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nalin [4]
3 years ago
6

An isosceles triangle has area of 105 ft2. If the base is 12 ft, what is the length of each leg? Round your answer to the neares

t tenth.
A) 450.3 ft
B) 14.5 ft
C) 21.2 ft
D) 18.5 ft
Mathematics
1 answer:
e-lub [12.9K]3 years ago
3 0
Answer: Option D) 18.5 ft

Solution:
Area of the isosceles triangle: A=105 ft^2
Base of the isosceles triangle: B=12 ft
Length of each leg of the isosceles triangle: L=?

A=BH/2
Height of the isosceles triangle (perpendicular to the base): H=?

Replacing the known values in the formula of Area:
105 ft^2=(12 ft)H/2
105 ft^2=(6 ft)H

Solving for H: Dividing both sides by 6 ft:
(105 ft^2)/(6 ft)=(6 ft)H/(6 ft)
17.5 ft=H
H=17.5 ft

Using the Pytagoras Theorem:
c^2=a^2+b^2
with:
c=L=?
a=B/2=(12 ft)/2→a=6 ft
b=H→b=17.5 ft

L^2=(6 ft)^2+(17.5)^2
L^2=36 ft^2+306.25 ft^2
L^2=342.25 ft^2
sqrt(L^2)=sqrt(342.25 ft^2)
L=18.5 ft

Answer: The length of each leg is 18.5 ft
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