Question

find to the three places o decimals 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 centimetres ( use π= 22/7 )​

Answer 1

Hope it will help you!!!!!

Answer 2

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${ \tt \: 14 \sqrt{3} }$

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we have to find the areas of both triangles by heron's formula,

$\bold{ \tt if \: a, \: b \: and \: c \: are \: sides \: of \triangle} \\ \bold{ \tt area = \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \bold{ \tt \: where ,\: \: s = \frac{a + b + c}{2} }$

for the first triangle,we have

a = 35, b = 53, and c = 66

$\bold{ \tt s = \frac{35 + 53 + 66}{2} } \\ \\ \bold{ \tt s = \frac{154}{2} = 77cm }$

Let , A1 be the area of the first triangle. Then,

${ \tt A1 = \sqrt{s(s - a)(s- b)(s - c)} } \\ \\ { \tt \implies A1 = \sqrt{77(77 - 35)(77 - 53)(77 - 66)} } \\ \\ { \tt \implies A1 = \sqrt{77 \times 42 \times 24 \times 11} } \\ \\ { \tt \implies A1 = \sqrt{7 \times 11 \times 7 \times 6 \times 6 \times 4 \times 11} } \\ \\ { \tt \implies A1 = \sqrt{ {7}^{2} \times {11}^{2} \times {6}^{2} \times {2}^{2} } } \\ \\ { \tt \implies A1 = 7 \times 11 \times 6 \times 2} \\ \\ { \tt \implies A1 = 924 {cm}^{2} ...........(i)}$

For second triangle, we have

a = 33 , b = 56 and c = 65

${ \tt s = \frac{33 + 56 + 65}{2} } \\ \\ { \tt \implies s = \frac{154}{2} = 77cm }$

Let, A2 be the area of second triangle. Then,

${ \tt A2 = \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ { \tt \implies A2 = \sqrt{77(77 - 33)(77 - 56)(77 - 65)} } \\ \\ { \tt \implies A2 = \sqrt{77 \times 44 \times 21 \times 12} } \\ \\ { \tt \implies A2 = \sqrt{7 \times 11 \times 4 \times 11 \times 3 \times 7 \times 3 \times 4} } \\ \\ { \tt \implies A2 = \sqrt{ {7}^{2} \times {11}^{2} \times {4}^{2} \times {3}^{2} } } \\ \\ { \tt \implies A2 = 7 \times 11 \times 4 \times 3} \\ \\ { \tt \implies A2 = 924 {cm}^{2}........(ii) }$

now, Let r be the radius of the circle. Then,

according to question,

Area of the circle = sum of areas of two triangles

${ \tt \implies \pi {r}^{2} =A1 + A2 } \\ \\ { \tt by \: using \: (i) \: and \: (ii) } \\ \\ { \tt \implies \pi {r}^{2} = 924 + 924 } \\ \\ { \tt \implies \frac{22}{7} \times {r}^{2} = 1848 } \\ \\ { \tt \implies \: {r}^{2} = 1848 \times \frac{7}{22} } \\ \\ { \tt \implies {r}^{2} = 3 \times 4 \times 7 \times 7 } \\ \\ { \tt \implies \: r = \sqrt{3 \times {2}^{2} \times {7}^{2} } } \\ \\ { \tt \implies \: r = 2 \times 7 \times \sqrt{3} } \\ \\ { \tt \implies r = 14 \sqrt{3} }$

hence, radius of circle is 14√3cm