# ML Aggarwal Solutions for Chapter 12 Pythagoras Theorem Class 9 Maths ICSE

**Exercise 12**

**1. Lengths of sides of triangles 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**

**Solution**

We use the Pythagoras theorem to check whether the triangles are right triangles.

We have h^{2} = b^{2}+a^{2} **[Pythagoras theorem]**

Where h is the hypotenuse, b is the base and a is the altitude.

**(i)** Given sides are 3 cm, 8 cm and 6 cm

b^{2}+a^{2} = 3^{2}+ 6^{2} = 9+36 = 45

⇒ h^{2} = 8^{2} = 64

Here, 45 ≠ 64

Hence the given triangle is not a right triangle.

**(ii)** Given sides are 13 cm, 12 cm and 5 cm

b^{2}+a^{2} = 12^{2}+ 5^{2} = 144+25 = 169

⇒ h^{2} = 13^{2} = 169

Here, b^{2}+a^{2} = h^{2}

Hence the given triangle is a right triangle.

Length of the hypotenuse is 13 cm.

**(iii)** Given sides are 1.4 cm, 4.8 cm and 5 cm

b^{2}+a^{2} = 1.4^{2}+ 4.8^{2} = 1.96+23.04 = 25

⇒ h^{2} = 5^{2} = 25

Here, b^{2}+a^{2} = h^{2}

Hence the given triangle is a right triangle.

Length of the hypotenuse is 5 cm.

**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.**

**Solution**

Let PR be the ladder and QR be the vertical wall.

Length of the ladder PR = 10 m

PQ = 6 m

Let height of the wall, QR = h

According to Pythagoras theorem,

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

⇒ 10^{2} = 6^{2}+QR^{2}

⇒ 100 = 36+QR^{2}

⇒ QR^{2} = 100-36

⇒ QR^{2} = 64

Taking square root on both sides,

QR= 8

Hence the height of the wall where the top of the ladder reaches is 8 m.

**3. A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be tight?**

**Solution**

Let AC be the wire and AB be the height of the pole.

AC = 24 cm

AB = 18 cm

According to Pythagoras theorem,

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

⇒ 24^{2} = 18^{2}+BC^{2}

⇒ 576 = 324+BC^{2}

⇒ BC^{2} = 576-324

⇒ BC^{2} = 252

Taking square root on both sides,

BC = √252

= √(4×9×7)

= 2×3√7

= 6√7 cm

Hence, the distance **is 6√7 cm.**

**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.**

**Solution**

Let AB and CD be the poles which are 12 m apart.

AB = 6 m

CD = 11 m

BD = 12 m

Draw AE BD

CE = 11-6 = 5 m

AE = 12 m

According to Pythagoras theorem,

AC^{2} = AE^{2}+CE^{2}

⇒ AC^{2} = 12^{2}+5^{2}

⇒ AC^{2} = 144+25

⇒ AC^{2} = 169

Taking square root on both sides

AC = 13

Hence, the distance between their tops is 13 m.

**5.** **In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is 4:3, find the sides.**

**Solution**

Given hypotenuse, h = 20 cm

Ratio of other two sides, a:b = 4:3

Let altitude of the triangle be 4x and base be 3x.

According to Pythagoras theorem,

h^{2} = b^{2}+a^{2}

⇒ 20^{2} = (3x)^{2}+(4x)^{2}

⇒ 400 = 9x^{2}+16x^{2}

⇒ 25x^{2} = 400

⇒ x^{2} = 400/25

⇒ x^{2} = 16

Taking square root on both sides

x = 4

so base, b = 3x = 3×4 = 12

altitude, a = 4x = 4×4 = 16

Hence the other sides are 12 cm and 16 cm.

**6.** **If the sides of a triangle are in the ratio 3:4:5, prove that it is right-angled triangle.**

**Solution**

Given the sides are in the ratio 3:4:5.

Let ABC be the given triangle.

Let the sides be 3x, 4x and hypotenuse be 5x.

According to Pythagoras theorem,

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

BC^{2}+AB^{2}= (3x)^{2}+(4x)^{2}

= 9x^{2}+16x^{2}

= 25x^{2}

AC^{2} = (5x)^{2} = 25x^{2}

⇒ AC^{2} = BC^{2}+AB^{2}

Hence ABC is a right angled triangle.

**7.** **For going to a city B from 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.**

**Solution**

Given AC = 2x km

CB = 2(x+7)km

AB = 26

Given AC CB.

According to Pythagoras theorem,

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

⇒ 26^{2} = ( 2(x+7))^{2}+(2x)^{2}

⇒ 676 = 4(x^{2}+14x+49) + 4x^{2}

⇒ 4x^{2}+56x+196+4x^{2} = 676

⇒ 8x^{2}+56x+196 = 676

⇒ 8x^{2}+56x +196-676 = 0

⇒ 8x^{2}+56x -480 = 0

⇒ x^{2}+7x -60 = 0

⇒ (x-5)(x+12) = 0

⇒ (x-5) = 0 or (x+12) = 0

⇒ x = 5 or x = -12

Length cannot be negative. So x = 5

BC = 2(x+7) = 2(5+7) = 2×12 = 24 km

AC = 2x = 2×5 = 10 km

Total distance = AC + BC = 10+24 = 34 km

Distance saved = 34-26 = 8 km

Hence the distance saved is 8 km.

**8. The hypotenuse of right triangle is 6m more than twice the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.**

**Solution**

Let the shortest side be x.

Then hypotenuse = 2x+6

Third side = 2x+6-2 = 2x+4

According to Pythagoras theorem,

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

⇒ (2x+6)^{2} = x^{2}+(2x+4)^{2}

⇒ 4x^{2}+24x+36 = x^{2}+4x^{2}+16x+16

⇒ x^{2}-8x-20 = 0

⇒ (x-10)(x+2) = 0

⇒ x-10 = 0 or x+2 = 0

⇒ x = 10 or x = -2

x cannot be negative.

So shortest side is 10 m.

Hypotenuse = 2x+6

= 2×10 +6

= 20+6

= 26 m

Third side = 2x+4

= = 2×10 +4

= 20+4

= 24 m

Hence the shortest side, hypotenuse and third side of the triangle are 10 m, 26 m and 24 m respectively.

**9.** **ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².**

**Solution**

Let ABC be the isosceles right angled triangle .

C = 90˚

AC = BC** [isosceles triangle]**

According to Pythagoras theorem,

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

⇒ AB^{2} = AC^{2}+AC^{2 }**[∵AC = BC]**

⇒ AB^{2} = 2AC^{2}

Hence, proved.

**10. In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².**

**Solution**

Given AD BC.

So ADB and ADC are right triangles.

In ADB,

AB^{2} = AD^{2}+BD^{2} **[Pythagoras theorem]**

AD^{2} = AB^{2}– BD^{2}** …(i)**

In ADC,

AC^{2} = AD^{2}+CD^{2} **[Pythagoras theorem]**

AD^{2} = AC^{2}– CD^{2} **…(ii)**

Comparing (i) and (ii)

AB^{2}– BD^{2 }= AC^{2}– CD^{2}

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

Hence, proved.

**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).**

**Solution**

Given PQ = a, PR = b, QD = c and DR = d.

PD QR.

So PDQ and PDR are right triangles.

In PDQ,

PQ^{2} = PD^{2}+QD^{2} **[Pythagoras theorem]**

⇒ PD^{2 }= PQ^{2}– QD^{2}

⇒ PD^{2} = a^{2}– c^{2} **…(i) [∵ PQ = a and QD = c]**

In PDR,

PR^{2} = PD^{2}+DR^{2} **[Pythagoras theorem]**

⇒ PD^{2} = PR^{2}– DR^{2}

⇒ PD^{2} = b^{2}– d^{2} **…(ii) [∵ PR = b and DR = d]**

Comparing (i) and (ii)

a^{2}– c^{2}= b^{2}– d^{2}

⇒ a^{2}– b^{2}= c^{2}– d^{2}

⇒ (a+b)(a-b) = (c+d)(c-d)

Hence, proved.

**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.**

**Solution**

Let AD be the altitude of ABC.

Given AB = AC = 12 cm

BC = 8 cm

The altitude to the base of an isosceles triangle bisects the base.

So BD = DC

BD = 8/2 = 4 cm

DC = 4 cm

ADC is a right triangle.

AB^{2} = BD^{2} +AD^{2} **[Pythagoras theorem]**

⇒ AD^{2} = AB^{2} -BD^{2}

⇒ AD^{2} = 12^{2}-4^{2}

⇒ AD^{2} = 144-16

⇒ AD^{2} = 128

Taking square root on both sides,

AD = √128 = √(2×64) = 8√2 cm

Area of ABC = ½ ×base ×height

= ½ ×8×8√2

= 4×8√2

= 32√2 cm^{2}

Hence the area of triangle is 32√2 cm^{2}.

**13. Find the area and the perimeter of a square whose diagonal is 10 cm long.**

**Solution**

Given length of the diagonal of the square is 10 cm.

AC = 10

Let AB = BC = x** [Sides of square are equal in measure]**

B = 90° **[All angles of a square are 90°]**

ABC is a right triangle.

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

⇒ 10^{2} = x^{2}+x^{2}

⇒ 100 = 2x^{2}

⇒ x^{2} = 50

⇒ x = √50 = √(25×2)

⇒ x = 5√2

So area of square = x^{2}

= (5√2)^{2} = 50 cm^{2}

Perimeter = 4x

= 4×5√2

= 20√2 cm

Hence, area and perimeter of the square are 50 cm^{2} and 20√2 cm.

**14.** **(a) In fig. (i) 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.**

**(b) In fig. (ii) 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.**

**Solution**

**(i)** Given AD = 13 cm, DC = 12 m

BC = 3 cm

ABD = BCD = 90°

BCD is a right triangle.

BD^{2} = BC^{2}+DC^{2} **[Pythagoras theorem]**

⇒ BD^{2} = 3^{2}+12^{2}

⇒ BD^{2} = 9+144

⇒ BD^{2} = 153

ABD is a right triangle.

AD^{2} = AB^{2}+BD^{2} **[Pythagoras theorem]**

⇒ 13^{2} = AB^{2}+153

⇒ 169 = AB^{2}+153

⇒ AB^{2} = 169-153

⇒ AB^{2} = 16

Taking square root on both sides,

AB = 4 cm

Hence the length of AB is 4 cm.

**(ii)** Given AB = AD, A = 90° = C, BC = 8 cm and CD = 6 cm

BCD is a right triangle.

BD^{2} = BC^{2}+DC^{2} **[Pythagoras theorem]**

⇒ BD^{2} = 8^{2}+6^{2}

⇒ BD^{2} = 64+36

⇒ BD^{2} = 100

Taking square root on both sides,

BD = 10 cm

ABD is a right triangle.

BD^{2} = AB^{2}+AD^{2} **[Pythagoras theorem]**

⇒ 10^{2} = 2AB^{2} **[∵AB = AD]**

⇒ 100 = 2AB^{2}

⇒ AB^{2} = 100/2

⇒ AB^{2} = 50

Taking square root on both sides,

AB = √50

⇒ AB = √(2×25)

⇒ AB = 5√2 cm

Hence the length of AB is 5√2 cm.

**15.** **(a) In figure (i) given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm. Calculate the length of BD.**

**(b) In figure (ii) given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.**

**(c) In figure (iii) given below, ∠ D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.**

**Solution**

**(a)** Given AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm

ABC is a right triangle.

AC^{2} = AB^{2}+BC^{2} **[Pythagoras theorem]**

⇒ 13^{2} = 12^{2}+BC^{2}

⇒ BC^{2} = 13^{2}-12^{2}

⇒ BC^{2} = 169-144

⇒ BC^{2} = 25

Taking square root on both sides,

BC = 5 cm

CDE is a right triangle.

CE^{2} = CD^{2}+DE^{2}** [Pythagoras theorem]**

⇒ 10^{2} = CD^{2}+6^{2}

⇒ 100 = CD^{2}+36

⇒ CD^{2} = 100-36

⇒ CD^{2} = 64

Taking square root on both sides,

CD = 8 cm

BD = BC +CD

⇒ BD = 5+8

⇒ BD = 13 cm

Hence the length of BD is 13 cm.

**(b)** Given PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm

PSQ is a right triangle.

PQ^{2} = PS^{2}+QS^{2} **[Pythagoras theorem]**

⇒ 10^{2} = PS^{2}+6^{2}

⇒ 100 = PS^{2}+36

⇒ PS^{2} = 100-36

⇒ PS^{2} = 64

Taking square root on both sides,

PS = 8 cm

PSR is a right triangle.

RS = RQ+QS

⇒ RS = 9+6

⇒ RS = 15 cm

⇒ PR^{2} = PS^{2}+RS^{2} **[Pythagoras theorem]**

⇒ PR^{2} = 8^{2}+15^{2}

⇒ PR^{2} = 64+225

⇒ PR^{2} = 289

Taking square root on both sides,

PR = 17 cm

Hence the length of PR is 17 cm.

**(c)** D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm

ADC is a right triangle.

AC^{2} = AD^{2}+CD^{2} **[Pythagoras theorem]**

⇒ 6^{2} = AD^{2}+CD^{2}** …(i)**

ABD is a right triangle.

AB^{2} = AD^{2}+BD^{2} **[Pythagoras theorem]**

⇒ 16^{2} = AD^{2}+(BC+CD)^{2}

⇒ 16^{2} = AD^{2}+(12+CD)^{2}

⇒ 256 = AD^{2}+144+24CD+CD^{2}

⇒ 256-144 = AD^{2}+CD^{2}+24CD

⇒ AD^{2}+CD^{2} = 112-24CD

⇒ 6^{2} = 112-24CD** [from (i)]**

⇒ 36 = 112-24CD

⇒ 24CD = 112-36

24CD = 76

⇒ CD = 76/24 = 19/6

**16.** **(a) In figure (i) given below, BC = 5 cm, ****∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.**

**(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB ^{2} = 4AD² – 3AC².**

**Solution**

**(a)** Given BC = 5 cm,

B =90°, AB = 5AE,

CD = 2AE and AC = ED

ABC is a right triangle.

AC^{2} = AB^{2}+BC^{2 }**…(i) [Pythagoras theorem]**

BED is a right triangle.

ED^{2} = BE^{2}+BD^{2 }**[Pythagoras theorem]**

⇒ AC^{2} = BE^{2}+BD^{2 }…(ii) **[∵AC = ED]**

Comparing (i) and (ii)

AB^{2}+BC^{2} = BE^{2}+BD^{2}

⇒ (5AE)^{2}+5^{2} = (4AE)^{2}+(BC+CD)^{2} **[∵BE = AB-AE = 5AE-AE = 4AE]**

⇒ (5AE)^{2}+25 = (4AE)^{2}+(5+2AE)^{2} **…(iii)[∵BC = 5, CD = 2AE]**

Let AE = x.

So (iii) becomes,

(5x)^{2}+25 = (4x)^{2}+(5+2x)^{2}

⇒ 25x^{2}+25 = 16x^{2}+25+20x+4x^{2}

⇒ 25x^{2} = 20x^{2}+20x

⇒ 5x^{2} = 20x

⇒ x = 20/5 = 4

AE = 4 cm

CD = 2AE = 2×4 = 8 cm

AB = 5AE

AB = 5×4 = 20 cm

ABC is a right triangle.

AC^{2} = AB^{2}+BC^{2} **[Pythagoras theorem]**

⇒ AC^{2} = 20^{2}+5^{2}

⇒ AC^{2} = 400+25

⇒ AC^{2} = 425

Taking square root on both sides,

AC = √425 = √(25×17)

⇒ AC = 5√17 cm

Hence EA = 4 cm, CD = 8 cm, AB = 20 cm and AC = 5√17 cm.

**(b)** Given D is the midpoint of BC.

DC = ½ BC

ABC is a right triangle.

AB^{2} = AC^{2}+BC^{2} **…(i) [Pythagoras theorem]**

ADC is a right triangle.

AD^{2} = AC^{2}+DC^{2} **…(ii) [Pythagoras theorem]**

⇒ AC^{2} = AD^{2}-DC^{2}

⇒ AC^{2} = AD^{2}– (½ BC)^{2} **[∵DC = ½ BC]**

⇒ AC^{2} = AD^{2}– ¼ BC^{2}

⇒ 4AC^{2} = 4AD^{2}– BC^{2}

⇒ AC^{2}+3AC^{2} = 4AD^{2}– BC^{2}

⇒ AC^{2}+BC^{2} = 4AD^{2}-3AC^{2}

⇒ AB^{2} = 4AD^{2}-3AC^{2} **[from (i)]**

Hence, proved.

**17.** **In ∆ ABC, AB = AC = x, BC = 10 cm and the area of ∆ ABC is 60 cm². Find x.**

**Solution**

Given AB = AC = x

So ABC is an isosceles triangle.

AD ⟂ BC

The altitude to the base of an isosceles triangle bisects the base.

BD = DC = 10/2 = 5 cm

Given area = 60 cm^{2}

½ ×base ×height = ½ ×10×AD = 60

⇒ AD = 60×2/10

⇒ AD = 60/5

⇒ AD = 12cm

ADC is a right triangle.

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

⇒ x^{2} = 12^{2}+5^{2}

⇒ x^{2} = 144+25

⇒ x^{2} = 169

Taking square root on both sides

x = 13 cm

Hence, the value of x is 13 cm.

**18.** **In a rhombus, If diagonals are 30 cm and 40 cm, find its perimeter.**

**Solution**

Let ABCD be the rhombus.

Given AC = 30cm

BD = 40 cm

Diagonals of a rhombus are perpendicular bisectors of each other.

OB = ½ BD = ½ ×40 = 20 cm

OC = ½ AC = ½ ×30 = 15 cm

OCB is a right triangle.

BC^{2} = OC^{2}+OB^{2} **[Pythagoras theorem]**

⇒ BC^{2 }= 15^{2}+20^{2}

⇒ BC^{2 }= 225+400

⇒ BC^{2 }= 625

Taking square root on both sides

BC = 25 cm

So, side of a rhombus, a = 25 cm.

Perimeter = 4a = 4×25 = 100 cm

Hence the perimeter of the rhombus is 100 cm.

**19.** **(a) In figure (i) given below, AB || DC, BC = AD = 13 cm. AB = 22 cm and DC = 12cm. Calculate the height of the trapezium ABCD.****(b) In figure (ii) given below, AB || DC, ∠ A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.****(c) In figure (iii) 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.**

**Solution**

**(i)** Given AB || DC, BC = AD = 13 cm.

AB = 22 cm and DC = 12cm

Here DC = 12

MN = 12 cm

AM = BN

AB = AM+MN+BN

⇒ 22 = AM+12+AM **[∵AM = BN]**

⇒ 2AM = 22-12 = 10

⇒ AM = 10/2

⇒ AM = 5 cm

AMD is a right triangle.

AD^{2} = AM^{2}+DM^{2} **[Pythagoras theorem]**

⇒ 13^{2} = 5^{2}+DM^{2}

⇒ DM^{2} = 13^{2}-5^{2}

⇒ DM^{2} = 169-25

⇒ DM^{2} = 144

Taking square root on both sides,

DM = 12 cm

Hence the height of the trapezium is 12 cm.

**(b)** Given AB || DC, A = 90°, DC = 7 cm,

AB = 17 cm and AC = 25 cm

ADC is a right triangle.

AC^{2} = AD^{2}+DC^{2 }**[Pythagoras theorem]**

⇒ 25^{2} = AD^{2}+7^{2}

⇒ AD^{2} = 25^{2}-7^{2}

⇒ AD^{2} = 625-49

⇒ AD^{2} = 576

Taking square root on both sides

AD = 24 cm

CM = 24 cm **[∵ABCD]**

DC = 7 cm

AM = 7 cm

BM = AB-AM

⇒ BM = 17-7 = 10 cm

BMC is a right triangle.

BC^{2} = BM^{2}+CM^{2}

⇒ BC^{2} = 10^{2}+24^{2}

⇒ BC^{2} = 100+576

⇒ BC^{2} = 676

Taking square root on both sides

BC = 26 cm

Hence length of BC is 26 cm.

**(c) (i)**Proof:

Given ABCD is a square of side 7 cm.

So AB = BC = CD = AD = 7 cm

Also given AE = FC = CG = HA = 3 cm

BE = AB-AE = 7-3 = 4 cm

BF = BC-FC = 7-3 = 4 cm

GD = CD-CG = 7-3 = 4 cm

DH = AD-HA = 7-3 = 4 cm

A = 90˚ **[Each angle of a square equals 90˚]**

AHE is a right triangle.

HE^{2} = AE^{2}+AH^{2} **[Pythagoras theorem]**

⇒ HE^{2} = 3^{2}+3^{2}

⇒ HE^{2} = 9+9 = 18

⇒ HE = √(9×2) = 3√2 cm

Similarly GF = 3√2 cm

EBF is a right triangle.

EF^{2} = BE^{2}+BF^{2} **[Pythagoras theorem]**

⇒ EF^{2} = 4^{2}+4^{2}

⇒ EF^{2} = 16+16 = 32

Taking square root on both sides

EF = √(16×2) = 4√2 cm

Similarly,

HG = 4√2 cm

Now join EG

In EFG

EG^{2} = EF^{2}+GF^{2}

⇒ EG^{2} = (4√2)^{2}+(3√2)^{2}

⇒ EG^{2} = 32+18 = 50

⇒ EG = √50 = 5√2 cm **…(i)**

Join HF.

Also HF^{2} = EH^{2}+HG^{2}

= (3√2)^{2}+(4√2)^{2}

= 18+32 = 50

HF = √50 = 5√2 cm **…(ii)**

From (i) and (ii)

EG= HF

Diagonals of the quadrilateral are congruent. So EFGH is a rectangle.

Hence, proved.

**(ii)** Area of rectangle EFGH = length × breadth

= HE ×EF

= 3√2×4√2

= 24 cm^{2}

Perimeter of rectangle EFGH = 2(length+breadth)

= 2×(4√2+3√2)

= 2×7√2

= 14√2 cm

Hence, area of the rectangle is 24 cm^{2} and perimeter is 14√2 cm.

**20.** **AD is perpendicular to the side BC of an equilateral Î” ABC. Prove that 4AD² = 3AB².**

**Solution**

Given AD BC

D = 90˚

Proof:

Since ABC is an equilateral triangle,

AB = AC = BC

ABD is a right triangle.

According to Pythagoras theorem,

AB^{2} = AD^{2}+BD^{2}

BD = ½ BC

AB^{2} = AD^{2}+( ½ BC)^{2}

⇒ AB^{2} = AD^{2}+( ½ AB)^{2} [∵BC = AB]

⇒ AB^{2} = AD^{2}+ ¼ AB^{2}

⇒ AB^{2} = (4AD^{2}+ AB^{2})/4

⇒ 4AB^{2} = 4AD^{2}+ AB^{2}

⇒ 4AD^{2 }= 4AB^{2}– AB^{2}

⇒ 4AD^{2 }= 3AB^{2}

Hence, proved.

**21. In figure (i) given below, D and E are mid-points of the sides BC and CA respectively of a Î”ABC, right angled at C.**

**Prove that :**

**(i)4AD ^{2} = 4AC^{2}+BC^{2}**

**(ii)4BE ^{2} = 4BC^{2}+AC^{2}**

**(iii)4(AD ^{2}+BE^{2}) = 5AB^{2}**

**Solution**

Proof:

**(i)** C = 90°

So ACD is a right triangle.

AD^{2} = AC^{2}+CD^{2} **[Pythagoras theorem]**

Multiply both sides by 4, we get

4AD^{2} = 4AC^{2}+4CD^{2}

⇒ 4AD^{2} = 4AC^{2}+4BD^{2} **[∵D is the midpoint of BC, CD = BD = ½ BC]**

⇒ 4AD^{2} = 4AC^{2}+(2BD)^{2}

⇒ 4AD^{2} = 4AC^{2}+BC^{2}**…(i) [∵BC = 2BD]**

Hence proved.

**(ii)** BCE is a right triangle.

BE^{2} = BC^{2}+CE^{2} **[Pythagoras theorem]**

Multiply both sides by 4 , we get

4BE^{2} = 4BC^{2}+4CE^{2}

⇒ 4BE^{2} = 4BC^{2}+(2CE)^{2}

⇒ 4BE^{2} = 4BC^{2}+AC^{2} **…(ii) [∵E is the midpoint of AC, AE = CE = ½ AC]**

Hence proved.

**(iii) **Adding (i) and (ii)

4AD^{2}+4BE^{2} = 4AC^{2}+BC^{2}+4BC^{2}+AC^{2}

⇒ 4AD^{2}+4BE^{2} = 5AC^{2}+5BC^{2}

⇒ 4(AD^{2}+BE^{2} ) = 5(AC^{2}+BC^{2})

⇒ 4(AD^{2}+BE^{2} ) = 5(AB^{2}) **[∵ABC is a right triangle, AB ^{2} = AC^{2}+BC^{2}]**

Hence, proved.

**22.** **If AD, BE and CF are medians of ABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).**

**Solution**

Construction:

Draw APBC

Proof:

APB is a right triangle.

AB^{2} = AP^{2}+BP^{2} **[Pythagoras theorem]**

⇒ AB^{2} = AP^{2}+(BD-PD)^{2}

⇒ AB^{2} = AP^{2}+BD^{2}+PD^{2}-2BD×PD

⇒ AB^{2} = (AP^{2}+PD^{2})+BD^{2}-2BD×PD

⇒ AB^{2 }= AD^{2}+ (½ BC)^{2}-2×( ½ BC)×PD **[∵AP ^{2}+PD^{2} = AD^{2} and BD = ½ BC]**

⇒ AB^{2 }= AD^{2}+ ¼ BC^{2}– BC×PD **…(i)**

APC is a right triangle.

AC^{2} = AP^{2}+PC^{2} **[Pythagoras theorem]**

⇒ AC^{2} = AP^{2}+(PD^{2}+DC^{2})

⇒ AC^{2} = AP^{2}+PD^{2}+DC^{2}+ 2×PD×DC

⇒ AC^{2} = (AP^{2}+PD^{2})+ (½ BC)^{2}+2×PD×( ½ BC) [DC = ½ BC]

⇒ AC^{2} = (AD)^{2}+ ¼ BC^{2}+PD× BC **…(ii) [In APD, AP ^{2}+PD^{2} = AD^{2}]**

Adding (i) and (ii), we get

AB^{2}+AC^{2} = 2AD^{2}+ ½ BC^{2} **…(iii)**

Draw perpendicular from B and C to AC and AB respectively.

Similarly we get,

BC^{2}+CA^{2} = 2CF^{2}+ ½ AB^{2} **…(iv)**

AB^{2}+BC^{2} = 2BE^{2}+ ½ AC^{2} **…(v)**

Adding (iii), (iv) and (v), we get

2(AB^{2}+BC^{2}+CA^{2}) = 2(AD^{2}+BE^{2}+CF^{2})+ ½ (BC^{2}+AB^{2}+AC^{2})

⇒ 2(AB^{2}+BC^{2}+CA^{2}) = 2(AB^{2}+BC^{2}+CA^{2}) – ½ (AB^{2}+BC^{2}+CA^{2})

⇒ 2(AD^{2}+BE^{2}+CF^{2}) = (3/2)× (AB^{2}+BC^{2}+CA^{2})

⇒ 4(AD^{2}+BE^{2}+CF^{2}) = 3(AB^{2}+BC^{2}+CA^{2})

Hence, proved.

**23.(a)** **In fig. (i) given below, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that** **AB² + CD² = AD² + BC².**

**Solution**

Given diagonals of quadrilateral ABCD, AC and BD intersect at O at right angles.

Proof:

AOB is a right triangle.

AB^{2} = OB^{2}+OA^{2} **…(i) [Pythagoras theorem]**

COD is a right triangle.

CD^{2} = OC^{2}+OD^{2} **…(ii) [Pythagoras theorem]**

Adding (i) and (ii), we get

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

⇒ AB^{2}+ CD^{2} = (OA^{2}+OD^{2})+ (OC^{2}+OB^{2}) **…(iii)**

AOD is a right triangle.

AD^{2} = OA^{2}+OD^{2} **…(iv) [Pythagoras theorem]**

BOC is a right triangle.

BC^{2} = OC^{2}+OB^{2} **…(v) [Pythagoras theorem]**

Substitute (iv) and (v) in (iii), we get

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

Hence, proved.

**24. In a quadrilateral ABCD, B = 90° = D. Prove that 2 AC² – BC ^{2} = AB² + AD² + DC².**

**Solution**

Given B = D = 90˚

So ABC and ADC are right triangles.

In ABC,

AC^{2} = AB^{2}+BC^{2} **…(i) [Pythagoras theorem]**

In ADC,

AC^{2} = AD^{2}+DC^{2} …(ii) [Pythagoras theorem]

Adding (i) and (ii)

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

⇒ 2AC^{2} -BC^{2} = AB^{2}+AD^{2}+DC^{2}

Hence, proved.

**25.** **In a ∆ ABC, A = 90°, CA = AB and D is a point on AB produced. Prove that :** **DC² – BD² = 2AB×AD.**

**Solution**

Given A = 90°

CA = AB

Proof:

In ACD,

DC^{2} = CA^{2}+AD^{2} **[Pythagoras theorem]**

⇒ DC^{2} = CA^{2}+(AB+BD)^{2}

⇒ DC^{2} = CA^{2}+AB^{2}+BD^{2}+2AB×BD

⇒ DC^{2} -BD^{2 }= CA^{2}+AB^{2}+2AB×BD

⇒ DC^{2} -BD^{2} = AB^{2}+AB^{2}+2AB×BD **[∵CA = AB]**

⇒ DC^{2} -BD^{2} = 2AB^{2}+2AB×BD

⇒ DC^{2} -BD^{2} = 2AB(AB+BD)

⇒ DC^{2} -BD^{2} = 2AB×AD [A-B-D]

Hence, proved.

**26. In an isosceles triangle ABC, AB = AC and D is a point on BC produced.**

**Prove that AD² = AC²+BD.CD.**

**Solution**

Given ABC is an isosceles triangle.

AB = AC

Construction: Draw AP BC

Proof:

APD is a right triangle.

AD^{2} = AP^{2}+PD^{2} **[Pythagoras theorem]**

⇒ AD^{2} = AP^{2}+(PC+CD)^{2} **[PD = PC+CD]**

⇒ AD^{2} = AP^{2}+PC^{2}+CD^{2} +2PC×CD **…(i)**

APC is a right triangle.

AC^{2} = AP^{2}+PC^{2} **…(ii) [Pythagoras theorem]**

Substitute (ii) in (i)

AD^{2} = AC^{2} +CD^{2}+2PC×CD **…(iii)**

Since ABC is an isosceles triangle,

PC = ½ BC **[The altitude to the base of an isosceles triangle bisects the base]**

AD^{2} = AC^{2} +CD^{2}+2× ½ BC ×CD

⇒ AD^{2} = AC^{2} +CD^{2}+BC×CD

⇒ AD^{2} = AC^{2} +CD(CD+BC)

⇒ AD^{2} = AC^{2} +CD×BD **[CD+BC = BD]**

⇒ AD^{2} = AC^{2} +BD×CD

Hence, proved.

**Chapter test**

**1.** (**a) In fig. (i) given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.**

**(b) In figure (ii) 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.**

**(c) In figure (iii) 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.**

**Solution**

**(a)** Given AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm

ADC is a right triangle.

AC^{2} = AD^{2}+DC^{2} **[Pythagoras theorem]**

⇒ 17^{2} = 15^{2}+DC^{2}

⇒ 289 = 225+DC^{2}

⇒ DC^{2} = 289-225

⇒ DC^{2} = 64

Taking square root on both sides,

DC = 8 cm

ADB is a right triangle.

AB^{2} = AD^{2}+BD^{2} **[Pythagoras theorem]**

⇒ 25^{2} = 15^{2}+BD^{2}

⇒ 625 = 225+ BD^{2}

⇒ BD^{2} = 625-225 = 400

Taking square root on both sides,

BD = 20 cm

⇒ BC = BD+DC

= 20+8

= 28 cm

Hence the length of BC is 28 cm.

**(b)**** **Given BAC = 90°, ADC = 90°

AD = 6 cm, CD = 8 cm and BC = 26 cm.

**(i)** ADC is a right triangle.

AC^{2} = AD^{2}+DC^{2} **[Pythagoras theorem]**

⇒ AC^{2} = 6^{2}+8^{2}

⇒ AC^{2} = 36+64

⇒ AC^{2} = 100

Taking square root on both sides,

AC = 10 cm

Hence length of AC is 10 cm.

**(ii)** ABC is a right triangle.

BC^{2} = AC^{2}+AB^{2} **[Pythagoras theorem]**

⇒ 26^{2} = 10^{2}+AB^{2}

⇒ AB^{2} = 26^{2}-10^{2}

⇒ AB^{2} = 676-100

⇒ AB^{2} = 576

Taking square root on both sides,

AB = 24 cm

Hence length of AB is 24 cm.

**(iii) **Area of ABC = ½ ×AB×AC

= ½ ×24×10

= 120 cm^{2}

Area of ADC = ½ ×AD×DC

= ½ ×6×8

= 24 cm^{2}

Area of shaded region = area of ABC- area of ADC

= 120-24

= 96 cm^{2}

Hence, the area of shaded region is 96 cm^{2}.

**(c)** Given B = 90°

AB = 9 cm, AC = 15 cm .

D, E are mid-points of the sides AB and AC respectively.

**(i)** ABC is a right triangle.

AC^{2} = AB^{2}+BC^{2} **[Pythagoras theorem]**

⇒ 15^{2} = 9^{2}+BC^{2}

⇒ 225 = 81+BC^{2}

⇒ BC^{2} = 225-81

⇒ BC^{2} = 144

Taking square root on both sides,

BC = 12 cm

Hence, the length of BC is 12 cm.

**(ii)** AD = ½ AB **[D is the midpoint of AB]**

AD = ½ ×9 = 9/2

⇒ AE = ½ AC **[E is the midpoint of AC]**

⇒ AE = ½ ×15 = 15/2

ADE is a right triangle.

AE^{2} = AD^{2}+DE^{2} **[Pythagoras theorem]**

⇒ (15/2)^{2} = (9/2)^{2}+DE^{2}

⇒ DE^{2} = (15/2)^{2} – (9/2)^{2}

⇒ DE^{2} = 225/4 -81/4

⇒ DE^{2} = 144/4

Taking square root on both sides,

DE = 12/2 = 6 cm.

Area of ADE = ½ ×DE×AD

= ½ ×6×9/2

= 13.5 cm^{2}

Hence, the area of the ADE is 13.5 cm^{2}.

**2.** **If in ∆ ABC, AB > AC and AD BC, prove that AB² – AC² = BD² – CD²**

**Solution**

Given AD BC, AB>AC

So ADB and ADC are right triangles.

Proof:

In ADB,

AB^{2} = AD^{2}+BD^{2} **[Pythagoras theorem]**

⇒ AD^{2} = AB^{2}-BD^{2} **…(i)**

In ADC,

AC^{2} = AD^{2}+CD^{2} [Pythagoras theorem]

⇒ AD^{2} = AC^{2}-CD^{2} **…(ii)**

Equating (i) and (ii)

AB^{2}-BD^{2} = AC^{2}-CD^{2}

AB^{2}-AC^{2} = BD^{2}– CD^{2}

Hence, proved.

**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.**

**Solution**

Construction:

Join AQ and BP.

Given C = 90°

**Proof:**

**(i)** In ACQ,

AQ^{2} = AC^{2}+CQ^{2} **[Pythagoras theorem]**

Multiplying both sides by 9, we get

9AQ^{2} = 9AC^{2}+9CQ^{2}

⇒ 9AQ^{2 }= 9AC^{2}+(3CQ)^{2} **…(i)**

Given BQ: CQ = 1:2

⇒ CQ/BC = CQ/(BQ+CQ)

⇒ CQ/BC = 2/3

⇒ 3CQ = 2BC **…(ii)**

Substitute (ii) in (i)

9AQ^{2 }= 9AC^{2}+(2BC)^{2}

⇒ 9AQ^{2 }= 9AC^{2}+4BC^{2} **…(iii)**

Hence, proved.

**(ii)** In BPC,

BP^{2} = BC^{2}+CP^{2} **[Pythagoras theorem]**

Multiplying both sides by 9, we get

9BP^{2} = 9BC^{2}+9CP^{2}

⇒ 9BP^{2} = 9BC^{2}+(3CP)^{2}** …(iv)**

Given AP: PC = 1:2

CP/AC = CP/AP+PC

⇒ CP/AC = 2/3

⇒ 3CP = 2AC **…(v)**

Substitute (v) in (iv)

9BP^{2} = 9BC^{2}+(2AC)^{2}

⇒ 9BP^{2} = 9BC^{2}+4AC^{2}** ...(vi)**

Hence, proved.

**(iii) **Adding (iii) and (vi), we get

9AQ^{2}+9BP^{2} = 9AC^{2}+4BC^{2}+9BC^{2}+4AC^{2}

⇒ 9(AQ^{2}+BP)^{2} = 13AC^{2}+13BC^{2}

⇒ 9(AQ^{2}+BP)^{2} = 13(AC^{2}+BC^{2}) **…(vii)**

In ABC,

AB^{2} = AC^{2}+BC^{2} **…(viii)**

Substitute (viii) in (viii), we get

9(AQ^{2}+BP)^{2} = 13AB^{2}

Hence proved.

**4.** **In the given figure, ∆PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² = 3PR² + 5PS².**

**Solution**

Given Q = 90°

S and T are points on RQ such that these points trisect it.

So RT = TS = SQ

To prove : 8PT² = 3PR² + 5PS².

Proof:

Let RT = TS = SQ = x

In PRQ,

PR^{2} = RQ^{2}+PQ^{2} **[Pythagoras theorem]**

⇒ PR^{2} = (3x)^{2}+PQ^{2}

⇒ PR^{2} = 9x^{2}+PQ^{2}

Multiply above equation by 3

3PR^{2} = 27x^{2}+3PQ^{2} **…(i)**

Similarly in PTS,

PT^{2} = TQ^{2}+PQ^{2} **[Pythagoras theorem]**

⇒ PT^{2} = (2x)^{2}+PQ^{2}

⇒ PT^{2} = 4x^{2}+PQ^{2}

Multiply above equation by 8

8PT^{2} = 32x^{2}+8PQ^{2} **…(ii)**

Similarly in PSQ,

PS^{2} = SQ^{2}+PQ^{2} **[Pythagoras theorem]**

⇒ PS^{2} = x^{2}+PQ^{2}

Multiply above equation by 5

5PS^{2} = 5x^{2}+5PQ^{2} **…(iii)**

Add (i) and (iii), we get

3PR^{2} +5PS^{2} = 27x^{2}+3PQ^{2}+5x^{2}+5PQ^{2}

⇒ 3PR^{2} +5PS^{2} = 32x^{2}+8PQ^{2}

⇒ 3PR^{2} +5PS^{2} = 8PT^{2} **[From (ii)]**

8PT^{2} = 3PR^{2} +5PS^{2}

Hence, proved.

**5. In a quadrilateral ABCD, B = 90°. If AD² = AB² + BC² + CD², prove that ACD = 90°.**

**Solution**

Given : B = 90˚ in quadrilateral ABCD

AD² = AB² + BC² + CD²

To prove: ACD = 90°

**Proof:**

In ABC,

AC^{2} = AB^{2}+BC^{2} **…(i) [Pythagoras theorem]**

Given,

AD² = AB² + BC² + CD²

⇒ AD² = AC^{2}+CD^{2} [from (i)]

In ACD, ACD = 90° **[Converse of Pythagoras theorem]**

Hence, proved.

**6. In the given figure, find the length of AD in terms of b and c.**

**Solution**

Given : A = 90°

AB = c

AC = b

ADB = 90°

In ABC,

BC^{2} = AC^{2}+AB^{2} **[Pythagoras theorem]**

⇒ BC^{2} = b^{2}+c^{2}

⇒ BC = √( b^{2}+c^{2}) **…(i)**

Area of ABC = ½ ×AB×AC

= ½ ×bc **…(ii)**

Also, Area of ABC = ½ ×BC×AD

= ½ ×√( b^{2}+c^{2}) ×AD **…(iii)**

Equating (ii) and (iii)

½ ×bc = ½ ×√( b^{2}+c^{2}) ×AD

⇒ AD = bc /(√( b^{2}+c^{2})

Hence, AD is bc /(√( b^{2}+c^{2}).

**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.**

**Solution**

Let x be each side of the square ABCD.

FB = ½ AB **[∵ F is the midpoint of AB]**

FB = ½ x **…(i)**

BE = (1/3) BC

⇒ BE = (1/3) x **…(ii)**

AC = √2 ×side **[Diagonal of a square]**

⇒ AC = √2x

Area of FBE = ½ FB×BE

108 = ½ × ½ x ×(1/3)x **[given area of FBE = 108 cm ^{2}]**

⇒ 108 = (1/12)x^{2}

⇒ x^{2} = 108×12

⇒ x^{2} = 1296

Taking square root on both sides.

x = 36

AC = √2×36 = 36√2

Hence, length of AC is 36√2 cm.

**8.** **In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC × CD, Prove that BD = BC.**

**Solution**

Given : In ABC, AB = AC

D is a point on side AC such that BC² = AC×CD

To prove : BD = BC

Construction: Draw BEAC

**Proof:**

In BCE ,

BC^{2} = BE^{2}+EC^{2} **[Pythagoras theorem]**

⇒ BC^{2} = BE^{2}+(AC-AE)^{2}

⇒ BC^{2} = BE^{2}+AC^{2}+AE^{2}- 2 AC×AE

⇒ BC^{2} = BE^{2}+AE^{2}+AC^{2}- 2 AC×AE **…(i)**

In ABC,

AB^{2} = BE^{2}+AE^{2} **...(ii)**

Substitute (ii) in (i)

BC^{2} = AB^{2}+AC^{2}- 2 AC×AE

⇒ BC^{2} = AC^{2}+AC^{2}- 2 AC×AE **[∵AB = AC]**

⇒ BC^{2} = 2AC^{2}-2 AC×AE

⇒ BC^{2} = 2AC(AC-AE)

⇒ BC^{2} = 2AC×EC

Given,

BC² = AC × CD

⇒ 2AC×EC = AC × CD

⇒ 2EC = CD **...(ii)**

E is the midpoint of CD.

EC = DE **…(iii)**

In BED and BEC,

EC = DE **[From (iii)]**

BE = BE **[common side]**

BED = BEC **[By SAS congruency rule]**

BD = BD **[c.p.c.t]**

Hence, proved.