Answer:
Perimeter would be 23.57 ft ( approx)
Step-by-step 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 = [tex]\frac{1}{2}[/tex] BC
In triangle ADB,
tan 15° = [tex]\frac{AD}{BD}[/tex]
[tex]\implies \tan 15^{\circ}=\frac{AD}{\frac{1}{2}BC}=\frac{2AD}{BC}[/tex]
[tex]\implies AD = \frac{BC \tan 15^{\circ}}{2}[/tex] .............(1)
Now, area of triangle ABC = [tex]\frac{1}{2}\times AD\times BC[/tex]
If area = 9 square ft,
[tex]\frac{1}{2}\times AD\times BC = 9[/tex]
From equation (1),
[tex]\frac{1}{2}\times \frac{BC \tan 15^{\circ}}{2}\times BC = 9[/tex]
[tex]BC^2\tan 15^{\circ} = 36[/tex]
[tex]BC^2 =\frac{36}{\tan 15^{\circ}}=134.35[/tex]
[tex]\implies BC = 11.59\text{ ft}[/tex]
From equation (1),
[tex]AD = \frac{11.59 \tan 15^{\circ}}{2}=1.55[/tex]
Using Pythagoras theorem,
[tex]AB = \sqrt{AD^2 + BD^2}=\sqrt{1.55^2+(\frac{11.59}{2})^2}=5.99\text{ ft}[/tex]
Hence, perimeter of the triangle ABC= AB + BC + CA
= 5.99 + 11.59 + 5.99
= 23.57 ft
