ML Aggarwal Class 9 Maths Solutions Chapter 11 Mid Point Theorem

Exercise 11

Question 1. Lengths of sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:
(i) 3 cm, 8 cm, 6 cm
(ii) 13 cm, 12 cm, 5 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
Answer: For each set of sides, identify the longest length and check if its square equals the sum of squares of the other two lengths.
(i) The longest side is 8 cm, with the other two being 3 cm and 6 cm. We have \( 8^2 = 64 \) and \( 3^2 + 6^2 = 9 + 36 = 45 \). Since \( 64 \neq 45 \), this is not a right triangle.
(ii) The longest side is 13 cm, with the other two being 5 cm and 12 cm. We have \( 13^2 = 169 \) and \( 5^2 + 12^2 = 25 + 144 = 169 \). Since \( 13^2 = 5^2 + 12^2 \), this is a right triangle with hypotenuse 13 cm.
(iii) The longest side is 5 cm, with the other two being 1.4 cm and 4.8 cm. We have \( 5^2 = 25 \) and \( (1.4)^2 + (4.8)^2 = 1.96 + 23.04 = 25 \). Since \( 5^2 = (1.4)^2 + (4.8)^2 \), this is a right triangle with hypotenuse 5 cm.
In simple words: For each set, square the longest side. If it matches the sum of squares of the other two sides, it is a right triangle.

Exam Tip: Always compare the square of the longest side with the sum of squares of the two shorter sides - this is the quickest way to identify right triangles.

Question 2. Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
Answer: Let AB represent the ladder and BC represent the vertical wall. Since the wall is vertical, triangle ABC forms a right angle at C.

Using the Pythagorean theorem:
\( AB^2 = BC^2 + AC^2 \)
\( 10^2 = BC^2 + 6^2 \)
\( 100 = BC^2 + 36 \)
\( BC^2 = 100 - 36 = 64 \)
\( BC = \sqrt{64} = 8 \text{ m} \)

Therefore, the ladder touches the wall at a height of 8 m above the ground.
In simple words: The ladder, the wall, and the ground form a right triangle. Use the Pythagorean theorem to find the missing side, which is the height where the ladder touches the wall.

Exam Tip: Draw a clear diagram showing the right angle at the base of the wall - this helps identify which side is the hypotenuse (the ladder) and apply the theorem correctly.

Question 3. A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to other end. How far from the base of the pole should the stake be driven so that the wire will be taught?
Answer: Let BC represent the pole and AB represent the wire. Let x be the distance from the base of the pole to where the stake is driven. Since the pole stands vertically, triangle ABC is a right triangle.

Using the Pythagorean theorem:
\( AB^2 = BC^2 + AC^2 \)
\( 24^2 = 18^2 + x^2 \)
\( 576 = 324 + x^2 \)
\( x^2 = 576 - 324 = 252 \)
\( x = \sqrt{252} = \sqrt{36 \times 7} = 6\sqrt{7} \text{ m} \)

The stake should be positioned at a distance of \( 6\sqrt{7} \) m from the base of the pole.
In simple words: The wire, pole, and ground create a right triangle. Apply the Pythagorean theorem to find how far the stake must be from the pole base.

Exam Tip: Simplify surds by factoring out perfect squares - this makes your final answer cleaner and shows good mathematical practice.

Question 4. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Answer: Let AB be the shorter pole (6 m) and CD be the taller pole (11 m). We construct point E on CD such that CE equals AB (6 m), making AE parallel to BC. This creates a right triangle ADE where AE is horizontal and ED is vertical.

From the construction: \( CE = AB = 6 \text{ m} \) and \( AE = BC = 12 \text{ m} \)
\( ED = CD - CE = 11 - 6 = 5 \text{ m} \)

Triangle ADE is a right triangle. Using the Pythagorean theorem:
\( AD^2 = AE^2 + ED^2 \)
\( AD^2 = 12^2 + 5^2 \)
\( AD^2 = 144 + 25 = 169 \)
\( AD = \sqrt{169} = 13 \text{ m} \)

The distance between the tops of the poles is 13 m.
In simple words: Draw an imaginary line from the top of the shorter pole parallel to the ground until it meets the taller pole. This creates a right triangle, and you can use the Pythagorean theorem to find the distance between the tops.

Exam Tip: When finding distances between the tops of poles of different heights, the key is to construct a right triangle by drawing a horizontal line from the top of the shorter pole.

Question 5. In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other sides is 4:3, find the sides.
Answer: Given that the two sides (other than the hypotenuse) are in the ratio 4:3, let them be 4x cm and 3x cm respectively.

Using the Pythagorean theorem:
\( 20^2 = (4x)^2 + (3x)^2 \)
\( 400 = 16x^2 + 9x^2 \)
\( 400 = 25x^2 \)
\( x^2 = \frac{400}{25} = 16 \)
\( x = 4 \)

Therefore: First side = \( 4x = 4 \times 4 = 16 \text{ cm} \) and Second side = \( 3x = 3 \times 4 = 12 \text{ cm} \)

The two sides are 12 cm and 16 cm.
In simple words: When sides are in a specific ratio, multiply that ratio by an unknown value (x) and use the Pythagorean theorem to solve for x, then find the actual side lengths.

Exam Tip: When given a ratio of sides, always represent the sides using the ratio multiplied by a variable (like 4x and 3x) - this keeps the algebra manageable.

Question 6. If the sides of the triangle are in the ratio 3:4:5, prove that it is right-angled triangle.
Answer: Let the sides be 3x, 4x, and 5x cm respectively. The greatest length is 5x cm, with the other two being 3x cm and 4x cm.

Calculate the square of the greatest side: \( (5x)^2 = 25x^2 \)

Calculate the sum of squares of the other two sides: \( (3x)^2 + (4x)^2 = 9x^2 + 16x^2 = 25x^2 \)

Since \( (5x)^2 = (3x)^2 + (4x)^2 \), the condition for a right triangle is satisfied. By the converse of the Pythagorean theorem, any triangle with sides in the ratio 3:4:5 must be a right-angled triangle.
In simple words: When sides follow the 3:4:5 ratio, the largest side squared always equals the sum of squares of the other two sides, which proves it is a right triangle.

Exam Tip: The 3:4:5 ratio is a special Pythagorean triple - memorizing this and a few others (like 5:12:13) can save you time during exams.

Question 7. For going to a city B from the city A, there is route via city C such that AC ⊥ CB. AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of highway.
Answer: Since AC is perpendicular to CB, triangle ABC is a right triangle. Using the Pythagorean theorem:
\( AB^2 = AC^2 + BC^2 \)
\( 26^2 = (2x)^2 + [2(x + 7)]^2 \)
\( 676 = 4x^2 + (2x + 14)^2 \)
\( 676 = 4x^2 + 4x^2 + 56x + 196 \)
\( 676 = 8x^2 + 56x + 196 \)
\( 480 = 8x^2 + 56x \)
\( 8(x^2 + 7x) = 480 \)
\( x^2 + 7x = 60 \)
\( x^2 + 7x - 60 = 0 \)
\( (x + 12)(x - 5) = 0 \)
\( x = 5 \text{ or } x = -12 \)

Since distance cannot be negative, \( x = 5 \).

Distance via route AC + BC: \( 2x + 2(x + 7) = 2(5) + 2(12) = 10 + 24 = 34 \text{ km} \)

Distance via highway AB: \( 26 \text{ km} \)

Distance saved: \( 34 - 26 = 8 \text{ km} \)
In simple words: The old route has two legs that add up to 34 km. The new direct highway is only 26 km, so 8 km of travel is saved.

Exam Tip: Always eliminate negative solutions for distance problems - distances must be positive. Also, remember to compare the total path length (sum of two legs) with the direct distance.

Question 8. The hypotenuse of a right triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.
Answer: Let the shortest side be x meters. Then:
Hypotenuse = 2x + 6 meters
Third side = (2x + 6) - 2 = 2x + 4 meters

Using the Pythagorean theorem:
\( (2x + 6)^2 = x^2 + (2x + 4)^2 \)
\( 4x^2 + 24x + 36 = x^2 + 4x^2 + 16x + 16 \)
\( 4x^2 + 24x + 36 = 5x^2 + 16x + 16 \)
\( 0 = x^2 - 8x - 20 \)
\( 0 = (x - 10)(x + 2) \)
\( x = 10 \text{ or } x = -2 \)

Since length cannot be negative, \( x = 10 \).

Shortest side = 10 m
Hypotenuse = 2(10) + 6 = 26 m
Third side = 2(10) + 4 = 24 m

The three sides are 10 m, 24 m, and 26 m.
In simple words: Express the other two sides in terms of the shortest side using the given conditions, then apply the Pythagorean theorem to solve for the unknown.

Exam Tip: When sides are described in relationship to one another, express everything in terms of one variable and use the Pythagorean theorem to create a solvable equation.

Question 9. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Answer: Since triangle ABC has a right angle at C and is isosceles, the two legs must be equal: AC = BC.

The hypotenuse is AB (the side opposite the right angle). Applying the Pythagorean theorem:
\( AB^2 = AC^2 + BC^2 \)
\( AB^2 = AC^2 + AC^2 \)
\( AB^2 = 2AC^2 \)

This proves the required relationship.
In simple words: In an isosceles right triangle, the two equal sides are the legs. When you apply the Pythagorean theorem, the hypotenuse squared equals twice the square of each leg.

Exam Tip: For isosceles right triangles, remember that the two legs are equal, making the hypotenuse always equal to leg times √2 - this is a useful shortcut to remember.

Question 10. In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².
Answer: Since AD is perpendicular to BC, we can apply the Pythagorean theorem to the two right triangles formed.

In right triangle ABD:
\( AB^2 = AD^2 + BD^2 \) ......(1)

In right triangle ACD:
\( AC^2 = AD^2 + CD^2 \) ......(2)

Subtracting equation (2) from equation (1):
\( AB^2 - AC^2 = (AD^2 + BD^2) - (AD^2 + CD^2) \)
\( AB^2 - AC^2 = BD^2 - CD^2 \)
\( AB^2 + CD^2 = AC^2 + BD^2 \)

This proves the required relationship.
In simple words: The perpendicular from the top vertex to the opposite side creates two right triangles. Writing the Pythagorean theorem for both triangles and then comparing them gives the desired result.

Exam Tip: When dealing with perpendiculars in triangles, write the Pythagorean theorem separately for each right triangle formed, then use algebraic manipulation (like subtraction) to prove relationships.

Question 11. In ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b)(a - b) = (c + d)(c - d).
Answer: Since PD is perpendicular to QR, we form two right triangles: PDQ and PDR.

In right triangle PDQ, applying the Pythagorean theorem:
\( PQ^2 = PD^2 + QD^2 \)
\( a^2 = PD^2 + c^2 \)
\( PD^2 = a^2 - c^2 \) ........(i)

In right triangle PDR, applying the Pythagorean theorem:
\( PR^2 = PD^2 + DR^2 \)
\( b^2 = PD^2 + d^2 \)
\( PD^2 = b^2 - d^2 \) ........(ii)

Since both expressions equal PD², we can equate them:
\( a^2 - c^2 = b^2 - d^2 \)
\( a^2 - b^2 = c^2 - d^2 \)
\( (a + b)(a - b) = (c + d)(c - d) \)

This proves the required relationship.
In simple words: The perpendicular creates two right triangles. Finding PD² in two different ways and setting them equal leads to the required identity.

Exam Tip: When you have multiple expressions for the same unknown (like PD²), equating them is a powerful technique for proving algebraic identities in geometry.

Question 12. ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and hence calculate its area.
Answer: Let AD represent the altitude drawn to BC, with BD = x cm and CD = (8 - x) cm. In right triangle ABD, by the Pythagorean theorem: \( AB^2 = AD^2 + BD^2 \)
\( 12^2 = AD^2 + x^2 \)
\( AD^2 = 144 - x^2 \) .......(i)
In right triangle ADC, by the Pythagorean theorem: \( AC^2 = AD^2 + DC^2 \)
\( 12^2 = AD^2 + (8 - x)^2 \)
\( AD^2 = 144 - (8 - x)^2 \) .......(ii)
From equations (i) and (ii): \( 144 - x^2 = 144 - (64 + x^2 - 16x) \)
\( 144 - x^2 = 144 - 64 - x^2 + 16x \)
\( 64 = 16x \)
\( x = 4 \)
Substituting x = 4 into equation (i): \( AD^2 = 144 - 16 = 128 \)
\( AD = \sqrt{128} = 8\sqrt{2} \) cm
Area of triangle = \( \frac{1}{2} \times AD \times BC = \frac{1}{2} \times 8\sqrt{2} \times 8 = 32\sqrt{2} \) cm²
Hence, AD = \( 8\sqrt{2} \) cm and area = \( 32\sqrt{2} \) cm²
In simple words: Since the triangle has two equal sides, the altitude from A cuts the base BC into two equal parts. Use the Pythagorean theorem twice to find where this height meets the base, then calculate the area by multiplying half the base by the height.

Exam Tip: In an isosceles triangle, the altitude from the vertex angle bisects the base - use this fact to simplify your setup. Always leave radical answers in simplified form unless the question asks for decimals.

Question 13. Find the area and the perimeter of a square whose diagonal is 10 cm long.
Answer: Let each side of the square be a cm. Since the sides of a square meet at right angles, in right triangle ABC: \( AC^2 = AB^2 + BC^2 \)
\( 10^2 = a^2 + a^2 \)
\( 100 = 2a^2 \)
\( a^2 = 50 \)
\( a = \sqrt{50} = 5\sqrt{2} \) cm
Perimeter = 4a = \( 4 \times 5\sqrt{2} = 20\sqrt{2} \) cm
Area = \( a^2 = 50 \) cm²
Hence, perimeter = \( 20\sqrt{2} \) cm and area = 50 cm²
In simple words: The diagonal of a square forms a right angle with the sides. Use the Pythagorean theorem to find the side length, then multiply by 4 for perimeter and square it for area.

Exam Tip: Recognize that in a square with diagonal d, the side length is always \( \frac{d}{\sqrt{2}} \) - this saves calculation time. Leave answers with radicals in simplified form.

Question 14(a). In figure given below, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, ∠ABD = ∠BCD = 90°. Calculate the length of AB.
Answer: In right triangle DBC, by the Pythagorean theorem: \( DB^2 = DC^2 + BC^2 \)
\( DB^2 = 12^2 + 3^2 \)
\( DB^2 = 144 + 9 = 153 \)
\( DB = \sqrt{153} \) cm
In right triangle ABD, by the Pythagorean theorem: \( AD^2 = AB^2 + DB^2 \)
\( 13^2 = AB^2 + 153 \)
\( 169 - 153 = AB^2 \)
\( AB^2 = 16 \)
\( AB = \sqrt{16} = 4 \) cm
Hence, AB = 4 cm
In simple words: First find DB using the measurements of triangle DBC. Then use that result to find AB in triangle ABD.

Exam Tip: When working with compound figures, identify each right triangle separately and solve in stages - solving backwards from the given hypotenuse is often key to finding unknown sides.

Question 14(b). In figure given below, ABCD is a quadrilateral in which AB = AD, ∠A = 90° = ∠C, BC = 8 cm and CD = 6 cm. Find AB and calculate the area of △ABD.
Answer: Let AB = AD = x cm. In right triangle BCD, by the Pythagorean theorem: \( BD^2 = BC^2 + CD^2 \)
\( BD^2 = 8^2 + 6^2 \)
\( BD^2 = 64 + 36 = 100 \)
\( BD = \sqrt{100} = 10 \) cm
In right triangle ABD, by the Pythagorean theorem: \( BD^2 = AB^2 + AD^2 \)
\( 10^2 = x^2 + x^2 \)
\( 100 = 2x^2 \)
\( x^2 = 50 \)
\( x = \sqrt{50} = 5\sqrt{2} \) cm
Area of right triangle ABD = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times AD \)
\( = \frac{1}{2} \times 5\sqrt{2} \times 5\sqrt{2} \)
\( = \frac{1}{2} \times 25 \times 2 = 25 \) cm²
Hence, AB = \( 5\sqrt{2} \) cm and area of right triangle ABD = 25 cm²
In simple words: First work out BD from the right triangle BCD. Then use the fact that AB and AD are equal to find their length from right triangle ABD. For the area, multiply half the base by the height.

Exam Tip: When two sides of a right triangle are equal (as AB = AD), the Pythagorean theorem becomes \( c^2 = 2a^2 \), simplifying to \( a = \frac{c}{\sqrt{2}} \). Also, for isosceles right triangles, area = \( \frac{1}{2}a^2 \).

Question 15(a). In figure given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm. Calculate the length of BD.
Answer: In right triangle ABC, by the Pythagorean theorem: \( AC^2 = AB^2 + BC^2 \)
\( 13^2 = 12^2 + BC^2 \)
\( BC^2 = 169 - 144 = 25 \)
\( BC = \sqrt{25} = 5 \) cm
In right triangle CDE, by the Pythagorean theorem: \( CE^2 = CD^2 + DE^2 \)
\( 10^2 = CD^2 + 6^2 \)
\( CD^2 = 100 - 36 = 64 \)
\( CD = \sqrt{64} = 8 \) cm
From the figure, BD = BC + CD = 5 + 8 = 13 cm
Hence, BD = 13 cm
In simple words: Two separate right triangles are involved. Find BC from the first triangle and CD from the second triangle, then add them together.

Exam Tip: When points lie on a straight line, the total distance is the sum of individual segments. Always check the figure to confirm the arrangement of points before adding segments.

Question 15(b). In figure given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
Answer: In right triangle PQS, by the Pythagorean theorem: \( PQ^2 = PS^2 + QS^2 \)
\( 10^2 = PS^2 + 6^2 \)
\( PS^2 = 100 - 36 = 64 \)
\( PS = \sqrt{64} = 8 \) cm
From the figure, RS = RQ + QS = 9 + 6 = 15 cm
In right triangle PRS, by the Pythagorean theorem: \( PR^2 = RS^2 + PS^2 \)
\( PR^2 = 15^2 + 8^2 \)
\( PR^2 = 225 + 64 = 289 \)
\( PR = \sqrt{289} = 17 \) cm
Hence, PR = 17 cm
In simple words: Start by finding PS in the first right triangle. Then add the given segments to get RS. Finally, use the Pythagorean theorem again to find PR.

Exam Tip: In multi-step problems, solve for unknown segments in order - each result feeds into the next calculation. Keep intermediate results in exact form to avoid rounding errors.

Question 15(c). In figure given below, ∠D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.
Answer: Let CD = x cm. In right triangle ACD, by the Pythagorean theorem: \( AC^2 = AD^2 + CD^2 \)
\( 6^2 = AD^2 + x^2 \)
\( AD^2 = 36 - x^2 \) .......(i)
In right triangle ABD, by the Pythagorean theorem: \( AB^2 = AD^2 + BD^2 \)
\( 16^2 = (36 - x^2) + (12 + x)^2 \)
\( 256 = 36 - x^2 + 144 + x^2 + 24x \)
\( 256 = 180 + 24x \)
\( 24x = 76 \)
\( x = \frac{76}{24} = 3\frac{1}{6} \) cm
Hence, CD = \( 3\frac{1}{6} \) cm
In simple words: Set up the Pythagorean theorem for both right triangles using a variable for the unknown side. Eliminate the other variable by substituting the first equation into the second, then solve for CD.

Exam Tip: When a figure contains overlapping triangles sharing a common side, express that shared side (like AD) in terms of one unknown and substitute systematically into the second equation to eliminate it.

Question 16(a). In figure given below, BC = 5 cm, ∠B = 90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.
Answer: In right triangle ABC, by the Pythagorean theorem: \( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (5AE)^2 + 5^2 \)
\( AC^2 = 25AE^2 + 25 \)
\( AC^2 = 25(AE^2 + 1) \) .......(i)
From the figure, EB = AB - AE = 5AE - AE = 4AE. In right triangle BED: \( ED^2 = EB^2 + BD^2 \)
\( ED^2 = (4AE)^2 + (5 + 2AE)^2 \)
\( ED^2 = 16AE^2 + 25 + 4AE^2 + 20AE \)
\( ED^2 = 20AE^2 + 20AE + 25 \) .......(ii)
Given that AC = ED, from equations (i) and (ii): \( 25(AE^2 + 1) = 20AE^2 + 20AE + 25 \)
\( 25AE^2 + 25 = 20AE^2 + 20AE + 25 \)
\( 5AE^2 - 20AE = 0 \)
\( 5AE(AE - 4) = 0 \)
Since AE ≠ 0, AE = 4 cm
Therefore: CD = 2AE = 8 cm, AB = 5AE = 20 cm
Substituting AE = 4 into equation (i): \( AC^2 = 25(16 + 1) = 25 \times 17 = 425 \)
\( AC = \sqrt{425} = 5\sqrt{17} \) cm
In simple words: Express all lengths in terms of AE using the given relationships. Use the Pythagorean theorem twice to set up two equations for AC and ED. Since AC = ED, solve these equal equations to find AE, then calculate all other lengths.

Exam Tip: In problems with multiple relationships (AB = 5AE, CD = 2AE, AC = ED), use variables strategically to reduce unknowns. Setting up the two Pythagorean equations and equating them is the key - this creates an equation in a single variable that can be solved algebraically.

Question 16(b). In figure given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB² = 4AD² - 3AC².
Answer: Since D is the midpoint of BC, we have DC = BC/2. Triangle ABC is a right triangle. Applying the Pythagorean theorem gives AB² = AC² + BC² ... (i). Triangle ADC is also a right triangle. By the Pythagorean theorem, AD² = AC² + DC² ... (ii). From (ii), AC² = AD² - DC². Substituting DC = BC/2, we get AC² = AD² - (BC/2)². This simplifies to AC² = AD² - BC²/4. Multiplying by 4 yields 4AC² = 4AD² - BC². Rearranging: AC² + 3AC² = 4AD² - BC². Adding BC² to both sides gives AC² + BC² = 4AD² - 3AC². By (i), AB² = AC² + BC², therefore AB² = 4AD² - 3AC². Thus the result is proved.
In simple words: We use the Pythagorean theorem twice - once on the big triangle and once on a smaller triangle. By finding a link between the distances, we can show that AB² equals 4AD² minus 3AC².

Exam Tip: Always identify the right angles in the figure and apply the Pythagorean theorem to each triangle separately - then look for a way to combine the two equations by substitution.

Question 17. In △ABC, AB = AC = x, BC = 10 cm and the area of △ABC is 60 cm². Find x.
Answer: Let AD represent the altitude from A to the base BC. In an isosceles triangle, the altitude from the vertex angle to the base divides the base into two equal parts. Therefore, BD = CD = 5 cm. In the right triangle ABD, applying the Pythagorean theorem: x² = AD² + 5². This gives AD² = x² - 25, so AD = √(x² - 25). The area of triangle ABC is given by (1/2) × base × height = (1/2) × BC × AD. Substituting: 60 = (1/2) × 10 × √(x² - 25), which simplifies to 120 = 10√(x² - 25). Dividing by 10 gives √(x² - 25) = 12. Squaring both sides: x² - 25 = 144, therefore x² = 169. Taking the square root: x = 13 cm.
In simple words: The altitude of an isosceles triangle splits the base in half. Using the area formula and the Pythagorean theorem, we can find the equal sides.

Exam Tip: Remember that in an isosceles triangle, the altitude from the apex is also the median - use this property to find the base segments quickly.

Question 18. In rhombus if diagonals are 30 cm and 40 cm, find its perimeter.
Answer: Let AC = 30 cm and BD = 40 cm. A key property of a rhombus is that its diagonals are perpendicular to each other and bisect one another. Therefore, the diagonals intersect at point O, where OB = BD/2 = 20 cm and AO = AC/2 = 15 cm. In the right triangle AOB, we apply the Pythagorean theorem: AB² = AO² + OB² = 15² + 20² = 225 + 400 = 625. Thus AB = 25 cm. Since all sides of a rhombus are equal, each side measures 25 cm. The perimeter is 4 × 25 = 100 cm.
In simple words: The diagonals of a rhombus cross at right angles and cut each other exactly in half. This creates four identical right triangles, and we can find the side length using the Pythagorean theorem.

Exam Tip: Always use the perpendicular bisector property of rhombus diagonals - this lets you immediately find the half-lengths needed for the Pythagorean theorem.

Question 19(a). In figure given below, AB || DC, BC = AD = 13 cm, AB = 22 cm and DC = 12 cm. Calculate the height of the trapezium ABCD.
Answer: From the figure, EF = DC = 12 cm. Since AD = BC and the heights from D and C to AB are equal (both h), we have AE = FB = x. From the relationship AB = AE + EF + FB, we get 22 = x + 12 + x, which gives 2x = 10, so x = 5 cm. In the right triangle ADE, applying the Pythagorean theorem: AD² = DE² + AE², thus 13² = h² + 5². This gives 169 - 25 = h², so h² = 144, and h = 12 cm. Therefore, the height of the trapezium is 12 cm.
In simple words: Drop perpendiculars from both top corners to the base. The base is divided into the top length plus two equal side pieces. Use the Pythagorean theorem on one of the side triangles to find the height.

Exam Tip: Recognise that an isosceles trapezium has equal slant sides, which creates two congruent right triangles on either end - this symmetry simplifies your calculation.

Question 19(b). In figure given below, AB || DC, ∠A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.
Answer: Since AB || DC, it follows that ∠D = ∠A = 90°. In the right triangle ACD, applying the Pythagorean theorem: AC² = AD² + CD², so 25² = AD² + 7², giving 625 = AD² + 49, thus AD² = 576, and AD = 24 cm. From the figure, let E be the point where a perpendicular from C meets AB. Then CE = AD = 24 cm. In the right triangle ACE: AC² = AE² + CE², so 25² = AE² + 24², giving 625 = AE² + 576, thus AE² = 49, and AE = 7 cm. Therefore, EB = AB - AE = 17 - 7 = 10 cm. In the right triangle BCE: BC² = CE² + EB² = 24² + 10² = 576 + 100 = 676, so BC = 26 cm.
In simple words: Drop a perpendicular from C to the base AB. This creates two right triangles. Use the Pythagorean theorem step-by-step on each triangle to eventually find BC.

Exam Tip: When one angle of a trapezium is 90°, dropping a perpendicular from the other slant side creates a rectangle on one side - use this to simplify your working.

Question 19(c). In the figure given below, ABCD is a square of side 7 cm. If AE = FC = CG = HA = 3 cm, (i) prove that EFGH is a rectangle. (ii) find the area and perimeter of EFGH.
Answer:
(i) Given ABCD is a square of side 7 cm and AE = FC = CG = HA = 3 cm. From the figure: DH = DA - HA = 7 - 3 = 4 cm, GD = DC - CG = 7 - 3 = 4 cm, EB = AB - AE = 7 - 3 = 4 cm, FB = BC - FC = 7 - 3 = 4 cm. Since the square has perpendicular sides, applying the Pythagorean theorem to each of the four corner triangles: HE² = HA² + AE² = 3² + 3² = 18, so HE = 3√2 cm. GF² = GC² + FC² = 3² + 3² = 18, so GF = 3√2 cm. GH² = GD² + DH² = 4² + 4² = 32, so GH = 4√2 cm. EF² = EB² + FB² = 4² + 4² = 32, so EF = 4√2 cm. In isosceles triangle AEH, the angles at E and H are each 45° (since the triangle has a 90° angle at A and two equal legs). On the baseline at E, we have ∠AEH + ∠HEF + ∠FEB = 180° (angles on a straight line), giving 45° + ∠HEF + 45° = 180°, so ∠HEF = 90°. By similar logic, all angles of EFGH are 90°. Since HE = GF and GH = EF, quadrilateral EFGH is a rectangle.
(ii) Area of EFGH = EF × GF = 4√2 × 3√2 = 24 cm². Perimeter of EFGH = 2(EF + GF) = 2(4√2 + 3√2) = 2 × 7√2 = 14√2 cm.
In simple words: When you mark equal distances from each corner of a square and connect the points, you make a tilted rectangle inside. All angles of this inner shape are 90°, and opposite sides are equal.

Exam Tip: To prove a quadrilateral is a rectangle, show that all four angles are 90° AND that opposite sides are equal - do not skip either condition, as both are essential for full marks.

Question 20. AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².
Answer: Given that AD is perpendicular to BC and triangle ABC is equilateral, we have AD ⊥ BC and AB = BC = CA. In an equilateral triangle, the perpendicular from any vertex to the opposite side bisects that side. Therefore, BD = BC/2. From the figure, applying the Pythagorean theorem to the right triangle ABD: AB² = AD² + BD² = AD² + (BC/2)² = AD² + BC²/4. This gives AB² = (4AD² + BC²)/4. Multiplying both sides by 4: 4AB² = 4AD² + BC². Since the triangle is equilateral, AB = BC. Substituting BC = AB: 4AB² = 4AD² + AB². Subtracting AB² from both sides: 3AB² = 4AD², or equivalently, 4AD² = 3AB². Thus the result is proved.
In simple words: In an equilateral triangle, the height cuts the base exactly in half. Using the Pythagorean theorem and the fact that all sides are equal, we can relate the height squared to the side squared.

Exam Tip: Always recall that in an equilateral triangle, the altitude, median, and perpendicular bisector from any vertex coincide - this property is key to solving this type of problem quickly.

Question 21. In the adjoining figure, D and E are midpoints of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²
Answer:
(i) Since D is the midpoint of BC:

\( \therefore CD = \frac{BC}{2} \)

In right triangle ACD, by the Pythagorean theorem:

\( AD^2 = AC^2 + CD^2 \)

\( AD^2 = AC^2 + \left(\frac{BC}{2}\right)^2 \)

\( AD^2 = AC^2 + \frac{BC^2}{4} \)

\( AD^2 = \frac{4AC^2 + BC^2}{4} \)

\( 4AD^2 = 4AC^2 + BC^2 \) .....(1)

Therefore, \( 4AD^2 = 4AC^2 + BC^2 \) is proved.

(ii) Since E is the midpoint of AC:

\( \therefore CE = \frac{AC}{2} \)

\( AC = 2CE \)

Triangle BCE is a right triangle. By the Pythagorean theorem:

\( BE^2 = BC^2 + CE^2 \)

Multiplying both sides by 4:

\( 4BE^2 = 4BC^2 + 4CE^2 \)

\( 4BE^2 = 4BC^2 + (2CE)^2 \)

\( 4BE^2 = 4BC^2 + AC^2 \) .....(2)

Therefore, \( 4BE^2 = 4BC^2 + AC^2 \) is proved.

(iii) Since ABC is a right triangle, by the Pythagorean theorem:

\( AB^2 = AC^2 + BC^2 \)

Adding equations (1) and (2) from the above parts:

\( 4AD^2 + 4BE^2 = 4AC^2 + BC^2 + 4BC^2 + AC^2 \)

\( 4(AD^2 + BE^2) = 5AC^2 + 5BC^2 \)

\( 4(AD^2 + BE^2) = 5(AC^2 + BC^2) \)

\( 4(AD^2 + BE^2) = 5AB^2 \)

Therefore, \( 4(AD^2 + BE^2) = 5AB^2 \) is proved.
In simple words: When you find the distances from two midpoints on a right triangle to the opposite vertex, those distances follow special patterns linked to the sides of the triangle.

Exam Tip: Show all three proofs separately using the Pythagorean theorem step by step - examiners award marks for clarity and showing each substitution clearly.

Question 22. If AD, BE and CF are medians of △ABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).
Answer: Draw AP perpendicular to BC.

In right triangle APD, by the Pythagorean theorem:

\( AD^2 = AP^2 + PD^2 \) .....(i)

In right triangle APB, by the Pythagorean theorem:

\( AB^2 = AP^2 + BP^2 \)

\( AB^2 = AP^2 + (BD - PD)^2 \)

\( AB^2 = AP^2 + BD^2 + PD^2 - 2BD \cdot PD \)

Since D is the midpoint of BC (as AD is a median):

\( BD = \frac{BC}{2} \)

\( AB^2 = AP^2 + PD^2 + \left(\frac{BC}{2}\right)^2 - 2 \cdot \frac{BC}{2} \cdot PD \)

\( AB^2 = AD^2 + \frac{BC^2}{4} - BC \cdot PD \) .....(ii) (From i)

In right triangle APC, by the Pythagorean theorem:

\( AC^2 = AP^2 + PC^2 \)

\( AC^2 = AP^2 + (PD + DC)^2 \)

\( AC^2 = AP^2 + PD^2 + DC^2 + 2PD \cdot DC \)

Since DC = \( \frac{BC}{2} \):

\( AC^2 = AD^2 + DC^2 + 2PD \cdot DC \) [From (i)]

\( AC^2 = AD^2 + \left(\frac{BC}{2}\right)^2 + 2 \cdot PD \cdot \frac{BC}{2} \)

\( AC^2 = AD^2 + \frac{BC^2}{4} + PD \cdot BC \) .....(iii)

Adding (ii) and (iii):

\( AB^2 + AC^2 = AD^2 + \frac{BC^2}{4} - BC \cdot PD + AD^2 + \frac{BC^2}{4} + PD \cdot BC \)

\( AB^2 + AC^2 = 2AD^2 + \frac{BC^2}{2} \) .....(iv)

Draw BQ perpendicular to AC. In right triangle BQE, by the Pythagorean theorem:

\( BE^2 = BQ^2 + QE^2 \) .....(v)

In right triangle ABQ, by the Pythagorean theorem:

\( AB^2 = BQ^2 + AQ^2 \)

\( AB^2 = BQ^2 + (AE - QE)^2 \)

\( AB^2 = BQ^2 + QE^2 + AE^2 - 2AE \cdot QE \)

Since E is the midpoint of AC (as BE is a median):

\( AE = \frac{AC}{2} \)

\( AB^2 = BE^2 + \left(\frac{AC}{2}\right)^2 - 2 \cdot \frac{AC}{2} \cdot QE \)
(From v)

\( AB^2 = BE^2 + \frac{AC^2}{4} - AC \cdot QE \) .....(vi)

In right triangle BQC, by the Pythagorean theorem:

\( BC^2 = BQ^2 + QC^2 \)

\( BC^2 = BQ^2 + (QE + EC)^2 \)

\( BC^2 = BQ^2 + QE^2 + EC^2 + 2QE \cdot EC \)

Since EC = \( \frac{AC}{2} \):

\( BC^2 = BE^2 + EC^2 + 2QE \cdot EC \) [From (v)]

\( BC^2 = BE^2 + \left(\frac{AC}{2}\right)^2 + 2 \cdot QE \cdot \frac{AC}{2} \)

\( BC^2 = BE^2 + \frac{AC^2}{4} + QE \cdot AC \) .....(vii)

Adding (vi) and (vii):

\( AB^2 + BC^2 = BE^2 + \frac{AC^2}{4} - AC \cdot QE + BE^2 + \frac{AC^2}{4} + QE \cdot AC \)

\( AB^2 + BC^2 = 2BE^2 + \frac{AC^2}{2} \) .....(viii)

Draw CR perpendicular to AB. In right triangle CFR, by the Pythagorean theorem:

\( CF^2 = CR^2 + FR^2 \) .....(ix)

In right triangle CBR, by the Pythagorean theorem:

\( CB^2 = CR^2 + RB^2 \)

\( CB^2 = CR^2 + (BF - FR)^2 \)

\( CB^2 = CR^2 + FR^2 + BF^2 - 2BF \cdot FR \)

Since F is the midpoint of AB (as CF is a median):

\( BF = \frac{AB}{2} \)

\( CB^2 = CF^2 + \left(\frac{AB}{2}\right)^2 - 2 \cdot \frac{AB}{2} \cdot FR \)
(From ix)

\( CB^2 = CF^2 + \frac{AB^2}{4} - AB \cdot FR \) .....(x)

In right triangle ACR, by the Pythagorean theorem:

\( AC^2 = RC^2 + AR^2 \)

\( AC^2 = RC^2 + (AF + FR)^2 \)

\( AC^2 = RC^2 + FR^2 + AF^2 + 2AF \cdot FR \)

Since AF = \( \frac{AB}{2} \):

\( AC^2 = CF^2 + AF^2 + 2AF \cdot FR \) [From (ix)]

\( AC^2 = CF^2 + \left(\frac{AB}{2}\right)^2 + 2 \cdot FR \cdot \frac{AB}{2} \)

\( AC^2 = CF^2 + \frac{AB^2}{4} + FR \cdot AB \) .....(xi)

Adding (x) and (xi):

\( CB^2 + AC^2 = CF^2 + \frac{AB^2}{4} - AB \cdot FR + CF^2 + \frac{AB^2}{4} + FR \cdot AB \)

\( CB^2 + AC^2 = 2CF^2 + \frac{AB^2}{2} \) .....(xii)

Adding equations (iv), (viii), and (xii):

\( AB^2 + AC^2 + AB^2 + BC^2 + CB^2 + AC^2 = 2AD^2 + \frac{BC^2}{2} + 2BE^2 + \frac{AC^2}{2} + 2CF^2 + \frac{AB^2}{2} \)

\( 2(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2) + \frac{1}{2}(AB^2 + BC^2 + CA^2) \)

\( 2(AB^2 + BC^2 + CA^2) - \frac{1}{2}(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2) \)

\( \frac{3}{2}(AB^2 + BC^2 + CA^2) = 2(AD^2 + BE^2 + CF^2) \)

\( 3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2) \)

Therefore, \( 3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2) \) is proved.
In simple words: When you draw perpendiculars from the median endpoints and use the Pythagorean theorem on each resulting right triangle, then add all the equations, the side squares and median squares follow this exact relationship.

Exam Tip: This is a long proof requiring careful management of multiple perpendiculars and Pythagorean applications - write equations (i) through (xii) clearly and label each step to show the full reasoning path.

Question 23. In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB² + CD² = AD² + BC².
Answer: By the Pythagorean theorem:

In right triangle AOB:

\( AB^2 = OB^2 + OA^2 \) .....(i)

In right triangle COD:

\( CD^2 = OC^2 + OD^2 \) .....(ii)

In right triangle AOD:

\( AD^2 = AO^2 + OD^2 \) .....(iii)

In right triangle BOC:

\( BC^2 = OB^2 + OC^2 \) .....(iv)

Adding equations (i) and (ii):

\( AB^2 + CD^2 = OB^2 + OA^2 + OC^2 + OD^2 \)

\( AB^2 + CD^2 = (OA^2 + OD^2) + (OC^2 + OB^2) \)

Substituting values from equations (iii) and (iv):

\( AB^2 + CD^2 = AD^2 + BC^2 \)

Therefore, \( AB^2 + CD^2 = AD^2 + BC^2 \) is proved.
In simple words: When two diagonals cross at right angles, you can use the Pythagorean theorem on the four right triangles they make, and the sum of opposite corner distances always equals the other opposite corner sum.

Exam Tip: Clearly identify all four right triangles formed at the intersection point - this makes it straightforward to apply Pythagorean theorem and group the terms correctly.

Question 24. In a quadrilateral ABCD, ∠B = 90° = ∠D. Prove that 2AC² - BC² = AB² + AD² + DC².
Answer: Given: ∠B = 90° = ∠D

By the Pythagorean theorem:

In right triangle ABC:

\( AC^2 = AB^2 + BC^2 \) .....(i)

By the Pythagorean theorem:

In right triangle ADC:

\( AC^2 = AD^2 + DC^2 \) .....(ii)

Adding equations (i) and (ii):

\( 2AC^2 = AB^2 + BC^2 + AD^2 + DC^2 \)

\( 2AC^2 - BC^2 = AB^2 + AD^2 + DC^2 \)

Therefore, \( 2AC^2 - BC^2 = AB^2 + AD^2 + DC^2 \) is proved.
In simple words: When opposite angles of a quadrilateral are right angles, the diagonal creates two right triangles, and applying the Pythagorean theorem to both then combining gives this relationship.

Exam Tip: Use the fact that both angles B and D are 90 degrees to identify two separate right triangles that share the diagonal AC as the hypotenuse.

Question 25. In a △ABC, ∠A = 90°, CA = AB and D is a point on AB produced. Prove that DC² - BD² = 2AB × AD.
Answer: In right triangle ACD:

\( DC^2 = CA^2 + AD^2 \) (by Pythagorean theorem)

\( DC^2 = CA^2 + (AB + BD)^2 \)

\( DC^2 = CA^2 + AB^2 + BD^2 + 2AB \cdot BD \)

Since CA = AB:

\( DC^2 - BD^2 = AB^2 + AB^2 + 2AB \cdot BD \)

\( DC^2 - BD^2 = 2AB^2 + 2AB \cdot BD \)

\( DC^2 - BD^2 = 2AB(AB + BD) \)

From the figure, AB + BD = AD:

\( DC^2 - BD^2 = 2AB \cdot AD \)

Therefore, \( DC^2 - BD^2 = 2AB \cdot AD \) is proved.
In simple words: The right angle at A and the equal sides CA and AB allow you to use the Pythagorean theorem to connect the distances from the extended point D to the other vertices.

Exam Tip: Note that D lies outside the original triangle on the extension of AB - this affects how you express AD as AB + BD.

Question 26. In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD × CD.
Answer: Draw AP perpendicular to BC.

Triangle APD is a right triangle. By the Pythagorean theorem:

\( AD^2 = AP^2 + PD^2 \)

\( AD^2 = AP^2 + (PC + CD)^2 \)

\( AD^2 = AP^2 + PC^2 + CD^2 + 2PC \cdot CD \) .....(i)

Triangle APC is a right triangle. By the Pythagorean theorem:

\( AC^2 = AP^2 + PC^2 \) .....(ii)

Substituting the value of AP² + PC² from equation (ii) into equation (i):

\( AD^2 = AC^2 + CD^2 + 2PC \cdot CD \) .....(iii)

Since ABC is an isosceles triangle, the altitude from A to the base BC bisects the base:

\( PC = \frac{BC}{2} \)

\( AD^2 = AC^2 + CD^2 + 2 \cdot \frac{1}{2} \cdot BC \cdot CD \)

\( AD^2 = AC^2 + CD^2 + BC \cdot CD \)

\( AD^2 = AC^2 + CD(CD + BC) \)

From the figure, CD + BC = BD:

\( AD^2 = AC^2 + CD \cdot BD \)

Therefore, \( AD^2 = AC^2 + BD \times CD \) is proved.
In simple words: In an isosceles triangle, the perpendicular to the base from the equal sides creates a relationship where the distance to an outside point on the extended base depends on the equal side and the segments created on the base.

Exam Tip: The key property is that the altitude in an isosceles triangle bisects the base - use this to express PC in terms of BC, then substitute back carefully into equation (iii).

Multiple Choice Questions

Question 1. In a △ABC, if AB = 6√3 cm, BC = 6 cm and AC = 12 cm, then ∠B is
(a) 120°
(b) 90°
(c) 60°
(d) 45°
Answer: (b) 90°
In simple words: When the square of the longest side equals the sum of squares of the other two sides, the angle opposite the longest side is always a right angle (90°).

Exam Tip: Recognize the Pythagorean relationship - if \( c^2 = a^2 + b^2 \), then the angle opposite side c is 90°. Check this relationship before assuming any angle measure.

Question 2. If the sides of a rectangular plot are 15 m and 8 m, then the length of its diagonal is
(a) 17 m
(b) 23 m
(c) 21 m
(d) 17 cm
Answer: (a) 17 m
In simple words: In a rectangle, the two sides meet at right angles. Use the Pythagorean theorem to find the diagonal by squaring both sides and adding them together.

Exam Tip: Always check that your final answer is in the correct unit (metres, not centimetres). The diagonal is the hypotenuse of the right triangle formed by two sides of the rectangle.

Question 3. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of the side of rhombus is
(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm
Answer: (b) 10 cm
In simple words: The diagonals of a rhombus split each other at right angles. Each diagonal cuts the other in half, creating four right triangles. Use Pythagoras on one of these triangles to find the side.

Exam Tip: Remember that the diagonals of a rhombus bisect each other perpendicularly. This creates four congruent right triangles, each with legs that are half the lengths of the diagonals.

Question 4. If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, then the length of the other diagonal is
(a) 6 cm
(b) 12 cm
(c) 20 cm
(d) 12 cm
Answer: (b) 12 cm
In simple words: The diagonals of a rhombus meet at right angles and cut each other in half. Knowing one diagonal and the side length, you can find the half-length of the other diagonal using Pythagoras, then double it.

Exam Tip: Set up a right triangle using half of the known diagonal, the unknown half-diagonal, and the side of the rhombus. The side is the hypotenuse of this triangle.

Question 5. If a ladder 10 m long reaches a window 8 m above the ground, then the distance of the foot of the ladder from the base of the wall is
(a) 18 m
(b) 8 m
(c) 6 m
(d) 4 m
Answer: (c) 6 m
In simple words: The ladder, the wall, and the ground form a right triangle. The ladder is the longest side, the wall height is one leg, and the distance from the wall is the other leg. Use Pythagoras to find it.

Exam Tip: Recognize this as a real-world application of the Pythagorean theorem - the ladder is the hypotenuse. Rearrange the formula to find the missing leg: \( a^2 = c^2 - b^2 \).

Question 6. A girl walks 200 m towards East and then she walks 150 m towards North. The distance of the girl from starting point is
(a) 350 m
(b) 250 m
(c) 300 m
(d) 225 m
Answer: (b) 250 m
In simple words: East and North are perpendicular directions. The girl's path creates a right angle. The direct distance from her starting point is found by treating this as a right triangle where the two walks are the legs.

Exam Tip: Always visualize direction-based problems as geometric shapes. Perpendicular directions (North-South, East-West) form right angles, making the Pythagorean theorem applicable.

Question 7. A ladder reaches a window 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. If the length of the ladder is 15 m, then the width of the street is
(a) 30 m
(b) 24 m
(c) 21 m
(d) 18 m
Answer: (c) 21 m
In simple words: The ladder stays the same length but reaches different heights on opposite sides of the street. Apply Pythagoras separately for each side to find the horizontal distance from the ladder's base to each building, then add them together.

Exam Tip: This problem requires two applications of the Pythagorean theorem - one for each side of the street. The street width is the sum of the two horizontal distances calculated separately.

Question 8. Consider the following two statements: Statement 1: The area of a square whose diagonal is 6 cm is 36 cm². Statement 2: A diagonal of a square divides it into two right angled isosceles triangle. Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (d) Statement 1 is false, and Statement 2 is true.
In simple words: A diagonal cuts a square into two equal triangles. Each triangle has one right angle (from the square's corner) and two equal sides (the square's sides), making them right isosceles triangles. For the area: if the diagonal is 6, the side is not 6 but smaller, so the area is 18 cm², not 36 cm².

Exam Tip: For any square with side a, the diagonal is \( a\sqrt{2} \). Use this to check Statement 1: if diagonal = 6, then \( a\sqrt{2} = 6 \), so \( a = \frac{6}{\sqrt{2}} \), and area = \( a^2 = 18 \) cm². Always verify both statements independently before selecting the answer.

Question. Assertion (A): If the sides of a rectangular plot are 80 m and 60 m, then the length of its diagonal is 100 m. Reason (R): In a rectangle, all angles are 90°.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: A rectangle has all right angles, which allows us to use the Pythagorean theorem on the triangle formed by two sides and a diagonal. Squaring and adding the sides (80² + 60²) gives 10,000, and the square root is 100 m. The reason explains why the assertion is correct.

Exam Tip: For Assertion-Reason questions, check: (1) Is the Assertion true? (2) Is the Reason true? (3) Does the Reason explain the Assertion? All three must be "yes" for option (c). The 90° angles are the foundation for applying Pythagoras.

Question. Assertion (A): If triangle ABC is isosceles with AC = BC and AB² = 2AC², then triangle ABC is right angled.
Reason (R): If in a triangle ABC, we have AB² = 2AC² then we can conclude that it is right angled triangle.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: Since triangle ABC is isosceles with AC = BC, we can write AB² = 2AC² as AB² = AC² + AC². This becomes AB² = AC² + BC² (since AC = BC). This expression matches the form of the Pythagorean Theorem where AB is the hypotenuse and AC and BC are the other two sides. Therefore, triangle ABC is a right angle triangle, so Assertion (A) holds true. However, the Reason given is incomplete - from the relation AB² = 2AC² alone, without knowing about the third side BC, we cannot reach the conclusion that the triangle is right angled. Thus Reason (R) is false.
In simple words: The Assertion is correct because when you know AC = BC, the equation AB² = 2AC² matches Pythagoras' rule. But the Reason is wrong because that rule by itself does not tell us the triangle is right angled.

Exam Tip: For Assertion-Reason questions, always check if both statements are true separately, and then verify if the Reason actually explains the Assertion - they must both be true AND logically connected for option (3) to be correct.

Question. Assertion (A): A triangle with sides 2 cm, 3 cm, 4 cm is not a right angled triangle.
Reason (R): A triangle with sides 2 cm, 3 cm, 4 cm is a scalene triangle.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: To check whether this is a right angled triangle, we examine the longest side (c = 4 cm) and apply Pythagoras' rule. The left side gives c² = 4² = 16. The right side gives a² + b² = 2² + 3² = 4 + 9 = 13. Since 16 does not equal 13, the Pythagorean relation fails, confirming that the triangle is not a right angled triangle. Therefore, Assertion (A) is true. A scalene triangle is defined as one in which all three sides have different lengths. Since the sides are 2 cm, 3 cm, and 4 cm - all distinct from each other - Reason (R) is also true. However, the fact that a triangle is scalene has no bearing on whether it is right angled or not. A scalene triangle can be right angled, or it can be non-right angled. Thus both statements are true, but the Reason does not explain the Assertion.
In simple words: The triangle is not right angled because the sides do not satisfy Pythagoras' rule. It is also scalene because all three sides are different. But being scalene does not tell us anything about whether it is right angled.

Exam Tip: Always verify Pythagorean relations by checking if the square of the longest side equals the sum of the squares of the other two sides - this is the definitive test for a right angled triangle.

Chapter Test

Question 1. In the figure given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.
Answer: Apply Pythagoras' theorem to find BD from the right triangle ADB. We have AB² = AD² + BD², so 25² = 15² + BD², giving 625 = 225 + BD². Thus BD² = 400 and BD = 20 cm. Similarly, in right triangle ADC, AC² = AD² + DC², so 17² = 15² + DC², giving 289 = 225 + DC². Thus DC² = 64 and DC = 8 cm. From the figure, BC = BD + DC = 20 + 8 = 28 cm.
In simple words: Use Pythagoras' theorem twice - once for each right triangle formed by the perpendicular - to find how far B and C are from point D. Then add these distances to get the total length BC.

Exam Tip: When a perpendicular is drawn from a vertex to the opposite side, it creates two right triangles. Apply Pythagoras' theorem to each separately, then combine the results to find the full length.

Question 1(b). In the figure given below, ∠BAC = 90°, ∠ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find (i) AC (ii) AB (iii) area of the shaded region.
Answer:
(i) In right angle triangle ADC, apply Pythagoras' theorem: AC² = AD² + DC² = 6² + 8² = 36 + 64 = 100. Therefore, AC = 10 cm.
(ii) In right angle triangle ABC, apply Pythagoras' theorem: BC² = AB² + AC². Substituting the known values: 26² = AB² + 10², so 676 = AB² + 100. This gives AB² = 576, so AB = 24 cm.
(iii) The shaded region is the area of triangle ABC minus the area of triangle ADC. Area of triangle ABC = \( \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 24 \times 10 = 120 \) cm². Area of triangle ADC = \( \frac{1}{2} \times AD \times DC = \frac{1}{2} \times 6 \times 8 = 24 \) cm². Therefore, area of shaded region = 120 - 24 = 96 cm².
In simple words: First find AC using Pythagoras in triangle ADC. Then find AB using Pythagoras in triangle ABC. Finally, subtract the smaller triangle's area from the larger triangle's area to get the shaded region.

Exam Tip: When finding areas of overlapping regions, clearly identify which triangle is the larger one and which is inside it, then subtract carefully to avoid mistakes.

Question 1(c). In figure given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate (i) the length of BC (ii) the area of △ADE.
Answer:
(i) In right angle triangle ABC, by Pythagoras' theorem: AC² = AB² + BC². Substituting: 15² = 9² + BC², so 225 = 81 + BC². Thus BC² = 144 and BC = 12 cm.
(ii) Since D and E are midpoints of AB and AC, the mid-point theorem tells us that DE = \( \frac{1}{2} \)BC = 6 cm. Since angle A is the right angle in the original triangle ABC, angle A remains the right angle in the smaller triangle ADE (because DE is parallel to BC). Therefore, AD = \( \frac{1}{2} \)AB = 4.5 cm. The area of triangle ADE = \( \frac{1}{2} \times AD \times DE = \frac{1}{2} \times 4.5 \times 6 = 13.5 \) cm².
In simple words: First find BC using Pythagoras. Then use the fact that D and E are midpoints to find that DE is half of BC. Since the angle at A is still a right angle, multiply half of AB by half of BC and divide by 2 to get the area.

Exam Tip: The mid-point theorem states that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length - this is a key shortcut that avoids extra calculations.

Question 2. If in △ABC, AB > AC and AD ⊥ BC, prove that AB² - AC² = BD² - CD².
Answer: In right angle triangle ADB, by Pythagoras' theorem: AB² = AD² + BD² ... (i). In right angle triangle ADC, by Pythagoras' theorem: AC² = AD² + CD² ... (ii). Subtracting equation (ii) from equation (i): AB² - AC² = (AD² + BD²) - (AD² + CD²) = BD² - CD². Hence proved.
In simple words: When you drop a perpendicular from A to BC, you create two right triangles. Use Pythagoras on both, then subtract the two equations - the AD² terms cancel out, leaving you with the required result.

Exam Tip: For proof questions, always write out the two Pythagoras equations clearly before subtracting - this shows your working and makes it easy to see which terms cancel.

Question 3. In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2 - 1. Prove that (i) 9AQ² = 9AC² + 4BC² (ii) 9BP² = 9BC² + 4AC² (iii) 9(AQ² + BP²) = 13AB²
Answer:
(i) In right angle triangle ACQ, by Pythagoras' theorem: AQ² = AC² + CQ². Multiplying both sides by 9: 9AQ² = 9AC² + 9CQ² = 9AC² + (3CQ)². Since Q divides CB in the ratio 1 - 2 from B, we have BQ - CQ = 1 - 2, so CQ/BC = 2/3, giving 3CQ = 2BC. Substituting: 9AQ² = 9AC² + (2BC)² = 9AC² + 4BC². Hence proved.
(ii) In right angle triangle BCP, by Pythagoras' theorem: BP² = BC² + CP². Multiplying both sides by 9: 9BP² = 9BC² + 9CP² = 9BC² + (3CP)². Since P divides CA in the ratio 2 - 1 from A, we have AP - PC = 2 - 1, so CP/AC = 2/3, giving 3CP = 2AC. Substituting: 9BP² = 9BC² + (2AC)² = 9BC² + 4AC². Hence proved.
(iii) In right angle triangle ABC, by Pythagoras' theorem: AB² = AC² + BC² ... (iii). Adding the results from (i) and (ii): 9AQ² + 9BP² = 9AC² + 4BC² + 9BC² + 4AC² = 13AC² + 13BC² = 13(AC² + BC²) = 13AB² [from (iii)]. Therefore, 9(AQ² + BP²) = 13AB². Hence proved.
In simple words: Apply Pythagoras to two different right triangles formed by the points P and Q. Use the ratio information to express the segment lengths in terms of the original sides. Then add the two equations and use Pythagoras on the main triangle to finish.

Exam Tip: When a point divides a side in a given ratio, express that relationship as a fraction of the whole side - this makes substitution straightforward and reduces algebraic errors.

Question 4. In the adjoining figure, △PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² = 3PR² + 5PS².
Answer: Let RT = TS = SQ = x, so QR = 3x. In right angle triangle PQR, by Pythagoras' theorem: PR² = PQ² + QR² = PQ² + (3x)² = PQ² + 9x². In right angle triangle PQS, by Pythagoras' theorem: PS² = PQ² + QS² = PQ² + x². In right angle triangle PQT, by Pythagoras' theorem: PT² = PQ² + QT² = PQ² + (2x)² = PQ² + 4x². Now, 3PR² + 5PS² = 3(PQ² + 9x²) + 5(PQ² + x²) = 3PQ² + 27x² + 5PQ² + 5x² = 8PQ² + 32x² = 8(PQ² + 4x²) = 8PT². Hence proved.
In simple words: Since S and T divide QR into three equal parts, express the distances QS, QT, and QR in terms of a common length x. Apply Pythagoras to three different right triangles, then show that a combination of PR² and PS² produces PT² with the right coefficients.

Exam Tip: When points trisect a side, use equal segments (like x, x, x) as your building blocks - this keeps the algebra simple and makes the final verification straightforward.

Question 5. In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², Prove that ∠ACD = 90°.
Answer: In the right triangle ABC, using the Pythagorean theorem, we find that AC² = AB² + BC². We are given that AD² = AB² + BC² + CD². Substituting the expression for AB² + BC² into this, we obtain AD² = AC² + CD². Since AD² equals AC² + CD², triangle ACD must be a right-angled triangle. By the converse of the Pythagorean theorem, the angle opposite the hypotenuse AD must be 90°. Therefore, ∠ACD = 90°.
In simple words: When AD² = AC² + CD², the angle at C in triangle ACD has to be 90 degrees. This is because of the reverse form of Pythagoras's rule.

Exam Tip: Always state which triangle is right-angled and identify the hypotenuse clearly. Using the converse of Pythagoras's theorem requires showing that the sum of squares of two sides equals the square of the third side.

Question 6. In the adjoining figure, find the length of AD in terms of b and c.
Answer: In the right triangle ABC, by Pythagoras's theorem, BC² = AB² + AC² = c² + b². Therefore, BC = \( \sqrt{b^2 + c^2} \). The total area of triangle ABC may be expressed as the sum of the areas of triangles ABD and ADC. This gives us: \( \frac{1}{2} \times AB \times AC = \frac{1}{2} \times AD \times BD + \frac{1}{2} \times AD \times CD \). Simplifying, we get AB·AC = AD·(BD + CD). Since BD + CD = BC, we have AB·AC = AD·BC. Solving for AD: \( AD = \frac{AB \times AC}{BC} = \frac{c \times b}{\sqrt{b^2 + c^2}} = \frac{bc}{\sqrt{b^2 + c^2}} \).
In simple words: The line AD, drawn perpendicular to BC, divides the triangle into two smaller triangles. The length of AD comes from equating the total area to the sum of the two smaller areas.

Exam Tip: Area-based approaches are powerful for finding altitudes. Always express BD + CD as BC and simplify the equation step-by-step to avoid algebraic errors.

Question 7. ABCD is a square, F is mid-point of AB and BE is one third of BC. If area of △FBE is 108 cm², find the length of AC.
Answer: Let the side length of the square be x cm. Since F is the midpoint of AB, we have FB = \( \frac{x}{2} \) cm. Since BE equals one third of BC, we have BE = \( \frac{x}{3} \) cm. The area of triangle FBE is given as 108 cm². Using the area formula: \( \frac{1}{2} \times FB \times BE = 108 \). Substituting the values: \( \frac{1}{2} \times \frac{x}{2} \times \frac{x}{3} = 108 \). This simplifies to \( \frac{x^2}{12} = 108 \), giving x² = 1296. Taking the square root, x = 36 cm. The diagonal of a square equals \( \sqrt{2} \) times the side length. Therefore, AC = \( 36\sqrt{2} \) cm.
In simple words: Find the side of the square using the area of triangle FBE. Then multiply the side by \( \sqrt{2} \) to get the diagonal length.

Exam Tip: Always use the formula for the area of a triangle carefully - a common mistake is forgetting to include the factor of \( \frac{1}{2} \). For a square, remember that the diagonal is \( \sqrt{2} \) times the side.

Question 8. In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC × CD. Prove that BD = BC.
Answer: Draw a perpendicular from B to AC, meeting AC at E. In the right triangle BEC, applying Pythagoras's theorem: BC² = BE² + EC². This can be rewritten as BC² = BE² + (AC - AE)², which expands to BC² = BE² + AC² + AE² - 2AC·AE ... (i). In the right triangle ABE, we have AB² = BE² + AE² ... (ii). Substituting the value of BE² + AE² from equation (ii) into equation (i): BC² = AB² + AC² - 2AC·AE. Since AB = AC (given), this becomes BC² = AC² + AC² - 2AC·AE = 2AC² - 2AC·AE = 2AC(AC - AE) = 2AC × EC ... (iii). From the given condition, BC² = AC × CD ... (iv). Comparing equations (iii) and (iv): 2AC × EC = AC × CD, which gives 2EC = CD, or EC = \( \frac{CD}{2} \). This means E is the midpoint of CD, so ED = EC. In triangles BDE and BCE, we have BE = BE (common side), ED = EC (proven above), and ∠BED = ∠BEC = 90°. By the SAS axiom, △BDE ≅ △BCE. Since corresponding parts of congruent triangles are equal, BD = BC.
In simple words: By dropping a perpendicular from B to AC and using Pythagoras's theorem twice, we show that E must be the midpoint of CD. This creates two congruent right triangles, which proves BD = BC.

Exam Tip: Perpendiculars are key constructions in geometry proofs. Once you establish that two triangles are congruent, state which axiom (SAS, ASA, SSS, RHS) you are using and identify the three pairs of equal parts clearly.