Question

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

Answer 1

Let equal sides be x
unequal side = 3x/2

now given
$x + x + \frac{3x}{2} = 42 \\ = > \frac{2x}{2} + \frac{2x}{2} + \frac{ 3x}{2} = 42 \\ = > \frac{7x}{2} = 42 \\ x = 42 \times \frac{2}{7} \\ x = 12$
Now equal sides are 12
so unequal side = 3/2 × 12
=> 3 × 6
= 18



For finding Height drop a perpendicular.
The perpendicular will cut the base 18 cm into half that is 9cm

Now by Pythagoras theorem

Height² + 9² = 12² (since 12 becomes hypotenuse of the triangle formed by perpendicular)

height² = 12² - 9²
height ² = 144 - 81
height² = 63
height = √63
height = 7.93



Hope it helps.

Answer 2

✬ 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.