PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR

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Q) PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semi-circles are drawn such that PQ = QR = RS. Semicircles are drawn on PQ and QS as diameters as shown in figure. Find the perimeter and area of the shaded region.

Ans:

Given that the radius = 6 cm

∴ Diameter PQRS = 2 x 6 = 12 cm

Since PQRS = PQ + QR + RS and its given that PQ = QR = RS

∴ 12 = 3 PQ

∴ PQ = $\frac{12}{3}$ = 4 cm

∴ PQ = QR = RS = 4 cm

Let’s start with calculating Perimeter of Shaded region:

∵ We know that the perimeter of a circle is given by = 2 π r or π d

∴ the perimeter of a half circle is given by: π r or π $\frac{d}{2}$

Now the Perimeter of shaded region = Perimeter of Semi-circle of dia. PS + Perimeter of Semi-circle of dia. PQ + Perimeter of Semi-circle of dia. QS

= π $\frac{PS}{2}$ + π $\frac{PQ}{2}$ + π $\frac{QS}{2}$

Here PS = 12 cm, PQ = 4 cm, QS = QR + RS = 4 + 4 = 8 cm

∴ Perimeter of shaded region = π $\frac{12}{2}$ + π $\frac{4}{2}$ + π $\frac{8}{2}$

= 6 π + 2 π + 4 π = 12 π

= 12 x $\frac{22}{7}$ = $\frac{264}{7}$ = 37.714

∴ Perimeter of shaded region = 37.714 cm

Next, we will calculate Area of the Shaded region:

∵ We know that the area of a circle is given by = π r² or π $\frac{d^2}{4}$

∴ the area of a half circle is given by: π $\frac{r^2}{2}$ or π $\frac{d^2}{8}$

Area of shaded region = Area of Semi-circle of dia. PS + Area of Semi-circle of dia. PQ – Area of Semi-circle of dia. QS

∴ Area of shaded region = π $\frac{PS^2}{8}$ + π $\frac{PQ^2}{8}$ – π $\frac{QS^2}{8}$

Here PS = 12 cm, PQ = 4 cm, QS = 8 cm

∴ Area of shaded region = π $\frac{12^2}{8}$ + $\frac{4^2}{8}$ – π $\frac{8^2}{8}$

= π $\frac{(12^2 + 4^2 - 8^2)}{8}$

= π $\frac{(144 + 16 - 64)}{8}$

= 12 π = 12 x $\frac{22}{7}$ = $\frac{264}{7}$ = 37.714

∴ Area of shaded region = 37.714 cm²

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