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
Angles
Triangles
Integer
Answer: is 47.
PRMO, 2019, Question 29
Geometry Vol I to IV by Hall and Stevens
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
AB=BD then\(\frac{AB}{BC}=\frac{AB}{BD+CD}=\frac{AB}{2BD}=\frac{1}{2}\) where AB=BD \(\frac{AE}{EC}\)=\(\frac{AB}{BC}\)=\(\frac{1}{2}\)
\(\frac{area triangle ABC}{area triangle ABE}=\frac{3}{1}\) 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.
