Question

The cross-section of a canal is in the form of a trapezium. If the canal top is 10 m wide and the bottom is 6 m wide, and the area of cross-section is 72 sq. m, then the depth of the canal is
A) 10 m
B) 7 m
C) 6 m
D) 9 m

Answer 1

Hint:
It is given that; the cross-section of a canal is in the form of a trapezium and the canal top is 10 m wide and the bottom is 6 m wide and the area of the well is 72 sq. m.
Let d be the depth of the well.
Thus, by using the formula for area of trapezium,
$A={\displaystyle \frac{1}{2}}\times \left(sumoflengthsofparallelsides\right)\times \left(depth\right)$ , we need to make d as subject from the formula of area of trapezoid.
Hence, the depth of the canal can be found.

Complete step by step solution:
It is given that; the cross-section of a canal is in the form of a trapezium and the canal top is 10 m wide and the bottom is 6 m wide.
Let a be the length of canal top, b be the length of the bottom part and h be the depth of the canal.
Thus, we get $a=10$ and $b=6$ .

Also, the area of the cross-section is 72 sq. m. So, $A=72$ .
Now, area of trapezium is given by $A={\displaystyle \frac{1}{2}}\times \left(a+b\right)\times$ depth
Substituting $A=72$ , $a=10$ and $b=6$ in the above formula to get the depth h of the canal.
 $$\begin{gathered} \therefore 72={\displaystyle \frac{1}{2}}\times \left(10+6\right)\times h \\ \therefore 72={\displaystyle \frac{1}{2}}\times 16\times h \\ \therefore 8h=72 \\ \therefore h={\displaystyle \frac{72}{8}} \\ \therefore h=9 \end{gathered}$$

Thus, we get the depth of the canal as 9 m.

Note:
Alternate Method:
It is given that; the cross-section of a canal is in the form of a trapezium and the canal top is 10 m wide and the bottom is 6 m wide.
Let the depth of the canal be h m.

So, we can find the area of the trapezium from the above diagram as the sum of area of triangle ABC and area of triangle ADC.

From above diagram, $$AB=a,DC=b$$ and $$DE=h$$ .
Also, the area of the cross-section is 72 sq. m. So, $A=72$ .
 $$\begin{gathered} \therefore A=ABC+ADC \\ \therefore 72=\left({\displaystyle \frac{1}{2}}\times AB\times DE\right)+\left({\displaystyle \frac{1}{2}}\times DC\times DE\right) \\ \therefore 72=\left({\displaystyle \frac{1}{2}}\times a\times h\right)+\left({\displaystyle \frac{1}{2}}\times b\times h\right) \\ \therefore 72=\left({\displaystyle \frac{1}{2}}\times 10\times h\right)+\left({\displaystyle \frac{1}{2}}\times 6\times h\right) \\ \therefore 5h+3h=72 \\ \therefore 8h=72 \\ \therefore h={\displaystyle \frac{72}{8}} \\ \therefore h=9 \end{gathered}$$
Thus, we get the depth of the canal as 9 m.