p_1, p_2, p_3 are altitudes of a triangle ABC | vedclass

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

Question

(C) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. From this,we get $\frac{1}{p_1} = \fra

Vedclass · Accepted Answer

$\frac{s-c}{\Delta}$

Vedclass · Answer

(C) We know that the area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$. From this,we get $\frac{1}{p_1} = \frac{a}{2\Delta}$,$\frac{1}{p_2} = \frac{b}{2\Delta}$,and $\frac{1}{p_3} = \frac{c}{2\Delta}$. Substituting these into the expression: $\frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} - \frac{c}{2\Delta} = \frac{a+b-c}{2\Delta}$. Since $2s = a+b+c$,we have $a+b = 2s-c$. Therefore,$\frac{a+b-c}{2\Delta} = \frac{(2s-c)-c}{2\Delta} = \frac{2s-2c}{2\Delta} = \frac{s-c}{\Delta}$.

$p_1, p_2, p_3$ are altitudes of a triangle $ABC$ drawn from the vertices $A, B, C$ respectively. If $\Delta$ is the area of the triangle and $2s$ is the sum of the sides,then $\frac{1}{p_1} + \frac{1}{p_2} - \frac{1}{p_3} =$

Similar Questions

In $\Delta ABC$,if $a = 2$,$b = 4$,and $\angle C = 60^\circ$,then $\angle A$ and $\angle B$ are equal to:

If $A = 60^\circ$,$a = 5$,and $b = 4\sqrt{3}$ in $\Delta ABC$,then $B =$

In a $\Delta ABC$,if $2s = a + b + c$ and $(s - b)(s - c) = x \sin^2 \frac{A}{2}$,then $x =$

In a triangle $ABC$,if $\tan \frac{A}{2} : \tan \frac{B}{2} : \tan \frac{C}{2} = 15 : 10 : 6$,then $\frac{a}{b-c} =$

If $A, B, C$ are the angles of a triangle with $\tan \frac{A}{2}=\frac{1}{3}$ and $\tan \frac{B}{2}=\frac{2}{3}$,then the value of $\tan \frac{C}{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module