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.
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
Answers
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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Answer:
Yep
Step-by-step explanation:
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