Question

An isosceles trapezium has an area of \[300{\text{ }}c{m^2}\] , height \[h = 12cm\] and base \[b = \dfrac{4}{3}h\] .Find
(a) the perimeter
(b) the surface area obtained by revolving it around the major base
(c) the volume and mass ( \[7.5\dfrac{g}{{c{m^3}}}\] )

Answer 1

Hint: In order to solve this, first we will draw the figure. Draw \[BE \bot AD\] and \[CF \bot AD\] . Then we will find all sides of the trapezium using the given conditions and find the perimeter. After that we will revolve the figure around the major base, and we will see that the shape will look like it is made up of two cones and one cylinder. Then we will find the surface area. For this, we will add the surface area of two cones and a cylinder. Then we will find the volume of an isosceles trapezium by adding the volume of two cones and a cylinder. And finally, we will find the value of the mass.

Formulas used:
Perimeter of an isosceles trapezium = sum of all its sides
Surface area of cone \[ = \pi rl\]
Surface area of cylinder \[ = 2\pi rh\]
Volume of cone \[ = \dfrac{1}{3}\pi {r^2}h\]
Volume of cylinder \[ = \pi {r^2}h\]
Mass \[ = V \cdot \rho \]

Complete step by step answer:
First, let’s draw the figure. Draw \[BE \bot AD\] and \[CF \bot AD\].

We are given that the trapezium is isosceles, which means the opposite sides are equal.
Therefore, we can say that
\[AB = CD{\text{ }} - - - \left( i \right)\]
Here we have given, height \[h = 12cm\]
\[ \Rightarrow BE = CF = h = 12cm\]
Also, it is given that, base \[b = \dfrac{4}{3}h\]
\[ \Rightarrow BC = b = \dfrac{4}{3}h\]
\[ \Rightarrow BC = b = \dfrac{4}{3} \cdot 12 = 16cm{\text{ }} - - - \left( {ii} \right)\]
Now let \[AD = a\]. Now it is given that, area of trapezium is \[300{\text{ }}c{m^2}\]

Now the area of trapezium is
\[A = \dfrac{1}{2}\left( {BC + AD} \right) \cdot BE\]
\[ \Rightarrow A = \dfrac{1}{2}\left( {b + a} \right) \cdot h\]
On substituting the values, we get
\[ \Rightarrow 300 = \dfrac{1}{2}\left( {16 + a} \right) \cdot 12\]
On solving, we get
\[a + 16 = \dfrac{{600}}{{12}} = 50\]
\[ \Rightarrow a = 50 - 16 = 34\]
\[\therefore AD = 34cm\]

Now consider the right-angled triangle, \[\vartriangle AEB\].Here first, we have to find \[AE\].From the figure, we can say that:
\[BC = EF = 16cm\]
\[\Rightarrow AE = FD\]
Now we can write,
\[AE + EF + FD = 34cm\]
\[ \Rightarrow 2AE + 16 = 34\]
\[ \Rightarrow AE = \dfrac{{34 - 16}}{2} = 9cm\]

Now we have, \[BE = 12cm\] and \[AE = 9cm\]
Therefore, by Pythagoras we have
\[A{B^2} = A{E^2} + B{E^2}\]
On substituting the values, we get
\[ \Rightarrow A{B^2} = {9^2} + {12^2}\]
\[ \Rightarrow AB = \sqrt {81 + 144} = \sqrt {225} \]
\[ \Rightarrow AB = 15cm{\text{ }} - - - \left( {iii} \right)\]

Now, we will find the perimeter
Perimeter of an isosceles trapezium = sum of all its sides
\[ \Rightarrow P = AB + BC + CD + AD\]
Using equation \[\left( i \right),\left( {ii} \right),\left( {iii} \right)\] we get
\[ \Rightarrow P = 15 + 16 + 15 + 34\]
\[ \Rightarrow P = 80cm\]

Now we will find the surface area and volume.Now, on revolving the figure around the major base, we will see that the shape will look like it is made up of two cones and one cylinder.So, the surface area of an isosceles triangle will be equal to:
Surface area= Surface area of 2cones + surface area of one cylinder.
Now we know that
Surface area of cone \[ = \pi rl\]
Surface area of cylinder \[ = 2\pi rh\]
Therefore, surface area \[ = 2\pi rl + 2\pi rh\]
Here after revolving, In cone, \[r = h\] and in cylinder, \[r = b\]
\[\therefore S = 2\pi hl + 2\pi bh\]
\[ \Rightarrow S = 2\pi h\left( {l + b} \right)\]
Here, \[h = 12cm,{\text{ }}l = 15cm,{\text{ }}b = 16cm\]
\[ \Rightarrow S = 2\pi \cdot 12\left( {15 + 16} \right)\]
\[ \Rightarrow S = 744\pi c{m^2}\]

Now, the volume of an isosceles triangle will be equal to:
volume= volume of 2cones + volume of one cylinder.
Now we know that
Volume of cone \[ = \dfrac{1}{3}\pi {r^2}h\]
Volume of cylinder \[ = \pi {r^2}h\]
Therefore, volume \[ = 2 \cdot \dfrac{1}{3}\pi {r^2}h + \pi {r^2}h\]
Here after revolving, in cylinder, \[r = h\] and \[h = b\]
\[\therefore V = \dfrac{2}{3}\pi {r^2}h + \pi {h^2}b\]
Here, \[h = 12cm,{\text{ }}b = 16cm\]
\[ \Rightarrow V = \dfrac{2}{3}\pi \cdot 144 \cdot 9 + \pi \cdot 144 \cdot 16\]
On solving, we get
\[ \Rightarrow V = 1088\pi {\text{ }}c{m^3}\]

Now, we will find the mass
We know that
Mass \[ = V \cdot \rho \]
Here, \[V = 1088\pi ,{\text{ }}\rho = 7.5\]
\[ \Rightarrow m = 1088\pi \times 7.5\]
\[ \therefore m = 25.635{\text{ }}kg\]

Hence, the perimeter of the trapezium is \[80cm\]. The surface area is \[744\pi {\text{ }}c{m^2}\]. The volume is \[1088\pi {\text{ }}c{m^3}\]. The mass is \[25.635{\text{ }}kg\].

Note: Here we have two right-angled triangles \[\vartriangle AEB\] and \[\vartriangle DFC\] , \[BE\] and \[CF\] are the heights, where \[BE = CF\] .So, instead of \[\vartriangle AEB\] you can also consider \[\vartriangle DFC\] to find the height. Also, after revolution, take the measurements correctly and do the calculation properly.