Find to the three places of decimal the radius of the circles whose area is the sum of the areas - Maths - Areas Related to Circles

Question

Find to the three places of decimal the radius of the circles
whose area is the sum of the areas of two triangles whose sides are
35,53,66and 33,56,65 measured in centimeters

Priyanka Kedia · Accepted Answer

Let radius of the circle be r cm

We know that,

$A r e a o f t r i a n g l e = \sqrt{s (s - a) (s - b) (s - c)} a n d, s = \frac{a + b + c}{2}$

Where a, b and c are the lengths of the sides of the triangle and s is the semi-perimeter of the triangle.

Therefore, Area of the triangle whose sides are a = 35, b = 53, c = 66 is as,

$s = \frac{35 + 53 + 66}{2} = 77 s o, A r e a = \sqrt{77 (77 - 35) (77 - 53) (77 - 66)} = \sqrt{77 \times 42 \times 24 \times 11} = \sqrt{853776} = 924 {c m}^{2} \Rightarrow A r e a o f 1 s t t r i a n g l e = 924 {c m}^{2} ...... (1)$

And, Area of the triangle whose sides are a = 33, b = 56, c = 65 is as,

$s = \frac{33 + 56 + 65}{2} = 77 s o, A r e a = \sqrt{77 (77 - 33) (77 - 56) (77 - 65)} = \sqrt{77 \times 44 \times 21 \times 12} = \sqrt{853776} = 924 {c m}^{2} \Rightarrow A r e a o f 2 n d t r i a n g l e = 924 {c m}^{2} ...... (2)$

Now, according to question,

$A r e a o f c i r c l e = S u m o f a r e a o f b o t h t h e t r i a n g l e s π r^{2} = 924 {c m}^{2} + 924 {c m}^{2} r^{2} = \frac{2 \times 924 \times 7}{22} = 588 \Rightarrow r = 24.24871131 c m \Rightarrow r = 24.249 c m$

Hence, radius of the circle is 24.249 cm.

Find to the three places of decimal the radius of the circles whose area is the sum of the areas of two triangles whose sides are 35,53,66and 33,56,65 measured in centimeters.