ii)

Method 1

The formula for the area of a trapezium is

\frac{(a + b) h}{2}

, where

a

and $b$ are the parallel sides and h is the vertical height.

Find BC.

Triangle CDF is isosceles so CD = DF = 9 cm.

BC = BD - CD = 12 - 9 = 3 cm

[1]

Substitute into the formula for the area of a trapezium.

Area ABCF =

\frac{(12 + 3) \times 8.86326 . . .}{2} = 66.4744 . . .

Correct method [1]

Area of trapezium ABCF = 66.5 cm²[1]

Method 2

The area of trapezium ABCF is equal to the area of parallelogram ABDF minus the area of triangle CDF.

Find the area ABDF using the formula area of a parallelogram = base × vertical height.
Use the unrounded answer for the height found in part (i).

Area ABDF = 12 × 8.86326... = 106.3592...

Find CD.

Triangle CDF is isosceles so CD = DF = 9 cm.

[1]

Find the area of CDF using the area of a triangle

= \frac{1}{2} a b \sin θ

, where

θ

is angle between a and b.

Area CDF =

\frac{1}{2} (9) (9) \sin 100 ° = 39.8847 . . .

Find the area of the trapezium by subtracting area CDF from area ABDF .

Area ABCF = 106.3592... - 39.8847... = 66.4744...

Correct method [1]

Area of trapezium ABCF = 66.5 cm²[1]

Area & Perimeter (CIE IGCSE Maths: Extended)

Exam Questions