Math, asked by Anonymous, 1 day ago

In the given fig., ABCD is a trapezium with AB||CD and , ∠BCD = 60°, If BFEC is a sector of a circle with centre C and AB = BC = 7 cm and DE = 4 cm,then find the area of the shaded region. (use π = 22/7) (√3 =1.732).


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Answered by user0888
50

Edited (Sorry! Mistakes are fixed. Please check my answer again.)

\rm\Huge\underline{\text{Ideas}}

\rm\Large\underline{\text{First idea:-}}

\rm\triangle BCL is a right triangle, right-angled at \rm\angle L and \rm\angle C=60^{\circ}.

In the right triangle \rm\triangle BCL

\rm\rightarrow\overline{BC}\times\sin60^{\circ}=\overline{BL}\ \therefore\overline{BL}=\dfrac{7\sqrt{3}}{2}\text{ cm}

\rm\Large\underline{\text{Second idea:-}}

Since the point \rm E lies on \rm\overline{CD}

\rm\rightarrow\overline{CE}+\overline{DE}=\overline{CD}\ \therefore\overline{CD}=11\text{ cm}

\rm\Large\underline{\text{Third idea:-}}

\rm\rightarrow\text{(Area of the circular sector BCE)}=\dfrac{60^{\circ}}{360^{\circ}}\times\pi\times7^2

\rm\therefore\text{(Area of the circular sector BCE)}=\dfrac{49}{6}\pi\text{ cm}^2

\rm\Huge\underline{\text{Solution}}

Now let's find our answer.

\rm\rightarrow\text{(Area of the shaded region)}=\square ABCD-\text{(Area of the circular sector BCE)}

By area formula of a trapezium

\rm\rightarrow\square ABCD=\dfrac{1}{2}\times(\overline{AB}+\overline{CD})\times\overline{BL}

\rm\iff\square ABCD=\dfrac{1}{2}\times(7+11)\times\dfrac{7\sqrt{3}}{2}

\rm\iff\square ABCD=\dfrac{1}{2}\times18\times\dfrac{7\sqrt{3}}{2}\ \therefore\underline{\square ABCD=\dfrac{63\sqrt{3}}{2}\text{ cm}^2}

We already found the area of the circular sector

\rm\rightarrow\underline{\text{(Area of the circular sector BCE)}=\dfrac{49}{6}\pi\text{ cm}^2}

Hence

\rightarrow\text{(Area of the shaded region)}=(\dfrac{63\sqrt{3}}{2}-\dfrac{49}{6}\pi)\text{ cm}^2

Approximately

\rightarrow\text{(Area of the shaded region)}\approx(63\times1.732\times\dfrac{1}{2}-\dfrac{49}{6}\times\dfrac{22}{7})\text{ cm}^2

\iff\text{(Area of the shaded region)}\approx(54.558-\dfrac{7}{3}\times11)\text{ cm}^2

\iff\text{(Area of the shaded region)}\approx(54.558-25.6666\dots)\text{ cm}^2

\iff\text{(Area of the shaded region)}\approx28.8914\text{ cm}^2

Rounding off at the fourth decimal point

\therefore\text{(Area of the shaded region)}\approx\red{\underline{28.891\text{ cm}^2}\tiny\text{//}}

\rm\Huge\underline{\text{Answer}}

The approximated area is 28.891 cm².

Answered by 390comsyu2020shivang
6

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this is all answer

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