Mensuration

Shape	Perimeter	Area
Rectangle	Perimeter = 2(l + b)	Area = l x b = lb
Parallelogram	Perimeter = 2(a + b)	Area = b x h = ab sinθ
Trapezium	Perimeter = a + b + c + l	Area = \( \frac{1}{2} \scriptsize (a +b) \: \times \: h\)
Triangle	Perimeter = a + b + c	Area = \( \frac{1}{2} \scriptsize \: \times \:b \: \times \: h \) Area = \( \frac{1}{2} \scriptsize ab \: Sin C \) Area = \( \frac{1}{2} \scriptsize ac \: Sin B \) Area = \( \frac{1}{2} \scriptsize bc \: Sin A \)
Square	Perimeter = 4d	Area = \( \scriptsize d^2 \)
Circle	Perimeter = \( \scriptsize 2\pi r = \pi d \)	Area = \( \scriptsize \pi r^2 \)

Example 1:

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

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

Solution:

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

Area of Triangle in figure 2 is equal to the area of triangle in figure 3 which is the area of ΔPQR

Given ΔPQR

Example 2:

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

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

We know |HQ| = 10cm and |PQ| = 6cm. 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 x perpendicular height

The base = PQ = SR = 6cm (as opposite sides are equal)

Height = |HP| = 8cm

Therefore, Area of PQRS = 6cm x 8cm

Area of PQRS = 48cm²