Question

From a circular piece of cardboard of radius 1.47 m, a sector of angle 60°has been
removed. Find the area of the remaining cardboard.​

Answer 1

Given: A circular piece of cardboard having radius 1.47 m; a sector of 60° has been removed.

To find: The area of the remaining cardboard.

Answer:

Let's first find the area of the cardboard.

Area of a circle = πr²

Area = π*(1.47)²

Area = 2.1609π m²

Now, let's find the area of the sector.

$\bf Area\ of\ a\ sector\ =\ \dfrac{\theta}{{360}^{\circ}}\ \times\ \pi\ \times\ r^2\\\\\\ \sf Area\ =\ \dfrac{{60}^{\circ}}{{360}^{\circ}}\ \times\ \pi\ \times\ (1.47)^2\\\\\\Area\ =\ \dfrac{1}{6}\ \times\ \pi\ \times\ 2.1609\\\\\\Area\ =\ 0.36015\pi\ m^2$

Area of the remaining portion = Area of the circle - Area of the sector.

Area of the remaining portion = 2.1609π - 0.36015π

Area of the remaining portion = 1.80075π

Area of the remaining portion = 5.654355 m²

Answer 2

$\huge{\underline{\underline{\red{Answer}}}}$

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$\mathbb{\boxed{\pink{GIVEN}}}$

• Radius= 1.47m
• Angle of sector =60°

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$\mathbb{\boxed{\pink{NEED\:TO\:FIND}}}$

• Area of remaining cardboard=?

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$\bf{\underline{\underline{\red{SOLUTION}}}}$

Total area of the cardboard

$= \pi \: {r}^{2} \\ = \frac{22}{7} \times 1.47 \times 1.47 {m}^{2} \\ = 6.7914$

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Area of the cardboard removed

$= \pi \: {r}^{2} \frac{ \theta }{360} \\ = \frac{22}{7} \times 1.47 \times 1.47 \times \frac{60}{360} \\ = 1.1319m^{2}$

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Area of remaining part

=Total Area of cardboard- Area of cardboard removed

=6.7914-1.1319 $m^{2}$

=5.6595$m^{2}$

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$\large{\underline{\boxed{\red{Answer=5.6595m^2}}}}$