What is the altitude of a triangle with area 57 sq. mi. and a base 6 mi.?

Question

please show your work.

Answer 1

The formula for the area of a right triangle is given by:

A = b h / 2

where A is area, b is the length of the base, and h is the altitude. If we rearrange to solve for h, we obtain:

h = 2A / b

Plugging in the given values, we compute the answer:

h = 2(57 mi^2) / (6 mi)

h = 19 mi

So, the triangle has an altitude of 19 miles.

Answer 2

Area = 0.5*base*height

Thus:

height = Area*2/base = 57*2/6 = 19 mi

Answer 3

Area (Triangle)=1/2*base*Altitude

Alt=2*area/base

Therefore, Alt=19 mi

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