Given: ABCD ∥gram, BK ⊥ AD , AB ⊥ BD AB=6, AK=3 Find: m∠A, BK, AABCD
Answers
Hi,
Answer:
Given data:
ABCD is a //gm
BK ⊥ AD
AB ⊥ BD
AB = 6 units
AK = 3 units
To find: m∠A, BK, & Area of //gm ABCD
Case (i): measure of side BK
Since BK is perpendicular to AD, so ∆ AKB is a right-angled triangle.
Applying Pythagoras theorem to ∆ AKB, we get
AB² = AK²+ BK²
⇒ BK = √[AB² – AK²]
⇒ BK = √[6² – 3²] = √[27]
⇒ BK = 3√3 units
Case (ii): Measure of ∠A
Again, let's consider right-angled ∆ AKB,
sin (∠A) = [perpendicular(BK)]/[hypotenuse(AB)]
⇒ sin (∠A) = 3√3 / 6
⇒ ∠A = sin⁻¹ (√3 / 2) = 60°
Case (iii): measure of the Area of //gm ABCD
Let’s consider right-angled triangle ABD (since AB ⊥ BD) and applying angle sum property, we get
∠A + ∠B + ∠D = 180°
⇒ 60° + 90° + ∠D = 180°
⇒ ∠D = 180° - 150° = 30°
Since in triangle ABD angle B is 90°, so AD is hypotenuse, AB & BD are the other two sides.
We know that in case of a 30°-60°-90° triangle, the shortest side is half the hypotenuse and here angle D = 30° is the shortest angle, so the side opposite to it i.e., side AB is the shortest side.
∴ AB = ½ * AD
⇒ AD = 2 * AB = 2 * 6 = 12 units
Thus,
The area of the //gm ABCD,
= breadth(AD) * height(BK)
= 12 * 3√3
= 36√3 sq. units
Hope this is helpful!!!!!
