SOLUTION: How do I find the largest possible area of an isosceles triangle if the length of each of the two equal sides is 10 m?

Question 1138788: How do I find the largest possible area of an isosceles triangle if the length of each of the two equal sides is 10 m?
Found 3 solutions by josmiceli, ikleyn, greenestamps:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
The height bisects the angle between 10m sides
Call these angles both
The height is
one-half of the base is
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The area is:

Use trig identity

The rate of change is the cos, so

Max is where

m2
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The angles are , ,
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Get a 2nd opinion if needed.
Check my math, too

Answer by ikleyn(52780)   (Show Source): You can put this solution on YOUR website!
.

For any triangle, its area is half of the product of any its two sides' lengths by the sine of the angle between them. Under the given conditions, the area of a triangle is maximal when the sine is equal to 1. It is the case when the angle is the right angle and the triangle is a right angled triangle. In this case its area is = 50 square meters. ANSWER

It is REALLY SIMPLE problem/question, which should be solved / answered in 3 lines.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

You don't even have to get into trigonometry with the sine of an angle to solve this problem.

Given the lengths of two sides of a triangle, the maximum area of a triangle with those two side lengths is when those two sides form a right angle.

That should be easy to see by considering one of the two given side lengths as the base. The area of the triangle is one-half base times height; and clearly the maximum height of the triangle is when the second given side is at right angles to the first.

And, to answer the question, the maximum area of a right triangle is one-half the product of the two legs; so the maximum possible area with two legs of length 10m is (1/2)(10m)(10m) = 50m^2.
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