(D) The maximum capacity of the canal is equal to the volume of the canal,which is the product of the area of the cross-section and the length of the

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(D) The maximum capacity of the canal is equal to the volume of the canal,which is the product of the area of the cross-section and the length of the canal.The cross-section is an isosceles trapezium with parallel sides $a = 4 \ m$ and $b = 5 \ m$,and height (depth) $h = 2 \ m$.Area of trapezium $= \frac{1}{2} \times (a + b) \times h$Area $= \frac{1}{2} \times (4 + 5) \times 2 = 9 \ m^2$Volume (Capacity) $= \text{Area} \times \text{Length} = 9 \times 120 = 1080 \ m^3$.

Competitive Quantitative Aptitude Mensuration Measurement of Volume and Surface Area

Medium

The cross section of a canal is in the shape of an isosceles trapezium which is $4 \ m$ wide at the bottom and $5 \ m$ wide at the top. If the depth of the canal is $2 \ m$ and it is $120 \ m$ long,what is the maximum capacity of this canal? (in $m^3$)

A
$2160$
B
$3240$
C
$4320$
D
$1080$

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Mensuration

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Measurement of Volume and Surface Area

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The cross section of a canal is in the shape of an isosceles trapezium which is $4 \ m$ wide at the bottom and $5 \ m$ wide at the top. If the depth of the canal is $2 \ m$ and it is $120 \ m$ long,what is the maximum capacity of this canal? (in $m^3$)

Similar Questions

$A$ cone-shaped circular tent is $9 \, m$ high and the circumference of its circular base is $44 \, m$. How much air is contained in the tent? (in $m^3$) (Use $\pi = \frac{22}{7}$)

The area of the iron sheet required to prepare a cone $24 \text{ cm}$ high with base radius $7 \text{ cm}$ is .......... $\text{cm}^2$ (Take $\pi = \frac{22}{7}$)

The outer dimensions of a closed wooden box of $1 \text{ cm}$ thickness are $12 \text{ cm} \times 10 \text{ cm} \times 8 \text{ cm}$. Find the cost of the wood required to make the box if $1 \text{ cm}^3$ of wood costs $₹ 3.00$.

$A$ solid piece of iron of dimensions $49 \times 33 \times 24 \text{ cm}$ is moulded into a sphere. The radius of the sphere is......$\text{cm}$.

Find the length of the canvas $2 \ m$ in width required to make a conical tent $12 \ m$ in diameter and $6.3 \ m$ in slant height (in $m$).

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