[Math] Radius of a circle inscribed in a right triangle is given, the hypotenuse is given, find the legs

circlesgeometrytrianglestrigonometry

The radius of the circle inscribed in the right triangle is 4, the hypotenuse is 20cm. How do I find the legs? Any help would be appreciated!

Best Answer

HINT:$$r = \cfrac{b+p - h}2$$,

$$b^2 + p^2 = h^2$$
Now you can solve to get answer

Related Solutions

[Math] Find legs of a right triangle if radius of circumscribed circle is 15 and radius of inscribed circle is 6

Try labelling a right angled triangle A, B, C, with A as the right angle. BC is 30 because of the well known circle theorem which says the angle in a semicircle is a right angle. Now draw the incircle and label the tangent points as P on AB, Q on BC and R on CA. Observe that AP and RA are both equal to the radius of the incircle. Let AB be x, and now write down all the lengths of the 4 remaining line segments in terms of x. Now use Pythagoras to form a quadratic in x.

[Math] Find length of triangle as well as radius of inscribed circle

From the Cosine Law,

$$AB^2+AC^2-BC^2=2\cdot AB\cdot AC\cdot \cos A \implies AB = 5 \text{ or } 16$$ The area of $\triangle ABC$ can be written as the following two ways: $$S_{ABC}=\frac{1}{2}\cdot AB\cdot AC\cdot \sin A = \frac{1}{2}\cdot(AB+AC+BC)\cdot r$$ where $r$ is the radius of the incircle of $\triangle ABC$. This will give us $$r=\frac{AB\cdot AC\cdot \sin60^\circ}{AB+AC+BC}$$ By plugging in the values of $AB, AC, BC$ and $\sin A$, we can get $r$. Further, because $BP=BQ,\ CQ=CR,\ AR=AP$, and $$BP+AP=AB\\BQ+CQ=BC\\AR+RC=AC$$ we know that $$BQ = \frac{BA+BC-AC}{2}$$Therefore, the following two cases are:

Case 1: $AB=5$, then $r=\frac{7\sqrt{3}}{6}$ and $BQ=\frac{3}{2}$.

Case 2: $AB = 16$, then $r = 3\sqrt{3}$ and $BQ = 7$.

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Related Solutions

[Math] Find legs of a right triangle if radius of circumscribed circle is 15 and radius of inscribed circle is 6

[Math] Find length of triangle as well as radius of inscribed circle

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