ML Aggarwal Solutions for Chapter 12 Pythagoras Theorem Class 9 Maths ICSE
Exercise 12
1. Lengths of sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:
(i) 3 cm, 8 cm, 6 cm
(ii) 13 cm, 12 cm, 5 cm
(iii) 1.4 cm, 4.8 cm, 5 cm
Solution
We use the Pythagoras theorem to check whether the triangles are right triangles.
We have 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+AC2 [∵AC = BC]
⇒ AB2 = 2AC2
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,
AB2 = AD2+BD2 [Pythagoras theorem]
AD2 = AB2– BD2 …(i)
In ADC,
AC2 = AD2+CD2 [Pythagoras theorem]
AD2 = AC2– CD2 …(ii)
Comparing (i) and (ii)
AB2– BD2 = 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]
⇒ PD2 = 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
ADC is a right triangle.
AB2 = BD2 +AD2 [Pythagoras theorem]
⇒ 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
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.
AD2 = AB2+BD2 [Pythagoras theorem]
⇒ 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.
BD2 = AB2+AD2 [Pythagoras theorem]
⇒ 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
ADC is a right triangle.
AC2 = AD2+CD2 [Pythagoras theorem]
⇒ 62 = AD2+CD2 …(i)
ABD is a right triangle.
AB2 = AD2+BD2 [Pythagoras theorem]
⇒ 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
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+BC2 …(i) [Pythagoras theorem]
BED is a right triangle.
ED2 = BE2+BD2 [Pythagoras theorem]
⇒ AC2 = BE2+BD2 …(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]
ADC is a right triangle.
AD2 = AC2+DC2 …(ii) [Pythagoras theorem]
⇒ 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.
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 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
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]
⇒ 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.
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.
AD2 = AM2+DM2 [Pythagoras theorem]
⇒ 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 [Pythagoras theorem]
⇒ 252 = AD2+72
⇒ AD2 = 252-72
⇒ AD2 = 625-49
⇒ AD2 = 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.
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
Given AD BC
D = 90˚
Proof:
Since ABC is an equilateral triangle,
AB = AC = BC
ABD is a right triangle.
According to Pythagoras theorem,
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.
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
Solution
Proof:
(i) C = 90°
So ACD is a right triangle.
AD2 = AC2+CD2 [Pythagoras theorem]
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 [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.
(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.
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
⇒ 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+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(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.
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
AB2+ CD2 = AD2+BC2
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]
In ADC,
AC2 = AD2+DC2 …(ii) [Pythagoras theorem]
Adding (i) and (ii)
2AC2 = AB2+BC2+ AD2+DC2
⇒ 2AC2 -BC2 = AB2+AD2+DC2
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+AD2 [Pythagoras theorem]
⇒ DC2 = CA2+(AB+BD)2
⇒ DC2 = CA2+AB2+BD2+2AB×BD
⇒ DC2 -BD2 = 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.
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.
AD2 = AP2+PD2 [Pythagoras theorem]
⇒ 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)
AD2 = AC2 +CD2+2PC×CD …(iii)
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 +CD2+BC×CD
⇒ AD2 = AC2 +CD(CD+BC)
⇒ AD2 = AC2 +CD×BD [CD+BC = BD]
⇒ AD2 = AC2 +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.
AC2 = AD2+DC2 [Pythagoras theorem]
⇒ 172 = 152+DC2
⇒ 289 = 225+DC2
⇒ DC2 = 289-225
⇒ DC2 = 64
Taking square root on both sides,
DC = 8 cm
ADB is a right triangle.
AB2 = AD2+BD2 [Pythagoras theorem]
⇒ 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 = AD2+DC2 [Pythagoras theorem]
⇒ 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
Area of ADC = ½ ×AD×DC
= ½ ×6×8
= 24 cm2
Area of shaded region = area of ABC- area of ADC
= 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
ADE is a right triangle.
AE2 = AD2+DE2 [Pythagoras theorem]
⇒ (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.
Area of ADE = ½ ×DE×AD
= 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
Given AD BC, AB>AC
So ADB and ADC are right triangles.
Proof:
In ADB,
AB2 = AD2+BD2 [Pythagoras theorem]
⇒ AD2 = AB2-BD2 …(i)
In ADC,
AC2 = AD2+CD2 [Pythagoras theorem]
⇒ AD2 = AC2-CD2 …(ii)
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
⇒ 9AQ2 = 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)
9AQ2 = 9AC2+(2BC)2
⇒ 9AQ2 = 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
ADB = 90°
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.