Area of Quadrilaterals and Polygons

We have learnt that for a closed plane figure, the perimeter is the distance around its boundary and its area is the region covered by it. We found the area and perimeter of various plane figures such as triangles, rectangles, circles etc. We have also learnt to find the area of pathways or borders in rectangular shapes.

**In this chapter, we will try to solve problems related to perimeter and area of other plane closed figures like quadrilaterals.**

**We will also learn about surface area and volume of solids such as cube, cuboid and cylinder.**

Let us take an example to review our previous knowledge. This is a figure of a rectangular park (Fig 1.1) whose length is 30 m and width is 20 m.

** (i) **What is the total length of the fence surrounding it? To find the length of the fence we need to find the perimeter of this park, which is 100 m. (Check it)

**(ii)** How much land is occupied by the park? To find the land occupied by this park we need to find the area of this park which is 600 square meters (m^{2}) (How?).

**(iii)** There is a path of one metre width running inside along the perimeter of the park that has to be cemented. If 1 bag of cement is required to cement 4 m^{2} area, how many bags of cement would be required to construct the cemented path?

We can say that the number of cement bags used =

Area of cemented path = Area of park – Area of park not cemented.

Path is 1 m wide, so the rectangular area not cemented is (30 – 2) × (20 – 2) m^{2}. That is 28 × 18 m^{2}.

Hence number of cement bags used = ------------------

**(iv)** There are two rectangular flower beds of size 1.5 m × 2 m each in the park as shown in the diagram (Fig 1.1) and the rest has grass on it. Find the area covered by grass.

Area of rectangular beds = ------------------

Area of park left after cementing the path = ------------------

Area covered by the grass = ------------------

Nazma owns a plot near a main road (Fig 1.2). Unlike some other rectangular plots in her neighbourhood, the plot has only one pair of parallel opposite sides. So, it is nearly a trapezium in shape. Can you find out its area?

Let us name the vertices of this plot as shown in Fig 1.3.

By drawing EC || AB, we can divide it into two parts, one of rectangular shape and the other of triangular shape, (which is right angled at C), as shown in Fig 1.3.

Area of rectangle ABCE = h × a = 12 × 20 = 240 m^{2}.

Area of trapezium ABDE = area of Δ ECD + Area of rectangle ABCE = 60 + 240 = 300 m^{2}.

We can write the area by combining the two areas and write the area of trapezium as

By substituting the values of h, b and a in this expression, we find

So to find the area of a trapezium we need to know the length of the parallel sides and the perpendicular distance between these two parallel sides. Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the area of trapezium.

We know that all congruent figures are equal in area. Can we say figures equal in area need to be congruent too? Are these figures congruent?

Draw at least three trapeziums which have different areas but equal perimeters on a squared sheet.

A general quadrilateral can be split into two triangles by drawing one of its diagonals. This “triangulation” helps us to find a formula for any general quadrilateral. Study the Fig 1.4.

Area of quadrilateral

Find the area of quadrilateral PQRS shown in Fig 1.5.

We can use the same method of splitting into triangles (which we called “triangulation”) to find a formula for the area of a rhombus. In Fig 1.6. ABCD is a rhombus. Therefore, its diagonals are perpendicular bisectors of each other.

Area of rhombus

Find the area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm.

Area of the rhombus

We split a quadrilateral into triangles and find its area. Similar methods can be used to find the area of a polygon. Observe the following for a pentagon: (Fig 1.7., 1.8.)

By constructing two diagonals AC and AD thepentagon ABCDE is divided into three parts. So, area ABCDE = area of Δ ABC + area of Δ ACD + area of Δ AED.

By constructing one diagonal AD and two perpendiculars BF and CG on it, pentagon ABCDE is divided into four parts. So,

area of ABCDE = area of right angled Δ AFB + area of trapezium BFGC + area of right angled Δ CGD + area of Δ AED. (Identify the parallel sides of trapezium BFGC.)

The area of a trapezium shaped field is 480 m^{2}, the distance between two parallel sides is 15 m and one of the parallel side is 20 m. Find the other parallel side.

One of the parallel sides of the trapezium is a = 20 m, let another parallel side be b, height h = 15 m.

The given area of trapezium =

Area of a trapezium

Hence the other parallel side of the trapezium is 44 m.

The area of a rhombus is 240 cm^{2 }and one of the diagonals is 16 cm. Find the other diagonal.

Let length of one diagonal d_{1} = 16 cm

and length of the other diagonal = d_{2}

Area of the rhombus

Therefore,

Hence the length of the second diagonal is 30 cm.

There is a hexagon MNOPQR of side 5 cm (Fig 1.9.).

Aman and Ridhima divided it in two different ways (Fig 1.10).

Find the area of this hexagon using both ways.

Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. You can verify it by paper folding (Fig 1.11).

Now area of trapezium

Δ MNO and Δ RPQ are congruent triangles with altitude 3cm (Fig.1.12)

You can verify this by cutting off these two triangles and placing them on one another.

Area of rectangle MOPR = 8 × 5 = 40 cm^{2.}

Now, area of hexagon MNOPQR = 40 + 12 + 12 = 64 cm^{2}.

In your earlier classes you have studied that two dimensional figures can be identified as the faces of three dimensional shapes. Observe the solids which we have discussed so far (Fig 1.13).

**Observe that some shapes have two or more than two identical (congruent) faces. Name them. Which solid has all congruent faces?**

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