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Question Number 72362 by Maclaurin Stickker last updated on 27/Oct/19

$$Calculatethesidesofatriangle$$$$knowingtheheights{h}_{\mathrm{a}}=\frac{1}{9}$$$$h_{\mathrm{b}}=\frac{1}{7}and{h}_{\mathrm{c}}=\frac{1}{4}$$

Answered by mr W last updated on 27/Oct/19

$$let\mathrm{\Delta }=areaoftriangle$$$${ah}_a={bh}_b={ch}_c=2\mathrm{\Delta }=k\left(say\right)$$$$\Rightarrow a=\frac{k}{h_a}=9k$$$$\Rightarrow b=\frac{k}{h_b}=7k$$$$\Rightarrow c=\frac{k}{h_c}=4k$$$$s=\frac{a+b+c}{2}=10k$$$$\mathrm{\Delta }=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{10\times 1\times 3\times 6}{k}^2=6\sqrt{5}{k}^2$$$$2\mathrm{\Delta }=12\sqrt{5}{k}^2=k$$$$\Rightarrow k=\frac{1}{12\sqrt{5}}=\frac{\sqrt{5}}{60}$$$$\Rightarrow a=\frac{9\sqrt{5}}{60}=\frac{3\sqrt{5}}{20}$$$$\Rightarrow b=\frac{7\sqrt{5}}{60}$$$$\Rightarrow c=\frac{4\sqrt{5}}{60}=\frac{\sqrt{5}}{15}$$

Commented by Maclaurin Stickker last updated on 27/Oct/19

$$Whya=9k?$$

Commented by mr W last updated on 27/Oct/19

$$a\times h_a=2\mathrm{\Delta }=kwith\mathrm{\Delta }=area$$$$a=\frac{2\mathrm{\Delta }}{h_a}=\frac{k}{h_a}=9k$$

Commented by Maclaurin Stickker last updated on 27/Oct/19

$$Oh,thankyou.$$

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