# In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

**Solution:**

We use the concept of areas of circle and equilateral triangle in the problem.

Mark O as center of the circle. Join BO and CO.

Since we know that equal chords of a circle subtend equal angles at the center and all sides of an equilateral triangle are equal,

∴ Each side of triangle ABC will subtend equal angles at the center.

∴ ∠BOC = 360°/3 = 120°

Consider ΔBOC. Drop a perpendicular from OM to BC

We know perpendicular from the center of circle to a chord bisects it.

⇒ BM = MC

OB = OC (radii)

OM = OM (common)

∴ ΔOBM ≅ ΔOCM (by SSS congruency)

⇒ ∠BOM = ∠COM (by CPCT)

∴ 2∠BOM = ∠BOC = 120°

∠BOM = 120°/2 = 60°

sin 60° = BM/BO = √3/2

∴ BM = √3/2 × BO = √3/2 × 32 = 16√3

⇒ BC = 2BM = 32√3

Using the formula of area of equilateral triangle = √3/4 (side)^{2}

We can find the area of ΔABC since a side BC of ΔABC is known.

Visually from the figure, it’s clear that

Area of the design = Area of circle - Area of ΔABC = πr² - √3/4 (BC)²

This can be solved with ease as both the radius of the circle and BC are known.

Radius of circle (r) = 32 cm

From figure, we observe area of design = Area of circle - Area of ΔABC

= πr² - √3/4 (BC)²

= 22/7 × (32)² - √3/4 × (32√3)²

= 22/7 × 1024 - √3/4 × 1024 × 3

= 22528/7 - 768√3

Area of design = (22528/7 - 768√3) cm²

**Video Solution:**

## In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

### NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 6 - Chapter 12 Exercise 12.3 Question 6:

**Summary:**

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. The area of the design is (22528/7−768√3) cm^{2}.