Question

Given: ABCD ∥gram, BK ⊥ AD , AB ⊥ BD AB=6, AK=3 Find: m∠A, BK, Area of ABCD

Answer 1

1. Consider right triangle ABK. In this triangle AB is the hypotenuse, BK and AK are legs. By the Pythagorean theorem,

$AB^2=AK^2+BK^2,\\\\6^2=3^2+BK^2,\\\\BK^2=36-9=27,\\\\BK=3\sqrt{3}\ un.$

2. Use the definition of $\cos \angle A:$

$\cos \angle A=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{AK}{AB}=\dfrac{3}{6}=\dfrac{1}{2}.$

Then $m\angle A=60^{\circ}.$

3. Consider right triangle ABD. In this triangle AD is the hypotenuse, AB and BD are legs. Since  $m\angle A=60^{\circ},$ then

$m\angle BDA=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}.$

The leg that is opposite to the angle of 30° is half of the hypotenuse, so

$AD=2AB=12\ un.$

4. The area of parallelogram aBCD is

$A_{ABCD}=AD\cdot BK=12\cdot 3\sqrt{3}=36\sqrt{3}\ sq. un.$