Question

One angle of an isosceles triangle is 150 degrees. Ifthe

area of the triangle is 9 square feet, what is the perimeter ofthe
triangle?

Answer 1

Perimeter would be 23.57 ft ( approx)

Explanation:

Consider triangle ABC is an isosceles triangle,

In which AB = AC,

And, m∠A = 150°,

∵ AB = AC ⇒ m∠B = m∠C,

Now sum of all interior angles of a triangle is 180°,

i.e. m∠A + m∠B + m∠C = 180°,

150°+ m∠B + m∠B= 180°,

2m∠B + 150° = 180°

2m∠B = 30°

⇒ m∠B = 15°

Let D ∈ BC such that AD ⊥ BC,

∵ Altitude of an isosceles triangle is its median,

i.e, BD = DC or BD =
`(1)/(2)` BC

In triangle ADB,

tan 15° =
`(AD)/(BD)`

`\implies \tan 15^(\circ)=(AD)/((1)/(2)BC)=(2AD)/(BC)`

`\implies AD = (BC \tan 15^(\circ))/(2)` .............(1)

Now, area of triangle ABC =
`(1)/(2)* AD* BC`

If area = 9 square ft,

`(1)/(2)* AD* BC = 9`

From equation (1),

`(1)/(2)* (BC \tan 15^(\circ))/(2)* BC = 9`

`BC^2\tan 15^(\circ) = 36`

`BC^2 =(36)/(\tan 15^(\circ))=134.35`

`\implies BC = 11.59\text{ ft}`

From equation (1),

`AD = (11.59 \tan 15^(\circ))/(2)=1.55`

Using Pythagoras theorem,

`AB = √(AD^2 + BD^2)=\sqrt{1.55^2+((11.59)/(2))^2}=5.99\text{ ft}`

Hence, perimeter of the triangle ABC= AB + BC + CA

= 5.99 + 11.59 + 5.99

= 23.57 ft