Contents

**GSEB Solutions for Class 10 mathematics – Similarity and The Theorem of Pythagoras (English Medium)**

### Exercise-7.1

**Question 1:**

∠B is a right angle in ∆ABC. and D ∈ . If AD = 4DC, prove that BD = 2DC.

**Solution :**

**Question 2:**

5, 12, 13 are the lengths of the sides of a triangle. Show that the triangle is right angled. Find the length of altitude on the hypotenuse.

**Solution :**

**Question 3:**

In ∆PQR, is the altitude to hypotenuse . If PM = 8, RM = 12, find PQ, QR and QM.

**Solution :**

**Question 4:**

In ∆ABC, m∠B = 90, , M ∈ . If AM – MC = 7, AB^{2} – BC^{2} = 175, find AC.

**Solution :**

**Question 5:**

∠A is right angle in ∆ABC. is an altitude of the triangle. If AB = , BD = 2, find the length of the hypotenuse of the triangle.

**Solution :**

**Question 6:**

m∠B = 90 in ∆ABC. is altitude to .

- If AM = BM = 8, find AC.
- If BM = 15, AC = 34, find AB.
- If BM = 2, MC = 6, find AC.
- If AB = , AM = 2.5, find MC.

**Question 6(1):**

**Solution :**

**Question 6(2):**

**Solution :**

**Question 6(3):**

**Solution :**

**Question 6(4):**

**Solution :**

**Question 7:**

In APQR m∠Q = 90, PQ = x, QR = y and . D ∈ . Find PD, QD, RD in terms Of x and y.

**Solution :**

**Question 8:**

∠Q is a right angle in ∆PQR and , M ∈ . If PQ = 4QR, then prove that PM = 16RM.

**Solution :**

**Question 9:**

PQRS is a rectangle. If PQ + QR = 7 and PR + QS = 10, then find the area of PQRS.

**Solution :**

**Question 10:**

The diagonals of a convex ABCD intersect at right angles. Prove that AB^{2} + CD^{2} = AD^{2} + BC^{2}.

**Solution :**

Given: The diagonals of a convex ⃞ABCD intersect at right angles at O.

To prove : AB^{2 }+ CD^{2 }= AD^{2 }+ BC^{2}

Proof : In ∆AOD, m∠O = 90°

∴ AD^{2 }= OA^{2 }+ OD^{2} … … (1)

In ∆BOC, m∠O = 90°

∴ BC^{2 }= OB^{2 }+ OC^{2} … … (2)

In ∆AOB, m∠O = 90°

∴ AB^{2 }= OA^{2 }+ OB^{2} … … (3)

In ∆COD, m∠O = 90°

∴CD^{2 }= OC^{2 }+ OD^{2} … … (4)

Adding results (3) and (4),

AB^{2 }+ CD^{2 }= OA^{2 }+ OB^{2 }+ OC^{2 }+ OD^{2}

∴AB^{2 }+ CD^{2 }= (OA^{2 }+ OD^{2}) + (OB^{2 }+ OC^{2})

∴AB^{2 }+ CD^{2 }= AD^{2 }+ BC^{2} [from (1) and (2)]

**Question 11:**

In ∆PQR, m∠Q = 90, M ∈ and N ∈ . Prove that PM^{2} + RN^{2} = PR^{2} + MN^{2}.

**Solution :**

**Question 12:**

The sides Of a triangle have lengths a^{2} + b^{2}, 2ab, a^{2} – b^{2}, where a > b and a, b ∈ R^{+}. Prove that the angle opposite to the side having length a^{2} + b^{2} is a right angle.

**Solution :**

**Question 13:**

In ∆ABC, m∠B = 90 and is a median. Prove that AB^{2} + BC^{2} + AC^{2} = 8BE^{2}.

**Solution :**

**Question 14:**

AB = AC and ∠A is right angle in ∆ABC. If BC = a, then find the area of the triangle. (a ∈ R, a > 0)

**Solution :**

### Exercise-7.2

**Question 1:**

In rectangle ∆BCD, AB + BC = 23, AC + BD = 34. Find the area of the rectangle.

**Solution :**

□ ABCD is a rectangle.

∴ AC = BD and AC + BD = 34

∴ 2AC = 34∴ AC = 17

Also, AB + BC = 23

Let, AB = x

∴ BC = 23 – x

In ∆ABC, m∠B = 90°

∴ AB^{2 }+ BC^{2 }= AC^{2}

∴ x^{2 }+ (23 – x)^{2 }= 17^{2}

∴ x^{2 }+ 529 – 46x + x^{2 }= 289

∴ 2x^{2 }– 46x + 240 = 0

∴ x^{2 } – 23x + 120 = 0

∴ (x – 15)(x – 8) = 0

∴ x = 15 or x = 8

∴ AB = 15 or AB = 8

If AB = 15, then BC = 23 – 15 = 8

If AB = 8,then BC = 23 – 8 = 15

Hence, under both conditions, the lengths of two adjacent sides of rectangle ABCD are 15 and 8.

Area of rectangle ABCD

= Product of two adjacent sides

= 15 × 8

= 120

∴ The area of rectangle ABCD is 120.

**Question 2:**

In ∆ABC m∠A = m∠B + m∠C, AB = 7, BC = 25. Find the perimeter of ∆ABC.

**Solution :**

In ∆ABC, m∠A = m∠B + m∠C and

m∠A + m∠B + m∠C = 180°

∴ m∠A + m∠A = 180°

∴ 2m∠A = 180°

∴ m∠A = 90°

In ∆ABC, m∠A = 90°

∴ AB^{2 }+ AC^{2 }= BC^{2}

∴ (7)^{2 }+ AC^{2 }= (25)^{2}

∴ AC^{2 }= 625 – 49 = 576 = (24)^{2}

∴ AC= 24

Perimeter of ∆ABC = AB + BC + AC

= 7 + 25 + 24 = 56

∴ The perimeter of ∆ABC is 56.

**Question 3:**

A staircase of length 6.5 meters touches a wall at height of 6 meter. Find the distance of base of the staircase from the wall.

**Solution :**

**Question 4:**

In ∆ABC AB = 7, AC = 5, AD = 5. Find BC, if the mid-point of is D.

**Solution :**

**Question 5:**

In equilateral ∆ABC, D ∈ such that BD : DC = 1 : 2. Prove that 3AD = AB.

**Solution :**

**Question 6:**

In ∆ABC, AB = 17, BC = 15, AC = 8, find the length of the median on the largest side.

**Solution :**

**Question 7:**

is a median of ∆ABC. AB^{2} + AC^{2} = 148 and AD = 7. Find BC.

**Solution :**

**Question 8:**

In rectangle ABCD, AC = 25 and CD = 7. Find perimeter of the rectangle.

**Solution :**

□ABCD is a rectangle. ∴ m∠D = 90°

In ∆ADC, m∠D = 90°

∴ AC^{2 }= AD^{2 }+ CD^{2}

∴ 25^{2 }= AD^{2 }+ 7^{2}

∴ 625 = AD^{2 }+ 49

∴ AD^{2 }= 576 = 24^{2}

∴ AD = 24

Perimeter of rectangle ABCD = 2(AD + CD)

= 2 (24 + 7)

= 2 × 31 = 62

∴ The perimeter of rectangle ABCD is 62.

**Question 9:**

In rhombus XYZW, XZ = 14 and YW = 48. Find XY.

**Solution :**

**Question 10:**

In ∆PQR, m∠Q : m∠R : m∠P = 1 : 2 : 1 If PQ = 2, find PR.

**Solution :**

### Exercise-7

**Question 1:**

, , are the medians of ∆ ABC. If BE = 12, CF = 9 and AB^{2} + BC^{2} + AC^{2} = 600, BC = 10, find AD.

**Solution :**

**Question 2:**

is the altitude of ∆ABC such that B – D – C. If AD^{2} = BD ∙ DC, prove that ∠BAC is right angle. [Hint : . So, B – D – C is given. So, ∆ADB and ∆ADC are right angled triangles to which Pythagoras’ theorem can be applied. Same method can be applied to solve Ex. 3, 4, 5.]

**Solution :**

**Question 3:**

In ∆ ABC, , B – D – C. If AB^{2 }= BD ∙ BC, prove that ∠BAC is a right angle.

**Solution :**

**Question 4:**

In ∆ABC, , B – D – C. If AC^{2} = CD ∙ BC, prove that ∠BAC is a right angle.

**Solution :**

**Question 5:**

is a median of ∆ABC. If BD = AD, prove that ∠A is a right angle in ∆ ABC.

**Solution :**

**Question 6:**

In figure 7.25, AC is the length of a pole standing vertical on the ground. The pole is bent at point B, so that the top of the pole touches the ground at a point 15 meters away from the base of the pole. If the length of the pole is 25, find the length of the upper part of the pole.

**Solution :**

When the pole is bent at point B, its top C touches the ground at point C’.

∴ BC = BC’

So, AC’ = 15 m and AC = 25 m.

AC = AB + BC

∴ 25 = AB + BC’

Let, AB = x m

∴ 25 = x + BC’

∴ BC’ = (25 – x) m

In ∆ABC, m∠A = 90°

∴ BC^{2 }= AB^{2 }+ AC^{2}

∴ (25 – x)^{2 }= x^{2 }+ 15^{2}

∴ 625 – 50x + x^{2 }= x^{2 }+225

∴ 625 – 225 = 50x

∴ 50x = 400

∴ x = 8

∴ AB = 8 m

Length of the upper part of the pole = BC = BC’ = 25-x = 25 – 8 = 17 m

**Question 7:**

In ∆ABC, AB > AC, D is the mid-point of . such that B – M – C. Prove that AB^{2} – AC^{2} = 2BC ∙ DM.

**Solution :**

**Question 8:**

In ∆ABC, , D ∈ and ∠B is right angle. If AC = SCD, prove that BD = 2CD.

**Solution :**

**Question 9:**

Select a proper option (a), (b), (c) or (d) from given options :

**Question 9(1):**

In ∆PQR, if m∠P + m∠Q = m∠R. PR = 7, QR = 24, then PQ = …….

**Solution :**

b. 25

In ∆PQR,

m∠P + m∠Q = m∠R

∴m∠R = 90°

∴PQ^{2 }= PR^{2 }+ QR^{2}

^{ }= 7^{2 }+ 24

= 49 + 576

= 625

= 25^{2}

∴PQ = 25

**Question 9(2):**

In ∆ABC, is an altitude and ∠A is right angle. If AB = BD = 4, CD = ………

**Solution :**

**Question 9(3):**

In ∆ABC, AB^{2} + AC^{2} = 50. The length of the median AD = 3. So, BC = ……..

**Solution :**

**Question 9(4):**

In ∆ABC, m∠B = 90, AB = BC. Then AB : AC = ………

**Solution :**

**Question 9(5):**

In ∆ABC, m∠B = 90, AC = 10. The length of the median BM = …….

**Solution :**

**Question 9(6):**

In ∆ABC, AB = BC = . m∠B ………

**Solution :**

**Question 9(7):**

In ∆ABC, if , then m∠B = ……….

**Solution :**

**Question 9(8):**

In ∆XYZ, m∠X : m∠Y : m∠Z = 1 : 2 : 3. If XY = 15, YZ = ……

**Solution :**

**Question 9(9):**

In ∆ABC, ∠B is a right angle and is an altitude. If AD = BD = 5, then DC = …….

**Solution :**

**Question 9(10):**

In ∆ABC, is median. If AB^{2} + AC^{2} = 130 and AD = 7, then BD = ……..

**Solution :**

**Question 9(11):**

The diagonal of a square is 5. The length of the side of the square is …….

**Solution :**

**Question 9(12):**

The length of a diagonal of a rectangle is 13. If one of the side of the rectangle is 5, the perimeter of the rectangle is ……….

**Solution :**

**
**b. 34

In any rectangle,

(Diagonal)

^{2 }= (Length)

^{2 }+ (Breadth)

^{2}

∴ 13

^{2 }= (Length)

^{2 }+ 5

^{2}

∴ 169 – 25 = (Length)

^{2}

∴ (Length)

^{2 }= 144

∴ (Length)

^{2 }= 12

Perimeter of rectangle = 2(Length + Breadth)

= 2(12 + 5) = 2(17) = 34

**Question 9(13):**

The length of a median of an equilateral triangle is . Length of the side of the triangle is .

**Solution :**

**Question 9(14):**

The perimeter of an equilateral triangle is 6. The length of the altitude of the triangle is ………

**Solution :**

**Question 9(15):**

In ∆ABC, m∠A = 90. is a median. If AD = 6, AB = 10, then AC = ……..

**Solution :**

**Question 9(16):**

In ∆PQR, m∠Q = 90 and PQ = QR. , M ∈ . If QM = 2, PQ = ………

**Solution :**

**Question 9(17):**

In ∆ABC, m∠A = 90, is an altitude. So AB^{2} =

**Solution :**

**Question 9(18):**

In ∆ABC, m∠A = 90, is an altitude. Therefore BD ∙ DC = …….

**Solution :**

Sosa Mahesh says

Very usful solutions sir

Aarya says

Very useful for practice