Two corners of an isosceles triangle are at $(8, 2)$ $(8 ,2 )$ and $(4, 3)$ $(4 ,3 )$ . If the triangle's area is $9$ $9$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(8, 2)$ $(8 ,2 )$ and $(4, 3)$ $(4 ,3 )$ . If the triangle's area is $9$ $9$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-06-10T09:11:00

$Isosceles triangle's sides are 4.12, 4.83, 4.83$ $color(indigo)("Isosceles triangle's sides are " 4.12, 4.83, 4.83$

Explanation:

![https://www.algebra.com/algebra/homework/Geometry-proofs/Geometry_proofshttp://.faq.question.209023.html](https://useruploads.socratic.org/AM7WAi4Tzyrvy0qBVAcz_isosceles%20triangle.jpg)

$A (8, 2), B (4, 3), A_{t} = 9$ $A(8,2), B(4,3), A_t = 9$

$c = {\sqrt{8 - 4}}^{2} + {(3 - 2)}^{2}) = 4.12$ $c = sqrt(8-4)^2 + (3-2)^2) = 4.12$

$h = \frac{2 \cdot A_{t}}{c} = \frac{2 \cdot 9}{4.12} = 4.37$ $h = (2 * A_t) / c = (2 * 9) / 4.12 = 4.37$

$a = b = \sqrt{{(\frac{4.12}{2})}^{2} + {4.37}^{2}} = 4.83$ $a = b = sqrt((4.12/2)^2 + 4.37^2) = 4.83$

Answer 2 · 2018-06-10T11:42:45

Base $\sqrt{17}$ $\sqrt{17}$ and common side $\sqrt{\frac{1585}{68}} .$ $sqrt{1585/68}.$

Explanation:

They're vertices, not corners. Why do we have the same bad wording of the question from all around the world?

Archimedes' Theorem says if $A, B and C$ $A,B and C$ are the squared sides of a triangle of area $S$ $S$ , then

$16 S^{2} = 4 A B - {(C - A - B)}^{2}$ $16S^2 = 4AB-(C-A-B)^2$

For an isosceles triangle, $A = B .$ $A=B.$

$16 S^{2} = 4 A^{2} - {(C - 2 A)}^{2} = 4 A C - C^{2}$ $16S^2 = 4A^2-(C-2A)^2 = 4AC-C^2$

We're not sure if the given side is $A$ $A$ (the duplicated side) or $C$ $C$ (the base). Let's work it out both ways.

$C = {(8 - 4)}^{2} + {(2 - 3)}^{2} = 17$ $C = (8-4)^2 + (2-3)^2 = 17$

$16 {(9)}^{2} = 4 A (17) - 17^{2}$ $16(9)^2 = 4A(17) - 17^2$

$A = \frac{1585}{68}$ $A = 1585/68$

If we started with $A = 17$ $A=17$ then

$16 {(9)}^{2} = 4 (17) C - C^{2}$ $16(9)^2 = 4(17)C - C^2$

$C^{2} - 68 C + 1296 = 0$ $C^2 - 68 C + 1296 = 0$

No real solutions for that one.

We conclude we have base $\sqrt{17}$ $\sqrt{17}$ and common side $\sqrt{\frac{1585}{68}} .$ $sqrt{1585/68}.$

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Two corners of an isosceles triangle are at $(8, 2)$ $(8 ,2 )$ and $(4, 3)$ $(4 ,3 )$ . If the triangle's area is $9$ $9$ , what are the lengths of the triangle's sides?

2 Answers

Explanation:

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Two corners of an isosceles triangle are at (8 ,2 )(8,2)(8 ,2 ) and (4 ,3 )(4,3)(4 ,3 ). If the triangle's area is 9 99 , what are the lengths of the triangle's sides?

2 Answers

Explanation:

Explanation:

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Two corners of an isosceles triangle are at $(8, 2)$ $(8 ,2 )$ and $(4, 3)$ $(4 ,3 )$ . If the triangle's area is $9$ $9$ , what are the lengths of the triangle's sides?