##### Class 10^{th} Mathematics Gujarat Board Solution

**Exercise 7.1**- ∠B is a right angle in ΔABC. bar bd perpendicular bar ac and D bar ac . If AD =…
- 5, 12, 13 are the lengths of the sides of a triangle. Show that the triangle is…
- In ΔPQR, bar qm is the altitude to hypotenuse bar pr ? If PM = 8, RM = 12, find…
- In ΔABC, m∠B = 90, bar bm perpendicular bar ac , M bar ac . If AM — MC = 7, AB^2…
- ∠A is right angle in ΔABC. AD is an altitude of the triangle. If AB = √5, BD =…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If AM = BM = 8, find AC.…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If BM = 15, AC = 34, find AB.…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If BM = 2√30, MC = 6, find AC.…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If AB = √10, AM = 2.5, find…
- In ΔPQR m∠Q = 90, PQ = x, QR = y and bar qd perpendicular bar pr . D in bar pr .…
- ∠Q is a right angle in ΔPQR and bar qm perpendicular bar pr , M in bar pr . If…
- âPQRS is a rectangle. If PQ + QR = 7 and PR + QS = 10, then find the area of â…
- The diagonals of a convex â ABCD intersect at right angles. Prove that AB^2 +…
- In ΔPQR, m∠Q = 90, M in bar qr and N in bar pq . Prove that PM^2 + RN^2 = PR^2…
- The sides of a triangle have lengths a^2 + b^2 , 2ab, a^2 — b^2 , where a b and…
- In ΔABC, m∠B = 90 and bar be is a median. Prove that AB^2 + BC^2 + AC^2 = 8BE^2…
- AB = AC and ∠A is right angle in ΔABC. If BC = √2 a, then find the area of the…

**Exercise 7.2**- In rectangle ABCD, AB + BC = 23, AC + BD = 34. Find the area of the rectangle.…
- In ΔABC m∠A = m∠B + m∠C, AB = 7, BC = 25. Find the perimeter of ΔABC.…
- A staircase of length 6.5 meters touches a wall at height of 6 meter. Find the…
- In ΔABC AB = 7, AC = 5, AD = 5. Find bar bc , if the mid - point of BC is D.…
- In equilateral ΔABC, D in bar bc such that BD : DC = 1 : 2. Prove that 3AD = √7…
- In ΔABC, AB = 17, BC = 15, AC = 8, find the length of the median on the largest…
- bar ad is a median of ΔABC. AB^2 + AC^2 = 148 and AD = 7. Find BC.…
- In rectangle ABCD, AC = 25 and CD = 7. Find perimeter of the rectangle.…
- In rhombus XYZW, XZ = 14 and YW = 48. Find XY.
- In ΔPQR, m∠Q : m∠R : m∠P = 1 : 2 : 1. If PQ = 2√6, find PR.

**Exercise 7**- bar ad , bar be , bar cf are the medians of ΔABC. If BE = 12, CF = 9 and AB^2 +…
- bar ad is the altitude of Δ ABC such that B—D—C. If AD^2 = BD ⋅ DC, prove that…
- In ΔABC, bar ad perpendicular bar bc , B—D—C. If AB^2 = BD ⋅ BC, prove that ∠…
- In ΔABC, bar ad perpendicular bar bc , B—D—C. If AC^2 = CD ⋅ BC, prove that ∠BAC…
- bar ad is a median of ΔABC. If BD = AD, prove that ∠A is a right angle in ΔABC.…
- In figure 7.25, AC is the length of a pole standing vertical on the ground. The…
- In ΔABC, AB AC, D is the mid-point of bar bc . bar am perpendicular vector bc…
- In ΔABC, bar bd perpendicular bar ac , D e bar ac and ∠B is right angle. If AC =…
- In ΔPQR, if m∠P + m∠Q = m∠R. PR = 7, QR = 24, then PQ =A. 31 B. 25 C. 17 D. 15…
- In ΔABC, bar ad is an altitude and ∠A is right angle. If AB = root 20 , BD = 4,…
- In ΔABC, AB^2 + AC^2 = 50. The length of the median AD = 3. So, BC =A. 4 B. 24…
- In ΔABC, m∠B = 90, AB = BC. Then AB: AC =A. 1:3 B. 1:2 C. 1 : √2 D. √2:1…
- In ΔABC, m∠B = 90 and AC = 10. The length of the median BM =A. 5 B. 5√2 C. 6 D.…
- In ΔABC, AB = BC = dc/root 2 . m∠BA. Is acute B. Is obtuse C. Is right angle D.…
- In ΔABC, if ab/1 = ac/2 = bc/root 3 then m ∠C = ……A. 90 B. 30 C. 60 D. 45…
- In ΔXYZ, m∠X : m∠Y : m∠Z = 1 : 2 : 3. If XY = 15, YZ = …A. 15 root 3/2 B. 17 C.…
- In Δ ABC, ∠B is a right angle and bar bd is an altitude. If AD = BD = 5, then…
- In Δ ABC, bar ad is median. If AB^2 + AC^2 = 130 and AD = 7, then BD =A. 4 B. 8…
- The diagonal of a square is 5√2. The length of the side of the square isA. 10…
- The length of a diagonal of a rectangle is 13. If one of the side of the…
- The length of a median of an equilateral triangle is √3. Length of the side of…
- The perimeter of an equilateral triangle is 6. The length of the altitude of…
- In ΔABC, m∠A = 90. bar ad is a median. If AD = 6, AB = 10, then AC =A. 8 B. 7.5…
- In Δ PQR, m∠Q = 90 and PQ = QR. bar qm perpendicular bar pr , in bar pr . If QM…
- In Δ ABC, m∠A = 90, bar ad is an altitude. So AB^2 = ……A. BD.BC B. BD.DC C.…
- In Δ ABC, m∠A = 90, bar ad is an altitude. Therefore BD.DC = ……A. AB^2 B. BC^2…

**Exercise 7.1**

- ∠B is a right angle in ΔABC. bar bd perpendicular bar ac and D bar ac . If AD =…
- 5, 12, 13 are the lengths of the sides of a triangle. Show that the triangle is…
- In ΔPQR, bar qm is the altitude to hypotenuse bar pr ? If PM = 8, RM = 12, find…
- In ΔABC, m∠B = 90, bar bm perpendicular bar ac , M bar ac . If AM — MC = 7, AB^2…
- ∠A is right angle in ΔABC. AD is an altitude of the triangle. If AB = √5, BD =…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If AM = BM = 8, find AC.…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If BM = 15, AC = 34, find AB.…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If BM = 2√30, MC = 6, find AC.…
- m∠B = 90 in ΔABC. bar bm is altitude to bar ac . If AB = √10, AM = 2.5, find…
- In ΔPQR m∠Q = 90, PQ = x, QR = y and bar qd perpendicular bar pr . D in bar pr .…
- ∠Q is a right angle in ΔPQR and bar qm perpendicular bar pr , M in bar pr . If…
- âPQRS is a rectangle. If PQ + QR = 7 and PR + QS = 10, then find the area of â…
- The diagonals of a convex â ABCD intersect at right angles. Prove that AB^2 +…
- In ΔPQR, m∠Q = 90, M in bar qr and N in bar pq . Prove that PM^2 + RN^2 = PR^2…
- The sides of a triangle have lengths a^2 + b^2 , 2ab, a^2 — b^2 , where a b and…
- In ΔABC, m∠B = 90 and bar be is a median. Prove that AB^2 + BC^2 + AC^2 = 8BE^2…
- AB = AC and ∠A is right angle in ΔABC. If BC = √2 a, then find the area of the…

**Exercise 7.2**

- In rectangle ABCD, AB + BC = 23, AC + BD = 34. Find the area of the rectangle.…
- In ΔABC m∠A = m∠B + m∠C, AB = 7, BC = 25. Find the perimeter of ΔABC.…
- A staircase of length 6.5 meters touches a wall at height of 6 meter. Find the…
- In ΔABC AB = 7, AC = 5, AD = 5. Find bar bc , if the mid - point of BC is D.…
- In equilateral ΔABC, D in bar bc such that BD : DC = 1 : 2. Prove that 3AD = √7…
- In ΔABC, AB = 17, BC = 15, AC = 8, find the length of the median on the largest…
- bar ad is a median of ΔABC. AB^2 + AC^2 = 148 and AD = 7. Find BC.…
- In rectangle ABCD, AC = 25 and CD = 7. Find perimeter of the rectangle.…
- In rhombus XYZW, XZ = 14 and YW = 48. Find XY.
- In ΔPQR, m∠Q : m∠R : m∠P = 1 : 2 : 1. If PQ = 2√6, find PR.

**Exercise 7**

- bar ad , bar be , bar cf are the medians of ΔABC. If BE = 12, CF = 9 and AB^2 +…
- bar ad is the altitude of Δ ABC such that B—D—C. If AD^2 = BD ⋅ DC, prove that…
- In ΔABC, bar ad perpendicular bar bc , B—D—C. If AB^2 = BD ⋅ BC, prove that ∠…
- In ΔABC, bar ad perpendicular bar bc , B—D—C. If AC^2 = CD ⋅ BC, prove that ∠BAC…
- bar ad is a median of ΔABC. If BD = AD, prove that ∠A is a right angle in ΔABC.…
- In figure 7.25, AC is the length of a pole standing vertical on the ground. The…
- In ΔABC, AB AC, D is the mid-point of bar bc . bar am perpendicular vector bc…
- In ΔABC, bar bd perpendicular bar ac , D e bar ac and ∠B is right angle. If AC =…
- In ΔPQR, if m∠P + m∠Q = m∠R. PR = 7, QR = 24, then PQ =A. 31 B. 25 C. 17 D. 15…
- In ΔABC, bar ad is an altitude and ∠A is right angle. If AB = root 20 , BD = 4,…
- In ΔABC, AB^2 + AC^2 = 50. The length of the median AD = 3. So, BC =A. 4 B. 24…
- In ΔABC, m∠B = 90, AB = BC. Then AB: AC =A. 1:3 B. 1:2 C. 1 : √2 D. √2:1…
- In ΔABC, m∠B = 90 and AC = 10. The length of the median BM =A. 5 B. 5√2 C. 6 D.…
- In ΔABC, AB = BC = dc/root 2 . m∠BA. Is acute B. Is obtuse C. Is right angle D.…
- In ΔABC, if ab/1 = ac/2 = bc/root 3 then m ∠C = ……A. 90 B. 30 C. 60 D. 45…
- In ΔXYZ, m∠X : m∠Y : m∠Z = 1 : 2 : 3. If XY = 15, YZ = …A. 15 root 3/2 B. 17 C.…
- In Δ ABC, ∠B is a right angle and bar bd is an altitude. If AD = BD = 5, then…
- In Δ ABC, bar ad is median. If AB^2 + AC^2 = 130 and AD = 7, then BD =A. 4 B. 8…
- The diagonal of a square is 5√2. The length of the side of the square isA. 10…
- The length of a diagonal of a rectangle is 13. If one of the side of the…
- The length of a median of an equilateral triangle is √3. Length of the side of…
- The perimeter of an equilateral triangle is 6. The length of the altitude of…
- In ΔABC, m∠A = 90. bar ad is a median. If AD = 6, AB = 10, then AC =A. 8 B. 7.5…
- In Δ PQR, m∠Q = 90 and PQ = QR. bar qm perpendicular bar pr , in bar pr . If QM…
- In Δ ABC, m∠A = 90, bar ad is an altitude. So AB^2 = ……A. BD.BC B. BD.DC C.…
- In Δ ABC, m∠A = 90, bar ad is an altitude. Therefore BD.DC = ……A. AB^2 B. BC^2…

###### Exercise 7.1

**Question 1.**∠B is a right angle in ΔABC. and D. If AD = 4DC, prove that BD = 2DC.

**Answer:**

m∠B = 90

AD = 4DC (Given) …Equation (i)

∴ BD^{2} = AD × DC

⇒ BD^{2} = 4DC^{2} (From Equation (i))

Taking Square Root both sides we get

∴ BD = 2DC

**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.

**Answer:**

Hypotenuse (AB) = 13

Perpendicular (AC) = 12

Base (BC) = 5

∴ AC^{2} + BC^{2} = 5^{2} + 12^{2}

⇒ AC^{2} + BC^{2} = 25 + 144

⇒ AC^{2} + BC^{2} = 169 = AB^{2}

Hence Pythagoras theorem is valid

m∠C = 90

Let AD = x, BD = 13 – x

Applying Pythagoras theorem

∴ Altitude CD^{2} = 12^{2} – x^{2} = 5^{2} – (13 – x) ^{2}

⇒ 144 – x^{2} = 25 – 169 – x^{2} + 26x

⇒ 26x = 288

⇒ x = 11.07 units

So Altitude CD = √21.43 = 4.63 units

**Question 3.**In ΔPQR, is the altitude to hypotenuse ? If PM = 8, RM = 12, find PQ, QR and QM.

**Answer:**

m∠Q = 90

RM = 12

MP = 8

∴ RP = (RM + MP) = 20

Let QM be x

Applying Pythagoras theorem in ΔQMP, we get

QP^{2} = x^{2} + 8^{2} = x^{2} + 64

Applying Pythagoras theorem in ΔQMR, we get

QR^{2} = x^{2} + 12^{2} = x^{2} + 144

Applying Pythagoras theorem in ΔPQR, we get

RP^{2} = (x^{2} + 64) + (x^{2} + 144)

⇒ 400 = 208 + 2x^{2}

⇒ x^{2} = 96

⇒ x = 4√6

⇒ QM = 4√6

PQ^{2} = x^{2} + 64

⇒ PQ^{2} = 160

⇒ PQ = 4√10

QR^{2} = x^{2} + 144

⇒ QR^{2} = 240

⇒ QR = 4√15

**Question 4.**In ΔABC, m∠B = 90, , M. If AM — MC = 7, AB^{2} — BC^{2} = 175, find AC.

**Answer:**

mB = 90

AM — MC = 7

Applying Pythagoras theorem in ΔABM and ΔCBM

AB^{2} = BM^{2} + AM^{2} …Equation (i)

BC^{2} = BM^{2} + MC^{2} …Equation (ii)

Subtracting Equation (ii) from Equation (i) we get

AB^{2} – BC^{2} = AM^{2} – MC^{2}

⇒ (AM – MC)(AM + MC) = AB^{2} – BC^{2}

Putting the values we get :

7 × (AM + MC) = 175

⇒ AM + MC = 25

⇒ AC = 25

**Question 5.**∠A is right angle in ΔABC. AD is an altitude of the triangle. If AB = √5, BD = 2, find the length of the hypotenuse of the triangle.

**Answer:**

m∠A = 90

AB = √5

BD = 2

Applying Pythagoras theorem in ΔADB we get

AD^{2} = AB^{2} – BD^{2}

⇒ AD^{2} = 5 – 4 = 1

⇒ AD = 1

Let AC = x

Applying Pythagoras theorem in ΔADC we get

DC^{2} = AC^{2} – AD^{2}

⇒ DC^{2} = x^{2} – 1

⇒ DC = √(x^{2} – 1)

Applying Pythagoras theorem in ΔABC we get

AB^{2} + AC^{2} = BC^{2}

⇒ 5 + x^{2} = (2 + √(x^{2} – 1))^{2}

⇒ 5 + x^{2} = 4 + x^{2} – 1 + 4√(x^{2} – 1)

Hypotenuse BC = BD + DC

⇒ Hypotenuse BC

**Question 6.**m∠B = 90 in ΔABC. is altitude to .

If AM = BM = 8, find AC.

**Answer:**

Applying Pythagoras theorem in ΔABM

AB^{2} = AM^{2} + BM^{2}

⇒ AB^{2} = 64 + 64 = 128

Let CM = x

Applying Pythagoras theorem in ΔCBM

⇒ BC^{2} = BM^{2} + CM^{2}

⇒ BC^{2} = x^{2} + 64

Applying Pythagoras theorem in ΔABC

AC^{2} = AB^{2} + BC^{2}

(x + 8)^{2} = x^{2} + 64 + 128

⇒ x^{2} + 64 + 16x = x^{2} + 192

⇒ 16x = 128

⇒ x = 8 units

AC = AM + CM = 16 units

**Question 7.**m∠B = 90 in ΔABC. is altitude to .

If BM = 15, AC = 34, find AB.

**Answer:**

Let AM = x, CM = 34 – x

BM^{2} = AM × MC

⇒ x^{2} – 34x = – 225

⇒ x^{2} – 34x + 225 = 0

⇒ x^{2} – 25x – 9x + 225 = 0

⇒ x(x – 25) – 9(x – 25) = 0

⇒ (x – 9)(x – 25) = 0

⇒ x = 9, 25

AB^{2} = 15^{2} + 9^{2}

⇒ AB^{2} = 306

AB = √306 units

**Question 8.**m∠B = 90 in ΔABC. is altitude to .

If BM = 2√30, MC = 6, find AC.

**Answer:**

BM = 2

MC = 6

Applying Pythagoras theorem :

BM^{2} = CM × AM

⇒ 120 = 6 × AM

⇒ AM = 20

AC = AM + MC = 26 units

**Question 9.**m∠B = 90 in ΔABC. is altitude to .

If AB = √10, AM = 2.5, find MC.

**Answer:**

AB =

AM = 2.5

Applying Pythagoras theorem in ΔABM

AB^{2} = AM^{2} + BM^{2}

⇒ BM^{2} = 10 – 6.25 = 3.75

Let CM = x

AC = x + 2.5

Applying Pythagoras theorem in ΔCBM

⇒ BC^{2} = BM^{2} + CM^{2}

⇒ BC^{2} = x^{2} + 3.75 …Equation (i)

Applying Pythagoras theorem in ΔABC

AC^{2} = AB^{2} + BC^{2}

(x + 2.5)^{2} = BC^{2} + 10

⇒ x^{2} + 6.25 + 5x = BC^{2} + 10

⇒ BC^{2} = x^{2} – 3.75 + 5x …Equation (ii)

Equating Equation (i) and Equation (ii)

x^{2} + 3.75 = x^{2} – 3.75 + 5x

⇒ 5x = 7.5

⇒ x = 1.5

So CM = 1.5

**Question 10.**In ΔPQR m∠Q = 90, PQ = x, QR = y and . D . Find PD, QD, RD in terms of x and y.

**Answer:**mQ = 90

Applying Pythagoras theorem

PR = √(x^{2} + y^{2})

PQ^{2} = PD × PR

QR^{2} = RD × PR

QD^{2} = PD × RD

⇒ QD^{2}

**Question 11.**∠Q is a right angle in ΔPQR and , M . If PQ = 4QR, then prove that PM = 16RM.

**Answer:**mQ = 90

PQ = 4QR

PQ^{2} = PR × PM …Equation(i)

QR^{2} = PR × RM …Equation(ii)

Dividing Equation (ii) by Equation(i) we get

Hence PM = 16RM

**Question 12.**âPQRS is a rectangle. If PQ + QR = 7 and PR + QS = 10, then find the area of â PQRS.

**Answer:**PQ + QR = 7

PR + QS = 10

Since Diagonals of a rectangle are of equal length, so

PR = QS = 5

Let PQ be x, QR = 7 – x

Applying Pythagoras theorem :

x^{2} + (7 – x)^{2} = 5^{2}

⇒ 2x^{2} – 14x + 24 = 0

⇒ x^{2} – 7x + 12 = 0

⇒ x^{2} – 4x – 3x + 12 = 0

⇒ x(x – 4) – 3(x – 4) = 0

⇒ (x – 3)(x – 4) = 0

x = 3, 4

∴ The two sides are 3 and 4

Area of the Rectangle = (3 × 4) = 12 sq. units

**Question 13.**The diagonals of a convex â ABCD intersect at right angles. Prove that AB^{2} + CD^{2} = AD^{2} + BC^{2}.

**Answer:**

Using Pythagoras theorem :

AO^{2} + OB^{2} = AB^{2} …Equation(i)

DO^{2} + OC^{2} = CD^{2} …Equation(ii)

AO^{2} + OD^{2} = AD^{2} …Equation(iii)

BO^{2} + OC^{2} = BC^{2} …Equation(iv)

Adding Equation(i) and Equation(ii)

AB^{2} + CD^{2} = AO^{2} + OB^{2} + DO^{2} + OC^{2} …Equation(v)

Adding Equation(iii) and Equation(iv)

AD^{2} + BC^{2} = AO^{2} + OB^{2} + DO^{2} + OC^{2} …Equation(vi)

The RHS of equation (v) and (vi) are similar

So we can say that

AB^{2} + CD^{2} = AD^{2} + BC^{2}

**Question 14.**In ΔPQR, m∠Q = 90, M and N . Prove that PM^{2} + RN^{2} = PR^{2} + MN^{2}.

**Answer:**

mQ = 90

Applying Pythagoras theorem

PM^{2} = QM^{2} + PQ^{2}

RN^{2} = QN^{2} + QR^{2}

Adding the above two equations we get

PM^{2} + RN^{2} = QM^{2} + PQ^{2} + QN^{2} + QR^{2}

⇒ PM^{2} + RN^{2} = (QM^{2} + QN^{2}) + (PQ^{2} + QR^{2})

⇒ PM^{2} + RN^{2} = MN^{2} + PR^{2}

**Question 15.**The sides of a triangle have lengths a^{2} + b^{2}, 2ab, a^{2} — b^{2}, where a > b and a, b ϵ R ^{+} . Prove that the angle opposite to the side having length a^{2} + b^{2} is a right angle.

**Answer:**Let a^{2} + b^{2} = p, 2ab = q, a^{2} — b^{2} = r

Since a, b ϵ R ^{+ ,} hence

p is the largest side of the triangle

p^{2} = (a^{2} + b^{2})^{2} = a^{4} + b^{4} + 2a^{2}b^{2}

q^{2} + r^{2} = 4a^{2}b^{2} + (a^{2} – b^{2})^{2}

⇒ q^{2} + r^{2} = a^{4} + b^{4} + 2a^{2}b^{2}

So p^{2} = q^{2} + r^{2}

Hence the Pythagoras theorem is established

Hence the angle opposite to the side p = a^{2} + b^{2} is a right angle

**Question 16.**In ΔABC, m∠B = 90 and is a median. Prove that AB^{2} + BC^{2} + AC^{2} = 8BE^{2}.

**Answer:**

mB = 90

is a median

∴ AE = EC

⇒ AC = 2AE

Applying Pythagoras Theorem :

AB^{2} + BC^{2} = AC^{2} …Equation (i)

Applying Appoloneous theorem

AB^{2} + BC^{2} = 2(AE^{2} + BE^{2})

⇒ AC^{2} = 2(AE^{2} + BE^{2})

Now since AC = 2AE

⇒ (2AE)^{2} = 2(AE^{2} + BE^{2})

⇒ 2AE^{2} = 2BE^{2}

⇒ AE = BE …Equation (ii)

∴ AB^{2} + BC^{2} + AC^{2} = 2AC^{2} (From Equation (i))

⇒ AB^{2} + BC^{2} + AC^{2} = 2 × (2AE)^{2} = 8AE^{2}

From Equation (ii) we can say that

⇒ AB^{2} + BC^{2} + AC^{2} = 8BE^{2}

**Question 17.**AB = AC and ∠A is right angle in ΔABC. If BC = √2 a, then find the area of the triangle. (a ∈ R, a > 0)

**Answer:**

mA = 90

Let AB = AC = x

BC = a

Applying Pythagoras Theorem :

∴ x^{2} + x^{2} = 2a^{2}

⇒ 2x^{2} = 2a^{2}

⇒ x^{2} = a^{2}

⇒ x = a

∴ Area of the triangle

⇒ Area of the triangle

**Question 1.**

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

**Answer:**

m∠B = 90

AD = 4DC (Given) …Equation (i)

∴ BD^{2} = AD × DC

⇒ BD^{2} = 4DC^{2} (From Equation (i))

Taking Square Root both sides we get

∴ BD = 2DC

**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.

**Answer:**

Hypotenuse (AB) = 13

Perpendicular (AC) = 12

Base (BC) = 5

∴ AC^{2} + BC^{2} = 5^{2} + 12^{2}

⇒ AC^{2} + BC^{2} = 25 + 144

⇒ AC^{2} + BC^{2} = 169 = AB^{2}

Hence Pythagoras theorem is valid

m∠C = 90

Let AD = x, BD = 13 – x

Applying Pythagoras theorem

∴ Altitude CD^{2} = 12^{2} – x^{2} = 5^{2} – (13 – x) ^{2}

⇒ 144 – x^{2} = 25 – 169 – x^{2} + 26x

⇒ 26x = 288

⇒ x = 11.07 units

So Altitude CD = √21.43 = 4.63 units

**Question 3.**

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

**Answer:**

m∠Q = 90

RM = 12

MP = 8

∴ RP = (RM + MP) = 20

Let QM be x

Applying Pythagoras theorem in ΔQMP, we get

QP^{2} = x^{2} + 8^{2} = x^{2} + 64

Applying Pythagoras theorem in ΔQMR, we get

QR^{2} = x^{2} + 12^{2} = x^{2} + 144

Applying Pythagoras theorem in ΔPQR, we get

RP^{2} = (x^{2} + 64) + (x^{2} + 144)

⇒ 400 = 208 + 2x^{2}

⇒ x^{2} = 96

⇒ x = 4√6

⇒ QM = 4√6

PQ^{2} = x^{2} + 64

⇒ PQ^{2} = 160

⇒ PQ = 4√10

QR^{2} = x^{2} + 144

⇒ QR^{2} = 240

⇒ QR = 4√15

**Question 4.**

In ΔABC, m∠B = 90, , M. If AM — MC = 7, AB^{2} — BC^{2} = 175, find AC.

**Answer:**

mB = 90

AM — MC = 7

Applying Pythagoras theorem in ΔABM and ΔCBM

AB^{2} = BM^{2} + AM^{2} …Equation (i)

BC^{2} = BM^{2} + MC^{2} …Equation (ii)

Subtracting Equation (ii) from Equation (i) we get

AB^{2} – BC^{2} = AM^{2} – MC^{2}

⇒ (AM – MC)(AM + MC) = AB^{2} – BC^{2}

Putting the values we get :

7 × (AM + MC) = 175

⇒ AM + MC = 25

⇒ AC = 25

**Question 5.**

∠A is right angle in ΔABC. AD is an altitude of the triangle. If AB = √5, BD = 2, find the length of the hypotenuse of the triangle.

**Answer:**

m∠A = 90

AB = √5

BD = 2

Applying Pythagoras theorem in ΔADB we get

AD^{2} = AB^{2} – BD^{2}

⇒ AD^{2} = 5 – 4 = 1

⇒ AD = 1

Let AC = x

Applying Pythagoras theorem in ΔADC we get

DC^{2} = AC^{2} – AD^{2}

⇒ DC^{2} = x^{2} – 1

⇒ DC = √(x^{2} – 1)

Applying Pythagoras theorem in ΔABC we get

AB^{2} + AC^{2} = BC^{2}

⇒ 5 + x^{2} = (2 + √(x^{2} – 1))^{2}

⇒ 5 + x^{2} = 4 + x^{2} – 1 + 4√(x^{2} – 1)

Hypotenuse BC = BD + DC

⇒ Hypotenuse BC

**Question 6.**

m∠B = 90 in ΔABC. is altitude to .

If AM = BM = 8, find AC.

**Answer:**

Applying Pythagoras theorem in ΔABM

AB^{2} = AM^{2} + BM^{2}

⇒ AB^{2} = 64 + 64 = 128

Let CM = x

Applying Pythagoras theorem in ΔCBM

⇒ BC^{2} = BM^{2} + CM^{2}

⇒ BC^{2} = x^{2} + 64

Applying Pythagoras theorem in ΔABC

AC^{2} = AB^{2} + BC^{2}

(x + 8)^{2} = x^{2} + 64 + 128

⇒ x^{2} + 64 + 16x = x^{2} + 192

⇒ 16x = 128

⇒ x = 8 units

AC = AM + CM = 16 units

**Question 7.**

m∠B = 90 in ΔABC. is altitude to .

If BM = 15, AC = 34, find AB.

**Answer:**

Let AM = x, CM = 34 – x

BM^{2} = AM × MC

⇒ x^{2} – 34x = – 225

⇒ x^{2} – 34x + 225 = 0

⇒ x^{2} – 25x – 9x + 225 = 0

⇒ x(x – 25) – 9(x – 25) = 0

⇒ (x – 9)(x – 25) = 0

⇒ x = 9, 25

AB^{2} = 15^{2} + 9^{2}

⇒ AB^{2} = 306

AB = √306 units

**Question 8.**

m∠B = 90 in ΔABC. is altitude to .

If BM = 2√30, MC = 6, find AC.

**Answer:**

BM = 2

MC = 6

Applying Pythagoras theorem :

BM^{2} = CM × AM

⇒ 120 = 6 × AM

⇒ AM = 20

AC = AM + MC = 26 units

**Question 9.**

m∠B = 90 in ΔABC. is altitude to .

If AB = √10, AM = 2.5, find MC.

**Answer:**

AB =

AM = 2.5

Applying Pythagoras theorem in ΔABM

AB^{2} = AM^{2} + BM^{2}

⇒ BM^{2} = 10 – 6.25 = 3.75

Let CM = x

AC = x + 2.5

Applying Pythagoras theorem in ΔCBM

⇒ BC^{2} = BM^{2} + CM^{2}

⇒ BC^{2} = x^{2} + 3.75 …Equation (i)

Applying Pythagoras theorem in ΔABC

AC^{2} = AB^{2} + BC^{2}

(x + 2.5)^{2} = BC^{2} + 10

⇒ x^{2} + 6.25 + 5x = BC^{2} + 10

⇒ BC^{2} = x^{2} – 3.75 + 5x …Equation (ii)

Equating Equation (i) and Equation (ii)

x^{2} + 3.75 = x^{2} – 3.75 + 5x

⇒ 5x = 7.5

⇒ x = 1.5

So CM = 1.5

**Question 10.**

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

**Answer:**

mQ = 90

Applying Pythagoras theorem

PR = √(x^{2} + y^{2})

PQ^{2} = PD × PR

QR^{2} = RD × PR

QD^{2} = PD × RD

⇒ QD^{2}

**Question 11.**

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

**Answer:**

mQ = 90

PQ = 4QR

PQ^{2} = PR × PM …Equation(i)

QR^{2} = PR × RM …Equation(ii)

Dividing Equation (ii) by Equation(i) we get

Hence PM = 16RM

**Question 12.**

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

**Answer:**

PQ + QR = 7

PR + QS = 10

Since Diagonals of a rectangle are of equal length, so

PR = QS = 5

Let PQ be x, QR = 7 – x

Applying Pythagoras theorem :

x^{2} + (7 – x)^{2} = 5^{2}

⇒ 2x^{2} – 14x + 24 = 0

⇒ x^{2} – 7x + 12 = 0

⇒ x^{2} – 4x – 3x + 12 = 0

⇒ x(x – 4) – 3(x – 4) = 0

⇒ (x – 3)(x – 4) = 0

x = 3, 4

∴ The two sides are 3 and 4

Area of the Rectangle = (3 × 4) = 12 sq. units

**Question 13.**

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

**Answer:**

Using Pythagoras theorem :

AO^{2} + OB^{2} = AB^{2} …Equation(i)

DO^{2} + OC^{2} = CD^{2} …Equation(ii)

AO^{2} + OD^{2} = AD^{2} …Equation(iii)

BO^{2} + OC^{2} = BC^{2} …Equation(iv)

Adding Equation(i) and Equation(ii)

AB^{2} + CD^{2} = AO^{2} + OB^{2} + DO^{2} + OC^{2} …Equation(v)

Adding Equation(iii) and Equation(iv)

AD^{2} + BC^{2} = AO^{2} + OB^{2} + DO^{2} + OC^{2} …Equation(vi)

The RHS of equation (v) and (vi) are similar

So we can say that

AB^{2} + CD^{2} = AD^{2} + BC^{2}

**Question 14.**

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

**Answer:**

mQ = 90

Applying Pythagoras theorem

PM^{2} = QM^{2} + PQ^{2}

RN^{2} = QN^{2} + QR^{2}

Adding the above two equations we get

PM^{2} + RN^{2} = QM^{2} + PQ^{2} + QN^{2} + QR^{2}

⇒ PM^{2} + RN^{2} = (QM^{2} + QN^{2}) + (PQ^{2} + QR^{2})

⇒ PM^{2} + RN^{2} = MN^{2} + PR^{2}

**Question 15.**

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

**Answer:**

Let a^{2} + b^{2} = p, 2ab = q, a^{2} — b^{2} = r

Since a, b ϵ R ^{+ ,} hence

p is the largest side of the triangle

p^{2} = (a^{2} + b^{2})^{2} = a^{4} + b^{4} + 2a^{2}b^{2}

q^{2} + r^{2} = 4a^{2}b^{2} + (a^{2} – b^{2})^{2}

⇒ q^{2} + r^{2} = a^{4} + b^{4} + 2a^{2}b^{2}

So p^{2} = q^{2} + r^{2}

Hence the Pythagoras theorem is established

Hence the angle opposite to the side p = a^{2} + b^{2} is a right angle

**Question 16.**

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

**Answer:**

mB = 90

is a median

∴ AE = EC

⇒ AC = 2AE

Applying Pythagoras Theorem :

AB^{2} + BC^{2} = AC^{2} …Equation (i)

Applying Appoloneous theorem

AB^{2} + BC^{2} = 2(AE^{2} + BE^{2})

⇒ AC^{2} = 2(AE^{2} + BE^{2})

Now since AC = 2AE

⇒ (2AE)^{2} = 2(AE^{2} + BE^{2})

⇒ 2AE^{2} = 2BE^{2}

⇒ AE = BE …Equation (ii)

∴ AB^{2} + BC^{2} + AC^{2} = 2AC^{2} (From Equation (i))

⇒ AB^{2} + BC^{2} + AC^{2} = 2 × (2AE)^{2} = 8AE^{2}

From Equation (ii) we can say that

⇒ AB^{2} + BC^{2} + AC^{2} = 8BE^{2}

**Question 17.**

AB = AC and ∠A is right angle in ΔABC. If BC = √2 a, then find the area of the triangle. (a ∈ R, a > 0)

**Answer:**

mA = 90

Let AB = AC = x

BC = a

Applying Pythagoras Theorem :

∴ x^{2} + x^{2} = 2a^{2}

⇒ 2x^{2} = 2a^{2}

⇒ x^{2} = a^{2}

⇒ x = a

∴ Area of the triangle

⇒ Area of the triangle

###### Exercise 7.2

**Question 1.**In rectangle ABCD, AB + BC = 23, AC + BD = 34. Find the area of the rectangle.

**Answer:**AB + BC = 23

AC + BD = 34

Since Diagonals of a rectangle are of equal length, so

AC = BD = 17

Let AB be x, BC = 23 – x

Applying Pythagoras theorem :

x^{2} + (23 – x)^{2} = 17^{2}

⇒ 2x^{2} – 46x + 240 = 0

⇒ x^{2} – 23x + 120 = 0

⇒ x^{2} – 15x – 8x + 120 = 0

⇒ x(x – 15) – 8(x – 15) = 0

⇒ (x – 8)(x – 15) = 0

x = 8, 15

∴ The two sides are 8 and 15

Area of the Rectangle = (8 × 15) = 120 sq. units

**Question 2.**In ΔABC m∠A = m∠B + m∠C, AB = 7, BC = 25. Find the perimeter of ΔABC.

**Answer:**Let m∠A = x

Sum of interior angles = mA = mB + mC

⇒ mA + mB + mC = 180 (Sum of interior angle of triangle is 180)

⇒ 2x = 180

⇒ x = 90

So ∠A is right angled

Applying Pythagoras theorem

AC^{2} = BC^{2} – AB^{2}

⇒ AC^{2} = 25^{2} – 7^{2}

⇒ AC^{2} = 576

⇒ AC = 24

Perimeter of ΔABC = (25 + 24 + 7) = 56 units.

**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.

**Answer:**Length of staircase = 6.5 metres

Height of Wall = 6 metres

Let the distance of base be d

Applying Pythagoras theorem

d^{2} = 6.5^{2} – 6^{2}

⇒ d^{2} = 6.25

⇒ d = 2.5 metres

Distance of base = 2.5 m

**Question 4.**In ΔABC AB = 7, AC = 5, AD = 5. Find , if the mid – point of BC is D.

**Answer:**

D is the mid – point of BC

AB = 7

AC = 5

AD = 5

Let BD = x

Applying Appoloneus theorem :

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 7^{2} + 5^{2} = 2(5^{2} + x^{2})

⇒ 74 = 50 + 2x^{2}

⇒ 2x^{2} = 24

⇒ x^{2} = 12

⇒ x = 2√3 units

BC = 2 × BD = 4√3 units

**Question 5.**In equilateral ΔABC, D such that BD : DC = 1 : 2. Prove that 3AD = √7 AB.

**Answer:**BD : DC = 1 : 2

Let BD = x, DC = 2x –

⇒ AB = BC = AC = 3x

Let M be the mid point of BC

BM

∴ DM

⇒ DM

Applying Pythagoras theorem :

AB^{2} = BM^{2} + AM^{2}

⇒ AM^{2}

⇒ AM^{2}

Applying Pythagoras theorem :

AD^{2} = AM^{2} + DM^{2}

⇒ AD^{2}

⇒ AD^{2} = 7x^{2}

⇒ AD = √7x

So

AB : AD = 3x : √7x

⇒ AB : AD = 3 : √7

So 3AD = AB

**Question 6.**In ΔABC, AB = 17, BC = 15, AC = 8, find the length of the median on the largest side.

**Answer:**AB = 17

BC = 15

AC = 8

Let the median meet the side AB at point D

AD = 9.5

∴ CD is the median on the largest side

Applying Appoloneus theorem :

AC^{2} + BC^{2} = 2(AD^{2} + CD^{2})

⇒ 8^{2} + 15^{2} = 2(9.5^{2} + CD^{2})

⇒ 144.5 = 90.25 + CD^{2}

⇒ CD^{2} = 54.25

⇒ CD = 7.36 units

**Question 7.** is a median of ΔABC. AB^{2} + AC^{2} = 148 and AD = 7. Find BC.

**Answer:**AB^{2} + AC^{2} = 148

AD = 7

Applying Apploneus theorem :

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 148 = 2(49 + BD^{2})

⇒ BD^{2} = 74 – 49 = 25

⇒ BD = 5 units

BC = 2 × BD = 10 units

**Question 8.**In rectangle ABCD, AC = 25 and CD = 7. Find perimeter of the rectangle.

**Answer:**AC = 25

CD = 7

Applying Pythagoras theorem :

AD^{2} = AC^{2} – CD^{2}

⇒ AD^{2} = 25^{2} – 7^{2} = 576

⇒ AD = 24 units

Perimeter of rectangle = 2(AD + CD) = 62 units

**Question 9.**In rhombus XYZW, XZ = 14 and YW = 48. Find XY.

**Answer:**

XZ = 14

⇒ XO = 7

YW = 48

⇒ YO = 24

Since Diagonals of a Rhombus bisect at Right angles,

So we apply Pythagoras theorem in ΔXOY

XY^{2} = XO^{2} + YO^{2}

⇒ XY^{2} = 49 + 576 = 625

⇒ XY = 25 units

**Question 10.**In ΔPQR, m∠Q : m∠R : m∠P = 1 : 2 : 1. If PQ = 2√6, find PR.

**Answer:****Given**: Ratio of angles of triangles = mQ : mR : mP = 1 : 2 : 1

Let mQ = x, mR = 2 x, mP = x

Sum of angles of triangle = 180°

mQ + mR + mP = 180°

⇒ x + 2x + x = 180°

⇒ 4x = 180

⇒ x = 45

So by putting value of x we get,

⇒ mR = 90

⇒ mQ = 45°

⇒ mP = 45°

It is an isosceles right angled triangle at R.

Let PR = QR = d

According to pythagoras theorem, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides

Applying Pythagoras theorem

PR^{2} + QR^{2} = PQ^{2}

2d^{2} = PQ^{2}

⇒ 2d^{2} = 24

⇒ d^{2} = 12

⇒ d = 2√3

So PR = 2√3 units

and QR = 2√3 units

**Question 1.**

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

**Answer:**

AB + BC = 23

AC + BD = 34

Since Diagonals of a rectangle are of equal length, so

AC = BD = 17

Let AB be x, BC = 23 – x

Applying Pythagoras theorem :

x^{2} + (23 – x)^{2} = 17^{2}

⇒ 2x^{2} – 46x + 240 = 0

⇒ x^{2} – 23x + 120 = 0

⇒ x^{2} – 15x – 8x + 120 = 0

⇒ x(x – 15) – 8(x – 15) = 0

⇒ (x – 8)(x – 15) = 0

x = 8, 15

∴ The two sides are 8 and 15

Area of the Rectangle = (8 × 15) = 120 sq. units

**Question 2.**

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

**Answer:**

Let m∠A = x

Sum of interior angles = mA = mB + mC

⇒ mA + mB + mC = 180 (Sum of interior angle of triangle is 180)

⇒ 2x = 180

⇒ x = 90

So ∠A is right angled

Applying Pythagoras theorem

AC^{2} = BC^{2} – AB^{2}

⇒ AC^{2} = 25^{2} – 7^{2}

⇒ AC^{2} = 576

⇒ AC = 24

Perimeter of ΔABC = (25 + 24 + 7) = 56 units.

**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.

**Answer:**

Length of staircase = 6.5 metres

Height of Wall = 6 metres

Let the distance of base be d

Applying Pythagoras theorem

d^{2} = 6.5^{2} – 6^{2}

⇒ d^{2} = 6.25

⇒ d = 2.5 metres

Distance of base = 2.5 m

**Question 4.**

In ΔABC AB = 7, AC = 5, AD = 5. Find , if the mid – point of BC is D.

**Answer:**

D is the mid – point of BC

AB = 7

AC = 5

AD = 5

Let BD = x

Applying Appoloneus theorem :

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 7^{2} + 5^{2} = 2(5^{2} + x^{2})

⇒ 74 = 50 + 2x^{2}

⇒ 2x^{2} = 24

⇒ x^{2} = 12

⇒ x = 2√3 units

BC = 2 × BD = 4√3 units

**Question 5.**

In equilateral ΔABC, D such that BD : DC = 1 : 2. Prove that 3AD = √7 AB.

**Answer:**

BD : DC = 1 : 2

Let BD = x, DC = 2x –

⇒ AB = BC = AC = 3x

Let M be the mid point of BC

BM

∴ DM

⇒ DM

Applying Pythagoras theorem :

AB^{2} = BM^{2} + AM^{2}

⇒ AM^{2}

⇒ AM^{2}

Applying Pythagoras theorem :

AD^{2} = AM^{2} + DM^{2}

⇒ AD^{2}

⇒ AD^{2} = 7x^{2}

⇒ AD = √7x

So

AB : AD = 3x : √7x

⇒ AB : AD = 3 : √7

So 3AD = AB

**Question 6.**

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

**Answer:**

AB = 17

BC = 15

AC = 8

Let the median meet the side AB at point D

AD = 9.5

∴ CD is the median on the largest side

Applying Appoloneus theorem :

AC^{2} + BC^{2} = 2(AD^{2} + CD^{2})

⇒ 8^{2} + 15^{2} = 2(9.5^{2} + CD^{2})

⇒ 144.5 = 90.25 + CD^{2}

⇒ CD^{2} = 54.25

⇒ CD = 7.36 units

**Question 7.**

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

**Answer:**

AB^{2} + AC^{2} = 148

AD = 7

Applying Apploneus theorem :

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 148 = 2(49 + BD^{2})

⇒ BD^{2} = 74 – 49 = 25

⇒ BD = 5 units

BC = 2 × BD = 10 units

**Question 8.**

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

**Answer:**

AC = 25

CD = 7

Applying Pythagoras theorem :

AD^{2} = AC^{2} – CD^{2}

⇒ AD^{2} = 25^{2} – 7^{2} = 576

⇒ AD = 24 units

Perimeter of rectangle = 2(AD + CD) = 62 units

**Question 9.**

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

**Answer:**

XZ = 14

⇒ XO = 7

YW = 48

⇒ YO = 24

Since Diagonals of a Rhombus bisect at Right angles,

So we apply Pythagoras theorem in ΔXOY

XY^{2} = XO^{2} + YO^{2}

⇒ XY^{2} = 49 + 576 = 625

⇒ XY = 25 units

**Question 10.**

In ΔPQR, m∠Q : m∠R : m∠P = 1 : 2 : 1. If PQ = 2√6, find PR.

**Answer:**

**Given**: Ratio of angles of triangles = mQ : mR : mP = 1 : 2 : 1

Let mQ = x, mR = 2 x, mP = x

Sum of angles of triangle = 180°

mQ + mR + mP = 180°

⇒ x + 2x + x = 180°

⇒ 4x = 180

⇒ x = 45

So by putting value of x we get,⇒ mR = 90

⇒ mQ = 45°

⇒ mP = 45°

It is an isosceles right angled triangle at R.

Let PR = QR = d

According to pythagoras theorem, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sidesApplying Pythagoras theorem

PR^{2} + QR^{2} = PQ^{2}

2d^{2} = PQ^{2}

⇒ 2d^{2} = 24

⇒ d^{2} = 12

⇒ d = 2√3

So PR = 2√3 units

and QR = 2√3 units

###### Exercise 7

**Question 1.** , , are the medians of ΔABC. If BE = 12, CF = 9 and AB^{2} + BC^{2} + AC^{2} = 600, BC = 10, find AD.

**Answer:**From the given info,

As AB^{2} + BC^{2} + AC^{2} = 600

and BC = 10,

⇒ AB^{2} + AC^{2} = 500 ...... (1)

As AD is median,

⇒ BD = 5 ...... (2)

In Δ ABC, using the theorem of Apolloneous, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

Substituting values for (1) and (2) in above equation,

⇒ 500 = 2 (AD^{2} + 5^{2})

⇒ 250 = AD^{2} + 25

⇒ AD^{2} = 225

⇒ AD^{2} = 15^{2}

⇒ AD = 15

**Question 2.** is the altitude of Δ ABC such that B—D—C. If AD^{2} = BD ⋅ DC, prove that ∠BAC is right angle.

**Answer:**

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB^{2} = BD^{2} + AD^{2} ...... (1)

And, AC^{2} = AD^{2} + DC^{2} ...... (2)

It is given that AD^{2} = BD.DC …… (3)

From equation (1) and (3), we get,

AB^{2} = BD^{2} + BD.DC

⇒ AB^{2} = BD(BD + DC) = BD. BC

Now, from equation (2) and (3),

AC^{2} = BD.DC + DC^{2}

⇒ AC^{2} = DC (BD + DC)

⇒ AC^{2} = DC(BC)

Also, BC = BD + DC,

Substituting the value of BD and DC in above equation, we get,

⇒ BC^{2} = AB^{2} + AC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**Question 3.**In ΔABC, , B—D—C. If AB^{2} = BD ⋅ BC, prove that ∠ BAC is a right angle.

**Answer:**

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB^{2} = BD^{2} + AD^{2} ...... (1)

And, AC^{2} = AD^{2} + DC^{2} ...... (2)

It is given that AB^{2} = BD.BC

From equation (1) and (2), we get,

AB^{2} – BD^{2} = AC^{2} – DC^{2}

⇒ AB^{2} – BD^{2} = AC^{2} – (BC – BD)^{2}

⇒ BC^{2} + BD^{2} – 2BC.BD = AC^{2} + BD^{2} – AB^{2}

⇒ BC^{2} – 2BC (BD) = AC^{2} – AB^{2}

Substituting the value of BD above, we get,

⇒ BC^{2} – 2(AB)^{2} = AC^{2} – AB^{2}

⇒ BC^{2} = AB^{2} + AC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**Question 4.**In ΔABC, , B—D—C. If AC^{2} = CD ⋅ BC, prove that ∠BAC is a right angle.

**Answer:**

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB^{2} = BD^{2} + AD^{2} ...... (1)

And, AC^{2} = AD^{2} + DC^{2} ...... (2)

It is given that AC^{2} = CD.BC

From equation (1) and (2), we get,

AB^{2} – BD^{2} = AC^{2} – CD^{2}

⇒ AB^{2} – (BC – CD)^{2} = AC^{2} – CD^{2}

⇒ AB^{2} – BC^{2} – CD^{2} + 2.BC.CD = AC^{2} – CD^{2}

⇒ AB^{2} – BC^{2} + 2.BC.CD = AC^{2}

Substituting the value of CD above, we get,

⇒ AB^{2} – BC^{2} + 2AC^{2} = AC^{2}

⇒ AB^{2} + AC^{2} = BC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**Question 5.**is a median of ΔABC. If BD = AD, prove that ∠A is a right angle in ΔABC.

**Answer:**

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

As it is given that AD = BD,

⇒ AB^{2} + AC^{2} = 4.BD^{2}

Since AD is the median to line BC, we have,

Substituting the value of BD in above equation, we get,

AB^{2} + AC^{2} = BC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**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.

**Answer:**Let the length of upper part of pole be x,

The length of AB becomes 25 – x.

Applying Pythagoras theorem in ΔABC,

(AC’)^{2} = AB^{2} + (BC’)^{2}

⇒ x^{2} = (25 – x)^{2} + 15^{2}

⇒ x^{2} = 625 + x^{2} – 50x + 225

⇒ 50x = 850

⇒ x = 17 m

∴ The length of the upper part of the pole is 17 meters.

**Question 7.**In ΔABC, AB > AC, D is the mid–point of . such that B—M—C. Prove that AB^{2} — AC^{2} = 2BC . DM.

**Answer:**From the given information, we construct the figure as:

As AM is perpendicular BC, ΔACM and ΔABM are right triangle.

By using Pythagoras Theorem, we get,

AC^{2} = CM^{2} + AM^{2} ...... (1)

Also, AB^{2} = BM^{2} + AM^{2} ...... (2)

Eliminating AM^{2} from equations (1) and (2), we get,

AC^{2} – CM^{2} = AB^{2} – BM^{2}

⇒ AC^{2} – AB^{2} = CM^{2} – BM^{2}

By using the identity a^{2} – b^{2} = (a + b)(a – b) on above equation,

⇒ AC^{2} – AB^{2} = (CM + BM)(CM – BM)

⇒ AC^{2} – AB^{2} = BC(CM – (BC – CM))

⇒ AB^{2} – AC^{2} = BC(2CM – BC)

As D is mid point of BC,

⇒ AB^{2} – AC^{2} = BC(2CM – 2BD)

⇒ AB^{2} – AC^{2} = 2BC(CM – BD)

⇒ AB^{2} – AC^{2} = 2BC(CM – CD)

⇒ AB^{2} – AC^{2} = 2BC(–DM)

⇒ AB^{2} — AC^{2} = 2BC.DM

Hence, proved.

**Question 8.**In ΔABC, , D and ∠B is right angle. If AC = 5CD, prove that BD = 2CD.

**Answer:**

ΔABC is a right triangle, right angle at B and BD is perpendicular from B vertex to hypotenuse AC.

So, as we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

⇒ BD^{2} = AD.CD

⇒ BD^{2} = (AC – CD).CD

As, AC = 5CD,

⇒ BD^{2} = (5CD – CD).CD

⇒ BD^{2} = 4CD^{2}

⇒ BD^{2} = (2CD)^{2}

⇒ BD = 2CD

Hence, proved.

**Question 9.**In ΔPQR, if m∠P + m∠Q = m∠R. PR = 7, QR = 24, then PQ =

A. 31

B. 25

C. 17

D. 15

**Answer:**

In ΔABC, we know that sum of all interior angles in triangle is equal to 180°

⇒ ∠P + ∠Q + ∠R = 180

Also, given that, ∠P + ∠Q = ∠R,

⇒ 2∠R = 180

⇒ ∠R = 90°

Using Pythagoras Theorem,

PQ^{2} = PR^{2} + QR^{2}

⇒ PQ^{2} = 7^{2} + (24)^{2}

⇒ PQ^{2} = 49 + 576

⇒ PQ^{2} = 625

⇒ PQ = 25

∴ Option (b) is correct.

**Question 10.**In ΔABC, is an altitude and ∠A is right angle. If AB = , BD = 4, then CD =

A. 5

B. 3

C. √5

D. 1

**Answer:**

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of each side other than the hypotenuse is the geometric mean of length of hypotenuse and segment of hypotenuse adjacent to the side.

⇒ AB^{2} = BD.BC

⇒ 20 = 4. BC

⇒ BC = 5

Also, CD = BC – BD

CD = 5 – 4

CD = 1

∴ Option (d) is correct.

**Question 11.**In ΔABC, AB^{2} + AC^{2} = 50. The length of the median AD = 3. So, BC =

A. 4

B. 24

C. 8

D. 16

**Answer:**

In Δ ABC, using the theorem of Apolloneous, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 50 = 2(3^{2} + BD^{2})

⇒ 50 = 2 (9 + BD^{2})

⇒ 25 = 9 + BD^{2}

⇒ BD^{2} = 16

⇒ BD = 4

⇒ BC = 2BD = 8

∴ Option (c) is correct.

**Question 12.**In ΔABC, m∠B = 90, AB = BC. Then AB: AC =

A. 1:3

B. 1:2

C. 1 : √2

D. √2:1

**Answer:**

In ΔABC, applying Pythagoras Theorem, we get,

AC^{2} = AB^{2} + BC^{2}

As AB = BC,

⇒ AC^{2} = 2AB^{2}

⇒ AB:AC = 1:√2

∴ Option (c) is correct.

**Question 13.**In ΔABC, m∠B = 90 and AC = 10. The length of the median BM =

A. 5

B. 5√2

C. 6

D. 8

**Answer:**

In Δ ABC, using the theorem of Apolloneous, we know that,

For BM to be the median, we have,

AB^{2} + AC^{2} = 2(BM^{2} + CM^{2})

Also, as Δ ABC is right angled at B,

⇒ AB^{2} + AC^{2} = AC^{2} = 10^{2} = 100

⇒ 100 = 2 (BM^{2} + 5^{2})

⇒ 50 = BM^{2} + 25

⇒ BM = 5

∴ Option (a) is correct.

**Question 14.**In ΔABC, AB = BC = . m∠B

A. Is acute

B. Is obtuse

C. Is right angle

D. Cannot be obtained

**Answer:**

Let us assume,

⇒ AB = BC = x and AC = √2x

As we see that, AC^{2} = AB^{2} + BC^{2}

So by converse of Pythagoras Theorem, ΔABC is right angled and right angle at B.

∴ Option (c) is correct.

**Question 15.**In ΔABC, if then m ∠C = ……

A. 90

B. 30

C. 60

D. 45

**Answer:**

Let us assume,

⇒ AB = x, AC = 2x and BC = √3x

As we see that, AC^{2} = AB^{2} + BC^{2}

So by converse of Pythagoras Theorem, ΔABC is right angled and right angle at B.

Let ∠C = θ,

⇒ θ = 30°

∴ Option (b) is correct.

**Question 16.**In ΔXYZ, m∠X : m∠Y : m∠Z = 1 : 2 : 3. If XY = 15, YZ = …

A.

B. 17

C. 8

D. 7.5

**Answer:**

Let ∠X = A,

⇒ ∠Y = 2A and ∠Z = 3A

As sum of all angles in a triangle is equal to 180°

∠X + ∠Y + ∠Z = 180

⇒ 6A = 180

⇒ A = 30

⇒ ∠Z = 90°

⇒ YZ = 7.5

∴ Option (d) is correct.

**Question 17.**In Δ ABC, ∠B is a right angle and is an altitude. If AD = BD = 5, then DC =

A. 1

B. √5

C. 5

D. 2.5

**Answer:**

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

BD^{2} = AD.CD

⇒ 5^{2} = 5.CD

⇒ CD = 5

∴ Option (c) is correct.

**Question 18.**In Δ ABC, is median. If AB^{2} + AC^{2} = 130 and AD = 7, then BD =

A. 4

B. 8

C. 16

D. 32

**Answer:**

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 130 = 2(7^{2} + BD^{2})

⇒ 65 = 49 + BD^{2}

⇒ 16 = BD^{2}

⇒ BD = 4

∴ Option (a) is correct.

**Question 19.**The diagonal of a square is 5√2. The length of the side of the square is

A. 10

B. 5

C. 3√2

D. 2√2

**Answer:**

Let the side of square be a.

Diagonal = √2a

⇒ 5√2 = √2a

⇒ a = 5

∴ Option (b) is correct.

**Question 20.**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 …

A. 36

B. 34

C. 48

D. 52

**Answer:**

Let the other side of rectangle be x

By Pythagoras Theorem,

AD^{2} = AC^{2} + CD^{2}

13^{2} = AC^{2} + 5^{2}

⇒ 169 – 25 = AC^{2}

⇒ 144 = AC^{2}

⇒ AC^{2} = 12^{2}

⇒ AC = 12

Perimeter = 2(AC + CD)

⇒ Perimeter = 2(12 + 5)

⇒ Perimeter = 34

∴ Option (b) is correct.

**Question 21.**The length of a median of an equilateral triangle is √3. Length of the side of the triangle is

A. 1

B. 2√3

C. 2

D. √3

**Answer:**

Median of an equilateral triangle is also a perpendicular to the base.

Let the side be a

⇒ a = 2

∴ Option (c) is correct.

**Question 22.**The perimeter of an equilateral triangle is 6. The length of the altitude of the triangle is …

A. 4

B. 2√3

C. 2

D. √3

**Answer:**

Let the length of altitude be x and side be a.

⇒ 3a = 6

⇒ a = 2

⇒ x = √3

∴ Option (d) is correct.

**Question 23.**In ΔABC, m∠A = 90. is a median. If AD = 6, AB = 10, then AC =

A. 8

B. 7.5

C. 16

D. 2√11

**Answer:**

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ BC^{2} = 2 (36 + BD^{2})

⇒ 4BD^{2} = 72 + 2BD^{2}

⇒ 2BD^{2} = 72

⇒ BD^{2} = 36

⇒ BD = 6

So, BC = 12 as BC = 2BD

By Pythagoras Theorem in ΔABC,

AC^{2} = BC^{2} – AB^{2}

⇒ AC^{2} = 144 – 100

⇒ AC = 2√11

∴ Option (d) is correct.

**Question 24.**In Δ PQR, m∠Q = 90 and PQ = QR. ,. If QM = 2, PQ =

A. 4

B. 2√2

C. 8

D. 2

**Answer:**

Since ∠Q = 90 and PQ = QR , therefore ∠QPR = ∠QRP = θ

As sum of all of triangle = 180°

⇒ 2θ + 90 = 180

⇒ θ = 45°

⇒ PQ = 2√2

∴ Option (b) is correct.

**Question 25.**In Δ ABC, m∠A = 90, is an altitude. So AB^{2} = ……

A. BD.BC

B. BD.DC

C.

D. BC.DC

**Answer:**

By Pythagoras Theorem in ΔABD,

AB^{2} = AD^{2} + BD^{2} ...... (1)

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

AD^{2} = BD.CD ...... (2)

From equation (1) and (2)

AB^{2} = BD.CD + BD^{2}

⇒ AB^{2} = BD(CD + BD)

⇒ AB^{2} = BD.BC

∴ Option (a) is correct.

**Question 26.**In Δ ABC, m∠A = 90, is an altitude. Therefore BD.DC = ……

A. AB^{2}

B. BC^{2}

C. AC^{2}

D. AD^{2}

**Answer:**

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

⇒ AD^{2} = BD.DC

∴ Option (d) is correct.

**Question 1.**

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

**Answer:**

From the given info,

As AB^{2} + BC^{2} + AC^{2} = 600

and BC = 10,

⇒ AB^{2} + AC^{2} = 500 ...... (1)

As AD is median,

⇒ BD = 5 ...... (2)

In Δ ABC, using the theorem of Apolloneous, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

Substituting values for (1) and (2) in above equation,

⇒ 500 = 2 (AD^{2} + 5^{2})

⇒ 250 = AD^{2} + 25

⇒ AD^{2} = 225

⇒ AD^{2} = 15^{2}

⇒ AD = 15

**Question 2.**

is the altitude of Δ ABC such that B—D—C. If AD^{2} = BD ⋅ DC, prove that ∠BAC is right angle.

**Answer:**

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB^{2} = BD^{2} + AD^{2} ...... (1)

And, AC^{2} = AD^{2} + DC^{2} ...... (2)

It is given that AD^{2} = BD.DC …… (3)

From equation (1) and (3), we get,

AB^{2} = BD^{2} + BD.DC

⇒ AB^{2} = BD(BD + DC) = BD. BC

Now, from equation (2) and (3),

AC^{2} = BD.DC + DC^{2}

⇒ AC^{2} = DC (BD + DC)

⇒ AC^{2} = DC(BC)

Also, BC = BD + DC,

Substituting the value of BD and DC in above equation, we get,

⇒ BC^{2} = AB^{2} + AC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**Question 3.**

In ΔABC, , B—D—C. If AB^{2} = BD ⋅ BC, prove that ∠ BAC is a right angle.

**Answer:**

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB^{2} = BD^{2} + AD^{2} ...... (1)

And, AC^{2} = AD^{2} + DC^{2} ...... (2)

It is given that AB^{2} = BD.BC

From equation (1) and (2), we get,

AB^{2} – BD^{2} = AC^{2} – DC^{2}

⇒ AB^{2} – BD^{2} = AC^{2} – (BC – BD)^{2}

⇒ BC^{2} + BD^{2} – 2BC.BD = AC^{2} + BD^{2} – AB^{2}

⇒ BC^{2} – 2BC (BD) = AC^{2} – AB^{2}

Substituting the value of BD above, we get,

⇒ BC^{2} – 2(AB)^{2} = AC^{2} – AB^{2}

⇒ BC^{2} = AB^{2} + AC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**Question 4.**

In ΔABC, , B—D—C. If AC^{2} = CD ⋅ BC, prove that ∠BAC is a right angle.

**Answer:**

Since AD is the altitude to BC,

⇒ ∠BDA = ∠CDA = 90°

So, Δ ABD and Δ ACD are right angle triangles.

Using Pythagoras theorem,

AB^{2} = BD^{2} + AD^{2} ...... (1)

And, AC^{2} = AD^{2} + DC^{2} ...... (2)

It is given that AC^{2} = CD.BC

From equation (1) and (2), we get,

AB^{2} – BD^{2} = AC^{2} – CD^{2}

⇒ AB^{2} – (BC – CD)^{2} = AC^{2} – CD^{2}

⇒ AB^{2} – BC^{2} – CD^{2} + 2.BC.CD = AC^{2} – CD^{2}

⇒ AB^{2} – BC^{2} + 2.BC.CD = AC^{2}

Substituting the value of CD above, we get,

⇒ AB^{2} – BC^{2} + 2AC^{2} = AC^{2}

⇒ AB^{2} + AC^{2} = BC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**Question 5.**

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

**Answer:**

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

As it is given that AD = BD,

⇒ AB^{2} + AC^{2} = 4.BD^{2}

Since AD is the median to line BC, we have,

Substituting the value of BD in above equation, we get,

AB^{2} + AC^{2} = BC^{2}

So, by the converse of Pythagoras Theorem, we can say that ∠BAC is a right angle.

**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.

**Answer:**

Let the length of upper part of pole be x,

The length of AB becomes 25 – x.

Applying Pythagoras theorem in ΔABC,

(AC’)^{2} = AB^{2} + (BC’)^{2}

⇒ x^{2} = (25 – x)^{2} + 15^{2}

⇒ x^{2} = 625 + x^{2} – 50x + 225

⇒ 50x = 850

⇒ x = 17 m

∴ The length of the upper part of the pole is 17 meters.

**Question 7.**

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

**Answer:**

From the given information, we construct the figure as:

As AM is perpendicular BC, ΔACM and ΔABM are right triangle.

By using Pythagoras Theorem, we get,

AC^{2} = CM^{2} + AM^{2} ...... (1)

Also, AB^{2} = BM^{2} + AM^{2} ...... (2)

Eliminating AM^{2} from equations (1) and (2), we get,

AC^{2} – CM^{2} = AB^{2} – BM^{2}

⇒ AC^{2} – AB^{2} = CM^{2} – BM^{2}

By using the identity a^{2} – b^{2} = (a + b)(a – b) on above equation,

⇒ AC^{2} – AB^{2} = (CM + BM)(CM – BM)

⇒ AC^{2} – AB^{2} = BC(CM – (BC – CM))

⇒ AB^{2} – AC^{2} = BC(2CM – BC)

As D is mid point of BC,

⇒ AB^{2} – AC^{2} = BC(2CM – 2BD)

⇒ AB^{2} – AC^{2} = 2BC(CM – BD)

⇒ AB^{2} – AC^{2} = 2BC(CM – CD)

⇒ AB^{2} – AC^{2} = 2BC(–DM)

⇒ AB^{2} — AC^{2} = 2BC.DM

Hence, proved.

**Question 8.**

In ΔABC, , D and ∠B is right angle. If AC = 5CD, prove that BD = 2CD.

**Answer:**

ΔABC is a right triangle, right angle at B and BD is perpendicular from B vertex to hypotenuse AC.

So, as we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

⇒ BD^{2} = AD.CD

⇒ BD^{2} = (AC – CD).CD

As, AC = 5CD,

⇒ BD^{2} = (5CD – CD).CD

⇒ BD^{2} = 4CD^{2}

⇒ BD^{2} = (2CD)^{2}

⇒ BD = 2CD

Hence, proved.

**Question 9.**

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

A. 31

B. 25

C. 17

D. 15

**Answer:**

In ΔABC, we know that sum of all interior angles in triangle is equal to 180°

⇒ ∠P + ∠Q + ∠R = 180

Also, given that, ∠P + ∠Q = ∠R,

⇒ 2∠R = 180

⇒ ∠R = 90°

Using Pythagoras Theorem,

PQ^{2} = PR^{2} + QR^{2}

⇒ PQ^{2} = 7^{2} + (24)^{2}

⇒ PQ^{2} = 49 + 576

⇒ PQ^{2} = 625

⇒ PQ = 25

∴ Option (b) is correct.

**Question 10.**

In ΔABC, is an altitude and ∠A is right angle. If AB = , BD = 4, then CD =

A. 5

B. 3

C. √5

D. 1

**Answer:**

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of each side other than the hypotenuse is the geometric mean of length of hypotenuse and segment of hypotenuse adjacent to the side.

⇒ AB^{2} = BD.BC

⇒ 20 = 4. BC

⇒ BC = 5

Also, CD = BC – BD

CD = 5 – 4

CD = 1

∴ Option (d) is correct.

**Question 11.**

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

A. 4

B. 24

C. 8

D. 16

**Answer:**

In Δ ABC, using the theorem of Apolloneous, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 50 = 2(3^{2} + BD^{2})

⇒ 50 = 2 (9 + BD^{2})

⇒ 25 = 9 + BD^{2}

⇒ BD^{2} = 16

⇒ BD = 4

⇒ BC = 2BD = 8

∴ Option (c) is correct.

**Question 12.**

In ΔABC, m∠B = 90, AB = BC. Then AB: AC =

A. 1:3

B. 1:2

C. 1 : √2

D. √2:1

**Answer:**

In ΔABC, applying Pythagoras Theorem, we get,

AC^{2} = AB^{2} + BC^{2}

As AB = BC,

⇒ AC^{2} = 2AB^{2}

⇒ AB:AC = 1:√2

∴ Option (c) is correct.

**Question 13.**

In ΔABC, m∠B = 90 and AC = 10. The length of the median BM =

A. 5

B. 5√2

C. 6

D. 8

**Answer:**

In Δ ABC, using the theorem of Apolloneous, we know that,

For BM to be the median, we have,

AB^{2} + AC^{2} = 2(BM^{2} + CM^{2})

Also, as Δ ABC is right angled at B,

⇒ AB^{2} + AC^{2} = AC^{2} = 10^{2} = 100

⇒ 100 = 2 (BM^{2} + 5^{2})

⇒ 50 = BM^{2} + 25

⇒ BM = 5

∴ Option (a) is correct.

**Question 14.**

In ΔABC, AB = BC = . m∠B

A. Is acute

B. Is obtuse

C. Is right angle

D. Cannot be obtained

**Answer:**

Let us assume,

⇒ AB = BC = x and AC = √2x

As we see that, AC^{2} = AB^{2} + BC^{2}

So by converse of Pythagoras Theorem, ΔABC is right angled and right angle at B.

∴ Option (c) is correct.

**Question 15.**

In ΔABC, if then m ∠C = ……

A. 90

B. 30

C. 60

D. 45

**Answer:**

Let us assume,

⇒ AB = x, AC = 2x and BC = √3x

As we see that, AC^{2} = AB^{2} + BC^{2}

So by converse of Pythagoras Theorem, ΔABC is right angled and right angle at B.

Let ∠C = θ,

⇒ θ = 30°

∴ Option (b) is correct.

**Question 16.**

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

A.

B. 17

C. 8

D. 7.5

**Answer:**

Let ∠X = A,

⇒ ∠Y = 2A and ∠Z = 3A

As sum of all angles in a triangle is equal to 180°

∠X + ∠Y + ∠Z = 180

⇒ 6A = 180

⇒ A = 30

⇒ ∠Z = 90°

⇒ YZ = 7.5

∴ Option (d) is correct.

**Question 17.**

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

A. 1

B. √5

C. 5

D. 2.5

**Answer:**

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

BD^{2} = AD.CD

⇒ 5^{2} = 5.CD

⇒ CD = 5

∴ Option (c) is correct.

**Question 18.**

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

A. 4

B. 8

C. 16

D. 32

**Answer:**

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ 130 = 2(7^{2} + BD^{2})

⇒ 65 = 49 + BD^{2}

⇒ 16 = BD^{2}

⇒ BD = 4

∴ Option (a) is correct.

**Question 19.**

The diagonal of a square is 5√2. The length of the side of the square is

A. 10

B. 5

C. 3√2

D. 2√2

**Answer:**

Let the side of square be a.

Diagonal = √2a

⇒ 5√2 = √2a

⇒ a = 5

∴ Option (b) is correct.

**Question 20.**

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 …

A. 36

B. 34

C. 48

D. 52

**Answer:**

Let the other side of rectangle be x

By Pythagoras Theorem,

AD^{2} = AC^{2} + CD^{2}

13^{2} = AC^{2} + 5^{2}

⇒ 169 – 25 = AC^{2}

⇒ 144 = AC^{2}

⇒ AC^{2} = 12^{2}

⇒ AC = 12

Perimeter = 2(AC + CD)

⇒ Perimeter = 2(12 + 5)

⇒ Perimeter = 34

∴ Option (b) is correct.

**Question 21.**

The length of a median of an equilateral triangle is √3. Length of the side of the triangle is

A. 1

B. 2√3

C. 2

D. √3

**Answer:**

Median of an equilateral triangle is also a perpendicular to the base.

Let the side be a

⇒ a = 2

∴ Option (c) is correct.

**Question 22.**

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

A. 4

B. 2√3

C. 2

D. √3

**Answer:**

Let the length of altitude be x and side be a.

⇒ 3a = 6

⇒ a = 2

⇒ x = √3

∴ Option (d) is correct.

**Question 23.**

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

A. 8

B. 7.5

C. 16

D. 2√11

**Answer:**

In Δ ABC, using Apolloneous Theorem, we know that,

For AD to be the median, we have,

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

⇒ BC^{2} = 2 (36 + BD^{2})

⇒ 4BD^{2} = 72 + 2BD^{2}

⇒ 2BD^{2} = 72

⇒ BD^{2} = 36

⇒ BD = 6

So, BC = 12 as BC = 2BD

By Pythagoras Theorem in ΔABC,

AC^{2} = BC^{2} – AB^{2}

⇒ AC^{2} = 144 – 100

⇒ AC = 2√11

∴ Option (d) is correct.

**Question 24.**

In Δ PQR, m∠Q = 90 and PQ = QR. ,. If QM = 2, PQ =

A. 4

B. 2√2

C. 8

D. 2

**Answer:**

Since ∠Q = 90 and PQ = QR , therefore ∠QPR = ∠QRP = θ

As sum of all of triangle = 180°

⇒ 2θ + 90 = 180

⇒ θ = 45°

⇒ PQ = 2√2

∴ Option (b) is correct.

**Question 25.**

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

A. BD.BC

B. BD.DC

C.

D. BC.DC

**Answer:**

By Pythagoras Theorem in ΔABD,

AB^{2} = AD^{2} + BD^{2} ...... (1)

As we know that,

If an altitude is drawn to hypotenuse of a right angled triangle, then the length of altitude is the geometric mean of lengths of segments of hypotenuse formed by the altitude.

AD^{2} = BD.CD ...... (2)

From equation (1) and (2)

AB^{2} = BD.CD + BD^{2}

⇒ AB^{2} = BD(CD + BD)

⇒ AB^{2} = BD.BC

∴ Option (a) is correct.

**Question 26.**

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

A. AB^{2}

B. BC^{2}

C. AC^{2}

D. AD^{2}

**Answer:**

As we know that,

⇒ AD^{2} = BD.DC

∴ Option (d) is correct.