Question

Find to the three places of decimals the radius of the circle whose area is the sum 41 , 58 , 85 and 54 , 58 , 72 measured in centimeters
( Use - π = 22/7 ).

Please read the question carefully.

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Answer 1

Answer:

For the first triangle, we have

$a = 35$

$b = 53$

And ,

$c = 66$

$∴s = \frac{a + b + c}{2} = \frac{35 + 53 + 66}{2}$

$= 77cm$

now ,

Δ¹ = Area of the first triangle

$</p><p>⇒Δ1 = \sqrt{77(77 - 35)(77 - 53)(77 - 66)}$

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

$\sqrt{7 \times 11 \times 7 \times 6 \times 6 \times 4 \times 11}$

$= \sqrt{ {7}^{2} \times {11}^{2} \times {6}^{2} \times {2}^{2} }$

⇒Δ 1 =7×11×6×2=924cm²

For the second triangle , we have

$a = 33$

$b = 56$

$c = 65$

$∴s = \frac{a + b + c}{2} = \frac{33 + 56 + 65}{2} = 77$

Area of the second triangle

$⇒Δ1 = \sqrt{s(s - a)(s - b)(s - c)}$

$⇒Δ1 = \sqrt{77(77 - 33)(77 - 56)(77 - 65)}$

$= \sqrt{7 \times 11 \times 4 \times 11 \times 3 \times 7 \times 3 \times 7}$

$⇒Δ1 = \sqrt{ {7}^{2} \times {11}^{2} \times {4}^{2} \times {3}^{2} }$

$= 7 \times 11 \times 4 \times 3 = 924 {cm}^{2}$

Let r be the radius of the circle . Then

Area of the circle = sum of the area of two triangles

Area related to circles

⇒πr² = Δ 1+Δ2

⇒πr² =924+924

$⇒ \frac{22}{7} \times {r}^{2} = 1848$

$⇒ {r}^{2} = 1848 \times \frac{7}{22} = 3 \times 4 \times \times 7 \times 7$

$⇒r = \sqrt{3 \times {2}^{2} \times {7}^{2} = 2 \times 7 \times \sqrt{3 } = 14 \sqrt{3cm} }$

Answer 2

Question:

Find to the three places of decimals the radius of the circle whose area is the sum of two triangles whose sides are 41 , 58 , 85 and 54 , 58 , 72 measured in centimeters.

Answer:

For the ∆ with sides 41cm,58cm and 85cm,

semi - perimeter

s = (a+b+c)/2

= (41cm + 58cm + 85cm)/2

= 184cm/2

= 92cm

Therefore,

Using Heron's formula,

Area

= √s(s-a)(s-b)(s-c)

= √92(92-41)(92-58)(92-85)

= √92*51*34*7

=√1116696

= 1056.738

Therefore,

area of 1st ∆ = 1056.738cm^2

For the ∆ with sides 54cm,58cm and 72cm,

semi-perimeter

s = (a+b+c)/2

= (54cm+58cm+72cm)/2

= 184cm/2

= 92cm

Therefore,

Using Heron's formula,

Area

= √s(s-a)(s-b)(s-c)

= √92(92-54)(92-58)(92-72)

= √92*38*34*20

= √2377280

= 1541.843

Therefore,

Area of 2nd ∆

= 1541.843cm^2

For the circle,

Let the radius be r cm.

Area of circle = Sum of areas of the two ∆s

=> πr^2 = 1056.738 + 1541.843

=> (22/7)*r^2 = 2598.581

=> r^2 = 2598.581*(7/22)

=> r^2 = 826.8212

=> r = √826.8212

=> r = 28.7544

=> r = 28.754

Therefore,

Radius of the circle = 28.754cm.