# Difficult Geometry & Mensuration Solved QuestionAptitude Discussion

 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.
 ✖ 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

Solution:
Option(B) is correct

$O$ 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
()

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.