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,
$A = \dfrac{1}{2} \times \left( {sum\,of\,lengths\,of\,parallel\,sides} \right) \times \left( {depth} \right)$ , 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 = \dfrac{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.
 $
  \therefore 72 = \dfrac{1}{2} \times \left( {10 + 6} \right) \times h \\
  \therefore 72 = \dfrac{1}{2} \times 16 \times h \\
  \therefore 8h = 72 \\
  \therefore h = \dfrac{{72}}{8} \\
  \therefore h = 9 \\
 $

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, $A = 72$ .
 $
  \therefore A = ABC + ADC \\
  \therefore 72 = \left( {\dfrac{1}{2} \times AB \times DE} \right) + \left( {\dfrac{1}{2} \times DC \times DE} \right) \\
  \therefore 72 = \left( {\dfrac{1}{2} \times a \times h} \right) + \left( {\dfrac{1}{2} \times b \times h} \right) \\
  \therefore 72 = \left( {\dfrac{1}{2} \times 10 \times h} \right) + \left( {\dfrac{1}{2} \times 6 \times h} \right) \\
  \therefore 5h + 3h = 72 \\
  \therefore 8h = 72 \\
  \therefore h = \dfrac{{72}}{8} \\
  \therefore h = 9 \\
 $
Thus, we get the depth of the canal as 9 m.