Question

A metal wire is used to form an equilateral triangle having the largest area that is 576√3cm².
If the same wire is used to form a semicircle of largest area, find the area of the circle.

Answer 1

Area of equilateral triangle = √3/4 (side)²

A wire is used to form an equilateral triangle having the largest area that is 576√3 cm².

→ 576√3 = √3/4 (side)²

→ (576√3 × 4)/√3 = (side)²

→ 2304 = (side)²

→ side = √2304

→ side = 48 cm

Perimeter of triangle = 3 × side

→ 3(48)

→ 144 cm

Perimeter of circle = 2πr

→ 2 × π × r = 144

→ r = 72/π

Area of circle = πr²

→ π × (72/π)²

→ 5184/π

→ (5184 × 7)/22

→ 36288/22

→ 1649.45 cm²

Answer 2

Answer:

${\bold{\therefore Area\:of\:circle=1649.45\:cm^{2}}}$

Step-by-step explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a wire which is bent to form an equilateral triangle whose area is given.

• We have to find the area of circle.

$\underline \bold{Given : } \\ \implies Area \: of \: equilateral \: triangle = 576 \sqrt{3} {cm}^{2} \\ \\ \underline\bold{To \: Find :} \\ \implies Area \: of \: circle = ?$

• According to given question :

$\implies Area \: of \: equilateral \: triangle = \frac{ \sqrt{3} {side}^{2} }{4} \\ \implies 576 \sqrt{3} = \frac{ \sqrt{3} \times {side}^{2} }{4} \\ \implies {side}^{2} = \frac{576 \sqrt{3} \times 4 }{ \sqrt{3} } \\ \implies {side}^{2} = 2304 \\ \implies side = \sqrt{2304} \\ \bold{\implies side = 48 \: cm }\\ \\ \bold{Circumference\: of \: circle = Perimeter \: of \: triangle} \\ \implies 2\pi r = 3 \times 48 \\ \implies r = \frac{144}{2\pi} \\ \bold{ \implies r = \frac{72}{\pi} } \\ \\ \implies \bold {Area \: of \: circle = \pi {r}^{2} } \\ \implies Area = \pi \times ({ \frac{72}{\pi} })^{2} \\ \implies Area = \frac{72 \times 72 \times 7}{22} \\ \bold{\implies Area = 1649.45 \: {cm}^{2}} \\ \\ \bold{\therefore Area \: of \: circle = 1649.45 \: {cm}^{2} }$