PRMO-2019

In a triangle ABC, the median AD(with D on BC) and the angular bisector BE(with E on AC) are perpendicular to each other. If AD=7 and BE=9, find the integer nearest to the area of triangle ABC.

#Geometry

Note by Jyothiraditya Chaitanya
1 year, 10 months ago

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Comments

The area comes out to be 47.25...….hence, the answer is 47

Aaghaz Mahajan - 1 year, 10 months ago

How was your paper??? Which class??

Aaghaz Mahajan - 1 year, 10 months ago

I am studying 10th class and i felt the paper to be tough. By the way, could you please post the solution.

Jyothiraditya Chaitanya - 1 year, 10 months ago

Let AD and BE meet at F. Drop a perpendicular from C to extended AD to meet at G. Then CG =BF=3EF. GF=AD=7=2AF. BE=9=BF+EF=CG+EF=4EF. So EF=2.25, CG=6.75 Area of triangle ABC is the sum of the areas of triangles ABD and ADC, or the sum of areas of triangles ABF, BFD and ADC, which is equal to the sum of areas of triangles AGC and ABF. Therefore the area of the triangle ABC is 12\dfrac{1}{2}GC.AG+12\dfrac{1}{2}BF.AF=47.25

A Former Brilliant Member - 1 year, 9 months ago
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