Question

$\huge{\mathtt{\underline{\underline{\color{lavender}{Question}}}}}$
Find the area of shaded region of the figure given above.

with explanation

please don't delete my question..​

Answer 1

Answer:

Hey Blinky !!

Plz refer the pic .

Hope help ..

Thnx!!

Will you be my friend ??

Answer 2

Answer:

$\pink{answer★}$

=Ar(ABCDEA)+ar(FGHF)+ar(HIDCJFH)

=Ar(semicircle ABC) - ar(semicircle AED)+ar(semicircle FGH)- ar (semicircle HID) - ar (semicircle FJC)

$( \frac{1}{2}\pi \times \frac{21}{4} \times \frac{21}{4} ) - \: ( \frac{1}{2}\pi \times \frac{7}{2} \times \frac{ 7}{2} ) \\ \\ + ( \frac{1}{2}\pi \times \frac{7}{2} \times \frac{7}{2}) + ( \frac{1}{2}\pi \times \frac{35}{4} \times \frac{35}{4}) - \\ \\ ( \frac{1}{2}\pi \times \frac{7}{2} \times \frac{7}{2} )m {}^{2}$

$( \frac{22}{7} \times \frac{441}{32}) + ( \frac{1}{2} \times \frac{22}{7} \times \frac{35}{4} \times \frac{35}{7} ) - \\ \\ ( \frac{22}{7} \times \frac{49}{8}) m {}^{2}$

$( \frac{11 \times 63}{16}) + ( \frac{55 \times 35}{16} ) - \frac{77}{4} ) {m}^{2} = \\ ( \frac{693}{16} + \frac{1925}{16} - \frac{77}{4}) {m}^{2}$

$( \frac{693 + 1925 - 308}{16} ) {m}^{2}$

$= \frac{1115}{8} {m}^{2} = 144.38 {m}^{2}$

Answer of required Area=144.38m²