Question

A trapezium ABCD has side AD parallel to BC, ∠BAD=90∘, BC=3 cm, and AD=8 cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is

Answer 1

Correct Answer: 66

The correct answer is 66:
Given the information:
- Side BC = 3 cm
- Side AD = 8 cm
- Perimeter of the trapezium = 36 cm
Let the other two sides of the trapezium be AB and CD. Since AD is parallel to BC, we have AB = CD.
The perimeter of a trapezium is given by the sum of all its four sides:
Perimeter = BC + CD + DA + AB
Given that BC = 3 cm and AD = 8 cm, we can substitute the values and simplify the equation:
36=3+CD+8+AB
Now, rearrange the equation to solve for AB + CD:
AB+CD=36-(3+8)
AB+CD=25
Since AB=CD, we can write:
2. AB=25
\(AB=CD=\text(\frac{25}{2})\)
AB=CD=12.5 cm
Now, let's find the height of the trapezium. Since ∠BAD=90°, we can use the Pythagorean theorem:
\(BD^2 = AB^2 - AD^2\)
\(BD^2 = (12.5)^2 - (8)^2\)
\(BD^2=156.25-64\)
\(BD^2=92.25\)
\(BD=\sqrt{92.25}\)
BD≈9.61 cm
The area of the trapezium can be calculated using the formula:
Area=\((1/2)\times(sum of parallel sides)\times{height}\)
Area = \((1/2)\times(AB+CD)\times{BD}\)
Area = \((1/2)\times(12.5+12.5)\times9.61\)
Area≈66 square cm
Hence, the area of the trapezium is approximately 66 square cm.

Answer 2

Apply Pythagorus Theorem,
\(CD^2= y^2+25\)
\(CD= \sqrt {y^2+25}\)
According to the question,
Perimeter of trapezium \(= 36\)
\(11+y+y^2+25 = 36\)
\(y^2+25 = 25-y\)
\(y^2+25 = 25^2+y^2-50y\)
\(2y=24\)
\(y= \frac {24}{2}\)
\(y=12\)
Area of trapezium \(= 3y+\frac {5y}{2}\)
\(= \frac {11y}{2}\)

\(= \frac {11}{2} \times 12\)
\(= 66\)

So, the answer is \(66 \ cm^2\).