Two corners of an isosceles triangle are at $(8, 7)$ $(8 ,7 )$ and $(2, 3)$ $(2 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at $(8, 7)$ $(8 ,7 )$ and $(2, 3)$ $(2 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

1 Answer

Impact of this question

1514 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-05T14:15:04

See a solution process below:

Explanation:

The formula for the area of an isosceles triangle is:

$A = \frac{b h_{b}}{2}$ $A = (bh_b)/2$

First, we must determine the length of the triangles base. We can do this by calculating the distance between the two points given in the problem. The formula for calculating the distance between two points is:

$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ $d = sqrt((color(red)(x_2) - color(blue)(x_1))^2 + (color(red)(y_2) - color(blue)(y_1))^2)$

Substituting the values from the points in the problem gives:

$d = \sqrt{{(2 - 8)}^{2} + {(3 - 7)}^{2}}$ $d = sqrt((color(red)(2) - color(blue)(8))^2 + (color(red)(3) - color(blue)(7))^2)$

$d = \sqrt{{(- 6)}^{2} + {(- 4)}^{2}}$ $d = sqrt((-6)^2 + (-4)^2)$

$d = \sqrt{36 + 16}$ $d = sqrt(36 + 16)$

$d = \sqrt{52}$ $d = sqrt(52)$

$d = \sqrt{4 \times 13}$ $d = sqrt(4 xx 13)$

$d = \sqrt{4} \sqrt{13}$ $d = sqrt(4)sqrt(13)$

$d = 2 \sqrt{13}$ $d = 2sqrt(13)$

The Base of the Triangle is: $2 \sqrt{13}$ $2sqrt(13)$

We are given the area is $64$ $64$ . We can substitute our calculation above for $b$ $b$ and solve for $h_{b}$ $h_b$ :

$64 = \frac{2 \sqrt{13} \times h_{b}}{2}$ $64 = (2sqrt(13) xx h_b)/2$

$64 = \sqrt{13} h_{b}$ $64 = sqrt(13)h_b$

$\frac{64}{\sqrt{13}} = \frac{\sqrt{13} h_{b}}{\sqrt{13}}$ $64/color(red)(sqrt(13)) = (sqrt(13)h_b)/color(red)(sqrt(13))$

$64/sqrt(13) = (color(red)(cancel(color(black)(sqrt(13))))h_b)/cancel(color(red)(sqrt(13)))$

$h_b = 64/sqrt(13)$

The Height of the Triangle is: $64/sqrt(13)$

To find the length of the triangles sides we need to remember the mid-line of an isosceles:
- bisects the base of the triangle into two equal parts
- forms a right angle with the base

enter image source here

Therefore, we can use the Pythagorean Theorem to find the length of the side of the triangle where the side is the hypotenuse and the height and $1/2$ the base are the sides.

$c^2 = a^2 + b^2$ becomes:

$c^2 = (1/2 xx 2sqrt(13))^2 + (64/sqrt(13))^2$

$c^2 = (sqrt(13))^2 + (64/sqrt(13))^2$

$c^2 = 13 + 4096/13$

$c^2 = 169/13 + 4096/13$

$c^2 = 4265/13$

$sqrt(c^2) = sqrt(4265/13)$

$c^2 = (sqrt(25)sqrt(185))/sqrt(13)$

$c^2 = (5sqrt(185))/sqrt(13)$

The Length of the Triangle's Side is: $(5sqrt(185))/sqrt(13)$

Science

Math

Humanities

... and beyond

Two corners of an isosceles triangle are at $(8, 7)$ $(8 ,7 )$ and $(2, 3)$ $(2 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Two corners of an isosceles triangle are at (8 ,7 )(8,7)(8 ,7 ) and (2 ,3 )(2,3)(2 ,3 ). If the triangle's area is 64 6464 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

Related questions

Impact of this question

Two corners of an isosceles triangle are at $(8, 7)$ $(8 ,7 )$ and $(2, 3)$ $(2 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?