Question

24. In figure. OA=7 cm, AB = 3.5 cm. Calculate
the perimeter and area of the shaded region.

The answer is 34.5cm , 48.125cm2

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Answer 1

Answer:

area:48.125

perimeter:12.5

Step-by-step explanation:

area of shaded region=area of sector OCB- area of sector ODA

area of shaded region= 90/360*pi*(7+3.5)^2 - 90/360*pi*(7)^2

=90/360*pi*(110.25-49)

=1/4*pi*

=1/4*pi*(61.25)

=48.125

perimeter of shaded region=perimeter of sector OCB- perimeter of sector ODA

perimeter of shaded region=2*10.5+(90/360*2*pi*10.5)- 2*7-(90/360*2*pi*7)

=

21+(90/360*2*pi*10.5)-14-(90/360*2*pi*7)

=21+(16.5)-14-11

=12.5

Answer 2

Given :

• OA = 7cm
• AB = 3.5 cm

To find :

• Perimeter of shaded region
• Area of shaded region

Formula used :

$\sf \bullet \: Area \: of \: quadrant \: = \dfrac{1}{4} \times \pi( {r}^{2} ) \\ \\ \sf \bullet \: Length \: of \: arc \: = \dfrac{ \theta}{360} \times 2\pi \: r$

Let :

• Radius of outer quadrant (OB) = R
• Radius of inner quadrant (OA) = r

Solution :

◈ OB = OA + AB

➠ OB = 7 + 3.5 = 10.5cm

◈ Perimeter of shaded region = = Length of arc BC + CD + length of arcDA + AB

$\sf \implies \: Perimeter \: of \: shaded \: region \: = (\dfrac{ \theta}{360} \times 2\pi \times R\: ) + \:BC + ( \dfrac{ \theta}{360} \times 2\pi \times r) + AB \\ \\ \sf \implies \: Perimeter \: of \: shaded \: region \: = (\dfrac{ 90}{360} \times 2 \times \dfrac{22}{7} \times 10.5\: ) + \:3.5 + ( \dfrac{ 90}{360} \times 2 \times \dfrac{22}{7} \times 7) + 3.5\\ \\ \sf \implies \: Perimeter \: of \: shaded \: region \: = (16.5 ) + (\:3.5 )+ ( 11) + \: (3.5) \: \\ \\ \sf \implies \: Perimeter \: of \: shaded \: region \: = 34.5cm$

◈ Area of shaded region = Area of quadrant (OBCO) - Area of quadrant (OADO)

$\sf \implies \: Area \: of \: shaded \: region \: = \dfrac{1}{4}\pi {R}^{2} - \dfrac{1}{4} \pi {r}^{2} \\ \\ \sf \implies \: Area \: of \: shaded \: region \: = \dfrac{1}{4} \pi( {R}^{2} - {r}^{2} ) \\ \\ \sf \implies \: Area \: of \: shaded \: region \: = \dfrac{1}{4} \pi( {10.5}^{2} - {7}^{2} ) \\ \\ \sf \implies \: Area \: of \: shaded \: region \: = \dfrac{1}{4} \pi(110.25 - 49) \\ \\ \sf \implies \: Area \: of \: shaded \: region \: = \dfrac{1}{4} \times \dfrac{22}{7} (61.25) \\ \\ \sf \implies \: Area \: of \: shaded \: region \: = 48.125 {cm}^{2}$

ANSWER :

• Perimeter = 34.5 cm
• Area = 48.125cm²