Question

In the figure below, the triangle is $a3-4-5$ sides triangle. Two equal circle are placed as in the figure. The radius of each of circle is

• A. $\frac{4}{5}$
• B. $\frac{5}{8}$
• C. $\frac{6}{11}$
• D. $\frac{5}{7}$

Answer 1

The correct option is D $\frac{5}{7}$

figure shows right angled triangle $△ABC$.

Let $AB=4$, $BC=3$ and $AC=5$

Points P and Q are centers of identical circles as shown in figure. Join them and construct right angled triangle PQR as shown in figure.

Let r= radius of each identical circle.

As circles are touching each other, we can say from the figure that,

$PR=r+r$

$\therefore PR=2r$

As $△ABC$ and $△PQR$ both are right angled triangles, both are similar triangles.

Thus, corresponding angles are congruent.

Now, Let $\angle ACB=\theta$

$sin\theta =\frac{AB}{AC}$

$\therefore sin\theta =\frac{4}{5}$

By property of similar triangles, $\angle PRQ=\theta$

$\therefore sin\theta =\frac{PQ}{PR}$

$\therefore \frac{4}{5}=\frac{PQ}{2r}$

$\therefore PQ=\frac{4}{5}\times 2r$

$\therefore PQ=\frac{8r}{5}$ Equation (1)

Now, In $△ABC$, $cos\theta =\frac{BC}{AC}$

$\therefore cos\theta =\frac{3}{5}$

By property of similar triangles, $cos\theta =\frac{QR}{PR}$

$\therefore \frac{3}{5}=\frac{QR}{2r}$

$\therefore QR=\frac{3}{5}\times 2r$

$\therefore QR=\frac{6r}{5}$ Equation (2)

Now, As shown in figure,

$R\acute{T}=r$

$\therefore \stackrel{`}{S}B=PQ+R\acute{T}$

$\therefore \stackrel{`}{S}B=\frac{8r}{5}+r$

$\therefore \stackrel{`}{S}B=\frac{13r}{5}$

Now, $A\stackrel{`}{S}=AB-\stackrel{`}{S}B$

$\therefore A\stackrel{`}{S}=4-\frac{13r}{5}$

$\therefore AS=A\stackrel{`}{S}=4-\frac{13r}{5}$ (Tangents from same point to same circle) Equation (3)

Similarly, $B\stackrel{`}{T}=\stackrel{`}{S}P+QR$

$\therefore B\stackrel{`}{T}=r+\frac{6r}{5}$

$\therefore B\stackrel{`}{T}=\frac{11r}{5}$

Now, $\therefore C\stackrel{`}{T}=BC-B\stackrel{`}{T}$

$\therefore C\stackrel{`}{T}=3-\frac{11r}{5}$

$\therefore CT=C\stackrel{`}{T}=3-\frac{11r}{5}$ (Tangents from same point to same circle) Equation (4)

Similarly, $ST=PR=2r$

Now, we can write from the figure that,

$AC=AS+ST+CT$

From equations (3) and (4), we get,

$AC=4-\frac{13r}{5}+2r+3-\frac{11r}{5}$

$\therefore 5=4-\frac{13r}{5}+2r+3-\frac{11r}{5}$

$\therefore -2=-\frac{13r}{5}+2r-\frac{11r}{5}$

$\therefore -2=\frac{-13r-11r+10r}{5}$

$\therefore -2=\frac{-14r}{5}$

$\therefore \frac{14r}{5}=2$

$\therefore r=\frac{10}{14}$

$\therefore r=\frac{5}{7}$

Thus, answer is option (D)