Two corners of an isosceles triangle are at $(1, 5)$ $(1 ,5 )$ and $(3, 7)$ $(3 ,7 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(1, 5)$ $(1 ,5 )$ and $(3, 7)$ $(3 ,7 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-11-04T07:10:07

The lengths of the sides are: $4 \sqrt{2}$ $4sqrt2$ , $\sqrt{10}$ $sqrt10$ , and $\sqrt{10}$ $sqrt10$ .

Explanation:

Let the given line segment be called $X$ $X$ . After using the distance formula $a^{2} + b^{2} = c^{2}$ $a^2+b^2=c^2$ , we get $X = 4 \sqrt{2}$ $X=4sqrt2$ .

Area of a triangle $= \frac{1}{2} b h$ $=1/2bh$
We are given the area is 4 square units, and the base is side length X.

$4 = \frac{1}{2} (4 \sqrt{2}) (h)$ $4=1/2(4sqrt2)(h)$
$4 = 2 \sqrt{2} h$ $4=2sqrt2h$
$h = \frac{2}{\sqrt{2}}$ $h=2/sqrt2$
Now we have the base and the height and the area. we can divide the isosceles triangle into 2 right triangles to find the remaining side lengths, which are equal to each other.
Let the remaining side length = $L$ $L$ . Using the distance formula:
${(\frac{2}{\sqrt{2}})}^{2} + {(2 \sqrt{2})}^{2} = L^{2}$ $(2/sqrt2)^2+(2sqrt2)^2=L^2$
$L = \sqrt{10}$ $L=sqrt10$

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Two corners of an isosceles triangle are at $(1, 5)$ $(1 ,5 )$ and $(3, 7)$ $(3 ,7 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(1, 5)$ $(1 ,5 )$ and $(3, 7)$ $(3 ,7 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?