ML Aggarwal Solution Class 9 Chapter 12 Pythagoras Theorem MCQs

 MCQs

Choose the correct answer from the given four options (1 to 7):

Question 1

In a ∆ABC, if AB = 6√3 cm, BC = 6 cm and AC = 12 cm, then ∠B is

(a) 120°

(b) 90°

(c) 60°

(d) 45°

Sol :
In ∆ABC, if AB = 6√3 cm, BC = 6 cm and AC = 12 cm 






∵AB2+BC2=(6√3)2+(6)2

=108+36=144

and AC2=122=144

∴√B=90°

Ans (b) (Converse of Pythagoras Theorem)


Question 2

If the sides of a rectangular plot are 15 m and 8 m, then the length of its diagonal is

(a) 17 m

(b) 23 m

(c) 21 m

(d) 17 cm

Sol :
Length of a rectangle (l)=15m

and breadth (b)=8m

∴Diagonal=√l2+b2

=√152+82

=√225+64=√289=17 m

Ans (a)


Question 3

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of the side of the rhombus is

(a) 9 cm

(b) 10 cm

(c) 8 cm

(d) 20 cm

Sol :
Lengths of diagonals of rhombus are 16cm and 12cm






∵Diagonals of rhombus bisects each other at right angles 

Length of side$=\sqrt{\left(\frac{\text { First diagonal }}{2}\right)^{2}+\left(\frac{\text { Second diagonal }}{2}\right)^{2}}$

$=\sqrt{\left(\frac{16}{2}\right)^{2}+\left(\frac{12}{2}\right)^{2}}$

$=\sqrt{8^{2}+6^{2}}$

=√64+36=√100=10 cm

Ans (b)


Question 4

If a side of a rhombus is 10 cm and one of the diagonals is 16 cm, then the length of the other diagonals is

(a) 6 cm

(b) 12 cm

(c) 20 cm

(d) 12 cm

Sol :
One diagonal of rhombus=16 cm
Side=10 cm






∵The diagonals of rhombus bisect each other at right angles.

∴In right ΔAOB

AO$=\frac{16}{2}=8$ , AB=10 cm

∴AB2=AO2+BO2

⇒102=82+BO2

⇒BO2=100-64=36=(6)2

∴BO=6 cm

∴Other diagonal BD=6×2 =12 cm

Ans (b)


Question 5

If a ladder 10 m long reaches a window 8 m above the ground, then the distance of the foot of the ladder from the base of the wall is

(a) 18 m

(b) 8 m

(c) 6 m

(d) 4 m

Sol :
Length of ladder=10 m
Height of window=8 m






∴Distance of ladder from the base of wall

$=\sqrt{A C^{2}-A B^{2}}=\sqrt{10^{2}-8^{2}}$

$=\sqrt{100-64}=\sqrt{36}$

=6 m

Ans (c)


Question 6

A girl walks 200 m towards East and then she walks ISO m towards North. The distance of the girl from the starting point is

(a) 350 m

(b) 250 m

(c) 300 m

(d) 225 m

Sol :
A girls walks 200m towards East and then 150 m towards North








Distance of girls from the starting point (OB)

$=\sqrt{\mathrm{OA}^{2}+\mathrm{AB}^{2}}=\sqrt{(200)^{2}+(150)^{2}}$

=√40000+22500

=√62500=250 m 

Ans (b)


Question 7

A ladder reaches a window 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. If the length of the ladder is 15 m, then the width of the street is

(a) 30 m

(b) 24 m

(c) 21 m

(d) 18 m

Sol :
Height of window=12 m
Length of ladder=15 m










In right ΔABC

⇒AC2=AB2+BC2

⇒BC2=AC2-AB2

⇒BC2=152-122

⇒BC2=225-144=81=(9)2

∴EC=12 m
∴Width of street EB=EC+CB
=9+12=21 m
Ans (c)

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