Question

35. In the given figure, ABC is a right triangle, right-angled at A. Find the area of the shaded regions if AB = 16 cm,
BC= 20 cm and O is the centre of the incircle of AABC
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Answer 1

Given :-

• A right triangle ABC right angled at A, such that AB = 16 cm and BC = 20 cm.

• A circle is inscribed in it touches the sides AC, AB and BC at R, P and Q respectively.

To Find :-

• Area of shaded region

Solution :-

In right-angled triangle ABC,

Using Pythagoras Theorem,

$\rm :\longmapsto\: {BC}^{2} = {AB}^{2} + {AC}^{2}$

$\rm :\longmapsto\: {20}^{2} = {16}^{2} + {AC}^{2}$

$\rm :\longmapsto\:400 = 256 + {AC}^{2}$

$\rm :\longmapsto\:400 - 256 = {AC}^{2}$

$\rm :\longmapsto\:144 = {AC}^{2}$

$\rm :\longmapsto\: {12}^{2} = {AC}^{2}$

$\bf\implies \:AC = 12 \: cm$

Now,

We know that Radius of inscribed circle is given by

$\rm :\longmapsto\:r = \dfrac{Area_{( \triangle \: ABC)}}{semi \: perimeter \: of \: \triangle \: ABC}$

$\rm :\longmapsto\:r = \dfrac{\cancel{\dfrac{1}{2}}\times AB \times AC }{ \: \: \: \cancel{\dfrac{1}{2}}(AB + BC + AC) \: \: \: }$

$\rm :\longmapsto\:r = \dfrac{16 \times 12}{16 + 12 + 20}$

$\rm :\longmapsto\:r = \dfrac{16 \times 12}{48}$

$\bf\implies \:r = 4 \: cm$

Now,

$\rm \: \:Area_{( \triangle \: ABC)} = \dfrac{1}{2} \times AB \times AC = \dfrac{1}{2} \times 16 \times 12 = 96 \: {cm}^{2}$

and

$\rm \: Area_{(circle)} = \pi {(r)}^{2} = \dfrac{22}{7} \times {4}^{2} = \dfrac{352}{7} = 50.28 {cm}^{2}$

Hence,

$\bf\:Area_{(shaded \: region)}= Area_{(\triangle\:ABC)}-Area_{(circle)}$

$\rm \: Area_{(shaded \: region)} =96 - 50.28$

$\bf\implies \:Area_{(shaded \: region)} =45.72 \: {cm}^{2}$

Additional Information :-

Let us consider a right-angle triangle ABC right-angled at A, such that AB = c, BC = a and AC = b and let a circle of radius 'r' is inscribed in triangle ABC, then

$\boxed{ \bf \: 2r = b + c - a}$