Question

in in the adjoining figure pqrs is a rectangle of length 36 CM and breadth 30 cm p u s and ptq quarter circles find the area of shaded part ​

Answer 1

$\bold {Total\ Area\ of\ shaded\ region= 627.30\ cm^2}$

Step-by-step explanation:

We have a rectangle PQRS

⇒∠P=∠Q=∠R=∠S=90°

In sector PRT,

PR=QT = 36 cm [Radius of same quadrant]

Now,$\bold {Area\ of\ sector\ PRT=\frac{\theta}{360} \times \pi r^2}$

                                        =$\frac{90}{360}\times \pi(36)^2$

                                        = $\frac{1}{4}\times\frac{22}{7}\times 36\times 36$

                                        = $1018.2 cm^2$

$\bold {Area\ of\ \Delta PTQ = \frac{1}{2}\times base\times height}$

                      =$\frac{1}{2}\times 36\times 36$

                      =$648 cm^2$

Area of shaded region 1 = Area of sector PRT-Area of ΔPTQ

                                        =$1018.2\ cm^2 - 648\ cm^2$

                                        = $370.2\ cm^2$

Now,in quadrant PSU,

Here PS=SU [Radius of same circle]

$\bold {Area\ of\ quadrant\ PSR = \frac{\theta}{360\degree}\pi r^2}$

                                    = $\frac{90}{360}\times \frac{22}{7}\times 30\times 30$

                                    =$\frac{1}{4}\times \frac{22}{7}\times 30\times 30$

                                    =$707.10\ cm^2$

$\bold {Area\ of\ \Delta PSU = \frac{1}{2}\times base\times height}$

                        =$\frac{1}{2}\times 30\times 30$

                       = $450\ cm^2$

Area of shaded region 2= Area of quadrant PSR-Area of ΔPSU

                                       =$707.10\ cm^2 - 450\ cm^2$

                                       =$257.10\ cm^2$

Total Area of shaded region = $370.2\ cm^2 + 257.10\ cm^2$

                                               = $\bold {627.30\ cm^2}$