Question

ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter.
Find the area of the shaded region.

Only Meigenis are allowed to answer.

Thanks @Akshita ,@ Ashu21sufu03 , @shariya112005 , @krisWuYifan , @KrisGalaxy


Love you all.
​

Answer 1

$\bf \fbox \purple { THE ANSWER IS 98 cm² }$

$\bf \fbox \green {Here you go with your answer}$

Step-by-step explanation:

We know that the quadrant is formed by the intersection of 2 diameters at 90⁰

Therefore ,

Angle BAC = 90⁰

IN ∆BAC

APPLYING PYTHAGORAS THEOREM

AB² + AC² = BC²

${(14)}^{2} + {(14)}^{2} = {BC}^{2} \\ \\ 196 + 196 = {BC}^{2} \\ \\ \sqrt{392} = BC \\ \\ 14 \sqrt{2} = BC$

$AREA \: OF \: ∆ABC = \: \frac{1}{2} \times b \: \times h \: \\ \\ = > \frac{1}{2} \times 14 \times 14 \\ \\ = > {98 \: cm \: }^{2}$

$AREA \: OF \: SEGMENT \: BDCE = \: area \: of \: quadrant \: - \: area \: of \: ABC \\ \\ = > \frac{1}{4} \pi {r}^{2} - 98 \\ \\ = > 154 - 98 \\ \\ = > 56 \: {cm \: }^{2}$

AREA OF SEMICIRCLE OF DIAMETER BC( 14√2)

$= > \frac{1}{2} \pi {r}^{2} \\ \\ = > \frac{1}{2} \times \frac{22}{7} \times \frac{14 \sqrt{2} }{2} \times \frac{ 14\sqrt{2} }{2} \\ \\ = > 154 \: {cm}^{2}$

AREA OF THE SHADED REGION

= AREA OF SEMICIRCLE - AREA OF BDCE

= 154 - 56

= 98 CM²

$\bf \fbox \red {HENCE THE ANSWER IS 98 cm² }$