A triangle ABC has area of P square units and | vedclass

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Question

(A) Let the sides of the triangle $ABC$ be $a, b, c$ units and the corresponding altitudes be $h_1, h_2, h_3$ units respectively.Then the perimeter is

Vedclass · Accepted Answer

$S^2 - 2P^2$

Vedclass · Answer

(A) Let the sides of the triangle $ABC$ be $a, b, c$ units and the corresponding altitudes be $h_1, h_2, h_3$ units respectively.Then the perimeter is $a + b + c = 2S$ and the area is $P = \frac{1}{2}(ah_1) = \frac{1}{2}(bh_2) = \frac{1}{2}(ch_3)$.This implies $h_1 = \frac{2P}{a}, h_2 = \frac{2P}{b}, h_3 = \frac{2P}{c}$.Now,consider the expression $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right]$.Substituting the values of $h_1, h_2, h_3$:$= P^2 \left[ \frac{(\frac{4P^2}{ab} + \frac{4P^2}{bc} + \frac{4P^2}{ca})^2}{(\frac{8P^3}{abc})^2} - 2 \right]$$= P^2 \left[ \frac{16P^4 (\frac{c+a+b}{abc})^2}{\frac{64P^6}{a^2b^2c^2}} - 2 \right]$$= P^2 \left[ \frac{16P^4 (\frac{2S}{abc})^2}{\frac{64P^6}{a^2b^2c^2}} - 2 \right]$$= P^2 \left[ \frac{16P^4 \cdot \frac{4S^2}{a^2b^2c^2}}{\frac{64P^6}{a^2b^2c^2}} - 2 \right]$$= P^2 \left[ \frac{64P^4S^2}{64P^6} - 2 \right]$$= P^2 \left[ \frac{S^2}{P^2} - 2 \right] = S^2 - 2P^2$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

$A$ triangle $ABC$ has area of $P$ square units and perimeter $2S$ units. If $h_1, h_2$ and $h_3$ are respectively the lengths of the altitudes of the triangle drawn from the vertices $A, B$ and $C$,then $P^2 \left[ \frac{(h_1 h_2 + h_2 h_3 + h_3 h_1)^2}{h_1^2 h_2^2 h_3^2} - 2 \right] =$

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