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

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

Angles

Triangles

Integer

But try the problem first...

Answer: is 47.

Source

Suggested Reading

PRMO, 2019, Question 29

Geometry Vol I to IV by Hall and Stevens

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

Content

[hide]

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

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

Angles

Triangles

Integer

But try the problem first...

Answer: is 47.

Source

Suggested Reading

PRMO, 2019, Question 29

Geometry Vol I to IV by Hall and Stevens

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.

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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