Samacheer Kalvi Class 10 Maths Solutions Chapter 4 Geometry Exercise 4.5

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.5

Question 1. If in triangles ABC and EDF, \( \frac { AB }{ DE } = \frac { BC }{ FD } \) then they will be similar, when ..........
(a) \( \angle B = \angle E \)
(b) \( \angle A = \angle D \)
(c) \( \angle B = \angle D \)
Answer: (c) \( \angle B = \angle D \)
In simple words: When two triangles have two pairs of sides that are in the same ratio, and the angles *between* those sides are equal, then the triangles are similar. This rule is called the Side-Angle-Side (SAS) similarity criterion.

🎯 Exam Tip: Remember the specific angle placement for SAS similarity; it must be the *included* angle between the two proportional sides.

Question 2. In \( \triangle LMN \), \( \angle L = 60^\circ \), \( \angle M = 50^\circ \). If \( \triangle LMN \sim \triangle PQR \) then the value of \( \angle R \) is ..........
(a) 40°
(b) 70°
(c) 30°
(d) 110°
Answer: (b) 70°
In simple words: First, find the third angle \( \angle N \) in the first triangle by subtracting the other two angles from 180 degrees. The sum of angles in any triangle is always 180 degrees. Because the triangles are similar, \( \angle N \) will be equal to \( \angle R \).

🎯 Exam Tip: Corresponding angles of similar triangles are equal, and the sum of angles in any triangle is 180°.

Question 3. If \( \triangle ABC \) is an isosceles triangle with \( \angle C = 90^\circ \) and \( AC = 5 \) cm, then \( AB \) is ..........
(a) 2.5 cm
(b) 5 cm
(c) 10 cm
(d) \( 5 \sqrt { 2 } \) cm
Answer: (d) \( 5 \sqrt { 2 } \) cm
In simple words: Since it's an isosceles right-angled triangle and \( \angle C \) is 90 degrees, the two shorter sides \( AC \) and \( BC \) must be equal. The Pythagorean theorem is a fundamental rule for right-angled triangles where the square of the longest side (hypotenuse) equals the sum of the squares of the other two sides.

🎯 Exam Tip: For an isosceles right-angled triangle, if the equal sides are 'x', the hypotenuse is \( x\sqrt{2} \).

Question 4. In a given figure \( ST \parallel QR \), \( PS = 2 \) cm and \( SQ = 3 \) cm. Then the ratio of the area of \( \triangle PQR \) to the area of \( \triangle PST \) is ..........
(a) 25:4
(b) 25: 7
(c) 25: 11
(d) 25: 13
Answer: (a) 25:4
In simple words: Since \( ST \) is parallel to \( QR \), the smaller triangle \( PST \) is similar to the larger triangle \( PQR \). First, find the full length of \( PQ \) by adding \( PS \) and \( SQ \). Then, use the rule that the ratio of their areas is equal to the square of the ratio of their corresponding sides. This theorem is a powerful tool for solving problems involving similar figures.

🎯 Exam Tip: The ratio of areas of similar triangles is the square of the ratio of their corresponding sides, not just the ratio itself.

Question 5. The perimeters of two similar triangles \( \triangle ABC \) and \( \triangle PQR \) are 36 cm and 24 cm respectively. If \( PQ = 10 \) cm, then the length of \( AB \) is .............
(a) \( 6 \frac { 2 }{ 3 } \) cm
(b) \( \frac{10 \sqrt{6}}{3} \)
(c) \( 66 \frac { 2 }{ 3 } \) cm
(d) 15 cm
Answer: (d) 15 cm
In simple words: When two triangles are similar, the ratio of their total outside lengths (perimeters) is the same as the ratio of any matching side lengths. This relationship holds true for any pair of similar polygons, not just triangles. We can use this rule to find the unknown side length \( AB \).

🎯 Exam Tip: Remember that for similar figures, the ratio of perimeters is equal to the ratio of corresponding sides, while the ratio of areas is the square of the ratio of sides.

Question 6. If in \( \triangle ABC \), \( DE \parallel BC \). \( AB = 3.6 \) cm, \( AC = 2.4 \) cm and \( AD = 2.1 \) cm then the length of \( AE \) is ..........
(a) 1.4 cm
(b) 1.8 cm
(c) 1.2 cm
(d) 1.05 cm
Answer: (a) 1.4 cm
In simple words: When a line inside a triangle is parallel to one side, it cuts the other two sides into parts that are in proportion. This property is a direct application of the Basic Proportionality Theorem (BPT). We can set up a simple ratio of corresponding sides to find the missing length \( AE \).

🎯 Exam Tip: Clearly identify the corresponding sides when applying the Basic Proportionality Theorem to avoid errors in setting up the ratio.

Question 7. In a \( \triangle ABC \), \( AD \) is the bisector of \( \angle BAC \). If \( AB = 8 \) cm, \( BD = 6 \) cm and \( DC = 3 \) cm. The length of the side \( AC \) is ..........
(a) 6 cm
(b) 4 cm
(c) 3 cm
(d) 8 cm
Answer: (b) 4 cm
In simple words: When a line splits an angle of a triangle exactly in half (an angle bisector), it also divides the opposite side into two parts. The lengths of these parts are in the same ratio as the other two sides of the triangle. This rule is called the Angle Bisector Theorem and helps relate the lengths of sides.

🎯 Exam Tip: Correctly identify the sides that form the ratio for the Angle Bisector Theorem: \( \frac{BD}{DC} = \frac{AB}{AC} \).

Question 8. In the adjacent figure \( \angle BAC = 90^\circ \) and \( AD \perp BC \) then ..........
(a) \( BD \cdot CD = BC^2 \)
(b) \( AB \cdot AC = BC^2 \)
(c) \( BD \cdot CD = AD^2 \)
(d) \( AB \cdot AC = AD^2 \)
Answer: (c) \( BD \cdot CD = AD^2 \)
In simple words: In a right-angled triangle, if you draw a line straight down from the right angle to the longest side (hypotenuse), that line's length squared is equal to the product of the two parts it divides the hypotenuse into. This relationship is often called the "altitude rule" or "geometric mean altitude theorem".

🎯 Exam Tip: This theorem is a special case derived from the similarity of the three triangles formed within the larger right-angled triangle.

Question 9. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?
(a) 13 m
(b) 14 m
(c) 15 m
(d) 12.8 m
Answer: (a) 13 m
In simple words: We can form a right-angled triangle by drawing a horizontal line from the top of the shorter pole to the taller pole. One short side of this triangle is the difference in the poles' heights (11m - 6m = 5m). The other short side is the distance between their bottoms (12m). The distance between their tops is the long side (hypotenuse) of this triangle, which we find using the Pythagoras rule. This is a classic application of the Pythagorean theorem in real-world scenarios.

🎯 Exam Tip: Always visualize or sketch the scenario to identify the right-angled triangle and the sides involved before applying the Pythagorean theorem.

Question 10. In the given figure, \( PR = 26 \) cm, \( QR = 24 \) cm, \( \angle PAQ = 90^\circ \), \( PA = 6 \) cm and \( QA = 8 \) cm. Find \( \angle PQR \).
(a) 80°
(b) 85°
(c) 75°
(d) 90°
Answer: (d) 90°
In simple words: First, we use the Pythagoras theorem in the smaller triangle \( PAQ \) to find the length of side \( PQ \). The converse of the Pythagorean theorem allows us to determine if a triangle is right-angled based on its side lengths. Then, we check if the square of the longest side \( PR \) in triangle \( PQR \) is equal to the sum of the squares of the other two sides \( PQ \) and \( QR \). If it is, then \( \angle PQR \) is 90 degrees.

🎯 Exam Tip: Remember to apply the Pythagorean theorem twice: first to find a missing side, then its converse to prove if an angle is 90 degrees.

Question 11. A tangent is perpendicular to the radius at the ..........
(a) centre
(b) point of contact
(c) infinity
(d) chord
Answer: (b) point of contact
In simple words: A tangent line to a circle always forms a right angle (90 degrees) with the radius that meets it at the exact point where the tangent touches the circle. This fundamental property is crucial for solving many circle-related geometry problems.

🎯 Exam Tip: This perpendicular relationship is a core definition in circle geometry and should be memorized for quick application.

Question 12. How many tangents can be drawn to the circle from an exterior point?
(a) one
(b) two
(c) three
(d) zero
Answer: (b) two
In simple words: From any point outside a circle, exactly two tangent lines can be drawn to that circle. These two tangents will have equal lengths from the external point to their respective points of contact on the circle. This property is why we often see two equal tangent segments originating from a single external point.

🎯 Exam Tip: Visualizing the external point and the circle helps confirm that only two distinct lines can touch the circle from that point.

Question 13. The two tangents from an external points \( P \) to a circle with centre at \( O \) are \( PA \) and \( PB \). If \( \angle APB = 70^\circ \) then the value of \( \angle AOB \) is ..........
(a) 100°
(b) 110°
(c) 120°
(d) 130°
Answer: (b) 110°
In simple words: The angles where the radius meets the tangent (\( \angle OAP \) and \( \angle OBP \)) are always 90 degrees. A shape with four sides (quadrilateral OAPB) has angles that add up to 360 degrees. This property shows the relationship between the angle formed by two tangents and the angle subtended by the chord of contact at the center. We know three angles, so we can find the fourth one by subtracting the known angles from 360.

🎯 Exam Tip: Remember that the quadrilateral formed by the two tangents and the two radii to the points of contact has a sum of angles equal to 360°, and the tangent-radius angles are 90°.

Question 14. In figure \( CP \) and \( CQ \) are tangents to a circle with centre at \( O \). \( ARB \) is another tangent to the circle at \( R \). If \( CP = 11 \) cm and \( BC = 7 \) cm, then the length of \( BR \) is ..........
(a) 6 cm
(b) 5 cm
(c) 8 cm
(d) 4 cm
Answer: (d) 4 cm
In simple words: Lines drawn from the same outside point that just touch a circle are always equal in length. So, \( PC \) is equal to \( QC \). Also, from point B, \( BQ \) is equal to \( BR \). This property of equal tangents from an external point is very useful in solving such problems. We know \( QC \) is made of \( QB \) and \( BC \). Using these facts, we can find the length of \( QB \), which is the same as \( BR \).

🎯 Exam Tip: Always apply the equal tangent property from each external point (C and B in this case) to simplify the problem.

Question 15. In figure if \( PR \) is tangent to the circle at \( P \) and \( O \) is the centre of the circle, then \( \angle POQ \) is .............
(a) 120°
(b) 100°
(c) 110°
(d) 90°
Answer: (a) 120°
In simple words: The line from the center to where the tangent touches is always at 90 degrees to the tangent. We can use this to find an angle in the triangle \( OPQ \). Since \( OP \) and \( OQ \) are both radii, they are equal, making \( \triangle OPQ \) an isosceles triangle. This means two of its angles are the same. This problem combines the tangent-radius property with the properties of isosceles triangles. Then, we use the fact that all angles in a triangle add up to 180 degrees to find the last angle.

🎯 Exam Tip: Don't forget that radii in the same circle are always equal, which often leads to isosceles triangles in circle geometry problems.