Question

Given: ∆AKL, AK = 9

Answer 1

Answer:

Given: In triangle AKL,  AK = 9 units,  $m\angle K = 90^{\circ}$  $m\angle A = 60^{\circ}$

In triangle AKL

Sum of interior angles of the triangle is 180 degree;

$m\angle A +m\angle K + m\angle L = 180^{\circ}$

or

$m\angle L = 180^{\circ}-(\angle A +\angle K)$

Substitute the given values we get;

$m\angle L = 180^{\circ}-(60^{\circ} + 90^{\circ}) = 180^{\circ} -150^{\circ} = 30^{\circ}$

Now, in a 30°-60°-90° triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is $\sqrt{3}$ times the length of the shorter leg.

Length of shorter leg (AK) = 9 units

From the above definition:

$\text{Length of hypotenuse(AL)}=2 \times \text{Length of Shorter leg}$

⇒ $AL = 2 \times 9 = 18$ units

and $\text{Length of longer leg(KL)}= \sqrt{3} \times \text{Length of Shorter leg}$

⇒ $KL = \sqrt{3} \times 9 = 9\sqrt{3}$ units.

To find the perimeter of ∆AKL;

Perimeter(P) is the sum of all the sides of a triangle.

⇒$P=AK+KL+AL$

Substitute the given values we get;

$P=9+9\sqrt{3}+18= 27+9\sqrt{3}$ units.

Now, to find the area of triangle AKL;

$\text{Area of triangle(AKL)} = \frac{1}{2}(KL)(AK)$

Substitute the given values we get;

$\text{Area of triangle(AKL)} = \frac{1}{2}(9\sqrt{3})(9)=\frac{81\sqrt{3}}{2}$ square units

Therefore, the perimeter of ∆AKL is, $27+9\sqrt{3}$ units anmd area of triangle is, $\frac{81\sqrt{3}}{2}$ square units