A triangle has sides with lengths of 5, 9, and 8. What is the radius of the triangles inscribed circle?

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A triangle has sides with lengths of 5, 9, and 8. What is the radius of the triangles inscribed circle?

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Answer 1 · 2016-01-21T22:42:29

$1.809$ $1.809$

Explanation:

Refer to the figure below

I created this figure using MS Excel

As the sides of the triangle are 5, 8 and 9:
$x + y = 9$ $x+y=9$
$x + z = 8$ $x+z=8$
$y + z = 5$ $y+z=5$ => $z = 5 - y$ $z=5-y$
$\to x + 5 - y = 8$ $-> x+5-y=8$ => $x - y = 3$ $x-y=3$

Adding the first and last equations
$2 x = 12$ $2x=12$ => $x = 6$ $x=6$

Using the Law of Cosines:
$5^{2} = 9^{2} + 8^{2} - 2 \cdot 9 \cdot 8 \cdot cos α$ $5^2=9^2+8^2-2*9*8*cos alpha$

$cos α = \frac{81 + 64 - 25}{144} = \frac{120}{144} = \frac{5}{6}$ $cos alpha=(81+64-25)/144=120/144=5/6$

$α = {33.557}^{\circ}$ $alpha=33.557^@$

In the right triangle with $x$ $x$ as cathetus, we can see that
$tan (\frac{α}{2}) = \frac{r}{x}$ $tan (alpha/2)=r/x$

$r = 6 \cdot tan (\frac{{33.557}^{\circ}}{2})$ $r=6*tan (33.557^@/2)$ => $r = 1.809$ $r=1.809$

Answer 2 · 2016-01-23T13:27:14

Radius of inscribed circle is $= \frac{6}{\sqrt{11}} ≅ 1.81$ $=6/sqrt(11) ~= 1.81$

Explanation:

The radius of a circle inscribed in a triangle is
$XXX r = \frac{{Area}_{△}}{s}$ $color(white)("XXX")r= ("Area"_triangle)/s$ where $s$ $s$ is the semi-perimeter of the triangle.

For a triangle with sides $5, 9, and 8$ $5, 9, and 8$
$XXX s = 11$ $color(white)("XXX")s=11$

Using Heron's formula
$XXX {Area}_{△} = \sqrt{s (s - a) (s - b) (s - c)}$ $color(white)("XXX")"Area"_triangle = sqrt(s(s-a)(s-b)(s-c))$

$XXXXXXX = \sqrt{11 (6) (2) (3)} = 6 \sqrt{11}$ $color(white)("XXXXXXX")=sqrt(11(6)(2)(3)) = 6sqrt(11)$

And the required radius is
$XXX \frac{6 \sqrt{11}}{11} = \frac{6}{\sqrt{11}}$ $color(white)("XXX")(6sqrt(11))/11 = 6/sqrt(11)$

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A triangle has sides with lengths of 5, 9, and 8. What is the radius of the triangles inscribed circle?

2 Answers

Explanation:

Explanation:

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