Question

✦✧✧ Question ✧✧✦ ➠ ᴅᴏɴᴛ sᴘᴀᴍ⚠️vertices abc of triangle ABC and centres are drawn with radius 5 cm if ab is equal to 14 cm BC is equal to 48 cm CA is equal to 58 cm find the area of shaded region​ Hint - use Angle a + angle b + angle c =180 , pi r square upon 360°​

Answer 1

Step-by-step explanation:

Answer :-

For area of remaining cardboard,

Area of remaining cardboard = Area of 2 circles + Area of 1 semicircle

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Area of Circle = πr²

For Area of 2 Circle = 2πr²

Area of 1 semicircle = \sf \frac{1}{2} πr²21πr²

Area of 1 semicircle = \sf \frac{πr²}{2}2πr²

___________

Now,

Area of rectangular cupboard = L × b

L = r + 2r + 2r = 5r

b = 2r

Area of rectangular cupboard = L × b

Area of rectangular cupboard = ( 5r × 2r )

Area of rectangular cupboard = 10 r²

___________

Now,

Area of remaining cardboard = Area of rectangular cupboard - ( Area of 2 circles + Area of 1 semicircle )

Area of remaining cardboard = \begin{gathered} \sf 10r² - \bigg( 2πr² + \frac{πr²}{2} \bigg) \\ \end{gathered}10r²−(2πr²+2πr²)

Area of remaining cardboard = \begin{gathered} \sf 10r² - \bigg(\frac{4πr² + πr²}{2} \bigg) \\ \end{gathered}10r²−(24πr²+πr²)

Area of remaining cardboard = \begin{gathered} \sf 10r² - \frac{5πr²}{2} \\ \end{gathered}10r²−25πr²

Area of remaining cardboard = \begin{gathered} \sf \bigg( \frac{20 {r}^{2} - 5 \frac{22}{7} {r}^{2} }{2} \bigg) \\ \end{gathered}(220r2−5722r2)

Area of remaining cardboard = \begin{gathered} \sf \cancel2 \bigg( \frac{10 {r}^{2} - 5 \times \frac{11}{7} {r}^{2}}{ \cancel2} \bigg) \\ \end{gathered}2(210r2−5×711r2)

Area of remaining cardboard = \begin{gathered} \sf \bigg(10 {r}^{2} - 5 \times \frac{11}{7} {r}^{2} \bigg) \\ \end{gathered}(10r2−5×711r2)

Area of remaining cardboard = \begin{gathered} \sf \bigg( \frac{70 {r}^{2} - 55 {r}^{2}}{7} \bigg) \\ \end{gathered}(770r2−55r2)

Area of remaining cardboard = \sf \bigg( \frac{15 {r}^{2} }{7} \bigg)(715r2)

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Now, finally Ratio between the area of the remaining cardboard and area of cardboard.

Ratio = \begin{gathered} \sf \frac{the \: \: area \: \: of \: \: the \: \: remaining \: \: cardboard}{area \: \: of \: \: cardboard} \\ \end{gathered}areaofcardboardtheareaoftheremainingcardboard

Ratio = \begin{gathered} \sf \frac{\bigg( \frac{15 {r}^{2} }{7} \bigg)}{10 \: \: r²}\\ \end{gathered}10r²(715r2)

Ratio = \begin{gathered} \sf \frac{15 \cancel{ {r}^{2}} }{7 \times 10 \cancel{ {r}^{2} }} \\ \end{gathered}7×10r215r2

Ratio = \begin{gathered} \sf \frac{ \cancel5 \times 3}{7 \times \cancel5 \times 2} \\ \end{gathered}7×5×25×3

Ratio = \begin{gathered} \sf \frac{3}{14} \\ \end{gathered}143

Ratio = \begin{gathered} \large \sf {\underline{ 3 : 14 }} \\ \end{gathered}3:14

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I hope it helps you ❤️✔️

I'm sorry I took too long time to calculate.

Answer 2

Answer:

The area of shaded region is 281.1 cm².

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

In figure, in △ABC,

AB = 14 cm

BC = 48 cm

AC = 58 cm

Three sectors are drawn with radius each 5 cm to the vertices of the triangle.

We have to find the area of the shaded region.

Now,

In △ABC,

• AB ( s₁ ) = 14 cm
• BC ( s₂ ) = 48 cm
• AC ( s₃ ) = 58 cm

Now,

$\displaystyle{\sf\:Semi\:perimeter\:of\:\triangle\:ABC\:=\:\dfrac{AB\:+\:BC\:+\:AC}{2}}$

$\displaystyle{\implies\sf\:s\:=\:\dfrac{s_1\:+\:s_2\:+\:s_3}{2}}$

$\displaystyle{\implies\sf\:s\:=\:\dfrac{14\:+\:48\:+\:58}{2}}$

$\displaystyle{\implies\sf\:s\:=\:\dfrac{62\:+\:58}{2}}$

$\displaystyle{\implies\sf\:s\:=\:\cancel{\dfrac{120}{2}}}$

$\displaystyle{\implies\boxed{\sf\:s\:=\:60\:cm\:}}$

Now, by Heron's formula,

$\displaystyle{\pink{\sf\:Area\:of\:\triangle\:ABC\:=\:\sqrt{s\:(\:s\:-\:s_1\:)\:(\:s\:-\:s_2\:)\:(\:s\:-\:s_3\:)}}}$

$\displaystyle{\implies\sf\:Area\:of\:\triangle\:ABC\:=\:\sqrt{60\:(\:60\:-\:14\:)\:(\:60\:-\:48\:)\:(\:60\:-\:58\:)}}$

$\displaystyle{\implies\sf\:Area\:of\:\triangle\:ABC\:=\:\sqrt{60\:\times\:46\:\times\:12\:\times\:2}}$

$\displaystyle{\implies\sf\:Area\:of\:\triangle\:ABC\:=\:\sqrt{4\:\times\:15\:\times\:23\:\times\:2\:\times\:4\:\times\:3\:\times\:2}}$

$\displaystyle{\implies\sf\:Area\:of\:\triangle\:ABC\:=\:\sqrt{4\:\times\:4\:\times\:4\:\times\:3\:\times\:3\:\times\:5\:\times\:23}}$

$\displaystyle{\implies\sf\:Area\:of\:\triangle\:ABC\:=\:2\:\times\:2\:\times\:2\:\times\:3\:\sqrt{5\:\times\:23}}$

$\displaystyle{\implies\sf\:Area\:of\:\triangle\:ABC\:=\:24\:\sqrt{115}}$

$\displaystyle{\implies\sf\:Area\:of\:\triangle\:ABC\:=\:24\:\times\:10.723}$

$\displaystyle{\implies\boxed{\blue{\sf\:Area\:of\:\triangle\:ABC\:=\:257.352\:cm^2\:}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:Area\:of\:sector\:=\:\dfrac{\theta}{360}\:\times\:\pi\:r^2}}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:A_{sector\:1}\:+\:A_{sector\:2}\:+\:A_{sector\:3}}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:\left(\:\dfrac{\theta_{1}}{360}\:\times\:\pi\:r^2\:\right)\:+\:\left(\:\dfrac{\theta_{2}}{360}\:\times\:\pi\:r^2\:\right)\:+\:\left(\:\dfrac{\theta_{3}}{360}\:\times\:\pi\:r^2\:\right)}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:\pi\:r^2\:\left(\:\dfrac{\angle\:a}{360}\:+\:\dfrac{\angle\:b}{360}\:+\:\dfrac{\angle\:c}{360}\:\right)}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:\pi\:r^2\:\left(\:\dfrac{\angle\:a\:+\:\angle\:b\:+\:\angle\:c}{360}\:\right)}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:\pi\:r^2\:\times\:\dfrac{\cancel{180}}{\cancel{360}}}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:\pi\:5^2\:\times\:\dfrac{1}{2}}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:\dfrac{25\:\times\:\cancel{3.14}}{\cancel{2}}}$

$\displaystyle{\implies\sf\:Area\:of\:3\:sectors\:=\:25\:\times\:1.57}$

$\displaystyle{\implies\boxed{\green{\sf\:Area\:of\:3\:sectors\:=\:39.25\:cm^2\:}}}$

Now,

$\displaystyle{\sf\:Area\:of\:shaded\:region\:=\:Area\:of\:\triangle\:ABC\:-\:Area\:of\:3\:sectors}$

$\displaystyle{\implies\sf\:Area\:of\:shaded\:region\:=\:257.352\:-\:39.25}$

$\displaystyle{\implies\sf\:Area\:of\:shaded\:region\:=\:218.102}$

$\displaystyle{\implies\:\underline{\boxed{\red{\sf\:Area\:of\:shaded\:region\:\approx\:218.1\:cm^2\:}}}}$

∴ The area of shaded region is 281.1 cm².