Question

1. Shankar is having a triangular open space in his plot. He divided the land into three parts by

drawing boundaries PQ and RS which are parallel to BC.

Other measurements are as shown in the figure.​

Answer 1

Given : Shankar is having a triangular open space in his plot. He divided the land into three parts by drawing boundaries PQ and RS which are parallel to BC.

To Find :

i. what is the are of the land?

a. 120m²  b. 60m²   c. 20m²    d. 30m²

ii. What is the length of PQ?

a. 2.5m   b. 5m    c. 6m     d. 8m

iii. the length of RS is

a. 5m     b. 6m       c. 8m       d. 4m

iv. Area of ∆APQ is

a. 7.5m²     b. 10m²     c. 3.75m²      d. 5m²

v. what is the area of ∆ARS?

a. 21.6m²    b. 10m²     c. 3.75m²     d. 6m²

​

Solution:

Area of the field = (1/2) * base * height

= (1/2) * 10 * 12

= 60 m²

Area of the field = 60 m²

ii.  length of PQ

ΔAPQ ≈ ΔABC     ( as ∠A=∠A  common , ∠P = ∠B  , ∠Q = ∠C  corresponding angles)

=> AP/AB = PQ/BC

AP = 5 , AB =AP + PR + RB =  5 + 7 + 8 = 20  , BC = 10

=> 5/20 = PQ/10  => PQ = 2.5

 length of PQ  = 2.5m  

iii. the length of RS

Similarly ΔARS ≈ ΔABC

AR/AB  = RS/BC

=> 12/20 = RS/10

=> RS = 6

the length of RS  =  6m

iv. Area of ∆APQ is

Ratio of area of similar triangle  = ( ratio of corresponding sides)²

Area of ∆APQ / Area of ∆ABC =  (1/4)²

=> Area of ∆APQ / 60 =  1/16

=>  Area of ∆APQ = 60/16  = 15/4 = 3.75 m²

Area of ∆APQ is 3.75m²      

v. area of ∆ARS

area of ∆ARS / 60 =  (3/5)²

=> area of ∆ARS = 21.6m²

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