ΔABC is an equilateral triangle of side 14 cm. A semi circle on BC as diameter

Question

Δ

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?

ΔABC Is An Equilateral Triangle Of Side 14 Cm. A Semi Circl...

Options:

A . 49[π2−√3] cm2

B . 49[π2−√32] cm2

C . 49 cm2

D . 49√2 cm2

Answer: Option B
:
B
Option (b)

O is the centre of the circle and the mid-point of BC. DO is parallel to AC. So,

∠

DOB = 60°
Area of Δ BDO =

\frac{3}{4}

* 49
Area of sector OBD =

\frac{49}{6}

Hence area of the shaded region
= 2[

\frac{49}{6}

–

\frac{3}{4}

*49]
= 49[

\frac{1}{3}

–

\frac{3}{2}

]
Shortcut
By graphical division, there are 3 equilateral triangles of areas of

\frac{\sqrt{3}}{4}

* 49
Area of interest = (area of semi circle [

\frac{r^{2}}{2}

] - area of three triangles) = (49*

\frac{1}{2}

- 3*3*

\frac{49}{4}

)*(

\frac{2}{3}

) = 49*(

\frac{1}{3}

–

\frac{3}{2}

)

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