A-triangle-has-area-15-and-circumradius-12-Find-the-product-of-its-heights-

Question Number 110782 by Aina Samuel Temidayo last updated on 30/Aug/20

A triangle has area 15 and

circumradius 12 . Find the product of

its heights .

Answered by som(math1967) last updated on 30/Aug/20

\frac{1}{2} bcsinA = 15

bc \times \frac{a}{24} = 30 [∵ sinA = \frac{a}{2 R}]

abc = 30 \times 24

let heights are h_{1}, h_{2}, h_{3}

∴ \frac{1}{2} \times h_{1} \times a = \frac{1}{2} \times bc \times sinA

\frac{1}{2} \times h_{2} \times b = \frac{1}{2} \times casinB

\frac{1}{2} \times h_{3} \times c = \frac{1}{2} \times absinC

∴ h_{1} \times h_{2} \times h_{3} = abcsinAsinBsinC

= abc \times \frac{a}{24} \times \frac{b}{24} \times \frac{c}{24}

= \frac{a^{2} b^{2} c^{2}}{24 \times 24 \times 24}

= \frac{30 \times 30 \times 24 \times 24}{24 \times 24 \times 24}

= \frac{900}{24} = \frac{75}{2}

Answered by mr W last updated on 30/Aug/20

R = \frac{a b c}{4 A} \Rightarrow a b c = 4 A R

A = \frac{1}{2} {a h}_{A}

A = \frac{1}{2} {b h}_{B}

A = \frac{1}{2} {c h}_{C}

\Rightarrow h_{A} h_{B} h_{C} = \frac{8 A^{3}}{a b c} = \frac{8 A^{3}}{4 A R} = \frac{2 A^{2}}{R} = \frac{2 \times 15^{2}}{12} = \frac{75}{2}

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Thanks .