Question

Work out the area of ABCD.
Give your answer to 1 decimal place.

Answer 1

Area of the quadrilateral ABCD = (21.5 + 23.4) cm² = 44.9 cm²

How to calculate the area of ABCD?

• Quadrilateral ABCD is formed from two triangles ΔABD and ΔBDC.

So, Area of quadrilateral ABCD= area of ΔABD and Δ BDC.

How to find area of triangle ABD?

ΔABD is a right angled triangle, right angled at A, BD is the hypotenuse.

• Area of ΔABD =  AD x AB/2

• We know that in a right angled triangle, cosФ= $\frac{base}{hypotenuse}$ and    sinФ=  $\frac{perpendicular}{hypotenuse}$, where Ф is the angle between base and hypotenuse.

Cos(55°) = $\frac{AD}{10}$

⇒ AD = 5.7 cm

Sin(55°)=$\frac{AB}{10}$

⇒AB=8.2 cm

Area of Δ ABD = ( 8.2 x 5.7/2) cm² =23.4 cm²

How to find area of triangle BDC?

• We can find the area of the triangle by the formula                           BD x CD x sin(44°)/2

• We know that sum of all the angles of a triangle is 180°

∠BDC+∠BCD+∠DBC=180°

⇒∠BCD=180°-44°-38°=99°

• We know according to sine rule,

in a triangle ABC,   $\frac{sinA}{a}=\frac{sinB}{b} =\frac{sinc}{c}$, where a,b,c are corresponding sides opposite to angle A,B,C

In ΔBDC

$\frac{sin99}{10}=\frac{sin38}{CD}$

⇒CD =6.2 cm

Area of Δ BDC = {6.2 x 10 x sin(44°)/2} cm² = 21.5 cm²

∴ Area of the quadrilateral ABCD = (21.5 + 23.4) cm² = 44.9 cm²

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