Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area .

Question

Homi · Accepted Answer

The idea is that a trapezium with parallel sides $a$ and $b$ , and height $h$ , has the same area as a rectangle with length $\frac{a+b}{2}$ and width $h$ .

Step 1 — Determine Dimensions of an Equivalent Rectangle

We need to construct a trapezium with an area of $144 \text{ cm}^2$ . Let us first find the dimensions of a rectangle that has this same area. Let the length of the rectangle be $L$ and its width be $W$ . The area of a rectangle is calculated as $L \times W$ . We are given that the area is $144 \text{ cm}^2$ . Let us choose a convenient width for our rectangle, say $W = 8 \text{ cm}$ .

$L \times W = 144$

$L \times 8 = 144$

$L = \frac{144}{8}$

$\boxed{L = 18 \text{ cm}}$

So, we will use a rectangle with length $18 \text{ cm}$ and width $8 \text{ cm}$ .

Diagram 1

Step 2 — Determine Dimensions of the Trapezium

Now, we will use the idea of converting a rectangle to a trapezium to find the dimensions of our trapezium. When a rectangle is converted into a trapezium of equal area, the width of the rectangle becomes the height of the trapezium. Also, the sum of the parallel sides of the trapezium becomes twice the length of the rectangle. Let the height of our trapezium be $h$ . Let the parallel sides of our trapezium be $a$ and $b$ . From Step 1, the width of the rectangle is $W = 8 \text{ cm}$ . So, the height of our trapezium will be $h = W$ .

$\boxed{h = 8 \text{ cm}}$

The length of the rectangle is $L = 18 \text{ cm}$ . So, the sum of the parallel sides of the trapezium will be $a+b = 2L$ .

$a+b = 2 \times 18$

$\boxed{a+b = 36 \text{ cm}}$

Now, we need to choose specific lengths for the parallel sides $a$ and $b$ such that their sum is $36 \text{ cm}$ . Let us choose $a = 10 \text{ cm}$ and $b = 26 \text{ cm}$ . (Many other combinations are possible, for example, $15 \text{ cm}$ and $21 \text{ cm}$ ). So, we will construct a trapezium with parallel sides $10 \text{ cm}$ and $26 \text{ cm}$ , and height $8 \text{ cm}$ .

Step 3 — Construct the Trapezium

Let us construct the trapezium with the dimensions we found: parallel sides $10 \text{ cm}$ and $26 \text{ cm}$ , and height $8 \text{ cm}$ . We will construct a right-angled trapezium for simplicity.

Draw a straight line segment AB of length $10 \text{ cm}$ . This will be one of the parallel sides.
At point A, draw a line perpendicular to AB. Let's call this line AX.
Mark a point D on AX such that AD is $8 \text{ cm}$ . This is the height of our trapezium.
Draw a line through D parallel to AB. Let's call this line DY. This line will contain the other parallel side.
At point B, draw a line perpendicular to AB. Let this line intersect DY at point E. Now, ABE D forms a rectangle, so DE is $10 \text{ cm}$ and BE is $8 \text{ cm}$ .
We need the total length of the second parallel side (on line DY) to be $26 \text{ cm}$ . Since DE is $10 \text{ cm}$ , we need to extend it by $26 - 10 = 16 \text{ cm}$ .
Extend line DY beyond E. Mark a point C on line DY such that EC is $16 \text{ cm}$ . Now, the length of the segment DC is $DE + EC = 10 \text{ cm} + 16 \text{ cm} = 26 \text{ cm}$ .
Join point B to point C.

The quadrilateral ABCD is the required trapezium. Its parallel sides are AB ( $10 \text{ cm}$ ) and DC ( $26 \text{ cm}$ ), and its height is AD ( $8 \text{ cm}$ ).

Let us verify its area: Area of trapezium = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$ Area = $\frac{1}{2} \times (10 + 26) \times 8$ Area = $\frac{1}{2} \times 36 \times 8$ Area = $18 \times 8$ Area = $144 \text{ cm}^2$ . This matches the required area.

Diagram 2

Question 34

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