OP Malhotra Class 9 Maths Solutions Chapter 13 Circle Chapter Test

Question 1. Which type of triangle can be constructed with a 50° angle between two 8 cm sides?
(a) Equilateral
(b) Isosceles
(c) Scalene
(d) Obtuse
Answer: (b) Isosceles
In simple words: When two sides of a triangle are equal (like both being 8 cm) and the angle between them is known (like 50°), the triangle will always have two equal sides, making it an isosceles triangle.

🎯 Exam Tip: Remember the definitions of different triangle types: Isosceles (two equal sides), Equilateral (all three equal sides), Scalene (all sides different lengths). The given information directly points to an isosceles triangle.

Question 2. Give all names that apply to figure.

Answer: The figure shown has four equal sides and four right angles. This means it is a quadrilateral, a rectangle, a parallelogram, a rhombus, and specifically, a square.
In simple words: The shape is a square because all its sides are equal and all its angles are 90 degrees. A square also fits the definition of a rectangle, rhombus, parallelogram, and quadrilateral.

🎯 Exam Tip: Understand the hierarchy of quadrilaterals: a square is a special type of rectangle and rhombus, which are both special types of parallelograms, and all are quadrilaterals. List all applicable names to score full marks.

Question 3. In the figure, ABCD is a rectangle. P is the midpoint of \( \overline{AB} \). Prove ADPC is isosceles.

Answer: In the given figure, ABCD is a rectangle, and P is the midpoint of AB. To prove that triangle ADPC is an isosceles triangle, we need to show that DP = CP. We will compare triangle ADP and triangle BCP.
(1) \( AD = BC \) (Opposite sides of a rectangle are equal)
(2) \( AP = PB \) (P is the midpoint of AB)
(3) \( \angle A = \angle B \) (Each angle in a rectangle is 90°)
So, by the SAS (Side-Angle-Side) axiom, \( \triangle ADP \cong \triangle BCP \).

\( \implies \) Because the triangles are congruent, their corresponding parts are equal (c.p.c.t.).

\( \implies \) Therefore, \( DP = CP \).

\( \implies \) Since two sides of triangle DPC are equal, \( \triangle DPC \) is an isosceles triangle. When two triangles are congruent, their corresponding sides and angles are equal.
In simple words: We showed that the two smaller triangles formed (ADP and BCP) are exactly the same size and shape. Because they are identical, the side DP must be the same length as CP. Since triangle DPC has two equal sides, it is an isosceles triangle.

🎯 Exam Tip: When proving triangles congruent, clearly state the three conditions (sides/angles) you are using and the congruence criterion (SAS, SSS, ASA, AAS, RHS). Then use c.p.c.t. to link to the required proof.

Question 4. In the figure, PQRS is a rhombus with diagonal \( \overline{PR} \). T is any point on \( \overline{PR} \) produced. Prove \( \overline{\mathrm{TS}}=\overline{\mathrm{TQ}} \).

Answer: In the given figure, PQRS is a rhombus with diagonal PR. T is a point on PR extended. We need to prove that TS = TQ.
In a rhombus, all sides are equal: \( PQ = QR = RS = SP \).
Also, the diagonal PR bisects the angles at P and R. This means \( \angle PRS = \angle PRQ \).
We know that angles on a straight line add up to 180°. So, \( \angle PRS + \angle SRT = 180° \) and \( \angle PRQ + \angle QRT = 180° \).
Since \( \angle PRS = \angle PRQ \), it follows that \( \angle SRT = \angle QRT \).
Now, let's look at \( \triangle SRT \) and \( \triangle QRT \):
(1) \( RT = RT \) (This is a common side to both triangles)
(2) \( SR = QR \) (Sides of a rhombus are equal)
(3) \( \angle SRT = \angle QRT \) (We just proved this)
So, by the SAS (Side-Angle-Side) axiom, \( \triangle SRT \cong \triangle QRT \).

\( \implies \) Because the triangles are congruent, their corresponding parts are equal (c.p.c.t.).

\( \implies \) Therefore, \( ST = QT \), or \( TS = TQ \). A rhombus has many symmetrical properties that help in proofs like this.
In simple words: We used the rules of a rhombus to show that two angles are equal. Then, we looked at two triangles (SRT and QRT) and saw that they share a side, have another pair of equal sides, and the angle between those sides is also equal. This means the triangles are identical, so their third sides (TS and TQ) must also be equal.

🎯 Exam Tip: When dealing with rhombuses, remember that all sides are equal, opposite angles are equal, and diagonals bisect angles and are perpendicular bisectors of each other. These properties are key to congruence proofs.

Question 5. In the figure, ABCD is a rhombus. Find :
(i) AB
(ii) m\( \angle \)ABC

Answer: ABCD is a rhombus. In a rhombus, all sides are equal, and diagonals bisect each other at right angles. Also, each diagonal bisects the opposite angles.
(i) To find AB:
Since all sides of a rhombus are equal, we have \( BC = CD = AB \).
So, \( 4x + 15 = 7x + 2 \).
Subtract \( 4x \) from both sides: \( 15 = 3x + 2 \).
Subtract \( 2 \) from both sides: \( 13 = 3x \).

\( \implies x = \frac{13}{3} \).
Now, substitute the value of \( x \) into the expression for AB (or BC, CD, DA):
\( AB = 4x + 15 = 4 \left( \frac{13}{3} \right) + 15 \)
\( AB = \frac{52}{3} + 15 \)
To add these, find a common denominator:
\( AB = \frac{52}{3} + \frac{15 \times 3}{3} = \frac{52 + 45}{3} = \frac{97}{3} \)

\( \implies AB = 32 \frac{1}{3} \). This calculation determines the length of each side of the rhombus.
(ii) To find m\( \angle \)ABC:
In a rhombus, diagonals bisect each other at right angles. Let the diagonals intersect at O.
So, \( \angle AOB = 90° \).
From the figure, the angle \( \angle DAO \) is labelled as \( 12y° \) and \( \angle CDO \) is \( (4y-1)° \). Diagonals bisect the angles, so \( \angle DAB = 2 \times 12y = 24y \) and \( \angle BCD = 2 \times (4y-1) \).
However, the provided solution uses \( 12y \) and \( 4y-1 \) directly in relation to the angles around the intersection, assuming \( \angle AOB = 90° \). This means we use the properties of a rhombus where the diagonals are perpendicular.
If \( \angle AOB = 90° \), then in \( \triangle AOB \), the sum of angles is 180°.
In a rhombus, adjacent angles sum to 180°. So \( \angle DAB + \angle ABC = 180° \).
The question asks for \( m\angle ABC \). Let's use the given angle expressions. The problem statement refers to \( 12y \) and \( (4y-1) \) as angles. The way they are placed suggests they are parts of angles, and perhaps the solution meant something else or is using specific properties. Let's re-evaluate based on how they appear in the solution for angle ADC.
The solution's step for \( \angle ADC \) is \( 180° - (8y-2) \). This comes from \( \angle BCA = \angle ACD = (4y-1)° \), which implies \( \angle BCD = 2(4y-1) \).
And \( \angle AOB = 90° \) implies \( 12y = 90° \).

\( \implies y = \frac{90°}{12} = \frac{15°}{2} \).
Now, \( \angle BCA = (4y-1)° = 4 \left( \frac{15}{2} \right) - 1 = 2 \times 15 - 1 = 30 - 1 = 29° \).
Since \( AC \) bisects \( \angle C \), \( \angle BCD = 2 \times \angle BCA = 2 \times 29° = 58° \).
In a rhombus, consecutive angles are supplementary, so \( \angle ABC + \angle BCD = 180° \).
\( \angle ABC + 58° = 180° \).

\( \implies \angle ABC = 180° - 58° = 122° \). This applies the rules of a rhombus to find the angle.
In simple words: First, we used the fact that all sides of a rhombus are equal to find the value of \( x \). This allowed us to calculate the length of side AB. Then, we used another rule of a rhombus: its diagonals cross each other at a 90-degree angle. This helped us find the value of \( y \). Using \( y \), we found part of angle C, then the full angle C, and finally, angle ABC since adjacent angles in a rhombus add up to 180 degrees.

🎯 Exam Tip: Clearly distinguish between properties of sides (all equal) and angles (opposite angles equal, consecutive angles supplementary) in a rhombus. The perpendicular intersection of diagonals is key for angle calculations.

Question 6. Tell whether each statement is sometimes, always, or never true.
(i) A rectangle is a parallelogram.
(ii) A rhombus is a square.
(iii) A parallelogram is a rhombus.
(iv) A rhombus is a rectangle.
(v) A square is a rectangle.
(vi) A rectangle is a quadrilateral.
(vii) A square is a parallelogram.
(viii)A rectangle is a square.
Answer:
(i) A rectangle is a parallelogram: **Always**. A rectangle is a special type of parallelogram where all angles are 90 degrees. A parallelogram always has opposite sides parallel.
(ii) A rhombus is a square: **Sometimes**. A rhombus has all four sides equal. It becomes a square only if its angles are all 90 degrees.
(iii) A parallelogram is a rhombus: **Sometimes**. A parallelogram has opposite sides parallel and equal. It becomes a rhombus only if all its sides are equal.
(iv) A rhombus is a rectangle: **Sometimes**. A rhombus has all sides equal. It becomes a rectangle only if all its angles are 90 degrees. This also means adjacent sides are unequal, which is confusing if referring to rhombus properties for rectangle.
(v) A square is a rectangle: **Always**. A square has four right angles and opposite sides are equal, which are the defining characteristics of a rectangle.
(vi) A rectangle is a quadrilateral: **Always**. A rectangle is a polygon with four sides, which is the definition of a quadrilateral.
(vii) A square is a parallelogram: **Always**. A square has opposite sides parallel and equal, which are the defining characteristics of a parallelogram.
(viii)A rectangle is a square: **Sometimes**. A rectangle has four right angles. It becomes a square only if all its sides are equal. Otherwise, it is just a rectangle.
In simple words: This question asks us to understand the differences and similarities between basic shapes like rectangles, rhombuses, squares, and parallelograms. "Always" means it's true for every shape of that type. "Sometimes" means it's true only for some specific examples, like when a rhombus also has right angles. "Never" means it's never true.

🎯 Exam Tip: Clearly know the definitions of each quadrilateral. Think of them as a family tree: A square is a type of rhombus AND a type of rectangle, and both of those are types of parallelograms, which are types of quadrilaterals.

Question 7. In the figure, AB = AC, CH = CB and HK is parallel to BC. If the exterior angle CAX is 137°, calculate the angle CHK.

Answer: In the given figure, we have \( AB = AC \) and \( CH = CB \). Also, \( HK \parallel BC \). The exterior angle \( \angle CAX = 137° \). We need to calculate \( \angle CHK \).
First, consider \( \triangle ABC \). Since \( AB = AC \), it is an isosceles triangle.

\( \implies \angle B = \angle C \).
The exterior angle of a triangle is equal to the sum of the two opposite interior angles. So, \( \angle XAC = \angle B + \angle C \).
Since \( \angle B = \angle C \), we can write \( 137° = \angle B + \angle B = 2\angle B \).

\( \implies \angle B = \frac{137°}{2} = 68.5° \).
Now, consider \( \triangle CBH \). Since \( BC = HC \), it is also an isosceles triangle.

\( \implies \angle B = \angle CHB = 68.5° \).
The sum of angles in \( \triangle CBH \) is 180°.
So, \( \angle BCH = 180° - (\angle B + \angle CHB) = 180° - (68.5° + 68.5°) = 180° - 137° = 43° \).
Finally, since \( HK \parallel BC \), \( \angle CHK \) and \( \angle BCH \) are alternate interior angles.

\( \implies \angle CHK = \angle BCH \).

\( \implies \angle CHK = 43° \). Understanding the properties of parallel lines and isosceles triangles is crucial here.
In simple words: We first found the base angles of the big triangle ABC using the exterior angle rule. Then, we used one of these base angles (angle B) in the smaller triangle CBH, which also has two equal sides. This let us find angle BCH. Because line HK is parallel to line BC, angle CHK is the same as angle BCH.

🎯 Exam Tip: Clearly identify isosceles triangles and use the exterior angle property. Remember that alternate interior angles are equal when two parallel lines are cut by a transversal. Show each step of calculation for angles clearly.

Question 8. D is a point on the base BC of an isosceles triangle ABC in which AB = AC. The triangle ADE is drawn such that AD = AE, \( \angle \)DAE = \( \angle \)BAC and D, E are on opposite sides of AC. Prove that
(i) \( \angle \)BAD = \( \angle \)CAE
(ii) triangle BAD and CAE are congruent.
(iii) AC bisects \( \angle \)BCE.

Answer: We are given an isosceles triangle ABC with \( AB = AC \). D is a point on BC. Another triangle ADE is drawn such that \( AD = AE \) and \( \angle DAE = \angle BAC \). D and E are on opposite sides of AC. We need to prove three things.
(i) To prove \( \angle BAD = \angle CAE \):
We are given that \( \angle BAC = \angle DAE \).
Subtract \( \angle DAC \) from both sides of the equation:
\( \angle BAC - \angle DAC = \angle DAE - \angle DAC \).

\( \implies \angle BAD = \angle CAE \). This is a foundational step, showing angle equality.
(ii) To prove \( \triangle BAD \cong \triangle CAE \):
Let's consider \( \triangle BAD \) and \( \triangle CAE \):
(1) \( AB = AC \) (Given that \( \triangle ABC \) is isosceles)
(2) \( AD = AE \) (Given)
(3) \( \angle BAD = \angle CAE \) (Proved in part (i))
So, by the SAS (Side-Angle-Side) axiom, \( \triangle BAD \cong \triangle CAE \). Congruent triangles have identical shape and size.
(iii) To prove AC bisects \( \angle BCE \):
Since \( \triangle BAD \cong \triangle CAE \) (proved in part (ii)), their corresponding parts are equal (c.p.c.t.).

\( \implies BD = CE \) and \( \angle B = \angle ACE \).
Also, in \( \triangle ABC \), since \( AB = AC \), we know that \( \angle B = \angle ACB \).
Therefore, \( \angle ACE = \angle ACB \).
This means that AC divides \( \angle BCE \) into two equal angles, which proves that AC bisects \( \angle BCE \).
In simple words: We started by showing that two angles are equal by subtracting a common part. Then, using two pairs of equal sides and the equal angles between them, we proved that triangle BAD and triangle CAE are identical. Because these triangles are identical, their matching angles must be equal, which helped us show that line AC cuts angle BCE exactly in half.

🎯 Exam Tip: When angles are subtracted or added, ensure you clearly name the angles. For congruence, meticulously list the three corresponding equal parts and the correct axiom. For bisection, demonstrate that the line divides the angle into two equal parts.

Question 9. In the given figure, which is not drawn accurately, \( \angle \)BAD = 59°, \( \angle \)DAC = 32° and AD = BD. of the angle ACB and state, giving reasons, which is greater, BD or DC.

Answer: In the given figure, \( \angle BAD = 59° \), \( \angle DAC = 32° \), and \( AD = BD \). We need to find \( \angle ACB \) and compare BD and DC.
(i) To find \( \angle ACB \):
In \( \triangle ABD \), we are given \( AD = BD \). This means \( \triangle ABD \) is an isosceles triangle.

\( \implies \angle ABD = \angle BAD \).
Since \( \angle BAD = 59° \), then \( \angle ABD = 59° \). So, \( \angle B = 59° \).
Now, consider \( \triangle ABC \). The sum of angles in a triangle is 180°.
\( \angle A = \angle BAD + \angle DAC = 59° + 32° = 91° \).
So, \( \angle A + \angle B + \angle C = 180° \).
\( 91° + 59° + \angle C = 180° \).
\( 150° + \angle C = 180° \).

\( \implies \angle C = 180° - 150° = 30° \).
Therefore, \( \angle ACB = 30° \). This applies the fundamental angle sum property of triangles.
(ii) To state which is greater, BD or DC:
In \( \triangle ADC \), we have \( \angle DAC = 32° \).
We just found \( \angle C = \angle ACD = 30° \).
To find \( \angle ADC \), use the angle sum property in \( \triangle ADC \):
\( \angle ADC = 180° - (\angle DAC + \angle ACD) = 180° - (32° + 30°) = 180° - 62° = 118° \).
Now, we compare the side lengths in \( \triangle ADC \). In a triangle, the side opposite a larger angle is longer.
Angle opposite to DC is \( \angle DAC = 32° \).
Angle opposite to AD is \( \angle ACD = 30° \).
Since \( \angle DAC > \angle ACD \) (32° > 30°), the side opposite \( \angle DAC \) must be greater than the side opposite \( \angle ACD \).

\( \implies DC > AD \).
We are given that \( AD = BD \).

\( \implies DC > BD \).
Thus, DC is greater than BD. This demonstrates the relationship between angles and side lengths in a triangle.
In simple words: First, we found angle B using the rule that angles opposite equal sides in a triangle are equal. Then, we found angle C in the big triangle ABC by subtracting the sum of angles A and B from 180 degrees. Next, in the smaller triangle ADC, we compared the angles opposite to sides AD and DC. The side opposite the bigger angle is longer, so DC is longer than AD. Since AD is the same length as BD, it means DC is also longer than BD.

🎯 Exam Tip: For problems involving side comparisons, always find all angles in the relevant triangle. The rule "the side opposite the greater angle is greater" is crucial. Remember to use given conditions like isosceles triangles to deduce angles or sides.

Question 10. A square ABCD and an equilateral triangle AXB are on the same base AB and on the same side of it. What is the size of the angle CBX? Calculate the angle CXB. The lines AC and BX cut each other at M. Prove that CX is equal to CM.

Answer: We have a square ABCD and an equilateral triangle AXB on the same base AB and on the same side. AC and BX cut each other at M. We need to find \( \angle CBX \), \( \angle CXB \) and prove \( CX = CM \).
In equilateral \( \triangle ABX \), each angle is 60°. So, \( \angle ABX = 60° \).
In square ABCD, each angle is 90°. So, \( \angle ABC = 90° \).
Now, we can find \( \angle CBX \):
\( \angle CBX = \angle ABC - \angle ABX = 90° - 60° = 30° \). This calculation is straightforward.
Next, we need to find \( \angle CXB \).
Since ABCD is a square, \( BC = AB \).
Since \( \triangle ABX \) is equilateral, \( AB = BX \).

\( \implies BC = BX \). So, \( \triangle BCX \) is an isosceles triangle.
In \( \triangle BCX \), \( \angle CXB = \angle BCX \).
The sum of angles in \( \triangle BCX \) is 180°.
So, \( \angle CBX + \angle CXB + \angle BCX = 180° \).
\( 30° + \angle CXB + \angle CXB = 180° \).
\( 30° + 2\angle CXB = 180° \).
\( 2\angle CXB = 180° - 30° = 150° \).

\( \implies \angle CXB = \frac{150°}{2} = 75° \). This applies isosceles triangle properties.
Finally, we need to prove that \( CX = CM \).
In a square, the diagonal AC bisects \( \angle C \). So \( \angle ACB = \frac{1}{2} \angle BCD = \frac{1}{2} (90°) = 45° \).
Now, consider \( \triangle BCX \). We found \( \angle BCX = 75° \).
Let's find \( \angle MCX \). \( \angle MCX = \angle BCX - \angle BCA = 75° - 45° = 30° \).
Now consider \( \triangle CMX \).
We know \( \angle CXB = 75° \), so \( \angle CXM = 75° \).
We just found \( \angle MCX = 30° \).
The sum of angles in \( \triangle CMX \) is 180°.
So, \( \angle CMX = 180° - (\angle CXM + \angle MCX) = 180° - (75° + 30°) = 180° - 105° = 75° \).
In \( \triangle CMX \), we have \( \angle CMX = 75° \) and \( \angle CXM = 75° \).
Since two angles are equal, the sides opposite those angles must also be equal.

\( \implies CX = CM \). This proves the last part by identifying an isosceles triangle.
In simple words: First, we used the angles of a square and an equilateral triangle to find angle CBX. Then, because triangle BCX has two equal sides, we found angle CXB. Lastly, we looked at triangle CMX. By calculating its angles, we found that two of its angles are equal, which means the sides opposite those angles (CX and CM) must also be equal.

🎯 Exam Tip: Break down complex figures into simpler triangles. Use properties of squares (all sides equal, angles 90°, diagonals bisect angles) and equilateral triangles (all sides equal, angles 60°). Look for isosceles triangles to find equal angles or sides.

Question 11. (i) In the figure, ABCD is an isosceles trapezium. Find the values of x and y.
(ii) Given, AC = 5x + 3 and BD = 9x – 12, find the value of x so that ABCD is isosceles.

Answer:
(i) In the given figure, ABCD is an isosceles trapezium, and \( AB \parallel DC \). This means \( AD = BC \).
In an isosceles trapezium, consecutive interior angles between parallel sides are supplementary. The sum of co-interior angles is 180°.
So, \( \angle A + \angle D = 180° \).
From the figure, \( \angle A = 70° \) and \( \angle D \) is given as \( 2x \).
\( 2x + 70° = 180° \).
\( 2x = 180° - 70° = 110° \).

\( \implies x = \frac{110°}{2} = 55° \).
Also, in an isosceles trapezium, the base angles are equal. So, \( \angle D = \angle C \) and \( \angle A = \angle B \).
We know \( \angle C = y \) from the figure.
So, \( \angle D = \angle C \implies 2x = y \).
Substitute the value of \( x \): \( 2(55°) = y \).

\( \implies y = 110° \).
So, the values are \( x = 55° \) and \( y = 110° \). These properties define an isosceles trapezium.
(ii) Given \( AC = 5x + 3 \) and \( BD = 9x - 12 \). ABCD is an isosceles trapezium.
In an isosceles trapezium, the diagonals are equal in length.
So, \( AC = BD \).
\( 5x + 3 = 9x - 12 \).
To solve for \( x \), move \( 5x \) to the right side and \( -12 \) to the left side:
\( 3 + 12 = 9x - 5x \).
\( 15 = 4x \).

\( \implies x = \frac{15}{4} \).

\( \implies x = 3.75 \). This is a direct application of the property that diagonals are equal.
In simple words: For the first part, we used the rules of an isosceles trapezium: angles between parallel lines add up to 180 degrees, and the base angles are equal. This helped us find x and y. For the second part, we used another rule for isosceles trapeziums: their diagonals are equal in length. We set the given expressions for the diagonals equal to each other and solved for x.

🎯 Exam Tip: Remember the two key properties of an isosceles trapezium: (1) Base angles are equal (\( \angle A = \angle B \) and \( \angle D = \angle C \)), (2) Diagonals are equal (\( AC = BD \)), and (3) Consecutive interior angles between parallel sides are supplementary. Use these directly in calculations.

Question 11. (iii) Find SR.

Answer: In the given figure, PQRS is a trapezium where \( PQ \parallel SR \). M and N are the midpoints of SP and RQ respectively.
The length of the line segment joining the midpoints of the non-parallel sides of a trapezium is half the sum of the lengths of the parallel sides.
So, \( MN = \frac{1}{2} (PQ + SR) \).
We are given \( MN = 35 \), \( PQ = 40 \). Let \( SR = x \).
\( 35 = \frac{1}{2} (40 + x) \).
Multiply both sides by 2: \( 35 \times 2 = 40 + x \).
\( 70 = 40 + x \).
Subtract 40 from both sides: \( x = 70 - 40 \).

\( \implies x = 30 \).
Therefore, \( SR = 30 \). This property simplifies finding the length of a parallel side.
In simple words: For a trapezium, if you draw a line connecting the middle points of the two non-parallel sides, the length of that line is exactly half the sum of the lengths of the two parallel sides. We used this rule, plugging in the numbers we knew, to find the length of side SR.

🎯 Exam Tip: Remember the midpoint theorem for trapeziums: the segment joining the midpoints of the non-parallel sides is parallel to the parallel sides and its length is the average of the lengths of the parallel sides. This is a common formula to apply.

Question 12. In the figure, ABCD is a kite whose diagonals intersect at O. If \( \angle \)BAD =110° and \( \angle \)BCD = 50°, find
(i) \( \angle \)OBC
(ii) \( \angle \)ODA

Answer: In the figure, ABCD is a kite, and its diagonals intersect at O. We are given \( \angle BAD = 110° \) and \( \angle BCD = 50° \). We need to find \( \angle OBC \) and \( \angle ODA \).
Properties of a kite: 1. Two pairs of adjacent sides are equal (\( AB = BC \) and \( AD = CD \)). 2. One diagonal (AC) is the perpendicular bisector of the other diagonal (BD). 3. The diagonal AC bisects the angles \( \angle A \) and \( \angle C \).
(i) To find \( \angle OBC \):
Since AC bisects \( \angle C \), \( \angle BCO = \angle DCO \).
\( \angle BCO = \frac{\angle BCD}{2} = \frac{50°}{2} = 25° \).
Also, the diagonals of a kite intersect at right angles, so \( \angle BOC = 90° \).
Now, consider \( \triangle OBC \). The sum of angles in a triangle is 180°.
\( \angle OBC + \angle BCO + \angle BOC = 180° \).
\( \angle OBC + 25° + 90° = 180° \).
\( \angle OBC + 115° = 180° \).

\( \implies \angle OBC = 180° - 115° = 65° \). This uses angle sum and perpendicular diagonals.
(ii) To find \( \angle ODA \):
Since AC bisects \( \angle A \), \( \angle BAO = \angle DAO \).
\( \angle DAO = \frac{\angle BAD}{2} = \frac{110°}{2} = 55° \).
In \( \triangle AOD \), we know \( \angle AOD = 90° \) (diagonals are perpendicular).
The sum of angles in \( \triangle AOD \) is 180°.
\( \angle ODA + \angle DAO + \angle AOD = 180° \).
\( \angle ODA + 55° + 90° = 180° \).
\( \angle ODA + 145° = 180° \).

\( \implies \angle ODA = 180° - 145° = 35° \). This applies the same logic to the other triangle.
In simple words: For a kite, one diagonal cuts the angles at its ends in half, and the diagonals cross each other at a perfect right angle. We used these rules to find the angles needed in the two triangles formed by the diagonals.

🎯 Exam Tip: Crucial properties of a kite are that one diagonal is the perpendicular bisector of the other and bisects two of the angles. Opposite angles are not necessarily equal, but angles between unequal sides are equal. Use these facts to set up angle sum equations in the right-angled triangles formed by the diagonals.

Question 13. Without using set square or protractor, contruct a rhombus in which the diagonals are of length 10 cm and 8 cm and measure one of its sides.

Answer: We need to construct a rhombus with diagonals of 10 cm and 8 cm. In a rhombus, the diagonals bisect each other at right angles.
Steps of construction:
(i) Draw a line segment AC = 10 cm. This will be one diagonal.
(ii) Draw the perpendicular bisector of AC. Let this bisector be XY, intersecting AC at point O. This point O is the midpoint of AC.
(iii) Since the diagonals bisect each other, half of the other diagonal (8 cm) is \( \frac{8}{2} = 4 \) cm. Mark points B and D on the perpendicular bisector XY such that \( OB = OD = 4 \) cm. Make sure B and D are on opposite sides of AC.
(iv) Join points A to B, B to C, C to D, and D to A. The figure ABCD is the required rhombus.
To measure one of its sides (e.g., AB), use a ruler. You will find that \( AB \approx 6.4 \) cm. (Using Pythagoras theorem in \( \triangle AOB \), \( AB^2 = AO^2 + OB^2 = 5^2 + 4^2 = 25 + 16 = 41 \), so \( AB = \sqrt{41} \approx 6.40 \) cm). The construction process allows you to visually create this specific geometric shape.
In simple words: First, draw a line 10 cm long. Find its exact middle and draw a line perfectly straight up and down through it. On this vertical line, mark two points, each 4 cm away from the middle point, one above and one below. Then, connect these four points with straight lines to form the rhombus. Finally, measure one of its sides with a ruler.

🎯 Exam Tip: Clearly state each step of construction. Remember that diagonals of a rhombus bisect each other at right angles – this is the crucial property for construction. Always indicate the final measurement requested.

Question 14. Draw parallelogram ABCD with the following data: AB = 6 cm, AD = 5 cm and \( \angle \)DAB = 45°. Let AC and DB meet in O and let E be the mid-point of BC. Join OE. Prove that
(i) OE \( \parallel \) AB
(ii) OE = \( \frac{1}{2} \) AB

Answer: First, let's list the steps to draw the parallelogram:
(i) Draw a line segment AB = 6 cm.
(ii) At point A, draw a ray AX making an angle of 45° with AB. Cut off AD = 5 cm on this ray.
(iii) With D as the center and radius 6 cm, draw an arc. With B as the center and radius 5 cm, draw another arc. Let these arcs intersect at point C.
(iv) Join DC and BC. ABCD is the required parallelogram.
Now, draw diagonals AC and BD intersecting at O. Let E be the midpoint of BC. Join OE.
In \( \triangle ABC \), O is the midpoint of AC (since diagonals of a parallelogram bisect each other), and E is the midpoint of BC (given).
According to the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
(i) To prove \( OE \parallel AB \):
Since O and E are midpoints of AC and BC respectively in \( \triangle ABC \), then by the Midpoint Theorem, \( OE \parallel AB \). This theorem is a direct consequence of geometric principles.
(ii) To prove \( OE = \frac{1}{2} AB \):
Similarly, by the Midpoint Theorem, the length of OE is half the length of the third side AB. So, \( OE = \frac{1}{2} AB \). This property holds true for any triangle.
In simple words: First, we drew the parallelogram using the given measurements. Then, we used a special rule called the Midpoint Theorem. This rule says that if you connect the middle points of two sides of a triangle (like O on AC and E on BC in triangle ABC), that connecting line (OE) will be parallel to the third side (AB) and also half its length.

🎯 Exam Tip: Clearly state the steps for construction. For the proof, correctly identify the triangle and the midpoints involved. Then directly apply the Midpoint Theorem, stating both parts (parallelism and half-length) to get full marks.

Question 15. In the figure, ABCD is a parallelogram in which angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are at right angles. Prove that XA = YC.

Answer: In the figure, ABCD is a parallelogram, with \( \angle A \) and \( \angle C \) being obtuse. X and Y are points on diagonal BD such that \( \angle XAD = 90° \) and \( \angle YCB = 90° \). We need to prove that \( XA = YC \).
To prove \( XA = YC \), we will show that \( \triangle ADX \) and \( \triangle CBY \) are congruent.
Let's consider \( \triangle ADX \) and \( \triangle CBY \):
(1) \( AD = BC \) (Opposite sides of a parallelogram are equal)
(2) \( \angle XAD = \angle YCB = 90° \) (Given that they are right angles)
(3) \( \angle ADX = \angle CBY \) (These are alternate interior angles because \( AD \parallel BC \) and BD is a transversal.)
So, by the ASA (Angle-Side-Angle) axiom, \( \triangle ADX \cong \triangle CBY \). This demonstrates the congruence of the two triangles.

\( \implies \) Because the triangles are congruent, their corresponding parts are equal (c.p.c.t.).

\( \implies XA = YC \). This shows that the two line segments are equal.
In simple words: We looked at two triangles, ADX and CBY. We know that in a parallelogram, opposite sides are equal. We were also told that two specific angles are 90 degrees. Finally, we identified that another pair of angles are equal because they are alternate interior angles when parallel lines are cut by a diagonal. Since we had two angles and a side between them equal in both triangles, it meant the triangles were identical. Therefore, their matching sides, XA and YC, must also be equal.

🎯 Exam Tip: For proofs in parallelograms, remember that opposite sides are parallel and equal, and opposite angles are equal. Identify transversals to find alternate interior angles. The ASA congruence criterion is often useful when you have angles and an included side.

Question 16. The sides PQ, PR of a triangle PQR are equal, and S, T are points on PR, PQ such that \( \angle \)PSQ and \( \angle \)PTR are right angles. Prove that the triangles PTR and PSQ are congruent. If QS and RT intersect at X, prove that the triangles PTX and PSX are congruent.

Answer: In \( \triangle PQR \), sides \( PQ = PR \). S is a point on PR, and T is a point on PQ such that \( \angle PSQ = 90° \) and \( \angle PTR = 90° \). QS and RT intersect at X. PX is joined.
Part 1: Prove \( \triangle PTR \cong \triangle PSQ \).
Let's consider \( \triangle PTR \) and \( \triangle PSQ \):
(1) \( \angle PTR = \angle PSQ = 90° \) (Given that they are right angles)
(2) \( \angle P = \angle P \) (Common angle to both triangles)
(3) \( PR = PQ \) (Given that \( \triangle PQR \) is an isosceles triangle)
So, by the AAS (Angle-Angle-Side) axiom, \( \triangle PTR \cong \triangle PSQ \). This is a direct application of the congruence criterion.

\( \implies \) Since the triangles are congruent, their corresponding parts are equal (c.p.c.t.).

\( \implies PT = PS \). This equality will be useful for the next part.
Part 2: Prove \( \triangle PTX \cong \triangle PSX \).
Let's consider \( \triangle PTX \) and \( \triangle PSX \):
(1) \( PT = PS \) (Proved above in Part 1)
(2) \( \angle PTX = \angle PSX = 90° \) (From \( \angle PTR = 90° \) and \( \angle PSQ = 90° \), these are parts of those angles.)
(3) \( PX = PX \) (Common side to both triangles)
So, by the SAS (Side-Angle-Side) axiom, \( \triangle PTX \cong \triangle PSX \). The presence of a common side and included angle is key here.
In simple words: First, we proved two larger triangles (PTR and PSQ) are identical because they share an angle, have a right angle each, and one matching side is equal. This helped us confirm that the sides PT and PS are equal. Then, we used this new fact along with another common side (PX) and the right angles to show that two smaller triangles (PTX and PSX) are also identical.

🎯 Exam Tip: When proving congruence, use the correct axiom (AAS, SAS, etc.). Clearly list the corresponding parts. Remember that if two triangles are congruent, their corresponding parts (sides and angles) are equal (c.p.c.t.). Identify common sides or angles carefully.

Question 18. In the rhombus PQRS the side PQ = 17 cm and the diagonal PR = 16 cm. Calculate the area of the rhombus.
Answer: In a rhombus PQRS, the side length PQ is 17 cm and one diagonal PR is 16 cm. We know that the diagonals of a rhombus bisect each other at right angles. Let O be the point where the diagonals intersect.
So, PO = \( \frac{1}{2} \)PR = \( \frac{16}{2} = 8 \) cm.
Now, consider the right-angled triangle ΔPOQ. We can use the Pythagoras Theorem:
PQ\( ^2 \) = PO\( ^2 \) + OQ\( ^2 \)
\( 17^2 = 8^2 + OQ^2 \)
\( 289 = 64 + OQ^2 \)
\( OQ^2 = 289 - 64 \)
\( OQ^2 = 225 \)
\( OQ = \sqrt{225} = 15 \) cm.
The length of the other diagonal QS is twice the length of OQ, since O is the midpoint of QS.
QS = 2 × OQ = 2 × 15 = 30 cm.
The area of a rhombus is calculated using the formula: Area = \( \frac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.
Area = \( \frac{1}{2} \times \) PR \( \times \) QS
Area = \( \frac{1}{2} \times 16 \times 30 \)
Area = \( 8 \times 30 \)
Area = 240 cm\( ^2 \).
In simple words: First, use the fact that rhombus diagonals cut each other in half at a right angle to find half of the other diagonal using the Pythagoras rule. Then, double that half to get the full length of the second diagonal. Finally, use the formula (half times diagonal 1 times diagonal 2) to find the total area.

🎯 Exam Tip: Always remember that the diagonals of a rhombus bisect each other perpendicularly. This allows you to use the Pythagorean theorem to find the missing parts of the diagonals before calculating the area.

Question 19. In the given figure, not drawn to scale, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm, RQ = 9 cm. Calculate the length of PR.
Answer: We need to find the length of PR using the given information and the figure. The figure shows a right-angled triangle ΔPQS at S and then a larger triangle ΔPRS also at S.
First, consider the right-angled triangle ΔPQS, where ∠PSR = 90° (which means ∠PSQ = 90°).
Using the Pythagoras Theorem:
PQ\( ^2 \) = PS\( ^2 \) + QS\( ^2 \)
\( 10^2 = PS^2 + 6^2 \)
\( 100 = PS^2 + 36 \)
\( PS^2 = 100 - 36 \)
\( PS^2 = 64 \)
\( PS = \sqrt{64} = 8 \) cm.
Next, we find the length of RS. From the figure, R, Q, and S are collinear points such that Q is between R and S. The given lengths are RQ = 9 cm and QS = 6 cm.
So, RS = RQ + QS = 9 + 6 = 15 cm.
Now, consider the larger right-angled triangle ΔPRS, where ∠PSR = 90°.
Using the Pythagoras Theorem again:
PR\( ^2 \) = RS\( ^2 \) + PS\( ^2 \)
\( PR^2 = 15^2 + 8^2 \)
\( PR^2 = 225 + 64 \)
\( PR^2 = 289 \)
\( PR = \sqrt{289} = 17 \) cm.
In simple words: First, use the Pythagoras rule in the smaller triangle to find the length of PS. Then, add the lengths RQ and QS to find the total length of RS. Finally, use the Pythagoras rule in the bigger triangle (PRS) with PS and RS to find PR.

🎯 Exam Tip: When a problem involves multiple right-angled triangles sharing a side, use the Pythagorean theorem in stages. Solve for a common side first, then use it in the larger triangle.

Question 20. In the given figure, AD is the median and DE || AB. Prove that BE is the median.
Answer: In the triangle ABC, AD is a median, which means that D is the midpoint of the side BC. This separates BC into two equal parts.
We are given that DE is parallel to AB (DE || AB).
Now, let's look at triangle ABC. We have AD as a median, making D the midpoint of BC. We also have a line segment DE starting from D and parallel to AB.
According to the converse of the Midpoint Theorem, if a line is drawn through the midpoint of one side of a triangle and is parallel to another side, then it bisects the third side. In this case, the line DE starts from D (the midpoint of BC) and is parallel to AB (another side of the triangle). This means that DE must bisect the third side, AC.
Therefore, E is the midpoint of AC.
If E is the midpoint of AC, then the line segment BE connects a vertex (B) to the midpoint of the opposite side (E on AC). By definition, such a line is a median of the triangle.
Hence, BE is a median of triangle ABC.
In simple words: Since AD is a median, D is the middle point of BC. Because DE is parallel to AB, and starts from the middle of BC, it must also cut AC exactly in the middle. If E is the middle of AC, then BE is also a median.

🎯 Exam Tip: To prove a line is a median, you must show that it connects a vertex to the midpoint of the opposite side. The Midpoint Theorem and its converse are very useful for this.

Question 21. Find:
(i) m\( \overparen{\mathrm{CDE}} \)
(ii) mRS if RS = PQ
(iii) AB
Answer:
(i) From the first figure, AD and BE are diameters of the circle with center O. We are given ∠AOB = 18°. Vertically opposite angles are equal, so ∠DOE = ∠AOB = 18°. Thus, the measure of arc DE is 18°. Also, the solution implies that ∠COD = 90°. A central angle's measure is equal to the measure of its intercepted arc. So, m\( \overparen{\mathrm{CD}} \) = ∠COD = 90°.
Now, m\( \overparen{\mathrm{CDE}} \) = m\( \overparen{\mathrm{CD}} \) + m\( \overparen{\mathrm{DE}} \) = 90° + 18° = 108°. The measure of an arc is equal to its central angle. This means the arc from C through D to E is 108 degrees.
(ii) From the second figure, RS and PQ are chords of a circle. We are given that RS = PQ. In a circle, if two chords are equal, then their corresponding arcs are also equal.
So, m\( \overparen{\mathrm{RS}} \) = m\( \overparen{\mathrm{PQ}} \).
We are given m\( \overparen{\mathrm{PQ}} \) = (2x + 27)° and m\( \overparen{\mathrm{RS}} \) = 3x°.
Therefore, \( 3x = 2x + 27 \).
Subtracting 2x from both sides gives \( x = 27° \).
So, m\( \overparen{\mathrm{RS}} \) = \( 3x = 3 \times 27° = 81° \).
(iii) From the third figure, O is the center of the circle, OP is the radius, and AB is a chord. OP is perpendicular to AB at Q. We are given OQ = 5 and QP = 8.
The radius of the circle is OP = OQ + QP = 5 + 8 = 13 cm. So, OA = OB = 13 cm.
In the right-angled triangle ΔOQB (since OP \( \perp \) AB), we can use the Pythagoras Theorem:
OB\( ^2 \) = OQ\( ^2 \) + QB\( ^2 \)
\( 13^2 = 5^2 + QB^2 \)
\( 169 = 25 + QB^2 \)
\( QB^2 = 169 - 25 \)
\( QB^2 = 144 \)
\( QB = \sqrt{144} = 12 \) cm.
Since the perpendicular from the center to a chord bisects the chord, AB = 2 × QB.
AB = 2 × 12 = 24 cm.
In simple words: (i) Use vertical angles to find one arc, and assume the given angle to find the other arc, then add them. (ii) Since the chords are equal, their arcs must be equal, so set the arc expressions equal to find x, then find the arc measure. (iii) Find the full radius of the circle. Then, use the Pythagoras rule in the small right triangle to find half of the chord. Finally, double that length to get the full chord.

🎯 Exam Tip: For circle problems, remember key theorems: vertically opposite angles, equal chords subtend equal arcs, and the perpendicular from the center to a chord bisects the chord. These are frequently used with Pythagoras theorem.

Question 22. Determine whether each statement is true or false. If false, explain why.
(i) The central angle of a minor arc is an acute angle.
(ii) Any two points on a circle determine a minor arc and a major arc.
(iii) In a circle, the perpendicular bisector of a chord must pass through the centre of the circle,
Answer:
(i) This statement is False. A minor arc is an arc that is smaller than a semicircle, meaning its measure is less than 180°. The central angle of a minor arc can be any angle between 0° and 180°. For example, it could be 90° (a right angle) or 120° (an obtuse angle), both of which correspond to a minor arc.
(ii) This statement is False. If the two points on the circle happen to be the endpoints of a diameter, then they divide the circle into two semicircles. These two arcs have equal measures (180° each), so neither is a minor arc nor a major arc. They are simply two equal arcs.
(iii) This statement is True. This is a fundamental theorem in circle geometry. The perpendicular bisector of any chord always passes through the center of the circle. This property is often used in constructions to find the center of a circle.
In simple words: (i) False, because a minor arc's central angle can be bigger than 90 degrees, like 100 degrees. (ii) False, because if the two points are exactly opposite on the circle, they make two equal half-circles, not one big and one small arc. (iii) True, this is a fixed rule for all circles.

🎯 Exam Tip: Clearly define and understand terms like 'minor arc,' 'major arc,' 'central angle,' and 'perpendicular bisector' to accurately evaluate such statements. Provide a counter-example for false statements.

Question 23. Consider the following statements in respect of any triangle.
I. The three medians of a triangle divide it into six triangles of equal area.
II. The perimeter of a triangle is greater than the sum of the lengths of its three medians.
Which of the statements given above is/are correct?
(a) I only
(b) II only
(c) Both I and II
(d) Neither I nor II
Answer: (c) Both I and II
Both statements are correct.
I. The medians of a triangle intersect at a single point called the centroid. The centroid divides the triangle into six smaller triangles, and all these six smaller triangles have equal areas. This property is an important result in geometry related to the medians. Each of these smaller triangles will have an area equal to one-sixth of the total area of the original triangle.
II. This is a known triangle inequality: the sum of the lengths of the three sides (perimeter) of any triangle is always greater than the sum of the lengths of its three medians. This means the outer boundary of the triangle is longer than all the internal medians added up.
In simple words: Statement I is true because the three lines that go from each corner to the middle of the opposite side (called medians) cut the triangle into six smaller parts that all have the same area. Statement II is also true because if you add up the lengths of all three sides of a triangle, the total length will always be more than if you add up the lengths of the three medians.

🎯 Exam Tip: Remember these fundamental properties of medians: they divide the triangle into six equal-area triangles, and the perimeter of the triangle is always greater than the sum of its medians.

Question 24. The three sides of a triangle are 10, 100 and x. Which one of the following is correct?
(a) 10 < x < 100
(b) 90 < x < 110
(c) 90 \( \le \) x \( \le \) 100
(d) 90 \( \le \) x < 110
Answer: (b) 90 < x < 110
To form a triangle with side lengths 10, 100, and x, we must follow the Triangle Inequality Theorem. This theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
Let's apply this rule to the given sides:
1. Sum of 10 and x must be greater than 100:
\( 10 + x > 100 \)
Subtract 10 from both sides: \( x > 90 \)
2. Sum of 10 and 100 must be greater than x:
\( 10 + 100 > x \)
\( 110 > x \), which can be written as \( x < 110 \)
3. Sum of x and 100 must be greater than 10:
\( x + 100 > 10 \)
Subtract 100 from both sides: \( x > -90 \). Since x is a side length, it must be positive anyway, so this condition is automatically satisfied by \( x > 90 \).
Combining the conditions \( x > 90 \) and \( x < 110 \), we get the range for x as \( 90 < x < 110 \). This ensures that a valid triangle can be constructed.
In simple words: For three side lengths to make a triangle, any two sides added together must be longer than the third side. Applying this rule to 10, 100, and x, we find that x must be bigger than 90 but smaller than 110.

🎯 Exam Tip: Always use the strict inequality (>) for the Triangle Inequality Theorem, as sides cannot be equal to the sum of the other two, nor can they be negative. Test all three possible pairs of sums.

Question 25.
(i) The median BD of the \( \triangle \)ABC meets AC at D. if BD = \( \frac{1}{2} \)AC, then which one of the following is correct?
(a) ∠ACB = 1 right angle
(b) ∠BAC = 1 right angle
(c) ∠ABC = 1 right angle
(d) None of these
(ii) ABC is an isosceles triangle such that AB = BC = 8 cm and ∠ABC = 90°. What is the length of the perpendicular drawn from B on AC?
(a) 4 cm
(b) \( 4\sqrt{2} \) cm
(c) \( 2\sqrt{2} \) cm
(d) 2 cm
Answer:
(i) The correct option is (c) ∠ABC = 1 right angle.
Given that BD is the median to AC, this means D is the midpoint of AC. Therefore, AD = DC = \( \frac{1}{2} \)AC.
We are also given that BD = \( \frac{1}{2} \)AC.
From these facts, we can conclude that AD = DC = BD. This means that the distance from the vertex B to the midpoint D of the opposite side AC is equal to half the length of AC itself.
In triangle ADB, since AD = BD, the angles opposite these sides are equal: ∠BAD = ∠ABD.
In triangle BDC, since DC = BD, the angles opposite these sides are equal: ∠DBC = ∠BCD.
Now, consider the angle at vertex B in the main triangle ABC: ∠ABC = ∠ABD + ∠DBC.
Substituting the equal angles, we get ∠ABC = ∠BAD + ∠BCD. Let's call ∠BAD as A and ∠BCD as C.
So, in ΔABC, the sum of angles is 180°: ∠BAC + ∠BCA + ∠ABC = 180°
This can be written as A + C + (A + C) = 180°
\( 2(A + C) = 180° \)
\( A + C = 90° \).
Since ∠ABC = A + C, it means ∠ABC = 90°. Therefore, triangle ABC is a right-angled triangle with the right angle at B. This is a specific property: if a median to a side is half the length of that side, then the angle opposite that side is a right angle.
(ii) The correct option is (b) \( 4\sqrt{2} \) cm.
Given triangle ABC is an isosceles right-angled triangle, with AB = BC = 8 cm and ∠ABC = 90°.
First, let's find the length of the hypotenuse AC using the Pythagoras Theorem in ΔABC:
AC\( ^2 \) = AB\( ^2 \) + BC\( ^2 \)
AC\( ^2 \) = \( 8^2 + 8^2 \)
AC\( ^2 \) = \( 64 + 64 \)
AC\( ^2 \) = 128
AC = \( \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2} \) cm.
Now, let BD be the perpendicular drawn from B to AC. We can find the length of BD by using the area of the triangle. The area of a right-angled triangle can be calculated in two ways:
Area = \( \frac{1}{2} \times \text{product of legs} \) = \( \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 8 \times 8 = 32 \) cm\( ^2 \).
Also, Area = \( \frac{1}{2} \times \text{base} \times \text{height} \) = \( \frac{1}{2} \times AC \times BD \).
Equating the two area expressions:
\( \frac{1}{2} \times AC \times BD = 32 \)
\( \frac{1}{2} \times 8\sqrt{2} \times BD = 32 \)
\( 4\sqrt{2} \times BD = 32 \)
\( BD = \frac{32}{4\sqrt{2}} = \frac{8}{\sqrt{2}} \).
To rationalize the denominator, multiply numerator and denominator by \( \sqrt{2} \):
\( BD = \frac{8\sqrt{2}}{2} = 4\sqrt{2} \) cm.
In simple words: (i) If the line from a corner to the middle of the opposite side (median) is exactly half the length of that side, then the angle at that corner must be 90 degrees. (ii) First, use the Pythagoras rule to find the length of the longest side (AC). Then, use the triangle area formula in two ways: once with the given sides and once with the longest side and the unknown perpendicular height. Set these two area calculations equal to each other to find the perpendicular height.

🎯 Exam Tip: (i) Memorize the property that if a median to a side is half the length of that side, the angle opposite that side is 90°. (ii) For finding altitude in a right-angled triangle, often calculating the area in two ways (using different base-height pairs) is the most efficient method.

Question 26. A ladder 34 m long is placed in a lane so as to reach window 30 m high and on turning the ladder to the other side of the lane and keeping its foot at the same place, reaches a point 16 m high. What is the breadth of the lane?
(a) 18 m
(b) 40 m
(c) 46 m
(d) 50 m
Answer: (c) 46 m
Imagine the ladder forming two right-angled triangles, with its foot as the common vertex at the bottom of the lane.
The ladder itself is the hypotenuse in both triangles, with a length of 34 m.
Let's find the width of the lane on each side of the ladder's foot.
**Side 1:**
Height of the window = 30 m (this is one leg of the right triangle).
Ladder length = 34 m (hypotenuse).
Let the base length on this side be LE. Using the Pythagoras Theorem:
\( \text{Ladder}^2 = \text{Height}^2 + LE^2 \)
\( 34^2 = 30^2 + LE^2 \)
\( 1156 = 900 + LE^2 \)
\( LE^2 = 1156 - 900 \)
\( LE^2 = 256 \)
\( LE = \sqrt{256} = 16 \) m.
**Side 2:**
Height reached = 16 m (this is one leg of the second right triangle).
Ladder length = 34 m (hypotenuse).
Let the base length on this side be BL. Using the Pythagoras Theorem:
\( \text{Ladder}^2 = \text{Height}^2 + BL^2 \)
\( 34^2 = 16^2 + BL^2 \)
\( 1156 = 256 + BL^2 \)
\( BL^2 = 1156 - 256 \)
\( BL^2 = 900 \)
\( BL = \sqrt{900} = 30 \) m.
The total breadth of the lane is the sum of the two base lengths (LE and BL).
Breadth of the lane = LE + BL = 16 m + 30 m = 46 m.
In simple words: Think of the ladder and the wall making a triangle on each side of the lane. Use the Pythagoras rule to find the ground distance for each side. The total width of the lane is simply these two ground distances added together.

🎯 Exam Tip: Visualize the problem as two distinct right-angled triangles sharing the ladder as a common hypotenuse. Carefully apply the Pythagorean theorem for each triangle and then sum the resulting base lengths.

Question 27. ABCD is a square in which AO = AX. What is ∠XOB?
(a) 22.5°
(b) 25°
(c) 30°
(d) 45°
Answer: (a) 22.5°
In a square ABCD, the diagonals bisect each other at right angles at point O. They also bisect the vertex angles of the square.
Therefore:
1. ∠AOB = 90° (diagonals are perpendicular).
2. ∠OAB = \( \frac{1}{2} \)∠DAB = \( \frac{1}{2} \times 90° = 45° \) (diagonals bisect vertex angles).
We are given that AO = AX. This means that ΔAOX is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are equal. So, ∠AOX = ∠AXO.
The sum of angles in ΔAOX is 180°:
∠OAX + ∠AOX + ∠AXO = 180°
Since ∠OAX is the same as ∠OAB, which is 45°:
\( 45° + ∠AOX + ∠AOX = 180° \)
\( 45° + 2∠AOX = 180° \)
\( 2∠AOX = 180° - 45° \)
\( 2∠AOX = 135° \)
\( ∠AOX = \frac{135°}{2} = 67.5° \).
Finally, we need to find ∠XOB. From the figure, we can see that ∠AOB is formed by ∠AOX and ∠XOB:
∠AOB = ∠AOX + ∠XOB
\( 90° = 67.5° + ∠XOB \)
\( ∠XOB = 90° - 67.5° \)
\( ∠XOB = 22.5° \).
In simple words: First, know that the lines inside a square (diagonals) cross at a 90-degree angle and cut the square's corners into 45-degree angles. Since AO equals AX, the triangle AOX is special, and two of its angles (AOX and AXO) are equal. Find these equal angles. Then, subtract the angle AOX from the 90-degree angle AOB to find the final angle XOB.

🎯 Exam Tip: This question requires combining properties of squares (diagonals bisect at 90°, bisect angles) and isosceles triangles (angles opposite equal sides are equal). Break down the larger angle into smaller, calculable parts.

Question 28. In the given figure, AB is a diameter of a circle and CD is perpendicular to AB. If AB = 10 cm and AE = 2 cm, then what is the length of ED?
(a) 5 cm
(b) 4 cm
(c) \( \sqrt{10} \) cm
(d) \( \sqrt{20} \) cm
Answer: (b) 4 cm
Given that AB is the diameter of the circle, and AB = 10 cm. The center of the circle, let's call it O, is the midpoint of the diameter.
So, the radius of the circle is OA = OB = OD = \( \frac{\text{AB}}{2} = \frac{10}{2} = 5 \) cm.
We are given AE = 2 cm. Since O is the center and A is an endpoint of the diameter, OA = 5 cm.
The distance from the center O to point E can be found as: OE = OA - AE = 5 cm - 2 cm = 3 cm.
We are also given that CD is perpendicular to AB. This means that in triangle ΔOED, the angle at E, ∠OED, is 90°. So, ΔOED is a right-angled triangle.
In ΔOED, we know the hypotenuse OD (which is the radius) = 5 cm, and one leg OE = 3 cm. We need to find the other leg ED.
Using the Pythagoras Theorem:
OD\( ^2 \) = ED\( ^2 \) + OE\( ^2 \)
\( 5^2 = ED^2 + 3^2 \)
\( 25 = ED^2 + 9 \)
\( ED^2 = 25 - 9 \)
\( ED^2 = 16 \)
\( ED = \sqrt{16} = 4 \) cm.
In simple words: First, figure out where the center of the circle is and what its radius is. Then, use the given lengths to find the distance from the center to point E. Since CD is straight up and down from AB, the triangle OED is a right-angle triangle. Use the Pythagoras rule in this triangle to find the length of ED.

🎯 Exam Tip: In circle problems, always identify the radius. When a perpendicular is involved, look for right-angled triangles to apply the Pythagorean theorem. Remember that the diameter is twice the radius.

Question 29. In the given figure, AD is a straight line. OP \( \perp \) AD and O is the centre of both circles. If OA = 20 cm, OB = 15 cm and OP = 12 cm, what is AB equal to?
(a) 7 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Answer: (a) 7 cm
We have two concentric circles (circles with the same center O). AD is a straight line that acts as a chord for the larger circle and BC for the smaller circle. OP is perpendicular to AD at P.
When a perpendicular is drawn from the center of a circle to a chord, it bisects the chord. This means P is the midpoint of AD and also the midpoint of BC.
Given:
Radius of the larger circle, OA = 20 cm.
Radius of the smaller circle, OB = 15 cm.
Distance from the center to the chord, OP = 12 cm.
First, consider the right-angled triangle ΔOPA (using the radius of the larger circle). Using the Pythagoras Theorem:
OA\( ^2 \) = AP\( ^2 \) + OP\( ^2 \)
\( 20^2 = AP^2 + 12^2 \)
\( 400 = AP^2 + 144 \)
\( AP^2 = 400 - 144 \)
\( AP^2 = 256 \)
\( AP = \sqrt{256} = 16 \) cm.
Next, consider the right-angled triangle ΔOPB (using the radius of the smaller circle). Using the Pythagoras Theorem:
OB\( ^2 \) = BP\( ^2 \) + OP\( ^2 \)
\( 15^2 = BP^2 + 12^2 \)
\( 225 = BP^2 + 144 \)
\( BP^2 = 225 - 144 \)
\( BP^2 = 81 \)
\( BP = \sqrt{81} = 9 \) cm.
From the figure, the length AB is the difference between AP and BP.
AB = AP - BP = 16 cm - 9 cm = 7 cm.
In simple words: First, use the Pythagoras rule for the bigger circle's radius and the distance OP to find the length of AP. Then, do the same for the smaller circle's radius and OP to find BP. Finally, subtract BP from AP to get the length of AB.

🎯 Exam Tip: Remember that a perpendicular from the center to a chord bisects the chord. In concentric circles, this perpendicular applies to all chords. Use the Pythagorean theorem for each right-angled triangle formed with the radii and the perpendicular distance.

Question 32. In the figure, YZ is parallel to MN, XY is parallel to LM and XZ is parallel to LN. Then MY is
(a) the median of \( \Delta LMN \)
(b) the angular bisector of \( \angle LMN \)
(c) perpendicular to LN
(d) perpendicular bisector of LN.
Answer: (a) the median of \( \Delta LMN \)
In simple words: When the segments YZ, XY, and XZ are parallel to the sides MN, LM, and LN respectively within triangle LMN, it means that X, Y, and Z are the midpoints of the sides LM, LN, and MN. Since MY connects vertex M to Y, which is the midpoint of the opposite side LN, MY is a median of the triangle LMN. A median divides the opposite side into two equal parts.

🎯 Exam Tip: Remember that a line segment joining a vertex to the midpoint of the opposite side in a triangle is defined as a median. This concept is often linked to the midpoint theorem and properties of parallelograms formed by joining midpoints.