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