Area of Polygons | IT

Question 22

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

Solution

We can find the area of each trapezium by dividing it into simpler shapes like rectangles and triangles, whose areas are easy to calculate.

Step 1 — Area of Trapezium ABCD

Let us find the area of the first trapezium, ABCD. We can divide it into a rectangle ABMD and a right-angled triangle BMC.

The height of the trapezium, AD, is 4 units. The length of the top base, AB, is 3 units. The length of the bottom base, DC, is 5 units (DM = 3 units, MC = 2 units).

The area of the rectangle ABMD is length multiplied by width. $\text{Area of rectangle ABMD} = \text{DM} \times \text{AD}$ $= 3 \times 4$ $= 12 \text{ square units}$

The area of the right-angled triangle BMC is half times base times height. $\text{Area of triangle BMC} = \frac{1}{2} \times \text{MC} \times \text{BM}$ $= \frac{1}{2} \times 2 \times 4$ $= 4 \text{ square units}$

The total area of trapezium ABCD is the sum of these areas. $\text{Area of trapezium ABCD} = \text{Area of ABMD} + \text{Area of BMC}$ $= 12 + 4$

$\boxed{16 \text{ square units}}$

Diagram 1

Step 2 — Area of Trapezium PQRS

Let us find the area of the second trapezium, PQRS. We can divide it into a right-angled triangle PST, a rectangle PTUQ, and a right-angled triangle QUR.

The height of the trapezium, PT, is 3 units. The length of the top base, PQ, is 5 units. The length of the bottom base, SR, is 9 units (ST = 2 units, TU = 5 units, UR = 2 units).

The area of the right-angled triangle PST is half times base times height. $\text{Area of triangle PST} = \frac{1}{2} \times \text{ST} \times \text{PT}$ $= \frac{1}{2} \times 2 \times 3$ $= 3 \text{ square units}$

The area of the rectangle PTUQ is length multiplied by width. $\text{Area of rectangle PTUQ} = \text{TU} \times \text{PT}$ $= 5 \times 3$ $= 15 \text{ square units}$

The area of the right-angled triangle QUR is half times base times height. $\text{Area of triangle QUR} = \frac{1}{2} \times \text{UR} \times \text{QU}$ $= \frac{1}{2} \times 2 \times 3$ $= 3 \text{ square units}$

The total area of trapezium PQRS is the sum of these areas. $\text{Area of trapezium PQRS} = \text{Area of PST} + \text{Area of PTUQ} + \text{Area of QUR}$ $= 3 + 15 + 3$

$\boxed{21 \text{ square units}}$

Diagram 2

Step 3 — Area of Trapezium WXYZ

Let us find the area of the third trapezium, WXYZ. We can divide it into a right-angled triangle WMZ, a rectangle WXNM, and a right-angled triangle XNY.

The height of the trapezium, WM, is 4 units. The length of the top base, WX, is 6 units. The length of the bottom base, ZY, is 11 units (ZM = 2 units, MN = 6 units, NY = 3 units).

The area of the right-angled triangle WMZ is half times base times height. $\text{Area of triangle WMZ} = \frac{1}{2} \times \text{ZM} \times \text{WM}$ $= \frac{1}{2} \times 2 \times 4$ $= 4 \text{ square units}$

The area of the rectangle WXNM is length multiplied by width. $\text{Area of rectangle WXNM} = \text{MN} \times \text{WM}$ $= 6 \times 4$ $= 24 \text{ square units}$

The area of the right-angled triangle XNY is half times base times height. $\text{Area of triangle XNY} = \frac{1}{2} \times \text{NY} \times \text{XN}$ $= \frac{1}{2} \times 3 \times 4$ $= 6 \text{ square units}$

The total area of trapezium WXYZ is the sum of these areas. $\text{Area of trapezium WXYZ} = \text{Area of WMZ} + \text{Area of WXNM} + \text{Area of XNY}$ $= 4 + 24 + 6$

$\boxed{34 \text{ square units}}$

Diagram 3

Answer

(i) The area of trapezium ABCD is 16 square units. (ii) The area of trapezium PQRS is 21 square units. (iii) The area of trapezium WXYZ is 34 square units.

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 Perimeter of Region 1 > Perimeter of Region 2, but Area of Region 1 < 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: $\Delta$ XDC or $\Delta$ YDC, if both the rectangles are identical?

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

Find the area of $\Delta$ 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 BC as the base?

Q11

Line $l \parallel \text{BC}$ . Consider the different triangles that have 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: $\text{Area of rhombus ABCD} = \text{Area}(\Delta\text{ADB}) + \text{Area}(\Delta\text{CDB}) = \frac{1}{2} \times \text{AO} \times \text{BD} + \frac{1}{2} \times \text{CO} \times \text{BD}$

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 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 cm and 29.7 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 in

(ii) 7.4 in

Q30

Express the following lengths in inches:

(i) 5.08 cm

(ii) 11.43 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, $161.29\text{ cm}^2 = \frac{161.29}{6.4516}\text{ in}^2$ .