Question

In given quadrilateral ABCD, BE = 3 cm and DF = 4 cm. Find the length of AC if area of ABCD is 28 \(cm^2\).

[3 marks]

Answer 1

Area of $△$ABC = $\frac{1}{2}$$\times$base$\times$height
= $\frac{1}{2}$$\times$AC$\times$BE
Similarly
Area of $△$ADC = $\frac{1}{2}$$\times$AC$\times$DF

Area of quadrilateral ABCD =
Area of $△$ABC
+ Area of $△$ADC
Area of quadrilateral ABCD
28 = $\frac{1}{2}$$\times$AC$\times$BE + $\frac{1}{2}$$\times$AC$\times$DF
28 = $\frac{1}{2}$$\times$AC$\times$(BE+DF)
28 = $\frac{1}{2}$$\times$AC$\times$(3+4) = $\frac{1}{2}$$\times$AC$\times$(7)
$\frac{28\times 2}{7}$ = AC = 8 cm

Length of AC is 8 cm.