Question

In fig,ABC is a quadrant of r=14cm
and a semicircle is drawn with BC as diameter.
Find the area of YELLOW region​

Answer 1

Ar. Of yellow region=(π(✓2r)^2)/2-(πr^2/2-r^2/2)

=πr^2-πr^2/2+ r^2/2

=(πr^2-r^2)/2

=(22*14*14/7-14*14)/2

=420/2

=210cm^2

Answer 2

In ∆BAC, AB _|_ AC

AB = AC = 14 cm (radii of same circle)

tanC =AB/AC = 14/14 = 1

tanC = 1 = tan45°

angle ACB = 45°

sinC = sin45° = AB/BC

$\frac{1}{ \sqrt{2} } = \frac{14}{ac }$

AC = 14√2 cm

for smaller circle;

r = 14 cm

$ar(quadrant) \\ = \frac{1}{4} \pi \: {r}^{2} \\ = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 \\ = \frac{1}{2} \times 22 \times 2 \times 7 = 1 \times 22 \times 7 \\ = 154 \: {cm}^{2}$

for bigger circle;

r = 14√3 cm

$ar(semicircle) \\ = \frac{1}{2} \times \pi \: {r}^{2} \\ = \frac{1}{2} \times \frac{22}{7} \times 14 \sqrt{3} \times 14 \sqrt{3} \\ = 1 \times 11 \times 2 \sqrt{3} \times 14 \sqrt{3} \\ = 11 \times 28 \times 3 \\ = 11 \times 84 \\ = 924 \: {cm}^{2}$

for ∆BAC;

ar(∆)

= 1/2 x AB x AC

= 1/2 x 14 x 14

= 98 cm²

ar(shaded portion)

= ar(triangle) + ar(semicircle) - ar(quadrant)

= 98 + 924 - 154

= 868 cm²

Hence, area of shaded portion 868 cm²