Question

if the altitude of a triangle is one-third of its base and area of the triangle is 54cm². then what is the value of the base.​

Answer 1

⛦ Given :

• The altitude of ∆ is one-third its base.
• Area of the triangle is 54cm².

⛦ To find :

• The base of the triangle∆.

⛦ Solution :

As it is given that the altitude of the triangle is one-third its base of the triangle.

Thus,

• Let the base of the triangle ∆ be x

• Then the altitude /height of the triangle will be $\sf\dfrac{1}{3}x$

$\bigstar{\boxed{\sf{Area \ of \ triangle =} \dfrac{1}{2}{(b\times{h})}}}$

[Here, b = base and h = height/altitude of the ∆]

According to Question,

$\mapsto\sf{54=}\dfrac{1}{2}\sf{(x\times\dfrac{1}{3}x)}$

$\mapsto\sf{54=}\dfrac{x\times{x}}{2\times3}$

$\mapsto\sf{54=}\dfrac{x^2}{6}$

$\mapsto\sf{54\times6=x^2}$

$\mapsto\sf{x^2=324}$

$\mapsto\sf{x=\sqrt{324}}$

$\mapsto\sf{x=\sqrt{(18^2)}}$

$\mapsto\sf{x=18cm}$

Therefore, the base of the triangle is 18cm.

___________________

$\underline{\sf{\green{Verification:}}}$

$\leadsto\sf{54=}\dfrac{1}{2}\sf{(x\times\dfrac{1}{3}x)}$

$\leadsto\sf{54=}\dfrac{1}{2}\sf{(x\times\dfrac{x}{3})}$

Putting the value of x.

$\mapsto\sf{54=}\dfrac{1}{\cancel{2}^1}\times\cancel{18}^{ \ 9}\times\dfrac{\cancel{18}^{ \ 6}}{\cancel{3}^1}$

$\mapsto\sf{54=9\times6}$

$\mapsto\sf{54=54}$

Thus, L.H.S. = R.H.S.

Hence, Checked !!

Answer 2

Question :

If the altitude of a triangle is one-third of its base and area of the triangle is 54cm². Then what is the value of the base.

To find:

• value of base

Given :

• Altitude of a triangle is 1/3 of its base
• Area of triangle = 54 cm²

Let:

• Base of triangle be x

• Lenth of triangle be $\sf{} \dfrac{1}{3}x$

How?

we know :

length of triangle = 1/3 of breadth

and we have supposed Breadth as x

=>length of triangle=1/3 of x

=>length of triangle=1/3 x

Solution:

So to find breadth first we should know value of x

We know:

$\boxed{ \boxed{ \rm{Area \: of \: triangle = \frac{1}{2} \times base \times heigth(altitude)}}}$

by using this formula we can find value of x

$: \implies\sf{}54 = \dfrac{1}{2} \times x \times \dfrac{x}{3}$

$: \implies\sf{}54 = \dfrac{x \times x}{2 \times 3}$

$: \implies\sf{}54 = \dfrac{ {x}^{2} }{6}$

$: \implies\sf{}54 \times 6 = {x}^{2}$

$: \implies\sf{} \sqrt{54 \times 6} = x$

$: \implies\sf{} \sqrt{(3 \times 3 \times 3 \times 2)\times (2 \times 3)} = x$

$: \implies\sf{} \sqrt{ {3}^{4} \times {2}^{2} } = x$

$: \implies\sf{} {3}^{2} \times 2 = x$

$: \implies\sf{} x = 9 \times 2$

$: \implies\underline {\boxed{\sf{} x = 18cm}}$

Now Let's Verify whether value of x is correct or not?

Verification:

$: \implies\sf{}54 = \dfrac{1}{2} \times x \times \dfrac{x}{3}$

put value of x in this equation:

$: \implies\sf{}54 = \dfrac{1}{2} \times 18\times \dfrac{18}{3}$

$: \implies\sf{}54 = \dfrac{18 \times 18}{2 \times 3}$

$: \implies\sf{}54 = \dfrac{324}{6}$

$: \implies\sf{}54 = \cancel \dfrac{324}{6}$

$: \implies \underline{ \boxed{\sf{}54 =54}}$

RHS=LHS

Hence verified!

.°. value of breadth = 18 cm