If p_1, p_2, p_3 are the altitudes of a trian | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$ and $\frac{1}{r} = \frac

Vedclass · Accepted Answer

$4\left(\frac{1}{p_1^2}+\frac{1}{p_2^2}+\frac{1}{p_3^2}\right)$

Vedclass · Answer

(D) We know that $\frac{1}{r_1} = \frac{s-a}{\Delta}, \frac{1}{r_2} = \frac{s-b}{\Delta}, \frac{1}{r_3} = \frac{s-c}{\Delta}$ and $\frac{1}{r} = \frac{s}{\Delta}$.Now,$\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{1}{\Delta^2} [(s-a)^2 + (s-b)^2 + (s-c)^2 + s^2]$.$= \frac{1}{\Delta^2} [4s^2 - 2s(a+b+c) + a^2 + b^2 + c^2]$.Since $a+b+c = 2s$,we have $4s^2 - 2s(2s) + a^2 + b^2 + c^2 = a^2 + b^2 + c^2$.So,the expression equals $\frac{a^2 + b^2 + c^2}{\Delta^2}$.Also,$\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$,which implies $a = \frac{2\Delta}{p_1}, b = \frac{2\Delta}{p_2}, c = \frac{2\Delta}{p_3}$.Substituting these,we get $\frac{1}{\Delta^2} [(\frac{2\Delta}{p_1})^2 + (\frac{2\Delta}{p_2})^2 + (\frac{2\Delta}{p_3})^2] = 4(\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2})$.

If $p_1, p_2, p_3$ are the altitudes of a triangle $ABC$ from the vertices $A, B, C$ respectively,then with the usual notation,$\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r^2}=$

Similar Questions

If $\alpha$ and $\beta$ are different values of $x$ satisfying $a \cos x + b \sin x = c,$ then $\tan \left( \frac{\alpha + \beta}{2} \right) = $

If $\alpha$ and $\beta$ are acute angles such that $\alpha+\beta$ and $\alpha-\beta$ satisfy the equation $\tan^2\theta - 4\tan\theta + 1 = 0$, then $\alpha$ and $\beta$ are respectively:

In a $\triangle ABC$,if $\angle C = 90^{\circ}$ and $\frac{a^2+b^2}{a^2-b^2} \sin(A-B) = 1$,then which of the following is true?

In $\triangle ABC$,with usual notations,if $a, b, c$ are in $A.P.$,then $a \cos^2\left(\frac{C}{2}\right) + c \cos^2\left(\frac{A}{2}\right) = $

If $a \cos A = b \cos B$,then $\Delta ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module