ML Aggarwal Pythagoras Theorem Exe-12 Class 9 ICSE Maths Solutions Ch-12. Step by Step Solutions of Exercise-12 on Pythagoras Theorem of ML Aggarwal for ICSE Class 9th Mathematics Questions. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Pythagoras Theorem Exe-12 Class 9 ICSE Maths Solutions Ch-12

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-12 | Pythagoras Theorem |

Topics | Solution of Exe-12 Questions |

Academic Session | 2024-2025 |

### Problems on Pythagoras Theorem

ML Aggarwal Exe-12 Class 9 ICSE Maths Solutions Ch-12.

**Question 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

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

**(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

Therefore, 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}

Therefore, 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}

Therefore, the given triangle is a right triangle.

Length of the hypotenuse is 5 cm.

**Question 2. ****Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wail. Find the height of the point on the wall where the top of the ladder reaches.**

**Answer : **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

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

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

**Question 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 taught?**

**Answer : **Let AC be the wire and AB be the height of the pole.

AC = 24 cm

AB = 18 cm

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

Therefore, the distance is 6√7 cm.

**Question 4. ****Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.**

**Answer : **Let AB 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

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

Therefore, the distance between their tops is 13 m.

**Question 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.**

**Answer : **Hypotenuse, h = 20 cm given

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

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

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

base, b = 3x = 3×4 = 12

altitude, a = 4x = 4×4 = 16

Therefore, the other sides are 12 cm and 16 cm

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

**Answer : **The sides are in the ratio 3:4:5. given

Let ABC be the given triangle.

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

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}

Therefore, ABC is a right angled triangle.

**Question 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.**

**Answer : **Given that AC = 2x km , CB = 2(x+7)km, AB = 26

Applying pythagoras themorem

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

Therefore, the distance saved is 8 km.

**Question 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.**

**Answer : **Let the shortest side be x.

Then hypotenuse = 2x+6

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

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

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

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

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

**Answer : **Let ABC be the isosceles right angled triangle .

C = 90˚

AC = BC** **

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

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

⇒ AB^{2} = 2AC^{2}

Hence proved.

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

**Answer : **Given AD is perpendiculat on BC.

So ADB and ADC are right triangles.

In ADB, AB^{2} = AD^{2}+BD^{2}

AD^{2} = AB^{2}– BD^{2}** …(i)**

In ADC, AC^{2} = AD^{2}+CD^{2}

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.

**Question 11. ****In ∆PQR, PD ⊥ QR, such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d).**

**Answer : ** PQ = a, PR = b, QD = c and DR = d.

PD is perpendicular QR.

So PDQ and PDR are right triangles.

In PDQ,

PQ^{2} = PD^{2}+QD^{2}

⇒ 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}

⇒ 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.**

**Question 12. ****ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and Hence, calculate its area**

**Answer : ** Let AD 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.

⇒ 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}

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

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

**Answer : **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°

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

area of square = x^{2}

= (5√2)^{2} = 50 cm^{2}

Perimeter = 4x

= 4×5√2

= 20√2 cm

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

**Question 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.**

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

BC = 3 cm

ABD = BCD = 90°

BCD is a right triangle.

⇒ BD^{2} = 3^{2}+12^{2}

⇒ BD^{2} = 9+144

⇒ BD^{2} = 153

ABD is a right triangle.

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

⇒ 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

Therefore, 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}

⇒ 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}

⇒ 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

Therefore, the length of AB is 5√2 cm.

**Question 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.**

**Answer : ****(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}

⇒ 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}** **

⇒ 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}

⇒ 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}

⇒ 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}

⇒ 6^{2} = AD^{2}+CD^{2}** …(i)**

ABD is a right triangle.

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

⇒ 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

CD = 3(1/6)

Hence, the length of CD is 3(1/6) cm

**Question 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 AB2 = 4AD² – 3AC².**

**Answer :**

**(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 }

BED is a right triangle.

ED^{2} = BE^{2}+BD^{2}

⇒ AC^{2} = BE^{2}+BD^{2 }…(ii)

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}

⇒ 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)

ADC is a right triangle.

AD^{2} = AC^{2}+DC^{2} …(ii)

⇒ 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.

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

**Answer : **AB = AC = x

So ABC is an isosceles triangle.

AD ⟂ BC

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

Therefore, the value of x is 13 cm.

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

**Answer : **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}

⇒ 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.

**Question 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.**

**Answer : ****(a) **AB || DC, BC = AD = 13 cm. (Given )

AB = 22 cm and DC = 12cm

Here DC = 12

MN = 12 cm

AM = BN

AB = AM+MN+BN

⇒ 22 = AM+12+AM

⇒ 2AM = 22-12 = 10

⇒ AM = 10/2

⇒ AM = 5 cm

AMD is a right triangle.

⇒ 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 }

⇒ 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

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˚

AHE is a right triangle.

HE^{2} = AE^{2}+AH^{2}

⇒ 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} = 4^{2}+4^{2}

⇒ EF^{2} = 16+16 = 32

Taking square root on both sides

EF = √(16×2) = 4√2 cm

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

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

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

**Answer : **Given AD BC, D = 90˚

**Proof:**

Since ABC is an equilateral triangle,

AB = AC = BC

ABD is a right triangle.

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.

**Question 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}

** **

**Answer : ****(i) C = 90°**

So ACD is a right triangle.

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}

Multiply both sides by 4

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.

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

**Answer : **Draw APBC

APB is a right triangle.

⇒ 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}+(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)

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.

**Question 23. ****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.**

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

**Proof: **AOB is a right triangle.

COD is a right triangle.

CD^{2} = OC^{2}+OD^{2} …(ii)

Adding (i) and (ii)

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

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

AOD is a right triangle.

AD^{2} = OA^{2}+OD^{2} …(iv)

BOC is a right triangle.

BC^{2} = OC^{2}+OB^{2} …(v)

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

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

Hence proved.

**Question 24. ****In a quadrilateral, ABCD∠B = 90° = ∠D. Prove that 2 AC² – BC2 = AB² + AD² + DC².**

** **

**Answer : **Given B = D = 90˚

So ABC and ADC are right triangles.

In ABC,

AC^{2} = AB^{2}+BC^{2} …(i)

In ADC,

AC^{2} = AD^{2}+DC^{2} …(ii)

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.

**Question 25. ****In a ∆ ABC, ∠ A = 90°, CA = AB and D is a point on AB produced. Prove that :**

DC² – BD² = 2AB. AD.

**Answer : Given A = 90° , CA = AB**

Proof: In ACD,

⇒ 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

⇒ 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.

**Question 26. ****In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD.CD.**

**Answer : **Given ABC is an isosceles triangle.

AB = AC

Draw AP BC

APD is a right triangle.

⇒ 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)

Substitute (ii) in (i)

AD^{2} = AC^{2} +CD^{2}+2PC×CD …(iii)

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.

— : End of ML Aggarwal Pythagoras Theorem Exe-12 Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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