Question

If the area of the base of a right circular cone
is 51 m² and volume is 68 m², then its
vertical height is​

Answer 1

Answer :

It is stated that the area of the base of a right circular cone is 51 m² and volume is 68 m³ respectively. So, in this question we are asked to find out its vertical height.

For that, we will use the formula for finding the volume of the right circular cone. Then, we will plug the value by making (1) and (2) equation. Thereafter, by taking the given output equal to the given volume we will get our answer.

So, Let's solve it now :

Formulae we are going to used :

$\bullet \: {\boxed{\bf{Volume \: of \: cone = \dfrac{1}{3} \pi \: r^{2} h}}}$

$\bullet \: {\boxed{\bf{Base \: Area \: of \: cone = \pi \: r^{2}}}}$

From given in question :-

• Area of the base of a right circular cone is 51 m²

• Volume of a right circular cone is 68 m³

Now,

According to the question :-

From given :

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{3} \pi \: r^{2} h \: \: \: .. (1) \: eqn$

$\dashrightarrow\:\:\sf Base \: Area = \pi \: r^{2} \: \: \: (2) \: eqn$

Putting the value of base area in eq (1) which is nothing but ' $\pi \: r^{2}$

So,

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{3} \times \pi \: r^{2} \: \times h$

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{3} \times 51 \: \times h$

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{\not{3}} \times {\not{51}^{ \: \: \: 17} } \: \times h$

$\dashrightarrow\:\:\sf Volume : 17h$

Now, we know from the question that the volume is 68 m³ .

So,.from this :-

$:\implies \sf 17 \times h = 68$

$:\implies \sf h = \dfrac{68}{17}$

$:\implies \sf h = \dfrac{{\not{68}^{ \: \: \: 4}} }{\not{17}}$

$:\implies \sf h = 4 \: m$

$\dashrightarrow\:\: \underline{ \boxed{\sf Height \: of \: Cone = {\red{4 \: m}}}}$

Therefore ,

The vertical height of the given cone is 4m.

Not sure about the answer ?

Let's verify it !

For confirmation , we will substitute the value of height , and let's check it whether it is satisfying the equation or not.

From above :-

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{3} \times 51 \: \times h$

Putting ' h = 4 m '

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{3} \times 51 \: \times {\red{4}}$

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{3} \times 204$

$\dashrightarrow\:\:\sf Volume = \dfrac{1}{ \not{3}^{\: \: \: 1} } \times {\cancel{204}^{ \: \: \: 68}}$

$\dashrightarrow\:\:\sf Volume = 68$

$\dashrightarrow\:\:\sf 68 = 68$

${\underline{\bf{L.H.S = R.HS}}}$

Hence,

Verified !

Answer 2

Answer:

4 m

Step-by-step explanation:

The area of the base of a right circular cone =51 m

2

Volume =68 m

1/3πr²h / πr² = 68/51

h=4 m

Hence, this is the answer is 4 m