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Q) Find upto three places of decimal the radius of the circle whose area is the sum of the areas of two triangles whose sides are 35, 53, 66 and 33, 56, 65 measured in centimeters (Use π = \frac{22}{7})

Ans: 

Let’s start with calculating area of triangles:

In 1st triangle: sides a = 35 cm, b = 53 cm, c = 66 cm

∵ s = \frac{ a + b + c}{2}

∴ s = \frac{35 + 53 + 66}{2} = \frac{154}{2} = 77 cm

Now Area of 1st triangle, Area1 = \sqrt {s (s - a) (s - b) (s - c)}

∴ Area1 = \sqrt {77 (77-35) (77 - 53) (77 - 66)}

= \sqrt {77 \times 42 \times 24 \times 11)}

= \sqrt {(7 \times 11) \times (2 \times 3 \times 7) \times (3 \times 2 \times 2^2) \times (11)}

= \sqrt {(7^2 \times 11^2 \times 2^4 \times 3^2)}

= (7 x 11 x 22 x 3)

= 84 x 11 = 924 cm2  

Similarly, in 2nd triangle: sides a = 33 cm, b = 56 cm, c = 65 cm

∵ s = \frac{ a + b + c}{2}

∴ s = \frac{33 + 56 + 65}{2} = \frac{154}{2} = 77 cm

Now Area of 2nd triangle, Area2 = \sqrt {s (s - a) (s - b) (s - c)}

∴ Area2 = \sqrt {77 (77-33) (77 - 56) (77 - 65)}

= \sqrt {77 \times 44 \times 21 \times 12)}

= \sqrt {(7 \times 11) \times (2 \times 2 \times 11) \times (3 \times 7) \times (2^2 \times 3)}

= \sqrt {(7^2 \times 11^2 \times 2^4 \times 3^2)}

= (7 x 11 x 22 x 3)

= 84 x 11 = 924 cm2  

Next, let’s consider the radius of circle be r ∴ Area of circle, Area = π r2

Its given that the Area of circle = Area of 1st Triangle + Area of 2nd triangle

By substituting, values from equation (i)  and (ii), we get:

π r2   = 924 + 924

∴ \frac{22}{7} r2   = 1848

∴  r2   = 1848 x \frac{7}{22}

∴  r2   = 84 x 7 = 3 x 4 x 7 x 7

∴  r   = \sqrt{3 \times 4 \times 7 \times 7} = 14√3

∴  r   = 14 x 1.732 = 24.248 cm

Therefore the radius of the circle is 24.248 cm.

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