Question

A circle of radius 6 cm is divided into 6 equal sectors.
An equilateral triangle is drawn on the chord of
each sector to lie outside the circle. Find the area of
resulting figure.​

Answer 1

The area of resulting figure is 183.52 sq miter.

Step-by-step explanation:

• We can understand that all the angle subtended at the center of circle have equal measures. Because all the triangles (MGH, MHL, MKL, MKJ, MJI, MIG) are made by radii of the circle and all the sectors are equal. So all the triangles are isosceles triangle.

• Here, we consider $\Delta MJK$first. (Refer the figure attached)

We draw a perpendicular MN from M to JK.

• From $\Delta MNK :$

Let, MN = h

• JK is the chord of circle.

• $\angle JMK = 60^o$

So, $\angle NMK = 30^o$

We know that,

• length of chord$JK = 2 \times radius \times sin \dfrac{ \angle JMK}{2}$

= $2 \times 6 \times \dfrac{1}{2}$

• JK = 6 cm

• NK = 3 cm

Now, $h^2 + 3^2 = 6^2$

$h^2 = 36 - 9$

$h = \pm 5 \ cm$

h = 5 cm,  Because h is the length.

• Now, the area of $\Delta MJK = \dfrac{1}{2} \times JK \times h$

= $\dfrac{1}{2} \times 6 \times 5$

= 15 sq miter (All the triangles as MJK, MKL, MLH, MHG, MGI, MIJ ) have same area.

• Now, the triangle out side \Delta JKD is an equilateral triangle. Its side length is equal to chord of the circle.

• So, the area of$\Delta JKD = \dfrac{\sqrt{3}}{4} \times a^2$

• $= \dfrac{\sqrt{3}}{4} \times 6^2$

$= 9 \sqrt{3} \ sq. \ miter$

All the outside triangles have the same area as $9 \sqrt{3}$ sq miter.

• Thus, the resulting figure area = All inner triangle's area + All outer triangle's area

• $= 6(15) + 6 (9 \sqrt{3})$

= 90 + 93.52

= 183.52 sq miter

Thus, the area of resulting figure is 183.52 sq miter.