Algebra in geometry?

Geometry Level 2

A B C ABC is a triangle with sides A B = 24 AB=24 , B C = 10 BC=10 and A C = 26 AC=26 . P P is point inside the triangle. Perpendiculars from P P are drawn to the sides B C BC , A B AB , and A C AC . These perpendiculars are named x x , y y and z z , respectively. Find the value of 5 x + 12 y + 13 z 5x+12y+13z .


The answer is 120.

Aaryan Maheshwari by Aaryan Maheshwari , 16, India

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1 solution

Chan Tin Ping
Nov 4, 2017

A r e a △ A B C = A r e a △ P A B + A r e a △ P B C + A r e a △ P A C 120 = 24 y × 1 2 + 10 x × 1 2 + 26 z × 1 2 \begin{aligned} Area \space\triangle ABC &= Area\space\triangle PAB + Area \space \triangle PBC + Area \space\triangle PAC \\120&=24y\times\frac{1}{2}+10x\times\frac{1}{2}+26z\times\frac{1}{2} \end{aligned}

Hence, 5x+12y+13z=120

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Amazing ! Didn`t thought of that. (+1)!

Rishu Jaar - 3 years, 7 months ago

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