Question

Base of a triangle is 3times as it altitude . If its area is 128cm^2. Find its altitude and base

Answer 1

Answer:

Altitude = 16/$\sqrt{3}$

Base = 16$\sqrt{3}$

Step-by-step explanation:

Let h = altitude of the given triangle

Let b = base of given triangle

We know that, b = 3h

Now,

Area of triangle = $\frac{1}{2}$*b*h

                           =$\frac{1}{2}$*3h*h

                           = $\frac{1}{2}$*3h²

ATQ

$\frac{1}{2}$*3h² = 128

3h² = 128*2

3h² = 256

h² = 256 / 3

h = $\sqrt{256}$ / $\sqrt{3}$

h = 16/$\sqrt{3}$

b = 3h

   = 3*16/$\sqrt{3}$

   = 16$\sqrt{3}$

Answer 2

Answer:

$\large\bold\red{Altitude = \frac{16}{\sqrt{3}}\:cm}\\\large\bold\red{Base=16\sqrt{3}\:cm}$

Step-by-step explanation:

Given,

A Triangle in which Base is b and it's Altitude is h.

Now,

According to Question,

b = 3h ........(i)

Also,

It's area ,$A = 128 {cm}^{2}$ ..........(ii)

Now,

We know that,

Are of a triangle is given by formula,

$A = \frac{1}{2} \times b \times h$

Now,

From Equation (i) and (ii),

Putting the respective values,

We get,

$= > 128 = \frac{1}{2} \times 3h \times h \\ = > 3 {h}^{2} = 128 \times 2 \\ = > 3 {h}^{2} = 256 \\ = > {h}^{2} = \frac{256}{3} \\ = > {h}^{2} = \frac{ {(16)}^{2} }{ {( \sqrt{3} )}^{2} } \\ = > h = \frac{16}{ \sqrt{3} }$

Therefore,

$= > b = 3h \\ = > b = 3 \times \frac{16}{ \sqrt{3} } \\ = > b = 16 \sqrt{3}$

Hence,

$\large\bold{Altitude = \frac{16}{\sqrt{3}}\:cm}\\\large\bold{Base=16\sqrt{3}\:cm}$