Area and Perimeter of Quadrilaterals

Find the perimeter of the parallelogram below.

Example 2, StudySmarter Originals

Solution

Recall that a parallelogram has opposite sides of equal length. This means that PQ = SR and PS = QR. Thus, SR = 16 cm and QR = 10 cm.

In order to find the perimeter of this given shape, we simply add the total length of each side as mentioned.

$$\begin{gathered} P=16+16+10+10 \\ \Rightarrow P=2\left(16\right)+2\left(10\right) \\ \Rightarrow P=32+20 \\ \Rightarrow P=52cm \end{gathered}$$

Thus, the perimeter of this parallelogram is 52 cm.

Find the length of the missing sides of the kite below given that the perimeter is equal to 98 cm.

Example 3, StudySmarter Originals

Solution

First, note that a kite has two pairs of equal adjacent sides. This means that WZ = WX (and YZ = YX = 32 cm).

By the formula for the perimeter of a quadrilateral, we obtain

$$\begin{gathered} P=32+32+WX+WZ \\ \Rightarrow 98=64+WX+WX \\ \Rightarrow 98=64+2WX \end{gathered}$$

Now rearranging this, we find that

$$\begin{gathered} 64+2WX=98 \\ \Rightarrow 2WX=98-64 \\ \Rightarrow 2WX=34 \end{gathered}$$

Further simplifying,

$$\begin{gathered} WX=\frac{34}{2} \\ \Rightarrow WX=17cm \end{gathered}$$

Thus, the length of WX and WZ is 17 cm.

Find the perimeter of a rectangle with vertices at A (1, 6), B (1, 2), C (4, 2) and D (4, 6).

Solution

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 4, StudySmarter Originals

Since we have a rectangle, AB = DC and AD = BC. Thus, we can use the distance formula for the lengths of AB and AD.

Distance AB, A (1, 6) and B (1, 2)

$$\begin{gathered} D_{AB}=\sqrt{{(1-1)}^2+{(2-6)}^2} \\ \Rightarrow D_{AB}=\sqrt{{\left(0\right)}^2+{(-4)}^2} \\ \Rightarrow D_{AB}=\sqrt{16} \\ \Rightarrow D_{AB}=4units \end{gathered}$$

Distance AD, A (1, 6) and D (4, 6)

$$\begin{gathered} D_{AD}=\sqrt{{(4-1)}^2+{(6-6)}^2} \\ \Rightarrow D_{AD}=\sqrt{{\left(3\right)}^2+{\left(0\right)}^2} \\ \Rightarrow D_{AD}=\sqrt{9} \\ \Rightarrow D_{AD}=3units \end{gathered}$$

Perimeter ABCD

$$\begin{gathered} P=AB+DC+AD+BC \\ \Rightarrow P=4+4+3+3 \\ \Rightarrow P=14units \end{gathered}$$

Deduce the perimeter of a quadrilateral with vertices at A (–2, 8), B (0, 8), C (1, 4) and D (–1, 6).

Solution

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 5, StudySmarter Originals

Looking at the sketch above, we need to find the distance of AB, BC, CD and AD to calculate the perimeter of ABCD.

Distance AB, A (–2, 8) and B (0, 8)

$$\begin{gathered} D_{AB}=\sqrt{{(-2-0)}^2+{(8-8)}^2} \\ \Rightarrow D_{AB}=\sqrt{{(-2)}^2+{\left(0\right)}^2} \\ \Rightarrow D_{AB}=\sqrt{4} \\ \Rightarrow D_{AB}=2units \end{gathered}$$

Distance BC, B (0, 8) and C (1, 4)

$$\begin{gathered} D_{BC}=\sqrt{{(1-0)}^2+{(4-8)}^2} \\ \Rightarrow D_{BC}=\sqrt{{\left(1\right)}^2+{(-4)}^2} \\ \Rightarrow D_{BC}=\sqrt{1+16} \\ \Rightarrow D_{BC}=\sqrt{17}units \end{gathered}$$

Distance CD, C (1, 4) and D (-1, 6)

$$\begin{gathered} D_{CD}=\sqrt{{(-1-1)}^2+{(6-4)}^2} \\ \Rightarrow D_{CD}=\sqrt{{(-2)}^2+{\left(2\right)}^2} \\ \Rightarrow D_{CD}=\sqrt{4+4} \\ \Rightarrow D_{CD}=\sqrt{8} \\ \Rightarrow D_{CD}=2\sqrt{2}units \end{gathered}$$

Distance AD, A (-2, 8) and D (-1, 6)

$$\begin{gathered} D_{AD}=\sqrt{{(-1-2)}^2+{(6-8)}^2} \\ \Rightarrow D_{AD}=\sqrt{{(-3)}^2+{(-2)}^2} \\ \Rightarrow D_{AD}=\sqrt{9+4} \\ \Rightarrow D_{AD}=\sqrt{13}units \end{gathered}$$

Perimeter ABCD

$$\begin{gathered} P=AB+DC+AD+BC \\ \Rightarrow P=2+\sqrt{17}+2\sqrt{2}+\sqrt{13} \\ \Rightarrow P=12.6units(correcttoonedecimalplace) \end{gathered}$$

Calculate the area of the rhombus below given that PO = 7 cm and SO = 4cm. Point O is the point at which the two diagonals PR and SQ perpendicularly bisect each other.

Example 6, StudySmarter Originals

Solution

Remember: We need the measures of the diagonals, PR and SQ, of the rhombus to calculate its area. Since the diagonals of a rhombus are perpendicular and bisect each other, we find that PO = OR and SO = OQ and thus,

$$\begin{gathered} PR=PO+OR=2PO \\ SQ=SO+OQ=2SO \end{gathered}$$

Solving this, we obtain

$$\begin{gathered} PR=2\left(7\right)=14cm \\ SQ=2\left(4\right)=8cm \end{gathered}$$

Thus, the vertical diagonal, PR is 14 cm and the horizontal diagonal, SQ is 8 cm. By the area formula for a rhombus,

$$\begin{gathered} A=\frac{1}{2}\times 14\times 8 \\ \Rightarrow A=\frac{112}{2} \\ \Rightarrow A=56c{m}^2 \end{gathered}$$

Thus, the area of this rhombus is 56 cm².

What is the height of the trapezium below given that the area is 330 cm²?

Example 7, StudySmarter Originals

Solution

Since AB is parallel to DC, the bases of this trapezium are given by AB = 13 cm and DC = 31 cm. The height is given by AD. By the formula for the area of a trapezium, we obtain

$$\begin{gathered} A=\frac{1}{2}\times \left(13+31\right)\times AD \\ \Rightarrow 330=\frac{1}{2}\times \left(44\right)\times AD \\ \Rightarrow 330=22\times AD \end{gathered}$$

Rearranging this and simplifying our expression, we obtain

$$\begin{gathered} 22\times AD=330 \\ \Rightarrow AD=\frac{330}{22} \\ \Rightarrow AD=15cm \end{gathered}$$

Thus, the height of this trapezium, AD is 15 cm.

Find the area of a kite with vertices at A (0, 4), B (1, 2), C (0, –4) and D (–1, 2).

Solution

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 8, StudySmarter Originals

Since we have a kite, we need the length of the diagonals to equate its area. The diagonals here are AC and BD.

Distance AC, A (0, 4) and C (0, –4)

$$\begin{gathered} D_{AC}=\sqrt{{(0-0)}^2+{(-4-4)}^2} \\ \Rightarrow D_{AC}=\sqrt{{\left(0\right)}^2+{(-8)}^2} \\ \Rightarrow D_{AC}=\sqrt{64} \\ \Rightarrow D_{AC}=8units \end{gathered}$$

Distance BD, B (1, 2) and D (–1, 2)

$$\begin{gathered} D_{BD}=\sqrt{{(-1-1)}^2+{(2-2)}^2} \\ \Rightarrow D_{BD}=\sqrt{{(-2)}^2+{\left(0\right)}^2} \\ \Rightarrow D_{BD}=\sqrt{4} \\ \Rightarrow D_{BD}=2units \end{gathered}$$

Area ABCD

$$\begin{gathered} A=\frac{1}{2}\times AC\times DB \\ \Rightarrow A=\frac{1}{2}\times 8\times 2 \\ \Rightarrow A=8unit{s}^2 \end{gathered}$$

Find the area of a square with vertices at A (2, 3), B (2, –3), C (–2, –3) and D (–2, 3).

Solution

Let us begin by sketching this quadrilateral on the Cartesian plane.

Example 9, StudySmarter Originals

As have a square, AB = BC = CD = AD. Thus, we can simply find one side to compute the area of this square. We shall choose to find AB.

Distance AB, A (2, 3) and B (2, –3)

$$\begin{gathered} D_{AB}=\sqrt{{(2-2)}^2+{(-3-3)}^2} \\ \Rightarrow D_{AB}=\sqrt{{\left(0\right)}^2+{(-6)}^2} \\ \Rightarrow D_{AB}=\sqrt{36} \\ \Rightarrow D_{AB}=6units \end{gathered}$$

Area ABCD

$$\begin{gathered} A=AB\times BC \\ \Rightarrow A={\left(AB\right)}^2 \\ \Rightarrow A=6^2 \\ \Rightarrow A=36unit{s}^2 \end{gathered}$$

Find the perimeter and area of the parallelogram MBND inscribed in the rectangle ABCD below. Here, AM = 6cm.

Example 10, StudySmarter Originals

Solution

The formula for the area of any parallelogram requires the length of its width and perpendicular height. The width is described by MB (or DN and MB = DN) while the perpendicular height is defined by MO. The length of MO is equal to the height of the rectangle ABCD. Thus, MO = AD = BC = 55 cm.

The width of the rectangle is AB = 84 cm. This is made up of both line segments AM and MB so

$$\begin{gathered} AB=AM+MB \\ \Rightarrow 84=48+MB \\ \Rightarrow MB=84-48 \\ \Rightarrow MB=36cm \end{gathered}$$

Thus, the length of MB is 36 cm. By the area formula for a parallelogram, we obtain

$$\begin{gathered} A_{MBND}=55\times 36 \\ \Rightarrow A_{MBND}=1980c{m}^2 \end{gathered}$$

Thus, the area of this parallelogram is 1980 cm².

Now we need to find the length of the side MD to calculate the perimeter of this parallelogram. Notice that MOD is a right-angle triangle. Since we have the lengths of MO = 55 cm and DO = AD = 48 cm, we can use Pythagoras' Theorem! Here, MD is the hypotenuse.

$$\begin{gathered} M{D}^2={55}^2+{48}^2 \\ \Rightarrow M{D}^2=3025+2304 \\ \Rightarrow M{D}^2=5329 \\ \Rightarrow MD=\sqrt{5329} \\ \Rightarrow MD=73cm \end{gathered}$$

Thus, the length of MD is 73 cm. Note that MD = BN. So the perimeter is equal to 218 cm since

$$\begin{gathered} P_{MBDN}=73+73+36+36 \\ \Rightarrow P_{MBDN}=218cm \end{gathered}$$