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Question Number 72734 by Mr Jor last updated on 01/Nov/19

A t r i a n g l e A B C i s i n s c r i b e d i n a

c i r c l e . A C = 10 c m, B C = 7 c m a n d

A B = 10 c m . F i n d t h e r a d i u s o f t h e

c i r c l e .

Answered by $@ty@m123 last updated on 01/Nov/19

a = 7, b = 10, c = 10

s = \frac{a + b + c}{2} = \frac{27}{2}

△ = \sqrt{s (s - a) (s - b) (s - c)}

= \sqrt{\frac{27}{2} (\frac{27}{2} - 7) (\frac{27}{2} - 10) (\frac{27}{2} - 10)}

= \sqrt{\frac{27}{2} \times \frac{13}{2} \times \frac{7}{2} \times \frac{7}{2}}

= \frac{21}{4} \sqrt{39} \dots \dots (1)

R = \frac{a b c}{4 △}

= \frac{7 \times 10 \times 10}{4 \times \frac{21 \sqrt{39}}{4}}

= \frac{100}{3 \sqrt{39}} A n s .

Answered by MJS last updated on 01/Nov/19

triangle with side lengths a, b, c

circumcircle R = \frac{a b c}{δ}

δ = \sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}

Commented by $@ty@m123 last updated on 01/Nov/19

a + b + c = 2 s

∴ δ = \sqrt{2 s . (2 s - 2 a) (2 s - 2 b) (2 s - 2 c)}

= 4 \sqrt{s (s - a) (s - b) (s - c)}

= 4 △

Answered by mr W last updated on 01/Nov/19

\sqrt{10^{2} - {(\frac{7}{2})}^{2}} - R = \sqrt{R^{2} - {(\frac{7}{2})}^{2}}

10^{2} - 2 R \sqrt{10^{2} - {(\frac{7}{2})}^{2}} = 0

\Rightarrow R = \frac{10^{2}}{\sqrt{{(2 \times 10)}^{2} - 7^{2}}} = \frac{100}{\sqrt{27 \times 13}} = \frac{100 \sqrt{39}}{117} \approx 5.338

Commented by mr W last updated on 01/Nov/19

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