If in Delta ABC,AB = 4,BC = 6 and AC = 5,and | vedclass

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(B) Let the sides opposite to vertices $A, B, C$ be $a, b, c$ respectively. Here $a = BC = 6$,$b = AC = 5$,and $c = AB = 4$.The area of the triangle $

Vedclass · Accepted Answer

$\frac{2\sqrt{7}}{15}$

Vedclass · Answer

(B) Let the sides opposite to vertices $A, B, C$ be $a, b, c$ respectively. Here $a = BC = 6$,$b = AC = 5$,and $c = AB = 4$.The area of the triangle $\Delta$ is given by $\Delta = \frac{1}{2} a h_1 = \frac{1}{2} b h_2 = \frac{1}{2} c h_3$.Thus,$h_1 = \frac{2\Delta}{a}$,$h_2 = \frac{2\Delta}{b}$,and $h_3 = \frac{2\Delta}{c}$.Substituting these into the expression: $\frac{1}{h_1} + \frac{1}{h_2} - \frac{1}{h_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta} = \frac{a+b-c}{2\Delta}$.Using Heron's formula,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+5+4}{2} = \frac{15}{2} = 7.5$.$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{7.5(7.5-6)(7.5-5)(7.5-4)} = \sqrt{7.5 	imes 1.5 	imes 2.5 	imes 3.5} = \sqrt{\frac{15}{2} 	imes \frac{3}{2} 	imes \frac{5}{2} 	imes \frac{7}{2}} = \frac{\sqrt{1575}}{4} = \frac{15\sqrt{7}}{4}$.Now,$\frac{a+b-c}{2\Delta} = \frac{6+5-4}{2 	imes \frac{15\sqrt{7}}{4}} = \frac{7}{\frac{15\sqrt{7}}{2}} = \frac{14}{15\sqrt{7}} = \frac{14\sqrt{7}}{15 	imes 7} = \frac{2\sqrt{7}}{15}$.

If in $\Delta ABC$,$AB = 4$,$BC = 6$ and $AC = 5$,and $h_1, h_2, h_3$ are the lengths of the altitudes from vertices $A, B, C$ respectively,then the value of $(\frac{1}{h_1} + \frac{1}{h_2} - \frac{1}{h_3})$ is equal to-

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