ML Aggarwal Solutions for Class 9 Maths Chapter 12 Pythagoras Theorem helps students to master the concept of Pythagoras theorem.
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Pythagoras theorem is the fundamental theorem in Mathematics, which defines the relationship between the hypotenuse, base and altitude of a right-angled triangle. According to this theorem, the square of the 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 different application questions.
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ML Aggarwal Solutions for Class 9 Maths Chapter 12 – Pythagoras Theorem
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Exercise 12
1. Lengths of the 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 the 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 the triangle is 32√2 cm2.
13. Find the area and the perimeter of a square whose diagonal is 10 cm long.
Solution:
Given the length of the diagonal of the square is 10 cm.
AC = 10
Let AB = BC = x [Sides of a 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 the 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 , and 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 a 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 the 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+AE 2 ..(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
BED BEC [By SAS congruency rule]
BD = BD [c.p.c.t]
Hence proved.
