Question

In the following figure, ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.​

Answer 1

Answer:

The area of the shaded region is 428.75 cm² .

Step-by-step explanation:

Given :  

∆ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm.  

In right ∆ABC , ∠B = 90°, by using Pythagoras theorem

AC² = AB² + BC²

AC² = 28² + 21²

AC²  = 784 + 441

AC² = 1225

AC = √1225

AC = 35 cm

Area of the shaded region ,A  = Area of Semi-circle with Diameter AC + Area of ∆ABC - Area of quadrant with Radius BC  

A = 1/2 x 22/7 x (35/2)² + 1/2 x 21 x 28 - 1/4 x 22/7 x 21²

[Area of Semi-circle = 1/2 πr² , Area of ∆ABC = 1/2  x base  x height , Area of quadrant = 1/4 πr²

A = 481.25 + 294 - 346.5

A = 775.25 - 346.5

A = 428.75 cm²

Area of the shaded region = 428.75 cm²

Hence, the area of the shaded region is 428.75 cm² .

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Answer 2

$\huge\underline\mathfrak \blue {Solution}$

Given that triangle ABC is a right angled at B , with AB = 28cm and BC = 21 cm .

In triangle ABC ....

${ac}^{2} = {ab}^{2} + {bc}^{2}$

$= {28}^{2} + {21}^{2}$

$= 784 + 441$

$= 1225$

$= > ac = 35cm$

From figure ,

Area of shaded region = Area of triangle ABC + area of semi circle - area of sector with radius BC .

$= \frac{1}{2} \times bc \times ab + \frac{1}{2} \pi {r}^{2} - \frac{1}{4}\pi {r}^{2}$

$= \frac{1}{2} \times 21 \times 28 + \frac{1}{2} \times \frac{22}{7} \times ( \frac{35}{2} ) \times ( \frac{35}{2}) - \frac{1}{4} \times \frac{22}{7} \times 21 \times 21$

$= 294 + 481.25 - 346.5$

$= 775.25 - 346.5$

$= 428.75 {cm}^{2}$

Hence , the area of shaded region is 428.75 cm^2 .