Two corners of an isosceles triangle are at $(4, 3)$ $(4 ,3 )$ and $(9, 3)$ $(9 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(4, 3)$ $(4 ,3 )$ and $(9, 3)$ $(9 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-08-04T10:26:32

Length of sides of triangle are $5, 25.72 (2 d p), 25.72 (2 d p)$ $5, 25.72(2dp) ,25.72(2dp)$ unit

Explanation:

The base of the isosceles triangle,

$b = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}} = \sqrt{{(4 - 9)}^{2} + {(3 - 3)}^{2}}$ $b=sqrt( (x_1-x_2)^2+(y_1-y_2)^2) = sqrt( (4-9)^2+(3-3)^2)$

$= \sqrt{25} = 5$ $= sqrt25 = 5$ unit .

The area of the isosceles triangle is $A_{t} = \frac{1}{2} \cdot b \cdot h = \frac{1}{2} \cdot 5 \cdot h$ $A_t = 1/2 * b *h =1/2*5 *h$

$A_t=64 :. h = (2* A_t )/b = (2*64)/5=128/5= 25.6$ unit.

Where $h$ is the altitude of triangle.

The legs of the isosceles triangle are $l_1=l_2= sqrt(h^2+(b/2)^2)=sqrt(25.6^2+(5/2)^2) ~~25.72(2dp)$ unit

Hence the length of three sides of triangle are

$5, 25.72(2dp) ,25.72(2dp)$ unit [Ans]

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Two corners of an isosceles triangle are at $(4, 3)$ $(4 ,3 )$ and $(9, 3)$ $(9 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

Two corners of an isosceles triangle are at (4 ,3 )(4,3)(4 ,3 ) and (9 ,3 )(9,3)(9 ,3 ). If the triangle's area is 64 6464 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(4, 3)$ $(4 ,3 )$ and $(9, 3)$ $(9 ,3 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?