the perimeter of an isosceles triangle is 42 cm and its base is 3.2 times each of the equal sides find the length of each side of the triangle area of the triangle and the height of the triangle
Answers
Base of isosceles triangle = (3x/2)
Given perimeter of isosceles triangle = 42 cm
That is (x + x +3x/2) = 42
7x / 2 = 42
Therefore x = 12 cm
The equal sides are 12 cm each.
Base is 18 cm
Using heron's formula find the area of the required triangle.
✬ Sides = 12 , 12 , 18 ✬
✬ Height = 7.94 ✬
Step-by-step explanation:
Given:
Perimeter of isosceles triangle is 42 cm.
Length of base of triangle is 3/2 times of equal sides.
To Find:
Length of each side of triangle and height of triangle.
Solution: Let ABC be a isosceles triangle where measure of each equal sides be x cm.
AB = AC = x
BC = 3/2 times of x
As we know that
★ Perimeter of ∆ = Sum of all sides ★
A/q
Perimeter (AB + BC + CA) = 42 cm
\implies{\rm }⟹ 42 = AB + BC + CA
\implies{\rm }⟹ 42 = x + 3x/2 + x
\implies{\rm }⟹ 42 = 2x + 3x/2
\implies{\rm }⟹ 42 = 4x + 3x/2
\implies{\rm }⟹ 42 × 2 = 7x
\implies{\rm }⟹ 84/7 = 12 = x
Hence, measure of sides of ∆ABC are ∼
AB = AC = 12 cm
BC = 3/2 × 12 = 18 cm
___________________
[ Now let's find height of the triangle ]
We will use Pythagoras Theorem here.
Let height of triangle be AZ i.e (AZ ⟂ BC, so BZ = ZC)
In right angled ∆AZC we have
AZ {perpendicular/height}
ZC {base} = 1/2 × 18 = 9
AC {hypotenuse} = 12 cm
★ H² = Perpendicular² + Base² ★
➼ AC² = AZ² + ZC²
➼ 12² = AZ² + 9²
➼ 144 – 81 = AZ²
➼ √63 = 7.937 = AZ
Hence, height of the triangle is 7.94 cm.
