Question

The area of a trapezium of height $40\,\text{cm$ is $1600\,\text{cm}^2$. One parallel side is $10\,\text{cm}$ longer than the other side. Find the ratio of the lengths of the parallel sides.}

• A. $7:9$
• B. $5:7$
• C. $3:5$
• D. $2:3$

Hint:

For trapezium problems, the sum of parallel sides comes directly from \(\text{Area}=\tfrac{1}{2}(a+b)h\). Combine with any linear relation (like a difference) to solve quickly.

Answer 1

The Correct Option is A

Step 1: Set variables and use area formula.
Let the shorter and longer parallel sides be \(a\) and \(b\) (in cm) with \(b=a+10\).
Area of a trapezium: \(\displaystyle \text{Area}=\frac{1}{2}(a+b)\times h\).
Given \(h=40\) and area \(=1600\): \[ \frac{1}{2}(a+b)\cdot 40 = 1600 \ \Rightarrow\ 20(a+b)=1600 \ \Rightarrow\ a+b=80. \] Step 2: Solve for the sides.
With \(b=a+10\): \[ a+(a+10)=80 \ \Rightarrow\ 2a=70 \ \Rightarrow\ a=35,\quad b=45. \] Step 3: Form the ratio.
\[ \text{Required ratio } a:b = 35:45 = \frac{35}{5}:\frac{45}{5}=7:9. \] \[ \boxed{7:9} \]