Selina Concise Class 9 Maths Chapter 13 Pythagoras Theorem Exercise 13A Solutions
EXERCISE – 13A
(1) A ladder 13m long rests against a vertical wall if the foot of the ladder is 5m from the foot of the wall, find the distance of the other end of the ladder from the ground.
=>
Given :
A ladder 13m long rest against a vertical wall.
∴ hypotenuse = AC = 13m
If the foot of the ladder is 5m from the foot of the wall.
∴ Base = BC = 5m
By Pythagoras Theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AC)2 = (BC)2 + (AB)2
(13)2 = (5)2 + (AB)2
169 = 25 + (AB)2
169 – 25 = (AB)2
144 = (AB)2
Taking square root on both sides,
√144 = √(AB)2
12cm = AB
∴ AB = 12cm
Go beyond traditional BBA College with NEXIS 3-year UG
∴ The distance of the other end of the ladder from the ground is 12m.
(2) A man goes 40m due north and then 50m due west. Find his distance from the starting point.
=>
Given, A man goes 40m due north.
Base = 40m
and 50m due west
height = 50m
By Pythagoras Theorem,
(hypotenuse)2 = (Base)2 + (height)2
H2 = (40)2 + (50)2
H2 = 1600 + 2500
H2 = 4100
Taking square root on both sides,
√H2 = √4100
√H2 = √41 × 100
H = 10√41 m
∴ His distance from the starting point is 10√41 m.
(3) In the figure : ∠PSQ = 90°, PQ = 10cm, QS = 6cm and RQ = 9cm. Calculate the length of PQ.
=>
Given :
∠PSQ = 90°
PQ = 10cm
QS = 6cm
and RQ = 9cm.
In △ PSQ,
By Pythagoras Theorem,
(hypotenuse)2 = (Base)2 + (height)2
(PQ)2 = (QS)2 + (PS)2
(10)2 = (6)2 + (PS)2
100 = 36 + (PS)2
100 – 36 = (PS)2
64 = (PS)2
Taking square root on both sides,
√64 = √(PS)2
8cm = PS
PS = 8cm
In △ PSR,
By Pythagoras theorem,
(PR)2 = (RS)2 + (PS)2
(PR)2 = (15)2 + (8)2
= 225 + 64
(PR)2 = 289
Taking square root on both sides,
√(PR)2 = √289
PR = 17cm
(4) The given figure shows a quadrilateral ABCD in which AD = 13cm, DC = 12cm, BC = 3cm and ABC = BCD = 90°. Calculate the length of AB.
=>
Given :
AD = 13cm
DC = 12cm
BC = 13cm
and ∠ABD = BCD = 90°
Now,
We have to find the length of AB.
In △ BCD
By Pythagoras Theorem,
(hypotenuse)2 = (Base)2 + (height)2
(BD)2 = (BC)2 + (CD)2
= (3)2 + (12)2
= 9 + 144
(BD)2 = 153
Taking Square root on both sides,
√(BD)2 = √153
BD = √153
In △ ABD,
by Pythagoras theorem,
(AD)2 = (BD)2 + (AB)2
(13)2 = (√153)2 + (AB)2
169 = 153 + (AB)2
169 – 153 = (AB)2
16 = (AB)2
Taking square root on both sides,
√16 = √(AB)2
4cm = AB
∴ AB = 4cm
∴ The length of AB is 4cm.
(5) AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10cm, find the length of AD, correct to 1 place of decimal.
=>
Given:
BC = 10cm
AD is draw n perpendicular to base BC of an equilateral triangle.
In △ ADB,
By Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AB)2 = (BD)2 + (AD)2
(10)2 = (5)2 + (AD)2
100 = 25 + (AD)2
100 – 25 = (AD)2
75 = (AD)2
Taking square root on both sides,
√75 = √(AD)2
√75 = AD
Now, we have to find the length of AD, correct to 1 place of decimal.
∴ AD = √75
AD = 8.66
AD = 8.7 cm
∴ The length of AD is 8.7 cm.
(6) In triangle ABC, given below, AB = 8cm, BC = 6cm and AC = 3cm. Calculate the length of OC.
=>
Given, AB = 8cm
BC = 6cm
AC = 3cm
Now,
we have to find the length of OC.
In △ AOC,
By Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AC)2 = (OC)2 + (AO)2
(3)2 = (AO)2 + x2
9 = (AO)2 + x2
9 – x2 = AO2 ……(i)
In △ AOB,
by Pythagoras theorem,
(AB)2 = (AO)2 + (BO)2
(8)2 = (AO)2 + (6 + x)2
64 = (AO)2 + (6 + x)2
64 – (6 + x)2 = (AO)2 …… (ii)
From equation (i) and (ii),
9 – x2 = 64 – (6 + x)2
9 – x2 = 64 – (36 + x2 + 12x) (∵ a2 + 2ab + b2)
9 – x2 = 64 – 36 – x2 – 12x
9 = 28 – 12x
9 – 28 = – 12x
– 19 = -12x
19 = 12x
19/12 = x
x = 19/12
x = 1 7/12 cm
∴ The length of OC is 1 7/12 cm.
(7) In triangle ABC, AB = AC = x, BC = 10cm and the area of the triangle is 60cm2. Find x.
=>
Given :
AB = AC = X
BC = 10cm and the area of triangle is 60cm2.
Now, we have to find the value of x.
We know that,
Area of triangle = 1/2 × Base × height
60 = 1/2 × x × x
60 × 2 = x2
120 = x2
Taking square root on both sides,
√120 = √x2
x = √120
Now, we have to draw isosceles triangle
Area of isosceles triangle =
base/4 √4a2 – b2
60 = 10/4 √4x2 – (10)2
60 = 5/2 √4x2 – 100
60 × 2 = 5 √4x2 – 100
120 = 5 √4x2 – 100
120/5 = √4x2 – 100
24 = √4x2 – 100
Taking square on both sides,
(24)2 = (√4x2 – 100)2
576 = 4x2 – 100
576 + 100 = 4x2
676 = 4x2
676/4 = x2
169 = x2
Taking square root on both sides,
√169 = √x2
13 = x
∴ x = 13
∴ The value of x = 13
(8) If the sides of triangle are in the ratio 1 : 2 : 1, show that is a right angled triangle.
=> Given, If the sides of triangle are in the ratio 1 : 2 : 1.
By Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(√2)2 = (1)2 + ( )2
2 = 1 + 1
2 = 2
R.H.S. = L.H.S.
Right hand side = Left hand side.
∴ This triangle is a right angle triangle.
(9) Two poles of heights 6m and 11m stand vertically on a plane ground. If the distance between their feet is 12m; find the distance between their tips.
=> Given,
Two poles of heights 6m and 11m stand 6m vertically on a plane ground.
If the distance between their feet is 12m.
Now, we have to find the distance between their tips.
In △ ADE,
By Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AE)2 = (AD)2 + (DE)2
(AE)2 = (12)2 + (5)2
= 144 + 25
AE2 = 169
Taking square root on both sides,
√169 = √AE2
13m = AE
∴ The distance between their tips is 13m.
(10) In the given figure, AB || CD, AB = 7cm, BD = 25cm and CD = 17cm; find the length of side BC.
=> Given,
AB parallel CD (AB || CD)
AB = 7cm
BD = 25cm and
CD = 17cm.
Now, we have to find the length of side BC.
In △ BAD,
by Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(BD)2 = (AB)2 + (AD)2
(25)2 = (7)2 + (AD)2
625 = 49 + (AD)2
625 – 49 = (AD)2
576 = AD2
Taking square root on both sides,
√576 = √AD2
24 = AD
In △ BMC,
by Pythagoras theorem,
(BC)2 = (BM)2 + (CM)2
(BC)2 = (24)2 + (10)2
= 576 + 100
(BC)2 = 676
Taking square root on both sides.
√BC2 = √676
BC = 26cm
Here is your solution of Selina Concise Class 9 Maths Chapter 13 Pythagoras Theorem Exercise 13A
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.