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Question Number 214220 by Ari last updated on 01/Dec/24

Answered by mr W last updated on 01/Dec/24

$$saya>b$$$$a^2+b^2=c^2...\left(i\right)$$$$a-b=6...\left(ii\right)$$$$h_c=\frac{ab}{c}=8...\left(iii\right)$$$${(a-b)}^2=36$$$$a^2+b^2-2ab=36$$$$c^2-16c-36=0$$$$(c+2)(c-18)=0$$$$\Rightarrow c=18✓$$$$ab=8\times 18=144$$$$a,-barerootsof{x}^2-6x-144=0$$$$x=3\pm 3\sqrt{17}$$$$\Rightarrow a=3(\sqrt{17}+1),b=3(\sqrt{17}-1)$$

Commented by Ari last updated on 01/Dec/24

$$excusemesirbutwhy$$$$h_c=\frac{a.b}{c}?$$

Commented by mr W last updated on 01/Dec/24

Commented by mr W last updated on 01/Dec/24

$$areaoftriangle=\frac{ab}{2}$$$$areaoftriangle=\frac{{ch}_c}{2}$$$$\Rightarrow \frac{{ch}_c}{2}=\frac{ab}{2}$$$$\Rightarrow h_c=\frac{ab}{c}$$

Commented by Ari last updated on 02/Dec/24

$$thankyouSir!$$

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