Question

In triangle RST; P, Q and U are midpoints of sides ST, RT and RS respectively. Show that area of PQRU is $\frac{1}{2}$ area of triangle RST

Answer 1

$\frac{UQ}{ST}=\frac{1}{2}$ (Mid point theorem)
Let the height be h.
Area of triangle $RUQ=\left(\frac{1}{2}\right)\times UQ\times \left(\frac{h}{2}\right)$
$=\left(\frac{1}{2}\right)\times \left(\frac{ST}{2}\right)\times \left(\frac{h}{2}\right)=\left(\frac{1}{2}\right)\times ST\times h\times \left(\frac{1}{4}\right)$
= Area of triangle $\frac{RST}{4}$
UP is parallel to RT and QP is parallel to RS using mid point theorem.
Thus RUPQ is a parallelogram with UQ as a diagonal which divides it into 2 triangles with equal areas.
Thus area of $RUPQ=2\times$ area of $RUQ=2\times$ Area of $\frac{RST}{4}$
= Area of $\frac{RST}{2}$