Question

Given: △AKL, AK=9 m∠K=90°, m∠A=60° Find: The perimeter of △AKL The area of △ AKL

Answer 1

Perimeter of ΔAKL is,
`27 +9 √(3)` units.

Area of ΔAKL is,
`(81√(3) )/(2)` square units

Explanation:

Given: In ΔAKL , AK = 9 units ,
`m\angle K = 90^(\circ)` and
`m\angle A= 60^(\circ)`.

In ΔAKL

`\tan (A) = (KL)/(AK)`

Substitute the value AK = 9 units and
`m\angle A =60^(\circ)` to solve for KL ;

`\tan(60^(\circ)) = (KL)/(9)`

`√(3) = (KL)/(9)`

⇒
`KL = 9√(3) units`

In right angle ΔAKL,

Using Pythagoras theorem;

`AL^2 = AK^2+KL^2` ......[1]

Substitute AK = 9 units and
`KL =9√(3) units` in [1] to solve for AL;

`AL^2 = 9^2+(9√(3))^2`

`AL^2 = 81+(81 \cdot 3)`

`AL^2 = 81+243 = 324`

`AL = √(324) = 18 units`

Perimeter of triangle is the sum of all the sides.

Perimeter of triangle AKL = AK +KL +AL =
`9 + 9√(3) + 18 = 27 +9 √(3)` units.

Formula for Area of right angle triangle is given by:

`A = (1)/(2) Base * Height`

Area of triangle AKL=
`(1)/(2) (9√(3)) * (9)`

=
`(81√(3) )/(2)` square units