Question

ABC is an equilateral triangle. AP and BQ are it's altitudes. find the radius of its circumcircle




no measurements given


don't give stup.id answers pls ​

Answer 1

Answer:

$r = \frac{9a {}^{2} + 4 }{24}$

Step-by-step explanation:

ABC is equilateral triangle..

AB = BC = CA = a (let)

$BP = \frac{a}{2}$

$In \: \: triangle, \: APB \\ {AB}^{2} = {AP}^{2} + {BP }^{2} \\ {a}^{2} =AP {}^{2} + ( \frac{a}{2} ) {}^{2} \\ {a}^{2} =AP {}^{2} + \frac{a {}^{2} }{4} \\ {a}^{2} = \frac{ 4AP {}^{2}}{4} + \frac{a {}^{2} }{4} \\ {a}^{2} = \frac{ 4AP {}^{2} + a {}^{2} }{4} \\ 4 {a}^{2} - a {}^{2} = 4AP \\ 3a {}^{2} = 4AP \\ \frac{3}{4} a {}^{2} = AP$

AO = BO = r (let)

OP = AP - OA

$OP = \frac{3}{4} a {}^{2} - r$

$In \: \: triangle , \: BOP \\ \\ OB {}^{2} = BP {}^{2} + OP {}^{2} \\ \\ r {}^{2} = (\frac{a}{2} ) {}^{2} + ( \frac{3}{4} a {}^{2} - r) {}^{2} \\ \\ r {}^{2} = \frac{a {}^{2} }{4} + \frac{9}{16} a {}^{4} + r {}^{2} - \frac{3}{2} a {}^{2} r \\ \\ 0 = \frac{a {}^{2} }{4} + \frac{9}{16} a {}^{4} - \frac{3}{2} a {}^{2}r \\ \\ 0 = \frac{4a {}^{2} + 9a {}^{4} -24a {}^{2} r }{16} \\ \\ 0 = 4a {}^{2} + 9a {}^{4} -24a {}^{2} r \\ \\ 24a {}^{2} r - 4a {}^{2} = 9a {}^{4} \\ \\ 4a {}^{2} (6r - 1) = 9a {}^{4} \\ \\ 6r - 1 = \frac{9a2}{4} \\ \\ 6r = \frac{9a2}{4} + 1 \\ \\ r = \frac{9a {}^{2} + 4 }{4 \times 6} \\ \\ r = \frac{9a {}^{2} + 4 }{24}$