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

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 h² = b²+a² [Pythagoras theorem]

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

b²+a² = 3²+ 6² = 9 + 36 = 45

⇒ h² = 8² = 64

Here, 45 ≠ 64

Therefore, the given triangle is not a right triangle.

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

b²+a² = 12²+ 5² = 144+25 = 169

⇒ h² = 13² = 169

Here, b²+a² = h²

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

b²+a² = 1.4²+ 4.8² = 1.96+23.04 = 25

⇒ h² = 5² = 25

Here, b²+a² = h²

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

PR² = PQ²+QR²

⇒ 10² = 6²+QR²

⇒ 100 = 36+QR²

⇒ QR² = 100-36

⇒ QR² = 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.

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

AC² = AE²+CE²

⇒ AC² = 12²+5²

⇒ AC² = 144+25

⇒ AC² = 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.

h² = b²+a²

⇒ 20² = (3x)²+(4x)²

⇒ 400 = 9x²+16x²

⇒ 25x² = 400

⇒ x² = 400/25

⇒ x² = 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.

AC² = BC²+AB²

BC²+AB²= (3x)²+(4x)²

= 9x²+16x²

= 25x²

AC² = (5x)² = 25x²

⇒ AC² = BC²+AB²

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

AB² = CB²+AC²

⇒ 26² = ( 2(x+7))²+(2x)²

⇒ 676 = 4(x²+14x+49) + 4x²

⇒ 4x²+56x+196+4x² = 676

⇒ 8x²+56x+196 = 676

⇒ 8x²+56x +196-676 = 0

⇒ 8x²+56x -480 = 0

⇒ x²+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

AB² = CB²+AC²

⇒ (2x+6)² = x²+(2x+4)²

⇒ 4x²+24x+36 = x²+4x²+16x+16

⇒ x²-8x-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 

AB² = BC²+AC²

⇒ AB² = AC²+AC² [∵AC = BC]

⇒ AB² = 2AC²

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,   AB² = AD²+BD²

AD² = AB²– BD² …(i)

In ADC,  AC² = AD²+CD²

AD² = AC²– CD² …(ii)

Comparing (i) and (ii)

AB²– BD² = AC²– CD²

AB²+ CD² = AC²+ BD²

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,

PQ² = PD²+QD²

⇒ PD² = PQ²– QD²

⇒ PD² = a²– c² …(i) [∵ PQ = a and QD = c]

In PDR,

PR² = PD²+DR²

⇒ PD² = PR²– DR²

⇒ PD² = b²– d² …(ii) [∵ PR = b and DR = d]

Comparing (i) and (ii)

a²– c²= b²– d²

⇒ a²– b²= c²– d²

⇒ (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.

⇒ AD² = AB² -BD²

⇒ AD² = 12²-4²

⇒ AD² = 144-16

⇒ AD² = 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 cm²

Therefore, the area of triangle is 32√2 cm².

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.

AC² = AB²+BC²

⇒ 10² = x²+x²

⇒ 100 = 2x²

⇒ x² = 50

⇒ x = √50 = √(25×2)

⇒ x = 5√2

area of square = x²

= (5√2)² = 50 cm²

Perimeter = 4x

= 4×5√2

= 20√2 cm

Therefore, area and perimeter of the square are 50 cm² 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.

⇒ BD² = 3²+12²

⇒ BD² = 9+144

⇒ BD² = 153

ABD is a right triangle.

AD² = AB²+BD²

⇒ 13² = AB²+153

⇒ 169 = AB²+153

⇒ AB² = 169-153

⇒ AB² = 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.

BD² = BC²+DC²

⇒ BD² = 8²+6²

⇒ BD² = 64+36

⇒ BD² = 100

Taking square root on both sides,

BD = 10 cm

ABD is a right triangle.

BD² = AB²+AD²

⇒ 10² = 2AB² [∵AB = AD]

⇒ 100 = 2AB²

⇒ AB² = 100/2

⇒ AB² = 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.

AC² = AB²+BC²

⇒ 13² = 12²+BC²

⇒ BC² = 13²-12²

⇒ BC² = 169-144

⇒ BC² = 25

Taking square root on both sides,

BC = 5 cm

CDE is a right triangle.

CE² = CD²+DE² 

⇒ 10² = CD²+6²

⇒ 100 = CD²+36

⇒ CD² = 100-36

⇒ CD² = 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.

PQ² = PS²+QS²

⇒ 10² = PS²+6²

⇒ 100 = PS²+36

⇒ PS² = 100-36

⇒ PS² = 64

Taking square root on both sides,

PS = 8 cm

PSR is a right triangle.

RS = RQ+QS

⇒ RS = 9+6

⇒ RS = 15 cm

⇒ PR² = PS²+RS²

⇒ PR² = 8²+15²

⇒ PR² = 64+225

⇒ PR² = 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.

AC² = AD²+CD²

⇒ 6² = AD²+CD² …(i)

ABD is a right triangle.

AB² = AD²+BD²

⇒ 16² = AD²+(BC+CD)²

⇒ 16² = AD²+(12+CD)²

⇒ 256 = AD²+144+24CD+CD²

⇒ 256-144 = AD²+CD²+24CD

⇒ AD²+CD² = 112-24CD

⇒ 6² = 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.

AC² = AB²+BC² 

BED is a right triangle.

ED² = BE²+BD²

⇒ AC² = BE²+BD² …(ii)

Comparing (i) and (ii)

AB²+BC² = BE²+BD²

⇒ (5AE)²+5² = (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

⇒ 25x²+25 = 16x²+25+20x+4x²

⇒ 25x² = 20x²+20x

⇒ 5x² = 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.

AC² = AB²+BC²

⇒ AC² = 20²+5²

⇒ AC² = 400+25

⇒ AC² = 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.

AB² = AC²+BC² (i)

ADC is a right triangle.

AD² = AC²+DC² …(ii)

⇒ AC² = AD²-DC²

⇒ AC² = AD²– (½ BC)² [∵DC = ½ BC]

⇒ AC² = AD²– ¼ BC²

⇒ 4AC² = 4AD²– BC²

⇒ AC²+3AC² = 4AD²– BC²

⇒ AC²+BC² = 4AD²-3AC²

⇒ AB² = 4AD²-3AC² [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 cm²

½ ×base ×height = ½ ×10×AD = 60

⇒ AD = 60×2/10

⇒ AD = 60/5

⇒ AD = 12cm

ADC is a right triangle.

AC² = AD²+DC²

⇒ x² = 12²+5²

⇒ x² = 144+25

⇒ x² = 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.

BC² = OC²+OB²

⇒ BC² = 15²+20²

⇒ BC² = 225+400

⇒ BC² = 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.

⇒ 13² = 5²+DM²

⇒ DM² = 13²-5²

⇒ DM² = 169-25

⇒ DM² = 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.

AC² = AD²+DC² 

⇒ 25² = AD²+7²

⇒ AD² = 25²-7²

⇒ AD² = 625-49

⇒ AD² = 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.

BC² = BM²+CM²

⇒ BC² = 10²+24²

⇒ BC² = 100+576

⇒ BC² = 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.

HE² = AE²+AH²

⇒ HE² = 3²+3²

⇒ HE² = 9+9 = 18

⇒ HE = √(9×2) = 3√2 cm

Similarly GF = 3√2 cm

EBF is a right triangle.

⇒ EF² = 4²+4²

⇒ EF² = 16+16 = 32

Taking square root on both sides

EF = √(16×2) = 4√2 cm

HG = 4√2 cm

Now join EG

In EFG

EG² = EF²+GF²

⇒ EG² = (4√2)²+(3√2)²

⇒ EG² = 32+18 = 50

⇒ EG = √50 = 5√2 cm …(i)

Join HF.

Also HF² = EH²+HG²

= (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 cm²

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 cm² 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.

AB² = AD²+BD²

BD = ½ BC

AB² = AD²+( ½ BC)²

⇒ AB² = AD²+( ½ AB)² [∵BC = AB]

⇒ AB² = AD²+ ¼ AB²

⇒ AB² = (4AD²+ AB²)/4

⇒ 4AB² = 4AD²+ AB²

⇒ 4AD² = 4AB²– AB²

⇒ 4AD² = 3AB²

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)4AD² = 4AC²+BC²

(ii)4BE² = 4BC²+AC²

(iii)4(AD²+BE²) = 5AB²

 

Answer :  (i) C = 90°

So ACD is a right triangle.

Multiply both sides by 4, we get

4AD² = 4AC²+4CD²

⇒ 4AD² = 4AC²+4BD² (∵D is the midpoint of BC, CD = BD = ½ BC)

⇒ 4AD² = 4AC²+(2BD)²

⇒ 4AD² = 4AC²+BC²…(i) [∵BC = 2BD]

Hence proved.

(ii) BCE is a right triangle.

BE² = BC²+CE²

Multiply both sides by 4

4BE² = 4BC²+4CE²

⇒ 4BE² = 4BC²+(2CE)²

⇒ 4BE² = 4BC²+AC² …(ii) (∵E is the midpoint of AC, AE = CE = ½ AC)

Hence proved.

(iii) Adding (i) and (ii)

4AD²+4BE² = 4AC²+BC²+4BC²+AC²

⇒ 4AD²+4BE² = 5AC²+5BC²

⇒ 4(AD²+BE² ) = 5(AC²+BC²)

⇒ 4(AD²+BE² ) = 5(AB²) (∵ABC is a right triangle, AB² = AC²+BC²)

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.

⇒ AB² = AP²+(BD-PD)²

⇒ AB² = AP²+BD²+PD²-2BD×PD

⇒ AB² = (AP²+PD²)+BD²-2BD×PD

⇒ AB² = AD²+ (½ BC)²-2×( ½ BC)×PD [∵AP²+PD² = AD² and BD = ½ BC]

⇒ AB² = AD²+ ¼ BC²– BC×PD …(i)

APC is a right triangle.

⇒ AC² = AP²+(PD²+DC²)

⇒ AC² = AP²+PD²+DC²+ 2×PD×DC

⇒ AC² = (AP²+PD²)+ (½ BC)²+2×PD×( ½ BC) [DC = ½ BC]

⇒ AC² = (AD)²+ ¼ BC²+PD× BC …(ii) [In APD, AP²+PD² = AD²]

Adding (i) and (ii), we get

AB²+AC² = 2AD²+ ½ BC² …(iii)

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

Similarly we get,

BC²+CA² = 2CF²+ ½ AB² …(iv)

AB²+BC² = 2BE²+ ½ AC² …(v)

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

2(AB²+BC²+CA²) = 2(AD²+BE²+CF²)+ ½ (BC²+AB²+AC²)

⇒ 2(AB²+BC²+CA²) = 2(AB²+BC²+CA²) – ½ (AB²+BC²+CA²)

⇒ 2(AD²+BE²+CF²) = (3/2)× (AB²+BC²+CA²)

⇒ 4(AD²+BE²+CF²) = 3(AB²+BC²+CA²)

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.

CD² = OC²+OD² …(ii)

Adding (i) and (ii)

AB²+ CD² = OB²+OA²+ OC²+OD²

⇒ AB²+ CD² = (OA²+OD²)+ (OC²+OB²) …(iii)

AOD is a right triangle.

AD² = OA²+OD² …(iv)

BOC is a right triangle.

BC² = OC²+OB² …(v)

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

AB²+ CD² = AD²+BC²

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,

AC² = AB²+BC² …(i)

In ADC,

AC² = AD²+DC² …(ii)

Adding (i) and (ii)

2AC² = AB²+BC²+ AD²+DC²

⇒ 2AC² -BC² = AB²+AD²+DC²

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,

⇒ DC² = CA²+(AB+BD)²

⇒ DC² = CA²+AB²+BD²+2AB×BD

⇒ DC² -BD² = CA²+AB²+2AB×BD

⇒ DC² -BD² = AB²+AB²+2AB×BD

⇒ DC² -BD² = 2AB²+2AB×BD

⇒ DC² -BD² = 2AB(AB+BD)

⇒ DC² -BD² = 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.

⇒ AD² = AP²+(PC+CD)² [PD = PC+CD]

⇒ AD² = AP²+PC²+CD² +2PC×CD …(i)

APC is a right triangle.

AC² = AP²+PC² …(ii)

Substitute (ii) in (i)

AD² = AC² +CD²+2PC×CD …(iii)

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

AD² = AC² +CD²+2× ½ BC ×CD

⇒ AD² = AC² +CD²+BC×CD

⇒ AD² = AC² +CD(CD+BC)

⇒ AD² = AC² +CD×BD [CD+BC = BD]

⇒ AD² = AC² +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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