![In the figure, PQR is a right angle triangle with angle Q = 90, QR = 17.5 cm and PQ = 42 cm, with PR as a diameter of a semicircle](https://www.gkseries.com/blog/wp-content/uploads/2024/04/In-the-figure-PQR-is-a-right-angle-triangle-with-angle-Q-90-QR-17.5-cm-and-PQ-42-cm-with-PR-as-a-diameter-of-a-semicircle.jpg)
In the figure, PQR is a right angle triangle with angle Q = 90, QR = 17.5 cm and PQ = 42 cm, with PR as a diameter of a semicircle and with QR as radius, a quarter circle is drawn. Find th e area of the shaded portion correct to two decimal places
![](https://www.gkseries.com/blog/wp-content/uploads/2024/04/image-249-823x1024.png)
(a) 3380.125 cm²
(b) 3540.251
(c) 3850.752 cm²
(d) 3284.25
Sol:
Area of shaded portion = area of semicircle of base PR + area of ∆PQR – area of quarter circle of radius
QR
![](https://www.gkseries.com/blog/wp-content/uploads/2024/04/image-250-1024x442.png)