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

Here, we are providing the solutions for Chapter 12 Pythagoras Theorem from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the twelve chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 12 Pythagoras Theorem ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on determining right triangles, solving questions based on Pythagoras theorem, finding distance of pole and height of it and proving perpendiculars and bisectors in a triangle.

### 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 h2 = b2+a2 [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

b2+a2 = 32+ 62 = 9+36 = 45

⇒ h2 = 82 = 64

Here, 45 ≠ 64

Hence 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

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

b2+a2 = 1.42+ 4.82 = 1.96+23.04 = 25

⇒ h2 = 52 = 25

Here, b2+a2 = h2

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,

PR2 = PQ2+QR2

⇒ 102 = 62+QR2

⇒ 100 = 36+QR2

⇒ QR2 = 100-36

⇒ QR2 = 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,

AC2 = AB2+BC2

⇒ 242 = 182+BC2

⇒ 576 = 324+BC2

⇒ BC2 = 576-324

⇒ BC2 = 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,

AC2 = AE2+CE2

⇒ AC2 = 122+52

⇒ AC2 = 144+25

⇒ AC2 = 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,

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

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,

AC2 = BC2+AB2

BC2+AB2= (3x)2+(4x)2

= 9x2+16x2

= 25x2

AC2 = (5x)2 = 25x2

⇒ AC2 = BC2+AB2

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,

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

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,

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

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,

AB2 = BC2+AC2

⇒ AB2 = AC2+AC[∵AC = BC]

⇒ AB2 = 2AC2

Hence, proved.

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

Solution

Comparing (i) and (ii)

AB2– BD= AC2– CD2

AB2+ CD2 = AC2+ BD2

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,

PQ2 = PD2+QD2 [Pythagoras theorem]

⇒ PD= PQ2– QD2

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

In PDR,

PR2 = PD2+DR2 [Pythagoras theorem]

⇒ 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.

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

AB2 = BD2 +AD2 [Pythagoras theorem]

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

Hence the area of triangle is 32√2 cm2.

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.

AC2 = AB2+BC2

⇒ 102 = x2+x2

⇒ 100 = 2x2

⇒ x2 = 50

⇒ x = √50 = √(25×2)

⇒ x = 5√2

So area of square = x2

= (5√2)2 = 50 cm2

Perimeter = 4x

= 4×5√2

= 20√2 cm

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

BD2 = BC2+DC2 [Pythagoras theorem]

⇒ BD2 = 32+122

⇒ BD2 = 9+144

⇒ BD2 = 153

ABD is a right triangle.

⇒ 132 = AB2+153

⇒ 169 = AB2+153

⇒ AB2 = 169-153

⇒ AB2 = 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.

BD2 = BC2+DC2 [Pythagoras theorem]

⇒ BD2 = 82+62

⇒ BD2 = 64+36

⇒ BD2 = 100

Taking square root on both sides,

BD = 10 cm

ABD is a right triangle.

⇒ 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

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.

AC2 = AB2+BC2 [Pythagoras theorem]

⇒ 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 [Pythagoras theorem]

⇒ 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 [Pythagoras theorem]

⇒ 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 [Pythagoras theorem]

⇒ 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

ABD is a right triangle.

⇒ 62 = 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 AB2 = 4AD² – 3AC².

Solution

(a) Given BC = 5 cm,

B =90°, AB = 5AE,

CD = 2AE and AC = ED

ABC is a right triangle.

AC2 = AB2+BC…(i) [Pythagoras theorem]

BED is a right triangle.

ED2 = BE2+BD[Pythagoras theorem]

⇒ AC2 = BE2+BD…(ii) [∵AC = ED]

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 [Pythagoras theorem]

⇒ 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) [Pythagoras theorem]

AD2 = AC2+DC2 …(ii) [Pythagoras theorem]

⇒ AC2 = AD2– (½ BC)2 [∵DC = ½ BC]

⇒ AC2 = AD2– ¼ BC2

⇒ AB2 = 4AD2-3AC2 [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.

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

BD = DC = 10/2 = 5 cm

Given area = 60 cm2

½ ×base ×height = ½ ×10×AD = 60

⇒ x2 = 122+52

⇒ x2 = 144+25

⇒ x2 = 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.

BC2 = OC2+OB2 [Pythagoras theorem]

⇒ BC= 152+202

⇒ BC= 225+400

⇒ BC= 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.

⇒ 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

Taking square root on both sides

CM = 24 cm [∵ABCD]

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˚ [Each angle of a square equals 90˚]

AHE is a right triangle.

HE2 = AE2+AH2 [Pythagoras theorem]

⇒ HE2 = 32+32

⇒ HE2 = 9+9 = 18

⇒ HE = √(9×2) = 3√2 cm

Similarly GF = 3√2 cm

EBF is a right triangle.

EF2 = BE2+BF2 [Pythagoras theorem]

⇒ EF2 = 42+42

⇒ EF2 = 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

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

Hence, area of the rectangle is 24 cm2 and perimeter is 14√2 cm.

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

Solution

D = 90˚

Proof:

Since ABC is an equilateral triangle,

AB = AC = BC

ABD is a right triangle.

According to Pythagoras theorem,

BD = ½ BC

⇒ AB2 = AD2+( ½ AB)2 [∵BC = AB]

⇒ AB2 = AD2+ ¼ AB2

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 :

(ii)4BE2 = 4BC2+AC2

Solution

Proof:

(i) C = 90°

So ACD is a right triangle.

Multiply both sides by 4, we get

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

⇒ 4AD2 = 4AC2+BC2…(i) [∵BC = 2BD]

Hence proved.

(ii) BCE is a right triangle.

BE2 = BC2+CE2 [Pythagoras theorem]

Multiply both sides by 4 , we get

4BE2 = 4BC2+4CE2

⇒ 4BE2 = 4BC2+(2CE)2

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

Hence proved.

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

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.

AB2 = AP2+BP2 [Pythagoras theorem]

⇒ AB2 = AP2+(BD-PD)2

⇒ AB2 = AP2+BD2+PD2-2BD×PD

⇒ AB2 = (AP2+PD2)+BD2-2BD×PD

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

⇒ AB= AD2+ ¼ BC2– BC×PD …(i)

APC is a right triangle.

AC2 = AP2+PC2 [Pythagoras theorem]

⇒ 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), we get

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

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.

AB2 = OB2+OA2 …(i) [Pythagoras theorem]

COD is a right triangle.

CD2 = OC2+OD2 …(ii) [Pythagoras theorem]

Adding (i) and (ii), we get

AB2+ CD2 = OB2+OA2+ OC2+OD2

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

AOD is a right triangle.

AD2 = OA2+OD2 …(iv) [Pythagoras theorem]

BOC is a right triangle.

BC2 = OC2+OB2 …(v) [Pythagoras theorem]

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

Hence, proved.

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

Solution

Given B = D = 90˚

So ABC and ADC are right triangles.

In ABC,

AC2 = AB2+BC2 …(i) [Pythagoras theorem]

AC2 = AD2+DC2 …(ii) [Pythagoras theorem]

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,

⇒ DC2 = CA2+(AB+BD)2

⇒ DC2 = CA2+AB2+BD2+2AB×BD

⇒ DC2 -BD= CA2+AB2+2AB×BD

⇒ DC2 -BD2 = AB2+AB2+2AB×BD [∵CA = AB]

⇒ DC2 -BD2 = 2AB2+2AB×BD

⇒ DC2 -BD2 = 2AB(AB+BD)

⇒ DC2 -BD2 = 2AB×AD [A-B-D]

Hence, proved.

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

Solution

Given ABC is an isosceles triangle.

AB = AC

Construction: Draw AP BC

Proof:

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) [Pythagoras theorem]

Substitute (ii) in (i)

Since ABC is an isosceles triangle,

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

AD2 = AC2 +CD2+2× ½ BC ×CD

⇒ AD2 = AC2 +CD×BD [CD+BC = BD]

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

⇒ 172 = 152+DC2

⇒ 289 = 225+DC2

⇒ DC2 = 289-225

⇒ DC2 = 64

Taking square root on both sides,

DC = 8 cm

⇒ 252 = 152+BD2

⇒ 625 = 225+ BD2

⇒ BD2 = 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.

⇒ AC2 = 62+82

⇒ AC2 = 36+64

⇒ AC2 = 100

Taking square root on both sides,

AC = 10 cm

Hence length of AC is 10 cm.

(ii) ABC is a right triangle.

BC2 = AC2+AB2 [Pythagoras theorem]

⇒ 262 = 102+AB2

⇒ AB2 = 262-102

⇒ AB2 = 676-100

⇒ AB2 = 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 cm2

= ½ ×6×8

= 24 cm2

= 120-24

= 96 cm2

Hence, the area of shaded region is 96 cm2.

(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.

AC2 = AB2+BC2 [Pythagoras theorem]

⇒ 152 = 92+BC2

⇒ 225 = 81+BC2

⇒ BC2 = 225-81

⇒ BC2 = 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

⇒ (15/2)2 = (9/2)2+DE2

⇒ DE2 = (15/2)2 – (9/2)2

⇒ DE2 = 225/4 -81/4

⇒ DE2 = 144/4

Taking square root on both sides,

DE = 12/2 = 6 cm.

= ½ ×6×9/2

= 13.5 cm2

Hence, the area of the ADE is 13.5 cm2.

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

Solution

Proof:

Equating (i) and (ii)

AB2-BD2 = AC2-CD2

AB2-AC2 = BD2– CD2

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,

AQ2 = AC2+CQ2 [Pythagoras theorem]

Multiplying both sides by 9, we get

9AQ2 = 9AC2+9CQ2

⇒ 9AQ= 9AC2+(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= 9AC2+(2BC)2

⇒ 9AQ= 9AC2+4BC2 …(iii)

Hence, proved.

(ii) In BPC,

BP2 = BC2+CP2 [Pythagoras theorem]

Multiplying both sides by 9, we get

9BP2 = 9BC2+9CP2

⇒ 9BP2 = 9BC2+(3CP)2 …(iv)

Given AP: PC = 1:2

CP/AC = CP/AP+PC

⇒ CP/AC = 2/3

⇒ 3CP = 2AC …(v)

Substitute (v) in (iv)

9BP2 = 9BC2+(2AC)2

⇒ 9BP2 = 9BC2+4AC2 ...(vi)

Hence, proved.

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

9AQ2+9BP2 = 9AC2+4BC2+9BC2+4AC2

⇒ 9(AQ2+BP)2 = 13AC2+13BC2

⇒ 9(AQ2+BP)2 = 13(AC2+BC2) …(vii)

In ABC,

AB2 = AC2+BC2 …(viii)

Substitute (viii) in (viii), we get

9(AQ2+BP)2 = 13AB2

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,

PR2 = RQ2+PQ2 [Pythagoras theorem]

⇒ PR2 = (3x)2+PQ2

⇒ PR2 = 9x2+PQ2

Multiply above equation by 3

3PR2 = 27x2+3PQ2 …(i)

Similarly in PTS,

PT2 = TQ2+PQ2 [Pythagoras theorem]

⇒ PT2 = (2x)2+PQ2

⇒ PT2 = 4x2+PQ2

Multiply above equation by 8

8PT2 = 32x2+8PQ2 …(ii)

Similarly in PSQ,

PS2 = SQ2+PQ2 [Pythagoras theorem]

⇒ PS2 = x2+PQ2

Multiply above equation by 5

5PS2 = 5x2+5PQ2 …(iii)

Add (i) and (iii), we get

3PR2 +5PS2 = 27x2+3PQ2+5x2+5PQ2

⇒ 3PR2 +5PS2 = 32x2+8PQ2

⇒ 3PR2 +5PS2 = 8PT2 [From (ii)]

8PT2 = 3PR2 +5PS2

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,

AC2 = AB2+BC2 …(i) [Pythagoras theorem]

Given,

AD² = AB² + BC² + CD²

⇒ AD² = AC2+CD2 [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

In ABC,

BC2 = AC2+AB2 [Pythagoras theorem]

⇒ BC2 = b2+c2

⇒ BC = √( b2+c2) …(i)

Area of ABC = ½ ×AB×AC

= ½ ×bc …(ii)

Also, Area of ABC = ½ ×BC×AD

= ½ ×√( b2+c2) ×AD …(iii)

Equating (ii) and (iii)

½ ×bc = ½ ×√( b2+c2) ×AD

⇒ AD = bc /(√( b2+c2)

Hence, AD is bc /(√( b2+c2).

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 cm2]

⇒ 108 = (1/12)x2

⇒ x2 = 108×12

⇒ x2 = 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 ,

BC2 = BE2+EC2 [Pythagoras theorem]

⇒ BC2 = BE2+(AC-AE)2

⇒ BC2 = BE2+AC2+AE2- 2 AC×AE

⇒ BC2 = BE2+AE2+AC2- 2 AC×AE …(i)

In ABC,

AB2 = BE2+AE2 ...(ii)

Substitute (ii) in (i)

BC2 = AB2+AC2- 2 AC×AE

⇒ BC2 = AC2+AC2- 2 AC×AE [∵AB = AC]

⇒ BC2 = 2AC2-2 AC×AE

⇒ BC2 = 2AC(AC-AE)

⇒ BC2 = 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.