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Algebra Arithmetic Geometry Math Olympiad PRMO

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

Area of Triangle and integer
  • is 107
  • is 47
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Triangles

Integer

Check the Answer


But try the problem first…

Answer: is 47.

Source
Suggested Reading

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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