[Math] Find the legs of isosceles triangle, given only the base

geometrytriangles

Is it possible to find the legs of isosceles triangle, given only the base length? I think that the info is insufficient. Am I right?

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The base length of an isosceles triangle is not enough to determine the triangle:

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[Math] How to find the base of an isosceles triangle with the angles and legs

Look at this picture http://mathworld.wolfram.com/images/eps-gif/Trigonometry_700.gif

Imagine that $\theta = 90-54.5= 35.5$.

This is a picture of your triangle cut in half, laid on its side, and shrunk to $1/43$ it's size.

We want to know what is the length of the side labeled "$\sin \theta $" is.

Your answer will be twice that multiplied by $43$.

So we plug it it into a calculator and get $\sin 35.5 = 0.58070295571093976105008.... $

So multiply that by $2$ and you will have the base of an isoceles triangle with legs of $1$.

Multiply that by $43$ and you will have the base of your tringle.

So the answer is $49.940454191140819450...... $.

[Math] Finding the area of an isosceles right triangle given its hypotenuse

Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.

However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.

In this case, $8^2 + 8^2 = (8\sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $\frac 12 (8)(8) = 32$.

(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8\sqrt 2)$, you can apply the converse to simplify your work a bit).

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Related Solutions

[Math] How to find the base of an isosceles triangle with the angles and legs

[Math] Finding the area of an isosceles right triangle given its hypotenuse

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