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# The volume generated by rotating the triangle with vertices at (0,0),(3,0) and (3,3) about axis is:

Last updated date: 13th Jul 2024
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Hint: After rotating through the axis, the slope formed will be cone. Observe radius and height of the cone to get volume.

We have given three coordinates of triangle as (0,0),(3,0) and (3,3): Let us represent these coordinates on coordinate plane:

Now, by representation only, we get that the given triangle is a right angled triangle at point (3,0). We can verify it by using Pythagoras formula as
$\left( Hypotenuse \right)={{\left( Base \right)}^{2}}+{{\left( Height \right)}^{2}}$
Distance formula between two points $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right)$ is
$\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
Hence, from the given triangle on coordinate axis we get sides of triangle as;
\begin{align} & AC=\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}}=\sqrt{18}=\sqrt{9\times 2} \\ & AC=3\sqrt{2} \\ \end{align}
And AB can be calculated as;
\begin{align} & AB=\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}}=\sqrt{9}=3 \\ & AB=3 \\ \end{align}
For BC we can write;
\begin{align} & BC=\sqrt{{{\left( 3-3 \right)}^{2}}+{{\left( 0-3 \right)}^{2}}}=\sqrt{9}=3 \\ & BC=3 \\ \end{align}
Now for using Pythagoras relation, we have three sides of triangle AB,BC and AC. As longest side of right angled triangle is hypotenuse, i.e. AC should be hypotenuse if the given triangle is the right angle.
Hence,
\begin{align} & {{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( 3 \right)}^{2}}+{{\left( 3 \right)}^{2}}=18 \\ & \And {{\left( AC \right)}^{2}}={{\left( 3\sqrt{2} \right)}^{2}}=18 \\ \end{align}
Therefore,
${{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}={{\left( AC \right)}^{2}}$
It is proved that a given triangle is a right angled triangle.
Now, we are rotating the given triangle about the axis i.e. X-axis on which base of the given triangle is lying.

We can observe that if we rotate right angle triangle about AB or X-axis, we will get a cone type structure with height 3 and diameter 6 and slant height of that cone should be $3\sqrt{2}$ from given diagram:

Now, we get height of cone=3
Slant height of cone= $3\sqrt{2}$
As we know volume of cone is $\dfrac{1}{3}\pi {{R}^{2}}H$
$\dfrac{1}{3}\times \pi \times {{\left( 3 \right)}^{2}}\times 3=9\pi uni{{t}^{3}}$
Hence, the volume of shape generated by rotating the given triangle about the axis is $9\pi uni{{t}^{3}}$.