ML Aggarwal Solutions For Class 9 Maths Chapter 12 Pythagoras Theorem helps students to master the concept of Pythagoras theorem.

These solutions provide students an advantage with practical questions.

This chapter deals with Pythagoras theorem and its different applications. ML Aggarwal Solutions can be downloaded from our website in PDF format for free. These solutions are prepared by our expert team. The concepts covered in this chapter are explained in a simple manner so that any student can easily understand.

Pythagoras theorem is the fundamental theorem in Mathematics, which defines the relation between the hypotenuse, base and altitude of a right angled triangle. According to this theorem, the square of hypotenuse is equal to the sum of squares of altitude and base of a right angled triangle.

In ML Aggarwal solutions For Class 9 Maths Chapter 12, we come across solving different application questions.

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**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 Δ 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 ΔABC, 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]

Multply 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

BED BEC [By SAS congruency rule]

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

Hence proved.