Question

find the area of triangle whose lenght of their all three altitudes are 20 28 and 35​

Answer 1

   

Let,

$h_a=20 \\ \\ h_b=28 \\ \\ h_c=35$

Let the sides corresponding to the altitudes  $h_a,\ h_b,\ h_c$  be a, b and c respectively.

Let the area be A.

According to  $h_a$  and a,

$A=\frac{1}{2} \times h_a \times a \\ \\ \\ \Rightarrow\ A=\frac{1}{2} \times 20a \\ \\ \\ \Rightarrow\ a=\frac{2A}{20} \\ \\ \\ \Rightarrow\ a=\frac{A}{10}$

According to  $h_b$  and b,

$A=\frac{1}{2} \times h_b \times b \\ \\ \\ \Rightarrow\ A=\frac{1}{2} \times 28b \\ \\ \\ \Rightarrow\ b=\frac{2A}{28} \\ \\ \\ \Rightarrow\ b=\frac{A}{14}$

According to  $h_c$  and c,

$A=\frac{1}{2} \times h_c \times c \\ \\ \\ \Rightarrow\ A=\frac{1}{2} \times 35c \\ \\ \\ \Rightarrow\ c=\frac{2A}{35}$

Now I'm using Heron's formula.

Semiperimeter,

$\Rightarrow\ s=\frac{a+b+c}{2} \\ \\ \Rightarrow\ s=\frac{\frac{A}{10}+\frac{A}{14}+\frac{2A}{35}}{2} \\ \\ \Rightarrow\ s=\frac{A(\frac{1}{10}+\frac{1}{14}+\frac{2}{35})}{2} \\ \\ \Rightarrow\ s=\frac{A(\frac{7}{70}+\frac{5}{70}+\frac{4}{70})}{2} \\ \\ \Rightarrow\ s=\frac{A \times \frac{8}{35}}{2} \\ \\ \Rightarrow\ s=\frac{4A}{35}$

Area = A,

$\Rightarrow\ A=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ \Rightarrow\ A=\sqrt{\frac{4A}{35}(\frac{4A}{35}-\frac{A}{10})(\frac{4A}{35}-\frac{A}{14})(\frac{4A}{35}-\frac{2A}{35})} \\ \\ \Rightarrow\ A=\sqrt{\frac{4A}{35} \times \frac{A}{70} \times \frac{3A}{70} \times \frac{2A}{35}} \\ \\ \Rightarrow\ A=\sqrt{A \times \frac{4}{35} \times A \times \frac{1}{70} \times A \times \frac{3}{70} \times A \times \frac{2}{35}} \\ \\ \Rightarrow\ A=\sqrt{A^4 \times \frac{4 \times 3 \times 2}{(35 \times 70)^2}}$

$\Rightarrow\ A=A^2\sqrt{\frac{2^2 \times 6}{2450^2}} \\ \\ \Rightarrow\ \frac{1}{A}=\sqrt{2^2 \times 6 \times \frac{1}{2450^2}} \\ \\ \Rightarrow\ \frac{1}{A}=\frac{2\sqrt{6}}{2450} \\ \\ \Rightarrow\ A=\frac{2450}{2\sqrt{6}} \\ \\ \Rightarrow\ A=\bold{\frac{1225}{6}\sqrt{6}}$

Hope this helps you. Plz mark it as the brainliest.

Plz ask me if you've any doubts.

Thank you. :-))