Question

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

Answer 1

We know, AC = r

In Δ ACB,

BC^2 = AC^2 + AB^2

⇒ BC = AC^2   (∵ AB = AC)

⇒ BC = r$\bf\huge\sqrt{2}$

Required area = ar (ΔACB) + ar (semicircle on  BC as diameter) – ar (quadrant ABPC)

= $\bf\huge\frac{1}{2}\times r\times r + \frac{1}{2}\times\frac{22}{7}\times\frac{r\sqrt{2} }{2})^2 - \frac{1}{4} \times\frac{22}{7}\times r^{2}$

= $\bf\huge\frac{r^{2} }{2} + \frac{pi r^2}{4} - \frac{pi r^2}{4}$

= $\bf\huge\frac{r^2}{2}$

= $\frac{(14)^{2} }{2}$

= $\bf\huge\frac{196}{2}$

= 98 cm^2

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Since, ABPC is quadrant.

ΔABC is right Δ at A n which AB = AC = 14 cm

In ΔABC

⇒ BC² = AB² + AC²

⇒ BC² = 14² + 14²

⇒ BC = 14√2 cm

Required Area = ar(ΔACB) + ar(semicircle on BC as diameter) - ar(quadrant ABPC)

Required Area = 1/2 × 14 × 14 + π/2× (14√2/2)² - π × 14² × 90/360

Required Area = 98 + 196π/4 - 196/4

Required Area = 98 m²

Hence, the area of the shaded region is 98 m².