Question

$\sf\small\orange{Given:-}$

✈︎In the figure, AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region. [Use π = 3.14].

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Answer 1

Answer:

Given:

• AB is the diameter =d of a circle.

ΔABC has the diameter AB as base & the point C is on the circumference.

• AC=6 cm and BC=8 cm.

$\red{ \tt{To find out:</p><p></p><p>Area of shaded portion in the given circle.}}</p><p></p><p>$

$\huge {\blue{\bigstar \: Solution:}}$

∠ACB=90* (* refers to Degree ex: x^0 degrees)

since ΔABC has been inscribed in a semicircle.

∴ΔABC is a right one with AB as hypotenuse . . . . .(i)

applying Pythagoras theorem, we have

$\pink{ \huge{AB= \sqrt{{(AC) }^{2} +{(BC)}^{2} }= \sqrt{{(6) }^{2} +{(8)}^{2}  cm}=10 cm=d. }}</u></em></strong></p><p></p><p></p><p><strong><em><u>[tex] \pink{ \huge{AB= \sqrt{{(AC) }^{2} +{(BC)}^{2} }= \sqrt{{(6) }^{2} +{(8)}^{2}  cm}=10 cm=d. }}$

∴ The radius of the given circle

$\huge \bf \green{= \frac d 2= \dfrac{10}{2} \: cm=5 cm. }$

$\tiny{ \orange{i.e \: \: \: The \: \: \: Area \: \: \: of \: \: \: circle ={πr }^{2} =3.14×{5 }^{2}{ cm }^{2} ={78.5cm }^{2} . }}$

$\red { \huge{Again, Area \: of  \: ΔABC= \frac12×AC×BC (by 1)}}$

$\large{ \color{cyan}= \frac12×6×{8cm }^{2} =24{cm }^{2} .}$

Now,

• Area of shaded region = Area of circle − area of ΔABC

=(78.5−24)cm²=54.5cm².

Step-by-step explanation:

Thank you

Answer 2

Answer:

Gɪᴠᴇɴ :

AB is a diameter of the circle

• ➛ AC = 6 cm
• ➛ BC = 8 cm

$\begin{gathered}\end{gathered}$

Tᴏ Fɪɴᴅ :

• ➛ The area of the shaded region.

$\begin{gathered}\end{gathered}$

Sᴏʟᴜᴛɪᴏɴ :

☼ Finding diameter of circle by applying Pythagoras theorem in right angled triangle ACB :-

$\longrightarrow{\underline{\boxed{\pmb{\sf{{(AB)}^{2} = {(AC)}^{2} + {(CB)}^{2}}}}}}$

$\longrightarrow\sf {(AB)}^{2} = {(AC)}^{2} + {(CB)}^{2}$

$\longrightarrow\sf {(AB)}^{2} = {(6)}^{2} + {(8)}^{2}$

$\longrightarrow\sf {(AB)}^{2} = {(6 \times 6)} + {(8 \times 8)}$

$\longrightarrow\sf {(AB)}^{2} = {(36)} + {(64)}$

$\longrightarrow\sf{{(AB)}^{2}=36 + 64}$

$\longrightarrow\sf{{(AB)}^{2}=100}$

$\longrightarrow\sf{(AB)= \sqrt{100} }$

$\longrightarrow\sf{(AB)= \sqrt{10 \times 10} }$

$\longrightarrow\sf{\underline{\underline{\red{(AB)= 10 \: cm}}}}$

∴ The diameter of circle is 10 cm.

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☼ Finding the radius of radius of circle :-

$\dashrightarrow{\underline{\boxed{\pmb{\sf{Radius = \dfrac{Diameter}{2}}}}}}$

$\dashrightarrow{\sf{Radius = \dfrac{Diameter}{2}}}$

$\dashrightarrow{\sf{Radius = \dfrac{10}{2}}}$

$\dashrightarrow{\sf{Radius = \cancel{\dfrac{10}{2}}}}$

$\dashrightarrow{\sf{\underline{\underline{\red{Radius = 5 \: cm}}}}}$

∴ The radius of circle is 5 cm.

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☼ Finding area of circle :-

$\longrightarrow\underline{\boxed{\pmb{\sf{Area \: of \: circle = \pi {r}^{2}}}}}$

$\longrightarrow{\sf{Area \: of \: circle = \pi {r}^{2}}}$

$\longrightarrow{\sf{Area \: of \: circle = 3.14{(5)}^{2}}}$

$\longrightarrow{\sf{Area \: of \: circle = 3.14{(5 \times 5)}}}$

$\longrightarrow{\sf{Area \: of \: circle = 3.14{(25)}}}$

$\longrightarrow{\sf{Area \: of \: circle = 3.14 \times 25}}$

$\longrightarrow{\sf{\underline{\underline{\red{Area \: of \: circle = 78.5 \: {cm}^{2}}}}}}$

∴ The area of circle is 78.5 cm².

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☼ Finding the area of right angled triangle :-

${\dashrightarrow{\underline{\boxed{\pmb{\sf{Area \: of \: triangle = \dfrac{1}{2} \times Base \times Height }}}}}}$

${\dashrightarrow{\sf{Area \: of \: triangle = \dfrac{1}{2} \times Base \times Height }}}$

${\dashrightarrow{\sf{Area \: of \: triangle = \dfrac{1}{2} \times AC \times CB}}}$

${\dashrightarrow{\sf{Area \: of \: triangle = \dfrac{1}{2} \times 6 \times 8}}}$

${\dashrightarrow{\sf{Area \: of \: triangle = \dfrac{1}{2} \times 48}}}$

${\dashrightarrow{\sf{Area \: of \: triangle = \dfrac{48}{2}}}}$

${\dashrightarrow{\sf{Area \: of \: triangle = \cancel{\dfrac{48}{2}}}}}$

${\dashrightarrow{\sf{\underline{\underline{\red{Area \: of \: triangle = 24 \: {cm}^{2}}}}}}}$

∴ The area of right angled triangle is 24 cm².

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☼ Now, finding the area of shaded region :-

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Area \: of \: shaded \: region = Area \: of \: circle - Area \: of \: triangle}}}}}}}$

${\longrightarrow{\small{\sf{Area \: of \: shaded \: region = Area \: of \: circle - Area \: of \: triangle}}}}$

${\longrightarrow{\sf{Area \: of \: shaded \: region = 78.5 - 24}}}$

${\longrightarrow{\sf{\underline{\underline{\red{Area \: of \: shaded \: region = 54.5 \: {cm}^{2}}}}}}}$

∴ The area of shaded region is 54.5 cm².

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Lᴇᴀʀɴ Mᴏʀᴇ :

$\begin{gathered}\begin{gathered}\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}\end{gathered}$

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