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ABC is similar to triangle PQR,PM is median of PQR. If ar triangle ABC =289cm2,BC=17cm,MR=6.5 cm then the area of triangle PQM is

given:$△ABC\approx △PQR$ and $area(△ABC)=289 c{m}^2$; 
PM is the median of PQR.
BC=17 cm and MR = 6.5 cm
since M is the mid-point of QR.
QR = 2*MR = 2*6.5 = 13 cm
the ratio of the areas of two similar triangles is equal to the ratio of the square of the corresponding sides.
$$\begin{gathered} \frac{area(△PQR)}{area(△ABC)}=\frac{Q{R}^2}{B{C}^2} \\ \frac{area(△PQR)}{289}=\frac{{13}^2}{{17}^2} \\ area(△PQR)=\frac{169}{289}*289=169 \end{gathered}$$
the area of the triangle PQR is 169 sq cm.
since the median divides the triangle into two equal areas.
area of the triangle PQM = 1/2* area of the triangle PQR
$$\begin{gathered} =\frac{1}{2}*169 \\ =84.5 sq cm \end{gathered}$$

hope this helps you