Area and Perimeter of Plane Shapes

Topic Content:

Rectangle:

Parallelogram:

Trapezium:

Triangle:

Square:

Circle:

Rhombus:

1.

Given lengths of diagonals, d₁ and d₂

2.

Given side and height

3.

Given side and angle

Example 7.1.1:

The diagram below shows the altitudes \( \scriptsize \overline{QT} \: and \: \overline{PM} \: of \: \Delta PQR \)

If |QR| = 8 cm, |PR| = 7 cm and |QT| = 4 cm, what is |PM|? (WAEC)

Solution:

There are two different triangles with different heights, and their areas are equal.

Area of the triangle in Fig. 7.1.2 is equal to the area of the triangle in Fig. 7.1.3, which is the area of ΔPQR

Given ΔPQR

Area of ΔPQR =

\( \normalsize \frac {1}{2} \scriptsize \: \times \: \: PR \: \times \:QT = \normalsize \frac {1}{2} \scriptsize \: \times \: QR \: \times \: PM \)

⇒ \( \normalsize \frac {1}{2} \scriptsize \: \times \: 7 \: \times \: 4 = \normalsize \frac {1}{2} \scriptsize \: \times \: 8 \: \times \: PM \)

\( \scriptsize 14 = 4PM \)

PM = \( \frac{14}{4} \\ = \frac{7}{2} \\ = \scriptsize 3.5 \: cm \)

|PM| = 3.5 cm

Example 7.1.2:

The figure below shows PQRS is a parallelogram, HSR is a straight line, and HPQ = 90^o . If |HQ| = 10 cm and |PQ| = 6 cm, what is the area of the parallelogram? (WAEC)

Solution

Consider right-angled \( \scriptsize \Delta HPQ \)

We know |HQ| = 10 cm and |PQ| = 6 cm. The first thing we do is calculate |HP|

Using Pythagoras theorem

\( \scriptsize HQ^2 = HP^2 \: + \: PQ^2 \)

\( \scriptsize HP^2 = HQ^2 \: – \: PQ^2 \)

\( \scriptsize HP = \sqrt{HQ^2 \: – \: PQ^2} \)

\( \scriptsize HP = \sqrt{10^2 \: – \: 6^2} \)

\( \scriptsize HP = \sqrt{100 \: – \: 36} \)

\( \scriptsize HP = \sqrt{64} \)

\( \scriptsize |HP| = 8 \: cm\)

Area of PQRS = base