Geometry & Mensuration
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Q.

\(\triangle ABC\) is an equilateral triangle of side 14 cm. A semi circle on \(BC\) as diameter is drawn to meet \(AB\) at \(D\),  and \(AC\) at \(E\). Find the area of the shaded region.

Image for Geometry and Mensuration, Aptitude:2189-1

 A.

\(49\left(\dfrac{\pi}{3}-\sqrt{3}\right)\text{ cm}^2\)

 B.

\(49\left(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{2}\right)\text{ cm}^2\)

 C.

\(49\text{ cm}^2\)

 D.

None

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Solution:
Option(B) is correct

Answer image for Geometry and Mensuration, Aptitude:2189-1

\(O\) is the centre of the circle and the mid-point of \(BC. DO\) is parallel to \(AC\).

So, \(\angle DOB=60^\circ\)

Area of \(\triangle BDO=\dfrac{49\sqrt{3}}{4}\)

Area of sector \(OBD=\dfrac{49\pi}{6}\)

Hence area of the shaded region

\(=2\left(\dfrac{49\pi}{6}-\dfrac{49\sqrt{3}}{4}\right)\)

\(=49\left(\dfrac{\pi}{3}-\dfrac{\sqrt{3}}{2}\right)\text{ cm}^2\)


(5) Comment(s)


Naveen Boora
 ()

how can we say the circle passes through the mid point of the two sides?



Vineet
 ()

trigo ques:

5cos(theta)+12 sin(theta)=13. find sin(theta)



Bhavesh
 ()

how can you directly take DO parallel to ac ? How will you prove that D is the mid point of AB or circle will cut this edge exactly at mid point ?



Vikas
 ()

Sir, can't I directly subtract the area the semicircle from the area of the equilateral triangle?

Because it has not given that $D$ and $E$ are the midpoints of the equilateral triangle.

The area left after subtracting may be equal to the area of $ADE$ and the area of two shaded regions


Ritika
 ()

Let me answer all the three points asked by you.

Point1: directly subtract the area the semicircle from the area of the equilateral triangle?

You can, but then you would not get the required area. You will also get an extra area that contains area $ADE$ too, which may become difficult to handle at the later stage.

Point 2: It is not given that $D$ and $E$ are the midpoints of the equilateral triangle.

Also, it is not mentioned in the question and you have to make it out that $D$ and $E$ are the midpoints.

Point 3: The area left after subtracting may be equal to the area of $ADE$ and the area of two shaded regions.

In mathematics we don't assume, you need to prove it to be sure if that's equal.