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Elementary Geometry For College Students, 7e

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter7: Locus And Concurrence

Section7.2: Concurrence Of Lines

Problem 19E: Draw an obtuse triangle and, by construction, find its orthocentre. HINT: You will have to extend...

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A triangle has side lengths of 5 and 8 and an area of 16. Show that there are actually two possible triangles that satisfy these conditions by finding the third side of each triangle AND the angle between the side with length 5 and the side with length 8. Round your answers to two decimal places. With the aid of a protractor and a ruler, draw a reasonably accurate diagram of each triangle.

Expert Solution

Step 1

Given that

the two sides of a triangle are 5 and 8

and area of that triangle is 16.

Let sides of the triangle are $a = 5, b = 8, c$

and area is $A = 16$ .

If $s = \frac{a + b + c}{2} = \frac{13 + c}{2}$ , then

Area will be

$A = \sqrt{s (s - a) (s - b) (s - c)}$

$\Rightarrow A = \sqrt{(\frac{13 + c}{2}) (\frac{13 + c}{2} - 5) (\frac{13 + c}{2} - 8) (\frac{13 + c}{2} - c)}$

$\Rightarrow 16 = \sqrt{(\frac{13 + c}{2}) (\frac{3 + c}{2}) (\frac{c - 3}{2}) (\frac{13 - c}{2})}$

$\Rightarrow (\frac{13^{2} - c^{2}}{4}) (\frac{c^{2} - 3^{2}}{4}) = 16^{2}$

$\Rightarrow (13^{2} - c^{2}) (c^{2} - 3^{2}) = 256 \times 16$

$\Rightarrow - c^{4} + (13^{2} + 3^{2}) c^{2} - 169 \times 9 = 4096$

$\Rightarrow c^{4} - 178 c^{2} + 5617 = 0$

$\Rightarrow c^{2} = \frac{178 \pm \sqrt{178^{2} - 4 (5617)}}{2} = \frac{178 \pm \sqrt{9216}}{2} = \frac{178 \pm 96}{2}$

Taking positive sign, we get

$c^{2} = \frac{178 + 96}{2} = \frac{274}{2} = 137$

$\Rightarrow c = \sqrt{137} = 11.70$

and taking negative sign, we get

$c^{2} = \frac{178 - 96}{2} = \frac{82}{2} = 41$

$\Rightarrow c = \sqrt{41} = 6.40$

So two possible triangles are $(5, 8, 11.70)$ and $(5, 8, 6.40)$ which has same area as 16.

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