One of the equal sides of an Isosceles triangle is 13cm and its perimeter is 50cm, the area of the triangle
Answers
Step-by-step explanation:
In isosceles ∆ABC
AB = AC = 13 cm But perimeter = 50 cm
∴ BC = 50 – (13 + 13) cm
= 50 – 26 = 24 cm
AD ⊥ BC
∴ AD = DC = 24/2 = 12 cm
In right ∆ ABD,
AB2 = AD2 + BD2 (Pythagoras Theorem)
(13)2 = AD2 + (12)2
⇒ 169 = AD2 + (144)
⇒ AD2 = 169 – 144
= 25 = (5)2
∴ AD = 5 cm
Now, area of ∆ABC = (1/2) Base × Altitude
= (1/2) × BC × AD
= (1/2) × 24 × 5 = 60 cm2
In isosceles ∆ABC
AB = AC = 13 cm and perimeter = 50 cm. Imagine a triangle with sides AB BC and CA and a line AD perpendicular to BC
BC = 50 – (13 + 13) cm
= 50 – 26 = 24 cm
AD ⊥ BC
so AD = DC = 24/2 = 12 cm
In ∆ ABD,
AB^2 = AD^2 + BD^2 (Pythagoras Theorem)
(13)^2 = AD^2 + (12)^2
169 = AD^2 + (144)
AD^2 = 169 – 144
AD = 5 cm
Area of ∆ABC = (1/2) Base × Height
= (1/2) × BC × AD
= (1/2) × 24 × 5 = 60 cm^2
Please mark as brainiest if you think the answer helped you