A circle is inscribed in a right triangle. The length

Ben Shaver 2021-12-19 Answered
A circle is inscribed in a right triangle. The length of the radius of the circle is 6 cm, and the length of the hypotenuse is 29 cm. Find the lengths of the two segments of the hypotenuse that are determined by the point of tangency.
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Expert Answer

scoollato7o
Answered 2021-12-20 Author has 158 answers
Step 1
Using Pythagorean theorem,
\(\displaystyle{A}{C}^{{{2}}}={A}{B}^{{{2}}}+{B}{C}^{{{2}}}\)
\(\displaystyle{A}{C}^{{{2}}}={\left({A}{E}+{E}{B}\right)}^{{{2}}}+{\left({B}{F}+{F}{C}\right)}^{{{2}}}\)
\(\displaystyle{29}^{{{2}}}={\left({X}+{6}\right)}^{{{2}}}+{\left({6}+{29}-{X}\right)}^{{{2}}}\)
\(\displaystyle{29}^{{{2}}}={\left({X}+{6}\right)}^{{{2}}}+{\left({35}-{X}\right)}^{{{2}}}\)
\(\displaystyle{841}={2}{X}^{{{2}}}-{58}{X}+{1261}\)
Step 2
On solving further
\(\displaystyle{2}{x}^{{{2}}}-{58}{x}+{420}={0}\)
\(\displaystyle{x}{\left({x}-{14}\right)}-{5}{x}+{210}={0}\)
\(\displaystyle{\left({x}-{15}\right)}{\left({x}-{14}\right)}={0}\)
Here
\(\displaystyle{x}={15}\) or \(\displaystyle{x}={14}\)
Therefore, the lengths of the two segments are as follows, 15 and 14 cm
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kalupunangh
Answered 2021-12-21 Author has 374 answers
Similar to circle sphere is a two dimensional space where the set of points that are at the same distance r from a given point in a three dimensional space. In analytical geometry with a center and radius is the locus of all points is called sphere Given: A circle is inscribed in a right triangle. The length of the radius of the circle is 6 cm, and the length of the hypotenuse is 29 cm. thus the given condition is can be drawn as follows, The objective is to find the lengths of the two segments of the hypotenuse that are determined by the point of the tangency Using Pythagorean theorem,
\(\displaystyle{A}{C}^{{{2}}}={A}{B}^{{{2}}}+{B}{C}^{{{2}}}\)
\(\displaystyle{A}{C}^{{{2}}}={\left({A}{E}+{E}{B}\right)}^{{{2}}}+{\left({B}{F}+{F}{C}\right)}^{{{2}}}\)
\(\displaystyle{29}^{{{2}}}={\left({X}+{6}\right)}^{{{2}}}+{\left({6}+{29}-{X}\right)}^{{{2}}}\)
\(\displaystyle{29}^{{{2}}}={\left({X}+{6}\right)}^{{{2}}}+{\left({35}-{X}\right)}^{{{2}}}\)
\(\displaystyle{841}={2}{X}^{{{2}}}-{58}{X}+{1261}\)
On solving further
\(\displaystyle{2}{x}^{{{2}}}-{58}{x}+{420}={0}\)
\(\displaystyle{x}{\left({x}-{14}\right)}-{5}{x}+{210}={0}\)
\(\displaystyle{\left({x}-{15}\right)}{\left({x}-{14}\right)}={0}\)
Here
\(\displaystyle{x}={15}\) or \(\displaystyle{x}={14}\)
Therefore, the lengths of the two segments are as follows, 15 and 14 cm
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