In the diagram above, TPS is a straight line, PQRS is a parallelogram with base QR and height 8cm. |QR| = 6cm and the area of triangle QST is 52cm2. Find the area of ΔQPT
Area of \(\Delta\)QST = 52cm\(^2\); height = 8cm
b=base length; but A = 1/2 b x h
=52=1/2 x b x 8cm = 104 =8b
b=104/8 = 13cm; ST= 13cm; PS = 6cm
TP = 13 - 6 = 7cm;
area of \(\Delta\)QST = 1/2 b x h
= 1/2 x 7 x 8/1 = 28cm\(^2\)
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