# 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

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 12

**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

**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}.

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