Question

There exist a tower near the house of Shankar. The top of the tower AB is tied with steel wire

and on the ground, it is tied with string support.

One day Shankar tried to measure the longest of the wire AC using Pythagoras theorem.

i. In the figure, the length of wire AC is: (take BC = 60 ft)

a. 75 ft

b. 100 ft

c. 120 ft

d. 90 ft

ii. What is the area of  ABC?

a. 2400 ft2

b. 4800 ft2

c. 6000 ft2

d. 3000 ft2

iii. What is the length of the wire PC?

a. 20 ft

b. 30 ft

c. 25 ft

d. 40 ft

iv. What is the length of the hypotenuse in  ABC?

a. 100 ft

b. 80 ft

c. 60 ft

d. 120 ft

v. What is the area of a  POC?

a. 100 ft2

b. 150 ft2

c. 200 ft2

d. 250 ft2​

Answer 1

Given : BC = 60 , AB = 80

Area of ΔABC

To Find : AC

Solution:

Applying Pythagoras theorem

AC² = AB² + BC²

=> AC² = 60² + 80²

=> AC² = 3600 + 6400

=> AC² = 10000

=> AC² = 100²

=> AC = 100  ft

Area of ΔABC = (1/2) AB * BC

= (1/2) 60 (80)

= 2400  sq ft

PC = √20² + 15² = 25 ft

length of hypotenuse in ΔABC = 100 ft

Area of  Δ POC  = (1/2) * 20 * 15

= 150 sq ft

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Answer 2

Answer:

In the figure, the length of wire AC is: (take BC = 60 ft)

a. 75 ft

b. 100 ft

c. 120 ft

d. 90 ft