Question

AB and DC are parallel.

Work out the area of the triangle BCD.
Give your answer correct to 3 significant figures.

Answer 1

583 cm²

Explanation:

To find the area of triangle BCD, we need to know the length of segment BD. That can be found from the law of sines:

BD/sin(A) = BA/sin(D)

BD = (sin(A)/sin(D))BA = sin(70°)/sin(74°)·(39 cm) ≈ 38.1249 cm

Then the area of BCD is ...

A = 1/2(DB)(DC)sin(BDC)

Angle BDC, together with the marked angles, makes a total of 180°.

∠BDC = 180° -74° -70° = 36°

A = 1/2)(38.1249 cm)(52 cm)sin(36°) ≈ 582.64 cm² ≈ 583 cm²

_____

Additional comment

The angle relation we used comes from the fact that consecutive interior angles where a transversal crosses parallel lines are supplementary. That means angle DAB and ADC are supplementary. Angle ADC is the sum of angles ADB and BDC, so we have ...

∠DAB +∠ADB +∠BDC = 180°

∠BDC = 180° -∠DAB -∠ADB