An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of $32$ $32$ and the triangle has an area of $16$ $16$ . What are the lengths of sides A and B?

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An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of $32$ $32$ and the triangle has an area of $16$ $16$ . What are the lengths of sides A and B?

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Answer 1 · 2017-11-13T21:10:46

$\sqrt{257} \approx 16.03$ $sqrt257~~16.03$

Explanation:

First things first - we start off with a diagram!
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Not the most elegant diagram in the world, but it does the job. By convention, the side opposite angle $A$ $A$ is $a$ $a$ , opposite angle $B$ $B$ is $b$ $b$ and $C$ $C$ is $c$ $c$ . $h$ $h$ is the perpendicular bisector of AB at M - it cuts AB in half at a right angle.

Let $a = b = x$ $a=b=x$ . Our job is to find $x$ $x$ .

$A r e a_{△} = \frac{1}{2} b h$ $Area_triangle=1/2bh$
$16 = \frac{1}{2} \times 32 h$ $16=1/2xx32h$
$16 = 16 h$ $16=16h$
$h = 1$ $h=1$

Now, we can apply Pythagoras' Theorem in one of the right-angled triangles. We have the height $h = 1$ $h=1$ and the base $= 16$ $=16$ , so we can work out x.

$1^{2} + 16^{2} = x^{2}$ $1^2+16^2=x^2$
$1 + 256 = x^{2}$ $1+256=x^2$
$x^{2} = 257$ $x^2=257$
$x = \sqrt{257} \approx 16.03$ $x=sqrt257~~16.03$

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An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of $32$ $32$ and the triangle has an area of $16$ $16$ . What are the lengths of sides A and B?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 3232 and the triangle has an area of 1616 . What are the lengths of sides A and B?

1 Answer

Explanation:

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Impact of this question

An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of $32$ $32$ and the triangle has an area of $16$ $16$ . What are the lengths of sides A and B?