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# Area of Triangle and Integer | PRMO 2019 | Question 29

Try this beautiful problem from the Pre-RMO, 2019 based on Area of Triangle and Integer. You may use sequential hints to solve the problem.

Try this beautiful problem from the PRMO, 2019 based on area of triangle and nearest integer.

## Area of triangle and integer – PRMO 2019

In a triangle ABC, the median AD (with D on BC) and the angle 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

• is 107
• is 47
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Triangles

Integer

PRMO, 2019, Question 29

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

let AD and BE meet at F angle ABF =angleFBD=$\frac{B}{2}$ angle AFB=angle BFD=90 (in degrees), BF is common with triangles ABF and BFD then triangle ABF is congruent to triangle BFD,AF=FD=$\frac{7}{2}$ where AF=FD

Second Hint

AB=BD then$\frac{AB}{BC}=\frac{AB}{BD+CD}=\frac{AB}{2BD}=\frac{1}{2}$ where AB=BD $\frac{AE}{EC}$=$\frac{AB}{BC}$=$\frac{1}{2}$

Final Step

$\frac{area triangle ABC}{area triangle ABE}=\frac{3}{1}$ then $area triangle ABC=3area triangle ABE$=$(3)(\frac{1}{2} \times AF \times BE)=\frac{3}{2} \times \frac{7}{2} \times 9$=47.25 then nearest integer=47.

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