TS EAMCET · Maths · Properties of Triangles

$p_{1}, p_{2}, p_{3}$ are altitudes of a triangle $A B C$ drawn from the vertices $A, B, C$ respectively. If $∆$ is the area of the triangle and $2 s$ is the sum of the sides, then $\frac{1}{p_{1}} + \frac{1}{p_{2}} - \frac{1}{p_{3}} =$

A $\frac{s - a}{∆}$
B $\frac{s - b}{∆}$
C $\frac{s - c}{∆}$
D $\frac{s}{∆}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\frac{s - c}{∆}$

Step-by-step Solution

Detailed explanation

We know ∆=12ap1=12bp2=12cp3 i.e. 1p1=a2∆, 1p2=b2∆, 1p3=c2∆ Now 1p1+1p2-1p3=a2∆+b2∆-c2∆=2s-2c2∆ =s-c∆

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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p1, p2, p3 are altitudes of a triangle ABC drawn from the vertices A, B, C respectively. If ∆ is the area of the triangle and 2s is the sum of the sides, then 1p1+1p2-1p3=

Step-by-step Solution

$p_{1}, p_{2}, p_{3}$ are altitudes of a triangle $A B C$ drawn from the vertices $A, B, C$ respectively. If $∆$ is the area of the triangle and $2 s$ is the sum of the sides, then $\frac{1}{p_{1}} + \frac{1}{p_{2}} - \frac{1}{p_{3}} =$