Area of Polygons | IT

Question 22

Find the areas of the following trapeziums by breaking them into figures whose areas can be computed.

Solution

Understand the Question

A trapezium can be split into simpler geometric shapes such as rectangles and right-angled triangles.
The total area is the sum of the areas of these individual components:
- $\text{Area of Rectangle} = \text{length} \times \text{breadth}$
- $\text{Area of Triangle} = \dfrac{1}{2} \times \text{base} \times \text{height}$

(i) Find the area of trapezium $\text{ABCD}$ .

Step 1 · Calculate Area of Trapezium ABCD

Trapezium $\text{ABCD}$ is divided into rectangle $\text{ABMD}$ and right-angled triangle $\text{BMC}$ . Diagram 1

Given dimensions: $\text{AD} = 4\text{ units}$ , $\text{AB} = \text{DM} = 3\text{ units}$ , $\text{MC} = \text{DC} - \text{DM} = 5 - 3 = 2\text{ units}$ , and $\text{BM} = \text{AD} = 4\text{ units}$ .

\begin{aligned} \text{Area of rectangle ABMD} &= \text{DM} \times \text{AD} \\ &= 3 \times 4 \\ &= 12 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of triangle BMC} &= \dfrac{1}{2} \times \text{MC} \times \text{BM} \\[0.6em] &= \dfrac{1}{2} \times 2 \times 4 \\[0.6em] &= 4 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of trapezium ABCD} &= \text{Area of ABMD} + \text{Area of BMC} \\ &= 12 + 4 \\ &= 16 \text{ square units} \end{aligned}

Answer

(i) $16\text{ square units}$

(ii) Find the area of trapezium $\text{PQRS}$ .

Step 1 · Calculate Area of Trapezium PQRS

Trapezium $\text{PQRS}$ is divided into two right-angled triangles $\text{PST}$ , $\text{QUR}$ , and a central rectangle $\text{PTUQ}$ . Diagram 2

Given dimensions: $\text{PT} = \text{QU} = 3\text{ units}$ , $\text{PQ} = \text{TU} = 5\text{ units}$ , $\text{ST} = 2\text{ units}$ , and $\text{UR} = 2\text{ units}$ .

\begin{aligned} \text{Area of triangle PST} &= \dfrac{1}{2} \times \text{ST} \times \text{PT} \\[0.6em] &= \dfrac{1}{2} \times 2 \times 3 \\[0.6em] &= 3 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of rectangle PTUQ} &= \text{TU} \times \text{PT} \\ &= 5 \times 3 \\ &= 15 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of triangle QUR} &= \dfrac{1}{2} \times \text{UR} \times \text{QU} \\[0.6em] &= \dfrac{1}{2} \times 2 \times 3 \\[0.6em] &= 3 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of trapezium PQRS} &= \text{Area of PST} + \text{Area of PTUQ} + \text{Area of QUR} \\ &= 3 + 15 + 3 \\ &= 21 \text{ square units} \end{aligned}

Answer

(ii) $21\text{ square units}$

(iii) Find the area of trapezium $\text{WXYZ}$ .

Step 1 · Calculate Area of Trapezium WXYZ

Trapezium $\text{WXYZ}$ is divided into two right-angled triangles $\text{WMZ}$ , $\text{XNY}$ , and a rectangle $\text{WXNM}$ . Diagram 3

Given dimensions: $\text{WM} = \text{XN} = 4\text{ units}$ , $\text{WX} = \text{MN} = 6\text{ units}$ , $\text{ZM} = 2\text{ units}$ , and $\text{NY} = 3\text{ units}$ .

\begin{aligned} \text{Area of triangle WMZ} &= \dfrac{1}{2} \times \text{ZM} \times \text{WM} \\[0.6em] &= \dfrac{1}{2} \times 2 \times 4 \\[0.6em] &= 4 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of rectangle WXNM} &= \text{MN} \times \text{WM} \\ &= 6 \times 4 \\ &= 24 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of triangle XNY} &= \dfrac{1}{2} \times \text{NY} \times \text{XN} \\[0.6em] &= \dfrac{1}{2} \times 3 \times 4 \\[0.6em] &= 6 \text{ square units} \end{aligned}

\begin{aligned} \text{Area of trapezium WXYZ} &= \text{Area of WMZ} + \text{Area of WXNM} + \text{Area of XNY} \\ &= 4 + 24 + 6 \\ &= 34 \text{ square units} \end{aligned}

Answer

(iii) $34\text{ square units}$

Common Mistakes

Forgetting the $\dfrac{1}{2}$ Factor: Missing the factor of $\dfrac{1}{2}$ when computing the area of the triangular components ( $\text{base} \times \text{height}$ instead of $\dfrac{1}{2} \times \text{base} \times \text{height}$ ).
Base Length Confusion: Using the entire bottom base of the trapezium instead of individual segment lengths (e.g. $\text{MC}$ , $\text{ST}$ , $\text{UR}$ ) for the triangle bases.
Formula Alternative Verification: You can cross-check using the standard trapezium formula $\text{Area} = \dfrac{1}{2}(a+b)h$ to ensure all component areas sum up correctly.

More questions in IT

Try to think of different creative ways to divide a square into 4 parts of equal area.

Why Can't Perimeter be a Measure of Area?

Why do we count the number of unit squares to assign measures for area? Couldn't we have just used the perimeter of a region, i.e., the length of its boundary as a measure of its area?

Context: Consider two regions, Region 1 and Region 2, such that $\text{Perimeter of Region 1} > \text{Perimeter of Region 2}$ , but $\text{Area of Region 1} < \text{Area of Region 2}$ .

Q. Find two rectangles that are examples of such regions. If needed, use a grid paper (given at the end of the book) for this.

Also give an example of two regions of other shapes, where the region with the larger perimeter has the smaller area! This property should be visually clear in your example.

In the given figure, which triangle has a greater area: $\Delta \text{XDC}$ or $\Delta \text{YDC}$ , if both the rectangles are identical?

In the given figure, which triangle has a greater area: $\Delta \text{XDC}$ or $\Delta \text{YBC}$ , if both the rectangles are identical?

Find the area of $\Delta \text{XDC}$ .

To find the area of a triangle, what measurements do we need?

How do we get the outer rectangle from the given triangle?

Q10

Will this formula hold for the kind of triangle, around which we cannot draw a rectangle with $\text{BC}$ as the base?

Q11

Line $l \parallel \text{BC}$ . Consider the different triangles that have $\text{BC}$ as their base, and with their third vertex lying anywhere on $l$ .

(i) Which of these triangles has the maximum area, and which has the minimum area?

(ii) Which of these triangles has the maximum perimeter, and which has the minimum perimeter?

Q12

Analyse whether $A$ lies on the perpendicular bisector of $BC$ .

Q13

Area of any Polygon

How do we find the area of this quadrilateral? What measurements do we need for this?

Q14

How do we find the area of this pentagon?

Q15

Can any polygon be divided into triangles?

Q16

Give a method to convert a parallelogram into a rectangle of equal area.

You can try this using a cut-out of a parallelogram.

Q17

Can $\Delta \text{AXD}$ and $\text{ABCX}$ fit together, as shown in the figure, to get a rectangle?

Q18

Try working this out!

Q19

What are the sidelengths of the rectangle $WXYZ$ ?

Q20

Area of rhombus ABCD can also be determined by finding the areas of $\Delta \text{ADB}$ and $\Delta \text{CDB}$ . What formula does this give us?

Q21

Context: The area of rhombus ABCD can be written as:

$\text{Area of rhombus } ABCD = \text{Area}(\Delta \text{ADB}) + \text{Area}(\Delta \text{CDB}) = \dfrac{1}{2} \times \text{AO} \times \text{BD} + \dfrac{1}{2} \times \text{CO} \times \text{BD}$

Q. Simplify the expression to show that we get the same formula for the area of a rhombus in terms of its diagonals.

Q22

Find the areas of the following trapeziums by breaking them into figures whose areas can be computed.

Q23

Will this formula hold for a trapezium that looks like this?

Q24

Will Approach 2 work for any type of trapezium?

Q25

What figure will we get when the two trapeziums are joined along $\text{BC}$ ?

Q26

What type of a quadrilateral is this?

Q27

What do you think is the area of an A4 sheet? Its sidelengths are $21\text{ cm}$ and $29.7\text{ cm}$ . Now find its area.

Q28

What do you think is the area of the tabletop that you use at school or at home? You could perhaps try to visualise how many A4 sheets can fit on your table.

Q29

Express the following lengths in centimeters:

(i) $5 \text{ in}$

(ii) $7.4 \text{ in}$

Q30

Express the following lengths in inches:

(i) $5.08 \text{ cm}$

(ii) $11.43 \text{ cm}$

Q31

How many $\text{cm}^2$ is $1 \text{ in}^2$ ?

Q32

Context: Convert $161.29 \text{ cm}^2$ to $\text{in}^2$ . Every $6.4516 \text{ cm}^2$ gives an $\text{in}^2$ . Hence, $161.29 \text{ cm}^2 = \dfrac{161.29}{6.4516} \text{ in}^2$ .