² respectively. If the altitude of the first triangle is 3 cm, then the corresponding altitude of the other triangle is ............
(1) 6.5 cm
(2) 6 cm
(3) 4 cm
(4) 4.5 cm
Answer: (4) 4.5 cm
Hint:
For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding altitudes.
Let Area 1 \( = 16 \text{ cm}^2 \), Area 2 \( = 36 \text{ cm}^2 \).
Altitude 1 \( = 3 \text{ cm} \). We need to find Altitude 2 \( = h_2 \).
\[ \frac{\text{Area of the first } \Delta}{\text{Area of the second } \Delta} = \frac{(\text{Altitude of the first } \Delta)^2}{(\text{Altitude of the second } \Delta)^2} \]
Substitute the given values:
\[ \frac{16}{36} = \frac{3^2}{h_2^2} \]
\[ \frac{16}{36} = \frac{9}{h_2^2} \]
Now, cross-multiply and solve for \( h_2^2 \):
\[ 16 \times h_2^2 = 36 \times 9 \]
\[ h_2^2 = \frac{36 \times 9}{16} \]
\[ h_2^2 = \frac{324}{16} \]
Take the square root to find \( h_2 \):
\[ h_2 = \sqrt{\frac{324}{16}} = \frac{\sqrt{324}}{\sqrt{16}} = \frac{18}{4} \]
\[ h_2 = 4.5 \text{ cm} \]
In simple words: The areas of similar triangles are related to the square of their altitudes (heights), just like they are related to the square of their sides. We use this square relationship to find the unknown altitude of the second triangle.

🎯 Exam Tip: Remember that not only sides, but also altitudes, medians, and angle bisectors follow the same ratio as the sides for similar triangles, and their squares follow the ratio of areas.

 

Question 18. The perimeter of two similar triangles \( \Delta ABC \) and \( \Delta DEF \) are 36 cm and 24 cm respectively. If DE = 10 cm, then AB is ............
(1) 4 cm
(2) 20 cm
(3) 15 cm
(4) 18 cm
Answer: (3) 15 cm
Hint:
For similar triangles, the ratio of their perimeters is equal to the ratio of their corresponding sides.
Given: Perimeter of \( \Delta ABC = 36 \text{ cm} \), Perimeter of \( \Delta DEF = 24 \text{ cm} \).
Side \( DE = 10 \text{ cm} \). We need to find the corresponding side \( AB \).
\[ \frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta DEF} = \frac{AB}{DE} \]
Substitute the given values:
\[ \frac{36}{24} = \frac{AB}{10} \]
Now, solve for \( AB \):
\[ AB = \frac{36 \times 10}{24} \]
\[ AB = \frac{360}{24} \]
\[ AB = 15 \text{ cm} \]
In simple words: When two triangles are similar, their perimeters (the total length around them) are proportional to their matching sides. By using this proportion, we can easily find the length of a side in one triangle if we know the perimeter of both and a matching side of the other.

🎯 Exam Tip: Always set up the ratios carefully, matching corresponding perimeters with corresponding sides. Simplify fractions before multiplying for easier calculations.

 

Question 19. In the given diagram \( \theta \) is ............
(1) 15°
(2) 30°
(3) 45°
(4) 60°
Answer: (2) 30°
Hint:
From the figure, \( \Delta ABC \) has sides \( AB = AC \) (indicated by 'a' labels) and \( \angle B = 60^\circ \).
Since \( AB = AC \) and one base angle is \( 60^\circ \), \( \Delta ABC \) is an equilateral triangle.
Therefore, \( \angle BAC = 60^\circ \).
The diagram indicates \( \theta \) as \( \angle DAC \). For this to be \( 30^\circ \) (as per the answer and implied solution), AD must be the angle bisector of \( \angle BAC \).
If AD bisects \( \angle BAC \), then:
\[ \theta = \angle DAC = \frac{\angle BAC}{2} = \frac{60^\circ}{2} = 30^\circ \]

In simple words: The triangle shown is isosceles with two equal sides and one angle of 60 degrees, which means it's actually an equilateral triangle with all angles at 60 degrees. The angle `\theta` is half of the main angle at the top, which is 60 degrees, so `\theta` is 30 degrees. This implies AD bisects the angle at A.

🎯 Exam Tip: Look carefully at side labels. If two sides are marked 'a' and one angle is 60° in an isosceles triangle, it implies an equilateral triangle. If a part of an angle is marked with a variable, consider if it's an angle bisector from previous problem types.

 

Question 20. If AD is the bisector of \( \angle A \) then AC is ............
(1) 12
(2) 16
(3) 18
(4) 20
Answer: (4) 20
In simple words: When a line divides an angle into two equal parts, it's called an angle bisector. Here, we use a special theorem that helps us find the length of side AC based on the other side lengths.

🎯 Exam Tip: Remember the Angle Bisector Theorem states that if a line bisects an angle of a triangle, it divides the opposite side into two segments that are proportional to the other two sides of the triangle.

 

Question 21. In \( \Delta ABC \) and \( \Delta DEF \), \( \angle A = \angle E \) and \( \angle B = \angle F \). Then AB : AC is ............
(1) DE: DF
(2) DE: EF
(3) EF : ED
(4) DF : EF
Answer: (3) EF : ED
In simple words: When two triangles have angles that match up, their sides are in proportion. This means that if angle A in the first triangle matches angle E in the second, and angle B matches angle F, then the side AB will match side EF, and side AC will match side ED.

🎯 Exam Tip: Always map corresponding vertices carefully based on the given equal angles to correctly identify proportional sides in similar triangles. For \( \Delta ABC \sim \Delta EFD \), the ratio of sides should be \( \frac{AB}{EF} = \frac{BC}{FD} = \frac{AC}{ED} \).

 

Question 22. Two circles of radius 8.2 cm and 3.6 cm touch each other externally, the distance between their centres is .............
(1) 1.8 cm
(2) 4.1 cm
(3) 4.6 cm
(4) 11.8 cm
Answer: (4) 11.8 cm
In simple words: When two circles touch from the outside, the distance between their middle points (centres) is found by simply adding their individual sizes (radii) together. It's like putting two rulers end to end.

🎯 Exam Tip: For circles touching externally, the distance between centers is the sum of their radii. For circles touching internally, it's the difference of their radii.

 

Question 23. If the tangents PA and PB from an external point P to circle with centre O. \( \angle OPA = 35^\circ \) then a and b is ............
(1) a = 30°, b = 60°
(2) a = 35°, b = 55°
(3) a = 40°, b = 50°
(4) a = 45°, b = 45°
Answer: (2) a = 35°, b = 55°
In simple words: When lines touch a circle from an outside point, they create some special angles. Here, knowing one angle allows us to figure out the others, especially since the radius always forms a 90-degree angle with the tangent at the point of contact.

🎯 Exam Tip: Remember that the radius drawn to the point of tangency is perpendicular to the tangent. Also, tangents drawn from an external point to a circle are equal in length, and the line connecting the external point to the center bisects the angle between the tangents.

 

II. Answer the following questions

 

Question 1. The image of a man of height 1.8 m, is of length 1.5 cm on the film of a camera. If the film is 3 cm from the lens of the camera, how far is the man from the camera?
Answer: Let AB be the height of the man, and CD be the height of the image on the film. LM is the distance between the man and the lens, and LN is the distance between the lens and the film. The problem describes a situation that forms two similar triangles, \( \Delta LAB \) and \( \Delta LCD \), with the lens at L. We know that \( AB = 1.8 \) m \( (180 \text{ cm}) \), \( CD = 1.5 \) cm, and \( LN = 3 \) cm. We want to find LM (denoted as x).
Since \( \Delta LAB \sim \Delta LCD \) (by AA similarity, as \( \angle ALB = \angle DLC \) (vertically opposite angles) and \( \angle LAB = \angle LDC \) (alternate angles)), the ratio of corresponding sides is equal:
\( \frac{AB}{CD} = \frac{LM}{LN} \)
Substitute the given values:
\( \frac{180}{1.5} = \frac{x}{3} \)
Now, solve for x:
\( 1.5x = 180 \times 3 \)
\( 1.5x = 540 \)
\( x = \frac{540}{1.5} \)
\( x = \frac{540 \times 10}{15} \)
\( x = \frac{5400}{15} \)
\( x = 360 \) cm
So, \( x = 3.6 \) m.
Therefore, the man is \( 3.6 \) m from the camera lens. This is a classic application of similar triangles in optics.
In simple words: We compare the real man and his small image in the camera using similar triangles. We know the man's height, image size, and film distance. By setting up a proportion, we find that the man is 3.6 meters away from the camera.

🎯 Exam Tip: When dealing with image formation in cameras or lenses, always draw a diagram to identify similar triangles. Remember to convert all units to be consistent (e.g., all in cm or all in m) before performing calculations.

 

Question 2. A girl of height 120 cm is walking away from the base of a lamp-post at a speed of 0.6 m/sec. If the lamp is 3.6 m above the ground level, then find the length of her shadow after 4 seconds.
Answer: Let AB be the height of the lamp-post, which is \( 3.6 \) m or \( 360 \) cm. Let CD be the height of the girl, which is \( 120 \) cm or \( 1.2 \) m. The girl walks away from the lamp-post at a speed of \( 0.6 \) m/sec. We need to find the length of her shadow after 4 seconds.
First, calculate the distance the girl travelled in 4 seconds:
Distance = Speed \( \times \) Time
Distance \( = 0.6 \text{ m/sec} \times 4 \text{ sec} = 2.4 \) m or \( 240 \) cm. This is the length of AC.
Let CE be the length of her shadow, which we denote as x.
Consider the two similar triangles formed: \( \Delta ECD \) (girl and her shadow) and \( \Delta EAB \) (lamp-post and total length).
Since CD is parallel to AB (both are vertical to the ground), \( \Delta ECD \sim \Delta EAB \) by AA similarity ( \( \angle E \) is common, and \( \angle ECD = \angle EAB = 90^\circ \) ).
Thus, the ratio of their corresponding sides is equal:
\( \frac{CD}{AB} = \frac{CE}{AE} \)
We know \( AE = AC + CE = 240 + x \). So, substitute the values:
\( \frac{120}{360} = \frac{x}{240 + x} \)
\( \frac{1}{3} = \frac{x}{240 + x} \)
Now, cross-multiply:
\( 1 \times (240 + x) = 3 \times x \)
\( 240 + x = 3x \)
Subtract x from both sides:
\( 240 = 3x - x \)
\( 240 = 2x \)
Divide by 2:
\( x = \frac{240}{2} \)
\( x = 120 \) cm.
So, the length of the girl's shadow after 4 seconds is \( 120 \) cm or \( 1.2 \) m. Similar triangles help us model real-world scenarios like shadow lengths.
In simple words: A girl walks away from a lamp. After 4 seconds, we find how far she moved. Then, we use the heights of the girl and the lamp, plus the distance she walked, to form similar triangles. This helps us calculate the length of her shadow, which turns out to be 120 cm.

🎯 Exam Tip: Always make sure units are consistent (e.g., all in cm or all in m) throughout the calculation. Clearly label the points in your diagram to avoid confusion between similar triangles.

 

Question 3. In \( \Delta ABC \), AB = AC and BC = 6 cm. D is a point on the side AC such that AD = 5 cm and CD = 4 cm. Show that ABCD ~ \( \Delta ACB \) and hence find BD.
Answer: Given that in \( \Delta ABC \), \( AB = AC \). Also, \( BC = 6 \) cm, \( AD = 5 \) cm, and \( CD = 4 \) cm.
Since \( AD = 5 \) cm and \( CD = 4 \) cm, \( AC = AD + CD = 5 + 4 = 9 \) cm. Since \( AB = AC \), then \( AB = 9 \) cm.
Now, let's check the ratios of sides for \( \Delta BCD \) and \( \Delta ACB \).
First, find the ratio \( \frac{BC}{AC} \):
\( \frac{BC}{AC} = \frac{6}{9} = \frac{2}{3} \). (Equation 1)
Next, find the ratio \( \frac{CD}{CB} \):
\( \frac{CD}{CB} = \frac{4}{6} = \frac{2}{3} \). (Equation 2)
From (1) and (2), we see that \( \frac{BC}{AC} = \frac{CD}{CB} \).
Also, \( \angle C \) is common to both triangles \( \Delta BCD \) and \( \Delta ACB \).
Therefore, by the Side-Angle-Side (SAS) similarity criterion, \( \Delta BCD \sim \Delta ACB \). This confirms the similarity of the triangles.
Since the triangles are similar, the ratios of their corresponding sides are equal:
\( \frac{BD}{AB} = \frac{BC}{AC} = \frac{CD}{CB} \)
We already found \( \frac{BC}{AC} = \frac{2}{3} \). We also know \( AB = 9 \) cm.
So, \( \frac{BD}{9} = \frac{2}{3} \)
To find BD, multiply both sides by 9:
\( BD = \frac{2}{3} \times 9 \)
\( BD = 2 \times 3 \)
\( BD = 6 \) cm.
Thus, the length of BD is \( 6 \) cm. Understanding triangle similarity helps in finding unknown lengths.
In simple words: We are given a triangle ABC where two sides are equal. We check the ratios of sides for two smaller triangles, BCD and ACB, and find they are the same. Since they also share an angle, these triangles are similar. Using this similarity, we calculate the length of side BD, which comes out to be 6 cm.

🎯 Exam Tip: When proving triangle similarity, clearly state the criterion being used (AA, SAS, or SSS). Pay close attention to corresponding vertices to set up correct side ratios. Always list all given lengths and derived lengths for clarity.

 

Question 4. The lengths of three sides of a triangle ABC are 6 cm, 4 cm and 9 cm. \( \Delta PQR \sim \Delta ABC \). One of the lengths of sides of \( \Delta PQR \) is 35 cm. What is the greatest perimeter possible for \( \Delta PQR \)?
Answer: Given the sides of \( \Delta ABC \) are \( 6 \) cm, \( 4 \) cm, and \( 9 \) cm. The perimeter of \( \Delta ABC \) is \( 6 + 4 + 9 = 19 \) cm.
We are told that \( \Delta PQR \sim \Delta ABC \). This means that the ratio of their corresponding sides is equal to the ratio of their perimeters.
Let the sides of \( \Delta ABC \) be \( a=4 \), \( b=6 \), \( c=9 \).
Let the sides of \( \Delta PQR \) be \( p, q, r \).
We have \( \frac{PQ}{AB} = \frac{QR}{BC} = \frac{PR}{AC} = \frac{\text{Perimeter of } \Delta PQR}{\text{Perimeter of } \Delta ABC} \).
The perimeter of \( \Delta ABC = 4 + 6 + 9 = 19 \) cm.
One side of \( \Delta PQR \) is given as \( 35 \) cm. To find the greatest possible perimeter for \( \Delta PQR \), this \( 35 \) cm side should correspond to the smallest side of \( \Delta ABC \). This would make the scaling factor between \( \Delta PQR \) and \( \Delta ABC \) as large as possible.
Smallest side of \( \Delta ABC = 4 \) cm.
So, let \( QR = 35 \) cm (corresponding to the side of 4 cm in \( \Delta ABC \)).
Then, \( \frac{QR}{BC} = \frac{35}{4} \). This ratio represents the scaling factor between the two triangles.
Now, we can find the perimeter of \( \Delta PQR \):
\( \frac{\text{Perimeter of } \Delta PQR}{\text{Perimeter of } \Delta ABC} = \frac{QR}{BC} \)
\( \frac{\text{Perimeter of } \Delta PQR}{19} = \frac{35}{4} \)
\( \text{Perimeter of } \Delta PQR = \frac{35}{4} \times 19 \)
\( \text{Perimeter of } \Delta PQR = \frac{665}{4} \)
\( \text{Perimeter of } \Delta PQR = 166.25 \) cm.
This perimeter is the greatest possible. If the \( 35 \) cm side corresponded to a larger side of \( \Delta ABC \), the scaling factor would be smaller, resulting in a smaller perimeter for \( \Delta PQR \).
In simple words: We know the sides of triangle ABC. Triangle PQR is similar to ABC. We want to find the biggest possible perimeter for PQR, given one of its sides is 35 cm. To make the perimeter largest, the 35 cm side must be linked to the smallest side of triangle ABC. We then use the ratio of perimeters to find the largest perimeter for PQR, which is 166.25 cm.

🎯 Exam Tip: For similar triangles, the ratio of corresponding sides is equal to the ratio of their perimeters. To maximize the perimeter of the larger triangle when one side is given, ensure that the given side corresponds to the smallest side of the smaller triangle. This yields the largest possible scaling factor.

 

Question 5. A man sees the top of a tower in a mirror which is at a distance of 87.6 m from the tower. The mirror is on the ground, facing upward. The man is 0.4 m away from the mirror, and the distance of his eye level from the ground is 1.5 m. How tall is the tower? (The foot of man, the mirror and the foot of the tower lie along a straight line).
Answer: Let AB be the height of the man's eye level from the ground, which is \( 1.5 \) m. Let ED be the height of the tower. Let C be the position of the mirror on the ground. The man is \( 0.4 \) m from the mirror, so \( BC = 0.4 \) m. The tower is \( 87.6 \) m from the mirror, so \( DC = 87.6 \) m.
According to the law of reflection, the angle of incidence is equal to the angle of reflection. This means that \( \angle BCA = \angle DCE \). Also, the man and the tower are perpendicular to the ground, so \( \angle ABC = 90^\circ \) and \( \angle EDC = 90^\circ \).
Thus, \( \Delta ABC \) and \( \Delta EDC \) are similar by AA similarity criterion.
Because these triangles are similar, the ratios of their corresponding sides are proportional:
\( \frac{AB}{ED} = \frac{BC}{DC} \)
We want to find ED (the height of the tower). Let's substitute the known values:
\( \frac{1.5}{ED} = \frac{0.4}{87.6} \)
Now, cross-multiply to solve for ED:
\( 0.4 \times ED = 1.5 \times 87.6 \)
\( ED = \frac{1.5 \times 87.6}{0.4} \)
\( ED = \frac{131.4}{0.4} \)
\( ED = 328.5 \) m.
Therefore, the height of the tower is \( 328.5 \) m. This problem showcases how geometric principles like similar triangles and reflection can be used to measure inaccessible heights indirectly.
In simple words: A man sees a tower's top in a mirror. We can make two right-angled triangles: one with the man and the mirror, and another with the tower and the mirror. Because light reflects at equal angles, these two triangles are similar. We use their side ratios to find the tower's height, which is 328.5 meters.

🎯 Exam Tip: Problems involving reflections in mirrors often create similar triangles. Always draw a clear diagram and identify the right-angled triangles. Remember that the angle of incidence equals the angle of reflection, which helps establish the similarity criterion.

 

Question 6. In \( \Delta PQR \), given that S is a point on PQ such that ST || QR and \( \frac{PS}{SQ} = \frac{3}{5} \). If PR = 5.6 cm, then find PT.
Answer: Given \( \Delta PQR \), with S on PQ and ST || QR. We are given the ratio \( \frac{PS}{SQ} = \frac{3}{5} \) and \( PR = 5.6 \) cm.
Since ST is parallel to QR, by the Basic Proportionality Theorem (also known as Thales's Theorem), the line ST divides the sides PQ and PR proportionally.
So, \( \frac{PS}{SQ} = \frac{PT}{TR} \).
We know \( \frac{PS}{SQ} = \frac{3}{5} \), so \( \frac{PT}{TR} = \frac{3}{5} \).
Let \( PT = x \). Then \( TR = PR - PT = 5.6 - x \).
Substitute these into the proportion:
\( \frac{x}{5.6 - x} = \frac{3}{5} \)
Now, cross-multiply:
\( 5x = 3(5.6 - x) \)
\( 5x = 16.8 - 3x \)
Add \( 3x \) to both sides:
\( 5x + 3x = 16.8 \)
\( 8x = 16.8 \)
Divide by 8:
\( x = \frac{16.8}{8} \)
\( x = 2.1 \) cm.
So, \( PT = 2.1 \) cm. This theorem is fundamental for understanding ratios in triangles.
In simple words: In triangle PQR, a line ST is parallel to QR. This means ST divides the sides PQ and PR in the same ratio. We are given the ratio for PQ and the total length of PR. By setting up an equation using these ratios, we find that the length of PT is 2.1 cm.

🎯 Exam Tip: The Basic Proportionality Theorem (BPT) is a key concept. Clearly identify parallel lines and the sides they divide. Practice setting up the ratios correctly: segment/segment = segment/segment. Also, remember that the lengths of the segments of the divided side add up to the total length of that side.

 

Question 7. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x - 3, BD = 3x - 1, AE = 8x - 7 and EC = 5x - 3, then find the value of x.
Answer: Given \( \Delta ABC \), with points D on AB and E on AC, such that DE is parallel to BC. We are given the lengths of the segments in terms of x:
\( AD = 4x - 3 \)
\( BD = 3x - 1 \)
\( AE = 8x - 7 \)
\( EC = 5x - 3 \)
According to the Basic Proportionality Theorem (BPT), since DE || BC, the line DE divides the sides AB and AC proportionally.
So, \( \frac{AD}{BD} = \frac{AE}{EC} \)
Substitute the given expressions for the lengths:
\( \frac{4x - 3}{3x - 1} = \frac{8x - 7}{5x - 3} \)
Now, cross-multiply to solve for x:
\( (4x - 3)(5x - 3) = (8x - 7)(3x - 1) \)
Expand both sides:
\( 20x^2 - 12x - 15x + 9 = 24x^2 - 8x - 21x + 7 \)
\( 20x^2 - 27x + 9 = 24x^2 - 29x + 7 \)
Move all terms to one side to form a quadratic equation:
\( 0 = 24x^2 - 20x^2 - 29x + 27x + 7 - 9 \)
\( 0 = 4x^2 - 2x - 2 \)
Divide the entire equation by 2 to simplify:
\( 0 = 2x^2 - x - 1 \)
Now, factor the quadratic equation. We look for two numbers that multiply to \( 2 \times -1 = -2 \) and add up to -1. These numbers are -2 and 1.
\( 2x^2 - 2x + x - 1 = 0 \)
Group the terms and factor:
\( 2x(x - 1) + 1(x - 1) = 0 \)
\( (2x + 1)(x - 1) = 0 \)
This gives two possible values for x:
\( 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \)
\( x - 1 = 0 \implies x = 1 \)
Since lengths cannot be negative, we must check if \( x = -\frac{1}{2} \) results in negative lengths. If \( x = -\frac{1}{2} \), then \( AD = 4(-\frac{1}{2}) - 3 = -2 - 3 = -5 \), which is not possible.
Therefore, we take the positive value, \( x = 1 \). This mathematical problem demonstrates the application of geometric theorems in algebraic contexts.
In simple words: We have a triangle with a line inside that is parallel to one side. This means the line divides the other two sides into equal ratios. We set up an equation using the given lengths (which include 'x') and solve it. We get two possible answers for 'x', but since lengths cannot be negative, we pick the positive answer, which is \( x = 1 \).

🎯 Exam Tip: Always verify that your solution for x yields positive lengths for all segments. A negative length is physically impossible in geometry. Be careful with algebraic manipulation and factoring quadratic equations.

 

Question 8. In the figure AC || BD and CE || DF. if OA = 12 cm, AB = 9cm, OC = 8 cm and EF = 4.5 cm, then find FO.
Answer: Given a figure with parallel lines and intersecting segments. We have \( AC \parallel BD \) and \( CE \parallel DF \).
Known lengths:
\( OA = 12 \) cm
\( AB = 9 \) cm
\( OC = 8 \) cm
\( EF = 4.5 \) cm
We need to find \( FO \).
First, consider \( \Delta OBD \). Since \( AC \parallel BD \), by Thales's Theorem (BPT), the ratio of segments on the transversals is equal.
From \( AC \parallel BD \), in \( \Delta OBD \), we have \( \frac{OA}{AB} = \frac{OC}{CD} \).
Substitute the given values:
\( \frac{12}{9} = \frac{8}{CD} \)
Cross-multiply to find CD:
\( 12 \times CD = 9 \times 8 \)
\( 12 \times CD = 72 \)
\( CD = \frac{72}{12} \)
\( CD = 6 \) cm.
Next, consider \( \Delta ODF \). Since \( CE \parallel DF \), again by Thales's Theorem (BPT), the ratio of segments is equal.
From \( CE \parallel DF \), in \( \Delta ODF \), we have \( \frac{OC}{CD} = \frac{OE}{EF} \).
Substitute the known values \( OC = 8 \), \( CD = 6 \), and \( EF = 4.5 \):
\( \frac{8}{6} = \frac{OE}{4.5} \)
Simplify the fraction \( \frac{8}{6} = \frac{4}{3} \):
\( \frac{4}{3} = \frac{OE}{4.5} \)
Cross-multiply to find OE:
\( 3 \times OE = 4 \times 4.5 \)
\( 3 \times OE = 18 \)
\( OE = \frac{18}{3} \)
\( OE = 6 \) cm.
Finally, we need to find \( FO \). From the figure, we can see that \( FO = FE + EO \).
\( FO = 4.5 + 6 \)
\( FO = 10.5 \) cm.
Therefore, the value of FO is \( 10.5 \) cm. This problem shows how applying BPT multiple times can solve for unknown lengths in complex figures.
In simple words: We are given a drawing with parallel lines and some lengths. We use a geometry rule (Thales's Theorem) twice. First, with one set of parallel lines, we find a missing length. Then, with another set of parallel lines and the length we just found, we find another missing length. Finally, we add two lengths together to get the answer for FO, which is 10.5 cm.

🎯 Exam Tip: Break down complex geometry problems into simpler steps. Clearly identify pairs of parallel lines and the triangles/transversals they affect to correctly apply the Basic Proportionality Theorem. Always double-check your calculations.

 

Question 9. Check whether AD is the bisector of \( \angle A \) of \( \Delta ABC \) in each of the following.
(i) AB = 4 cm, AC = 6 cm, BD 1.6 cm, and CD = 2.4 cm.
Answer: To check if AD is the bisector of \( \angle A \) in \( \Delta ABC \), we use the Converse of the Angle Bisector Theorem. This theorem states that if a line segment divides the opposite side in the ratio of the other two sides, then the line segment is the angle bisector.
Given:
\( AB = 4 \) cm
\( AC = 6 \) cm
\( BD = 1.6 \) cm
\( CD = 2.4 \) cm
First, find the ratio of the sides \( \frac{AB}{AC} \):
\( \frac{AB}{AC} = \frac{4}{6} = \frac{2}{3} \). (Equation 1)
Next, find the ratio of the segments of the opposite side \( \frac{BD}{DC} \):
\( \frac{BD}{DC} = \frac{1.6}{2.4} \)
To simplify this ratio, multiply the numerator and denominator by 10 to remove decimals:
\( \frac{16}{24} \)
Divide both by 8:
\( \frac{16 \div 8}{24 \div 8} = \frac{2}{3} \). (Equation 2)
Since \( \frac{AB}{AC} = \frac{BD}{DC} \) (both are \( \frac{2}{3} \)), by the Converse of the Angle Bisector Theorem, AD is indeed the internal bisector of \( \angle A \). This theorem is a powerful tool for verifying angle bisectors.
In simple words: We are checking if line AD cuts angle A exactly in half. A rule says if a line divides the opposite side in the same ratio as the other two sides, then it's an angle bisector. We compare the ratio of sides AB and AC with the ratio of segments BD and CD. Since both ratios are 2/3, line AD is indeed the bisector of angle A.

🎯 Exam Tip: Clearly state the theorem you are using (Angle Bisector Theorem or its Converse). Show all steps for calculating the ratios to ensure full marks. Remember, for AD to be an angle bisector, the ratio \( \frac{AB}{AC} \) must exactly equal \( \frac{BD}{DC} \).

 

(ii) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 3 cm
Answer: To check if AD is the bisector of \( \angle A \) in \( \Delta ABC \), we will again use the Converse of the Angle Bisector Theorem.
Given:
\( AB = 6 \) cm
\( AC = 8 \) cm
\( BD = 1.5 \) cm
\( CD = 3 \) cm
First, calculate the ratio of the sides \( \frac{AB}{AC} \):
\( \frac{AB}{AC} = \frac{6}{8} = \frac{3}{4} \). (Equation 1)
Next, calculate the ratio of the segments of the opposite side \( \frac{BD}{DC} \):
\( \frac{BD}{DC} = \frac{1.5}{3} \)
To simplify, we can write 1.5 as \( \frac{3}{2} \):
\( \frac{1.5}{3} = \frac{3/2}{3} = \frac{3}{2 \times 3} = \frac{1}{2} \). (Equation 2)
Comparing Equation 1 and Equation 2:
\( \frac{AB}{AC} = \frac{3}{4} \)
\( \frac{BD}{DC} = \frac{1}{2} \)
Since \( \frac{3}{4} \neq \frac{1}{2} \), the ratios are not equal. Therefore, by the Converse of the Angle Bisector Theorem, AD is not the bisector of \( \angle A \). This shows how the theorem can confirm or deny if a line segment bisects an angle.
In simple words: We are checking if line AD divides angle A into two equal parts. We compare the ratio of sides AB and AC (which is 3/4) with the ratio of segments BD and CD (which is 1/2). Since these ratios are not the same, AD is not the angle bisector of angle A.

🎯 Exam Tip: Be meticulous with calculations involving decimals or fractions when determining ratios. A small error can lead to an incorrect conclusion about whether a line segment is an angle bisector or not. State clearly whether the ratios are equal or not.

 

Question 10. In a \( \Delta ABC \), AD is the internal bisector of \( \angle A \), meeting BC at D. If AB 5.6 cm, AC = 6 cm, DC = 3 cm, find BC.
Answer: Given that AD is the internal bisector of \( \angle A \) in \( \Delta ABC \). The bisector meets BC at D.
We are given:
\( AB = 5.6 \) cm
\( AC = 6 \) cm
\( DC = 3 \) cm
We need to find the length of BC.
Let \( BD = x \). Then \( BC = BD + DC = x + 3 \).
According to the Angle Bisector Theorem, if AD is the bisector of \( \angle A \), then it divides the opposite side BC in the ratio of the other two sides:
\( \frac{BD}{DC} = \frac{AB}{AC} \)
Substitute the given values and \( x \):
\( \frac{x}{3} = \frac{5.6}{6} \)
Now, solve for x:
\( 6x = 3 \times 5.6 \)
\( 6x = 16.8 \)
\( x = \frac{16.8}{6} \)
\( x = 2.8 \) cm.
So, \( BD = 2.8 \) cm.
Now we can find BC:
\( BC = BD + DC \)
\( BC = 2.8 + 3 \)
\( BC = 5.8 \) cm.
Therefore, the length of BC is \( 5.8 \) cm. This demonstrates how the Angle Bisector Theorem is used to find unknown lengths in a triangle.
In simple words: In a triangle, a line AD cuts angle A in half and touches side BC. We are given some side lengths and part of BC. Using a rule called the Angle Bisector Theorem, we set up a proportion and solve for the missing part of BC. Then we add the two parts to find the total length of BC, which is 5.8 cm.

🎯 Exam Tip: Clearly state the Angle Bisector Theorem before applying it. Ensure you correctly identify which segments correspond to which sides in the ratio. Be careful with decimal calculations during multiplication and division.

 

Question 11. In the figure, tangents PA and PB are drawn to a circle with centre O from an external point P. If CD is a tangent to the circle at E and AP = 15 cm, find the perimeter of \( \Delta PCD \).
Answer: Given a circle with center O, and tangents PA and PB drawn from an external point P. Also, CD is another tangent to the circle at point E.
We are given \( AP = 15 \) cm. We need to find the perimeter of \( \Delta PCD \).
A key property of tangents from an external point to a circle is that their lengths are equal.
So, \( PA = PB \). Since \( AP = 15 \) cm, then \( PB = 15 \) cm.
Also, from point C, tangents \( CA \) and \( CE \) are drawn to the circle. So, \( CA = CE \).
Similarly, from point D, tangents \( DB \) and \( DE \) are drawn to the circle. So, \( DB = DE \).
The perimeter of \( \Delta PCD \) is the sum of its sides:
Perimeter \( = PC + CD + DP \)
We can rewrite CD as the sum of segments CE and ED:
Perimeter \( = PC + (CE + ED) + DP \)
Now, substitute \( CE = CA \) and \( ED = DB \):
Perimeter \( = PC + CA + DB + DP \)
Rearrange the terms to group the segments that form the longer tangents PA and PB:
Perimeter \( = (PC + CA) + (DB + DP) \)
Notice that \( PC + CA = PA \) and \( DB + DP = PB \).
So, Perimeter \( = PA + PB \).
Since \( PA = PB = 15 \) cm:
Perimeter \( = 15 + 15 \)
Perimeter \( = 30 \) cm.
Therefore, the perimeter of \( \Delta PCD \) is \( 30 \) cm. This property simplifies finding perimeters in such tangent configurations.
In simple words: We have a circle with lines touching it from outside. A rule says that lines from the same outside point touching a circle are equal in length. We use this rule for all the touching lines. The perimeter of triangle PCD is found by adding its three sides. When we replace parts of the sides using our rule, we find the perimeter is simply twice the length of one of the main tangents, making it 30 cm.

🎯 Exam Tip: Always remember the property that tangents from an external point to a circle are equal in length. This is crucial for solving problems like this. Clearly show how the perimeter of the inner triangle simplifies to the sum of the two main external tangents.

 

Question 12. ABCD is a quadrilateral such that all of its sides touch a circle. If AB = 6 cm, BC = 6.5 cm and CD = 7 cm, then find the length of AD.
Answer: Given a quadrilateral ABCD whose sides touch a circle. This is a special type of quadrilateral called a tangential quadrilateral or a circumscribed quadrilateral.
We are given the lengths of three sides:
\( AB = 6 \) cm
\( BC = 6.5 \) cm
\( CD = 7 \) cm
We need to find the length of AD.
A key property of a tangential quadrilateral is that the sums of the lengths of opposite sides are equal.
This means: \( AB + CD = BC + AD \).
Substitute the given values into this property:
\( 6 + 7 = 6.5 + AD \)
\( 13 = 6.5 + AD \)
To find AD, subtract 6.5 from both sides:
\( AD = 13 - 6.5 \)
\( AD = 6.5 \) cm.
Therefore, the length of AD is \( 6.5 \) cm. This property makes solving for unknown side lengths in tangential quadrilaterals straightforward.
In simple words: We have a four-sided shape (quadrilateral) where all its sides touch a circle. A special rule for such shapes is that if you add the lengths of opposite sides, they will be equal. We use this rule with the given side lengths to find the missing side AD, which is 6.5 cm.

🎯 Exam Tip: Clearly state the property of tangential quadrilaterals: the sum of opposite sides is equal (\( AB + CD = BC + AD \)). This is a direct application, and writing down the property ensures clarity and helps in solving similar problems quickly.

 

Question 13. A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.
Answer: Let the man's initial position be O. He first goes \( 10 \) m due east, reaching point A. Then, from point A, he goes \( 24 \) m due north, reaching point B.
We need to find the straight-line distance from his starting point O to his final position B.
When moving east and then north, the path forms a right-angled triangle. OA is the eastward displacement, AB is the northward displacement, and OB is the hypotenuse, representing the direct distance from start to finish.
So, in right-angled \( \Delta OAB \):
\( OA = 10 \) m (base)
\( AB = 24 \) m (height)
We can use the Pythagorean theorem to find OB, the hypotenuse:
\( OB^2 = OA^2 + AB^2 \)
Substitute the given values:
\( OB^2 = 10^2 + 24^2 \)
\( OB^2 = 100 + 576 \)
\( OB^2 = 676 \)
To find OB, take the square root of 676:
\( OB = \sqrt{676} \)
\( OB = 26 \) m.
Therefore, the man is \( 26 \) m from his starting point. This is a classic application of the Pythagorean theorem for displacement problems.
In simple words: A man walks east, then north. This creates a right-angled triangle. We can use the Pythagorean theorem (a² + b² = c²) to find the direct distance from where he started to where he finished. The distance is 26 meters.

🎯 Exam Tip: Always visualize or draw a diagram for direction-based problems. East-west movement is usually horizontal, and north-south movement is vertical, creating a right-angled triangle. Remember the Pythagorean theorem \( (a^2 + b^2 = c^2) \) to find the direct distance.

 

Question 14. Suppose AB, AC and BC have lengths 13, 16 and 20 respectively. If \( \frac{AF}{FB} = \frac{4}{5} \) and \( \frac{CE}{EA} = \frac{5}{12} \) Find BD and DC.
Answer: Given \( \Delta ABC \) with side lengths \( AB = 13 \), \( AC = 16 \), and \( BC = 20 \). Points F, E, D are on sides AB, AC, BC respectively. We are given ratios: \( \frac{AF}{FB} = \frac{4}{5} \) and \( \frac{CE}{EA} = \frac{5}{12} \). We need to find BD and DC.
This problem involves Ceva's Theorem, which applies when three cevians (lines from vertices to opposite sides) are concurrent (meet at a single point). While not explicitly stated that AD, BE, CF are concurrent, the problem implies a setup where such a theorem would be relevant, or it could be a simpler segment division problem.
Let's assume AD, BE, CF are cevians that intersect at a single point. Ceva's Theorem states:
\( \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = 1 \)
We are given \( \frac{AF}{FB} = \frac{4}{5} \) and \( \frac{CE}{EA} = \frac{5}{12} \). Let \( \frac{BD}{DC} = \frac{x}{y} \).
Substitute these ratios into Ceva's Theorem:
\( \frac{4}{5} \times \frac{x}{y} \times \frac{5}{12} = 1 \)
Simplify the expression:
\( \frac{4 \times 5 \times x}{5 \times 12 \times y} = 1 \)
\( \frac{20x}{60y} = 1 \)
\( \frac{x}{3y} = 1 \)
So, \( x = 3y \). (Equation 1)
We also know that \( BC = BD + DC \). Given \( BC = 20 \), and if \( BD = x \) and \( DC = y \), then:
\( x + y = 20 \). (Equation 2)
Now we have a system of two linear equations:
1) \( x = 3y \)
2) \( x + y = 20 \)
Substitute Equation 1 into Equation 2:
\( (3y) + y = 20 \)
\( 4y = 20 \)
\( y = \frac{20}{4} \)
\( y = 5 \)
So, \( DC = 5 \) cm.
Now, substitute \( y = 5 \) back into Equation 1 to find x:
\( x = 3 \times 5 \)
\( x = 15 \)
So, \( BD = 15 \) cm.
Thus, \( BD = 15 \) cm and \( DC = 5 \) cm. This problem illustrates how theorems like Ceva's can be used to solve for segment lengths.
In simple words: In a triangle, some lines connect corners to the opposite sides. We are given ratios of how these lines divide two sides. Using a rule called Ceva's Theorem, we set up an equation. We also know the total length of the third side. By solving these two pieces of information together, we find the lengths of BD (15 cm) and DC (5 cm).

🎯 Exam Tip: Ceva's Theorem is a powerful tool for problems involving concurrent cevians. Clearly define your variables for segments (e.g., \( BD = x, DC = y \)) and set up the system of equations carefully. Remember that the sum of the segments on a side equals the total length of that side.

 

Question 15. ABC is a right-angled triangle at B. Let D and E be any two points on AB and BC respectively. Prove that \( AE^2 + CD^2 = AC^2 + DE^2 \).
Answer: We need to prove that \( AE^2 + CD^2 = AC^2 + DE^2 \).
Since \( \triangle ABE \) is a right-angled triangle at B, by the Pythagoras theorem, we have:
\( AE^2 = AB^2 + BE^2 \) ...(1)
Similarly, \( \triangle DBC \) is a right-angled triangle at B, so by the Pythagoras theorem, we have:
\( CD^2 = BD^2 + BC^2 \) ...(2)
Now, let's add equations (1) and (2):
\( AE^2 + CD^2 = (AB^2 + BE^2) + (BD^2 + BC^2) \)
We can rearrange the terms as:
\( AE^2 + CD^2 = (AB^2 + BC^2) + (BE^2 + BD^2) \)
In the right-angled triangle \( \triangle ABC \) (right-angled at B), according to the Pythagoras theorem, \( AC^2 = AB^2 + BC^2 \).
Also, consider the triangle \( \triangle BDE \). If D is on AB and E is on BC, then \( \triangle BDE \) is a right-angled triangle at B. By Pythagoras theorem, \( DE^2 = BD^2 + BE^2 \).
Substituting these into our sum equation:
\( AE^2 + CD^2 = AC^2 + DE^2 \)
Thus, the statement is proven. This theorem shows how squared lengths of segments in a right triangle relate to its hypotenuse and other segments.
In simple words: When you have a right-angled triangle and add other points on its sides, the square of certain lines drawn from these points will equal the square of the main diagonal plus the square of another line connecting the points. It is like combining two Pythagoras theorems.

🎯 Exam Tip: When proving geometric relations, always clearly state the triangles you are considering and the theorems you are applying (like Pythagoras theorem) at each step. This makes your proof easy to follow and complete.

III. Answer the Following Questions

 

Question 1. A point O in the interior of a rectangle ABCD is joined to each of the vertices A, B, C and D. Prove that \( OA^2 + OC^2 = OB^2 + OD^2 \).
Answer: To prove this, we start with a rectangle ABCD and a point O inside it.

**Construction:** Draw a line segment EF through point O such that EF is parallel to AB. This line EF will also be parallel to DC because ABCD is a rectangle.
**Proof:**
Consider the right-angled triangle \( \triangle OEA \) (right-angled at E). By Pythagoras theorem:
\( OA^2 = OE^2 + AE^2 \) ...(1)
Consider the right-angled triangle \( \triangle OFC \) (right-angled at F). By Pythagoras theorem:
\( OC^2 = OF^2 + FC^2 \) ...(2)
Consider the right-angled triangle \( \triangle OFB \) (right-angled at F). By Pythagoras theorem:
\( OB^2 = OF^2 + FB^2 \) ...(3)
Consider the right-angled triangle \( \triangle OED \) (right-angled at E). By Pythagoras theorem:
\( OD^2 = OE^2 + ED^2 \) ...(4)
Now, let's add equations (1) and (2):
\( OA^2 + OC^2 = OE^2 + AE^2 + OF^2 + FC^2 \)
Next, let's add equations (3) and (4):
\( OB^2 + OD^2 = OF^2 + FB^2 + OE^2 + ED^2 \)
Since ABCD is a rectangle and EF is parallel to AB, it means EF is perpendicular to AD and BC. So, AE = FB and ED = FC.
Substituting these equal lengths into the sum of (3) and (4):
\( OB^2 + OD^2 = OF^2 + AE^2 + OE^2 + FC^2 \)
If we compare \( OA^2 + OC^2 \) and \( OB^2 + OD^2 \), we can see they are equal:
\( OA^2 + OC^2 = OE^2 + AE^2 + OF^2 + FC^2 \)
\( OB^2 + OD^2 = OE^2 + AE^2 + OF^2 + FC^2 \)
Therefore, \( OA^2 + OC^2 = OB^2 + OD^2 \). This property is a useful application of the Pythagorean theorem in coordinate geometry.
In simple words: If you pick any point inside a rectangle and connect it to all four corners, the sum of the squares of the distances to opposite corners will always be the same. For example, the sum of squares of distances to corners A and C will be equal to the sum of squares of distances to corners B and D.

🎯 Exam Tip: For geometry proofs, drawing a clear diagram and adding construction lines can often simplify the problem. Remember to apply theorems like Pythagoras theorem correctly to the right-angled triangles formed.

 

Question 2. A lotus is 20 cm above the water surface in a pond and its stem is partly below the water surface. As the wind blew, the stem is pushed aside so that the lotus touched the water 40 cm away from the original position of the stem. How much of the stem was below the water surface originally?
Answer: Let's set up a right-angled triangle to solve this problem.

Let OA be the length of the stem below the water surface. We can call this \( x \) cm.
The part of the stem above the water surface is given as 20 cm. Let this be AB.
So, the total length of the lotus stem is \( OA + AB = x + 20 \) cm.
When the wind pushes the stem, the lotus (point B) touches the water at point C, which is 40 cm away from the original position of the stem (point A).
This forms a right-angled triangle OAC, where:
- OA is the side representing the stem below the water (\( x \)).
- AC is the distance the lotus moved on the water surface (40 cm).
- OC is the hypotenuse, which is the total length of the stem (\( x + 20 \)).
Using the Pythagoras theorem in \( \triangle OAC \):
\( OC^2 = OA^2 + AC^2 \)
Substitute the values:
\( (x + 20)^2 = x^2 + 40^2 \)
Expand the equation:
\( x^2 + 40x + 400 = x^2 + 1600 \)
Subtract \( x^2 \) from both sides:
\( 40x + 400 = 1600 \)
Subtract 400 from both sides:
\( 40x = 1600 - 400 \)
\( 40x = 1200 \)
Divide by 40 to find \( x \):
\( x = \frac{1200}{40} \)
\( x = 30 \) cm
So, the stem was 30 cm below the water surface originally. The total stem length was 50 cm, a common length for water plants.
In simple words: We used a right-angled triangle to model the lotus stem. The part of the stem underwater, the distance the lotus moved, and the total stem length formed the sides. By using Pythagoras theorem, we found that the stem was 30 cm below the water.

🎯 Exam Tip: For word problems involving geometry, always draw a clear diagram and label the knowns and unknowns. This helps in correctly applying the relevant geometric theorems like the Pythagoras theorem.

 

Question 3. In the figure, DE || BC and \( \frac { AD }{ BD } = \frac {3}{5} \), calculate the value of
(i) \( \frac{\text{area of } \triangle ADE}{\text{area of } \triangle ABC} \)
(ii) \( \frac{\text{area of trapezium BCED}}{\text{area of } \triangle ABC} \)
Answer: Given that DE is parallel to BC, and the ratio \( \frac { AD }{ BD } = \frac {3}{5} \).

From the ratio \( \frac { AD }{ BD } = \frac {3}{5} \), we can say that \( AD = 3k \) and \( BD = 5k \) for some constant \( k \).
Then, the full length of side AB is \( AB = AD + BD = 3k + 5k = 8k \).
(i) Calculating \( \frac{\text{area of } \triangle ADE}{\text{area of } \triangle ABC} \):
Since DE is parallel to BC, \( \triangle ADE \) is similar to \( \triangle ABC \) (by AA similarity, as \( \angle A \) is common and \( \angle ADE = \angle ABC \) are corresponding angles).
For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
\( \frac{\text{area of } \triangle ADE}{\text{area of } \triangle ABC} = \frac{AD^2}{AB^2} \)
Substitute the values of AD and AB:
\( \frac{\text{area of } \triangle ADE}{\text{area of } \triangle ABC} = \frac{(3k)^2}{(8k)^2} = \frac{9k^2}{64k^2} = \frac{9}{64} \)
(ii) Calculating \( \frac{\text{area of trapezium BCED}}{\text{area of } \triangle ABC} \):
The area of trapezium BCED can be found by subtracting the area of \( \triangle ADE \) from the area of \( \triangle ABC \).
Area of trapezium BCED = Area of \( \triangle ABC \) - Area of \( \triangle ADE \)
Let's say Area of \( \triangle ADE = 9k' \) and Area of \( \triangle ABC = 64k' \) (from the ratio above).
Area of trapezium BCED = \( 64k' - 9k' = 55k' \)
Now, find the ratio:
\( \frac{\text{area of trapezium BCED}}{\text{area of } \triangle ABC} = \frac{55k'}{64k'} = \frac{55}{64} \)
The ratios are fixed because parallel lines create proportional segments and similar figures. This concept is fundamental in understanding scaling in geometry.
In simple words: When a line inside a triangle is parallel to its base, it creates a smaller triangle that is similar to the big one. We used the side ratios to find how their areas compare. Then, we subtracted the small triangle's area from the big one to find the area of the shape that looks like a table (trapezium) at the bottom.

🎯 Exam Tip: Remember that for similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides. This is a very common property used in competitive exams.

 

Question 4. A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?
Answer: In a kite, the diagonals are perpendicular to each other. So, let AC and BD be the diagonals intersecting at E, and \( \angle AEB = 90^\circ \). The problem assumes that the triangle ABC is a right-angled triangle at B, and BE is the altitude to the hypotenuse AC.

We are given \( AE = 16 \) cm and \( EC = 81 \) cm.
Let \( BE = z \). Since the diagonals of a kite are perpendicular, \( \triangle AEB \) and \( \triangle CEB \) are right-angled at E.
From \( \triangle AEB \), by Pythagoras theorem: \( AB^2 = AE^2 + BE^2 \implies AB^2 = 16^2 + z^2 \).
From \( \triangle CEB \), by Pythagoras theorem: \( BC^2 = CE^2 + BE^2 \implies BC^2 = 81^2 + z^2 \).
Also, in a right-angled triangle ABC (right-angled at B) with an altitude BE to the hypotenuse AC, a property states that \( BE^2 = AE \times EC \).
Using this property:
\( z^2 = 16 \times 81 \)
\( z^2 = 1296 \)
\( z = \sqrt{1296} \)
\( z = 36 \) cm
So, the length \( BE = 36 \) cm.
For a kite, one of the diagonals (in this case AC) is the perpendicular bisector of the other diagonal (BD). This means E is the midpoint of BD, so \( BE = ED \).
Therefore, the length of the cross bar BD is \( 2 \times BE \).
\( BD = 2 \times 36 = 72 \) cm.
Kites are often used for decorative purposes, with their crossbars providing structural stability.
In simple words: A kite has two sticks inside that cross each other straight. One stick cuts the other exactly in half. We used a special rule for right-angled triangles to find the length of half of the second stick. Since it cuts in half, we doubled that length to get the full length of the cross bar, which is 72 cm.

🎯 Exam Tip: Remember the properties of a kite: diagonals are perpendicular, and one diagonal bisects the other. Also, be aware of the geometric mean theorem for altitudes in right-angled triangles, as it is often useful in these types of problems.

 

Question 5. Find the unknown values in each of the following figures. All lengths are given in centimetres (All measures are not in scale).
(i)

(ii)

Answer:
(i) In \( \triangle ABC \) and \( \triangle ADE \):
\( \angle A \) is common to both triangles.
Since DE is parallel to BC, \( \angle ABC = \angle ADE \) (corresponding angles).
Therefore, \( \triangle ABC \sim \triangle ADE \) by AAA similarity.
According to the property of similar triangles, the ratio of corresponding sides is equal:
\( \frac{AC}{AE} = \frac{BC}{DE} \)
From the figure, we have AC = AE + EC = \( y + 6 \). BC = 24. DE = 8. AD = x, DB = 8.
Given \( \frac{AE}{AC} = \frac{DE}{BC} \). So, \( \frac{y}{y+6} = \frac{8}{24} \)
\( \frac{y}{y+6} = \frac{1}{3} \)
\( 3y = y + 6 \)
\( 2y = 6 \)
\( y = 3 \) cm.
Now, in \( \triangle EAG \) and \( \triangle ECF \):
\( \angle AEG = \angle CEF \) (vertically opposite angles).
Given CF || AG, so \( \angle EAG = \angle ECF \) (alternate interior angles).
Therefore, \( \triangle EAG \sim \triangle ECF \) by AA similarity.
The ratio of corresponding sides is equal:
\( \frac{EA}{EC} = \frac{AG}{CF} \)
From the figure, AE = x, EC = 8. AG = y, CF = 6.
\( \frac{x}{8} = \frac{y}{6} \)
Substitute \( y = 3 \) cm:
\( \frac{x}{8} = \frac{3}{6} \)
\( \frac{x}{8} = \frac{1}{2} \)
\( 2x = 8 \)
\( x = 4 \) cm.
Thus, the unknown values are \( x = 4 \) cm and \( y = 3 \) cm. The lengths correspond to proportional relationships created by parallel lines.
(ii) In \( \triangle HBC \) and \( \triangle HFG \):
\( \angle H \) is common to both triangles.
Given FG || BC, so \( \angle HFG = \angle HBC \) (corresponding angles).
Therefore, \( \triangle HBC \sim \triangle HFG \) by AA similarity.
The ratio of corresponding sides is equal:
\( \frac{HF}{HB} = \frac{FG}{BC} \)
From the figure, HF = 4. HB = HF + FB = \( 4 + 8 = 12 \). FG = x. BC = 10.
\( \frac{4}{12} = \frac{x}{10} \)
\( \frac{1}{3} = \frac{x}{10} \)
\( 3x = 10 \)
\( x = \frac{10}{3} \)
\( x \approx 3.33 \) cm.
This demonstrates how similar triangles formed by a parallel line within a larger triangle can be used to find unknown lengths. This technique is often used in surveying.
In simple words: For the first figure, we found two pairs of triangles that are similar because of parallel lines and shared angles. By setting up equations with the ratios of their sides, we solved for the unknown lengths x and y. For the second figure, we again used similar triangles to find the unknown length x.

🎯 Exam Tip: When dealing with figures involving parallel lines, always look for similar triangles. The ratio of corresponding sides is equal, which helps in finding unknown lengths. Clearly state the similar triangles and the reason for their similarity.

 

Question 6. The internal bisector of \( \angle A \) of \( \triangle ABC \) meets BC at D and the external bisector of \( \angle A \) meets BC produced at E. Prove that \( \frac { BD }{ BE } = \frac { CD }{ CE } \).
Answer: Given that AD is the internal bisector of \( \angle A \) of \( \triangle ABC \), meeting BC at D. According to the Angle Bisector Theorem (ABT):
\( \frac{BD}{CD} = \frac{AB}{AC} \) ...(1)
Also, AE is the external bisector of \( \angle A \) of \( \triangle ABC \), meeting BC produced at E. According to the External Angle Bisector Theorem:
\( \frac{BE}{CE} = \frac{AB}{AC} \) ...(2)
From equations (1) and (2), since both ratios are equal to \( \frac{AB}{AC} \), we can equate them:
\( \frac{BD}{CD} = \frac{BE}{CE} \)
To get the required form, we can rearrange this equation:
\( \frac{BD}{BE} = \frac{CD}{CE} \)
Thus, the statement is proven. This shows a powerful relationship between internal and external angle bisectors, revealing how they divide the opposite side proportionally.
In simple words: When you draw lines that cut an angle of a triangle in half (both inside and outside), these lines divide the opposite side of the triangle in a special way. The ratio of the parts created by the inside line is the same as the ratio of the parts created by the outside line.

🎯 Exam Tip: Remember both the internal and external angle bisector theorems. They are fundamental in geometry problems involving angle bisectors and side ratios. Make sure to apply them correctly to the respective segments.

 

Question 7. In a quadrilateral ABCD, the bisectors of \( \angle B \) and \( \angle D \) intersect on AC at E. Prove that \( \frac { AB }{ BC } = \frac { AD }{ DC } \).
Answer: Given that ABCD is a quadrilateral. BE is the bisector of \( \angle B \), intersecting AC at E. DE is the bisector of \( \angle D \), intersecting AC at E.
**To Prove:** \( \frac { AB }{ BC } = \frac { AD }{ DC } \).
**Proof:**
Consider \( \triangle ABC \). BE is the internal bisector of \( \angle B \). By the Angle Bisector Theorem:
\( \frac{AE}{EC} = \frac{AB}{BC} \) ...(1)
Consider \( \triangle ADC \). DE is the internal bisector of \( \angle D \). By the Angle Bisector Theorem:
\( \frac{AE}{EC} = \frac{AD}{DC} \) ...(2)
From equations (1) and (2), since both ratios are equal to \( \frac{AE}{EC} \), we can equate them:
\( \frac{AB}{BC} = \frac{AD}{DC} \)
Thus, the statement is proven. This result is a unique property of quadrilaterals where the angle bisectors of opposite angles meet on a diagonal. This principle is a cornerstone for understanding geometric symmetries.
In simple words: Imagine a four-sided shape (quadrilateral). If the lines that cut angles B and D in half meet on the diagonal AC, then the ratio of side AB to side BC will be the same as the ratio of side AD to side DC.

🎯 Exam Tip: For quadrilateral problems, look for triangles formed by diagonals or angle bisectors. Applying the Angle Bisector Theorem to these triangles is often the key to proving relationships between sides.

 

Question 8. ABCD is a quadrilateral with AB parallel to DC. A line drawn parallel to AB meets AD at P and BC at Q. Prove that \( \frac { AP }{ PD } = \frac { BQ }{ QC } \).
Answer: Given that ABCD is a quadrilateral with AB || DC. This means ABCD is a trapezium. A line PQ is drawn such that PQ || AB, and it meets AD at P and BC at Q.

**To Prove:** \( \frac { AP }{ PD } = \frac { BQ }{ QC } \).
**Proof:**
Draw the diagonal AC, which intersects PQ at O.
Since AB || DC and PQ || AB, it means PQ || DC.
Consider \( \triangle ADC \). We have PO || DC (because PQ || DC). By the Basic Proportionality Theorem (BPT, also known as Thales Theorem):
\( \frac{AP}{PD} = \frac{AO}{OC} \) ...(1)
Consider \( \triangle ABC \). We have OQ || AB (because PQ || AB). By the Basic Proportionality Theorem (BPT):
\( \frac{BQ}{QC} = \frac{AO}{OC} \) ...(2)
From equations (1) and (2), since both ratios are equal to \( \frac{AO}{OC} \), we can equate them:
\( \frac{AP}{PD} = \frac{BQ}{QC} \)
Thus, the statement is proven. This principle is very useful in engineering for dividing segments proportionally.
In simple words: If you have a trapezium (a shape with one pair of parallel sides) and draw another line parallel to those sides, this new line will cut the non-parallel sides in the same proportion. It's like a dividing line that keeps the balance.

🎯 Exam Tip: For problems involving parallel lines in a trapezium, drawing a diagonal often helps to create two triangles where the Basic Proportionality Theorem (BPT) can be applied effectively.

 

Question 9. D is the midpoint of the side BC of \( \triangle ABC \). If P and Q are points on AB and on AC such that DP bisects \( \angle BDA \) and DQ bisects \( \angle ADC \), then prove that PQ || BC.
Answer: Given that D is the midpoint of BC. So, \( BD = DC \). P is on AB and Q is on AC. DP bisects \( \angle BDA \) and DQ bisects \( \angle ADC \).

**To Prove:** PQ || BC.
**Proof:**
Consider \( \triangle ABD \). DP is the internal bisector of \( \angle BDA \). By the Angle Bisector Theorem:
\( \frac{AP}{PB} = \frac{AD}{BD} \) ...(1)
Consider \( \triangle ADC \). DQ is the internal bisector of \( \angle ADC \). By the Angle Bisector Theorem:
\( \frac{AQ}{QC} = \frac{AD}{DC} \) ...(2)
We are given that D is the midpoint of BC, so \( BD = DC \).
Substitute \( DC \) with \( BD \) in equation (2):
\( \frac{AQ}{QC} = \frac{AD}{BD} \) ...(3)
From equations (1) and (3), since both ratios are equal to \( \frac{AD}{BD} \), we can equate them:
\( \frac{AP}{PB} = \frac{AQ}{QC} \)
Now, in \( \triangle ABC \), the line segment PQ divides sides AB and AC such that \( \frac{AP}{PB} = \frac{AQ}{QC} \). According to the Converse of Basic Proportionality Theorem (Converse of Thales Theorem), if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Therefore, PQ || BC.
This proof beautifully links angle bisectors with parallel lines, a testament to the interconnectedness of geometric theorems. The points P and Q are key to establishing this proportionality.
In simple words: If you have a triangle, and a line is drawn from a midpoint that cuts an angle, and another line cuts another angle in a similar way, then the line connecting the two points P and Q will be parallel to the base of the triangle.

🎯 Exam Tip: This problem is a classic application of combining the Angle Bisector Theorem with the Converse of the Basic Proportionality Theorem. Always identify which theorem applies to which triangle or line segment to build a logical proof.

 

Question 10. ABCD is a trapezium with AB || DC. The diagonal AC and BD intersect at E. Prove that AD = BC.
Answer: Given that ABCD is a trapezium with AB parallel to DC. The diagonals AC and BD intersect at point E.

**To Prove:** AD = BC. (This means it's an isosceles trapezium).
**Proof:**
Consider \( \triangle ECD \) and \( \triangle EAB \).
Since AB || DC, we have:
\( \angle EDC = \angle EBA \) (Alternate interior angles)
\( \angle ECD = \angle EAB \) (Alternate interior angles)
Also, \( \angle DEC = \angle AEB \) (Vertically opposite angles).
Therefore, \( \triangle ECD \sim \triangle EAB \) by AAA similarity.
From the similarity, the ratio of corresponding sides is equal:
\( \frac{DE}{BE} = \frac{CE}{AE} = \frac{DC}{AB} \) ...(1)
Now, we are also given a condition (implied by the original solution attempting to prove equality) that leads to \( \triangle ADE \) being similar to \( \triangle BCE \). If \( \triangle ADE \sim \triangle BCE \), then:
\( \frac{DE}{CE} = \frac{AE}{BE} = \frac{AD}{BC} \) ...(2)
From equation (1), we have \( \frac{DE}{BE} = \frac{CE}{AE} \). Cross-multiplying gives \( DE \times AE = BE \times CE \).
From equation (2), we have \( \frac{DE}{CE} = \frac{AE}{BE} \). Cross-multiplying gives \( DE \times BE = CE \times AE \).
For both of these to hold true, it must be that \( AE = BE \) and \( DE = CE \).
If \( AE = BE \) and \( DE = CE \), substituting into equation (2) for \( \frac{AD}{BC} \):
\( \frac{AE}{BE} = \frac{AD}{BC} \)
Since \( AE = BE \), this simplifies to \( 1 = \frac{AD}{BC} \).
\( \implies AD = BC \).
Thus, it is proven that if the conditions for similar triangles \( \triangle ADE \) and \( \triangle BCE \) hold, then the non-parallel sides of the trapezium are equal, making it an isosceles trapezium. This property is crucial for analyzing architectural and structural designs.
In simple words: If you have a trapezium (a four-sided shape with two parallel sides) and you connect its opposite corners with diagonal lines, a special rule can help us prove that its non-parallel sides are equal. This happens because the triangles formed by the diagonals crossing each other become similar.

🎯 Exam Tip: When proving properties of trapeziums, remember that parallel lines create similar triangles. Look for vertically opposite angles and alternate interior angles to establish similarity, then use the ratios of corresponding sides.