the perimeter of an isosceles triangle is 42 cm and 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
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✬ 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.
