Question

1) The sides adjacent to the right angle of right angled triangular thin plate are 5 cm and 12cm .If the triangle is rotated with the longer side as the axis then find the volume of the shape formed by such rotation.


2) Ratio of the volumes of a right circular cylinder and a sphere is 1:3 . Ratio of the radii of the cylinder and sphere is 1:3 .If the sum of the height and radius of the base of the cylinder is 78cm cm, then what will be the height of the cylinder ?​

Answer 1

$\large\underline{\bold{Solution-1}}$

Given a right angled triangular thin plates of edges 5 cm and 12 cm Adjacent to right angles.

Let suppose that thin triangular plate be ABC right angled at B, such that AB = 5 cm and BC = 12 cm.

Now,

By pythagoras theorem,

⇛AC² = AB² + BC²

⇛AC² = 5² + 12²

⇛AC² = 25 + 144

⇛AC² = 169

⇛AC = 13 cm.

Now,

when a thin triangular plate is rotated with the longer side AC, then two cones generated as shown in attachment.

So,

Area of triangle ABC =

$\sf \: \dfrac{1}{2} \times AB \times BC = \dfrac{1}{2} \times AC \times OB$

$\sf \: \dfrac{1}{2} \times 5 \times 12 = \dfrac{1}{2} \times 13 \times OB$

$\therefore \: \bf \: OB \: = \: \dfrac{60}{13} \: cm$

Now,

we have two cones,

The dimensions of first cone,

Radius, r = OB = 60/13 cm

Height, h = OA cm

The dimensions of other cone

Radius, r = OB = 60/13 cm

Height, H = OC cm

Also,

AC = H + h = 13 cm

Therefore,

$\bf \: Volume \: of \: solid = \: V_1 + V_2$

$\sf \: V_{solid} \: = \: \dfrac{1}{3}\pi \: {r}^{2}h + \dfrac{1}{3} \pi \: {r}^{2} H$

$\sf \: V_{solid} \: = \: \dfrac{1}{3}\pi \: {r}^{2}(h \: + \: H)$

$\sf \: V_{solid} \: = \: \dfrac{1}{3} \times \dfrac{22}{7} \: {\bigg(\dfrac{60}{13} \bigg) }^{2} \times 15$

$\bf \: V_{solid} \: = \: 334.74 \: {cm}^{3}$

$\large\underline{\bold{Solution-2}}$

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{\dfrac{Volume_{(Cylinder)}}{Volume_{(sphere)}} = \dfrac{1}{3} } \\ &\sf{\dfrac{radius_{(Cylinder)}}{radius_{(sphere)}} = \dfrac{1}{3} }\\ &\sf{height_{(Cylinder)} + radius_{(Cylinder)} = 78 \: cm} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{height_{(Cylinder)}}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf \: Volume_{(Cylinder)} = \pi \: {r}^{2} h}$

$\boxed{ \bf \: Volume_{(sphere)} = \dfrac{4}{3} \pi \: {r}^{3} }$

Calculations :-

Given that

Ratio of the radii of the cylinder and sphere is 1:3.

Let radius of cylinder be 'r' cm

Let radius of sphere be 3r cm

Let height of cylinder be 'h' cm.

Now,

h + r = 78

So, h = 78 - r -----(1)

According to statement,

$\sf \: \dfrac{Volume_{(Cylinder)}}{Volume_{(sphere)}} = \dfrac{1}{3}$

$\sf \: Volume_{(sphere)} = 3 \: Volume_{(Cylinder)}$

$\sf \: \dfrac{4}{3} \pi \: {(3r)}^{3} = 3 \times \pi \: {(r)}^{2} h$

$\therefore \: 12r \: = \: h$

$\sf \: 12r = 78 - r \: \: (using \: 1)$

$\sf \: 13r = 78$

$\sf \: r \: = \: 6 \: cm$

So, put r = 6, in equation (1), we get

$\bf \: h \: = 78 - 6 = 72 \: cm$

Hence,

Height of cylinder, h = 72 cm.

More information :-

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length²+breadth²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

Answer 2

Question 1) :- The sides adjacent to the right angle of right angled triangular thin plate are 5 cm and 12cm .If the triangle is rotated with the longer side as the axis then find the volume of the shape formed by such rotation.

Answer :-

Points Used :--

→ When a right ∆ is revolve Around its Hypotenuse :--

• Here cone will be double .
• Height of cone = Hypotenuse of Right ∆.
• Radius of cone = ( Perpendicular * Base ) / Hypotenuse .

so, by pythagoras theorem we get,

→ Hypotenuse = √(5² + 12²)

→ Height of cone = √(25 + 144) = √(169) = 13 cm.

then,

→ Radius of cone = (P * B)/H = (5 * 12)/13 = (60/13) cm.

therefore,

→ volume of cone = (1/3) * π * (r)² * h

→ volume = (1/3) * 3.14 * (60/13)² * 13

→ volume = (1/3) * 3.14 * (3600/13)

→ volume = (3.14 * 1200)/13

→ volume ≈ 289.85 cm³ . (Ans.)

Hence, the volume of the shape formed by such rotation is 289.85 cm³.