Two corners of an isosceles triangle are at $(4, 8)$ $(4 ,8 )$ and $(5, 7)$ $(5 ,7 )$ . If the triangle's area is $5$ $5$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(4, 8)$ $(4 ,8 )$ and $(5, 7)$ $(5 ,7 )$ . If the triangle's area is $5$ $5$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-09-16T11:47:01

The length of three sides of triangle are $1.41, 7.11, 7.11$ $1.41 ,7.11 , 7.11$ unit.

Explanation:

Base length of the triangle is $B = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}})$ $B= sqrt((x_2-x_1)^2+(y_2-y_1)^2))$

$= \sqrt{{(5 - 4)}^{2} + {(7 - 8)}^{2}}) = \sqrt{1 + 1} = \sqrt{2} \approx 1.41 (2 d p)$ $= sqrt((5-4)^2+(7-8)^2)) =sqrt(1+1)=sqrt2 ~~ 1.41(2dp)$ unit

We know area of triangle is $A_{t} = \frac{1}{2} \cdot B \cdot H$ $A_t =1/2*B*H$ Where $H$ $H$ is altitude.

$:. 5=1/2*1.41*H or H= 10/1.41 ~~7.07(2dp)$ unit

Legs are $L = sqrt(H^2+(B/2)^2)$

$= sqrt( 7.07^2+(1.41/2)^2) ~~7.11 (2dp)$ unit

The length of three sides of triangle are $1.41 ,7.11 , 7.11$ unit [Ans]

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Two corners of an isosceles triangle are at $(4, 8)$ $(4 ,8 )$ and $(5, 7)$ $(5 ,7 )$ . If the triangle's area is $5$ $5$ , 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 ,8 )(4,8)(4 ,8 ) and (5 ,7 )(5,7)(5 ,7 ). If the triangle's area is 5 55 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(4, 8)$ $(4 ,8 )$ and $(5, 7)$ $(5 ,7 )$ . If the triangle's area is $5$ $5$ , what are the lengths of the triangle's sides?