Lesson 6, Topic 1
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Mensuration

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ShapePerimeter Area
Rectangle
Perimeter
= 2(l + b)
Area
= l x b
= lb
Parallelogram
Perimeter
= 2(a + b)
Area
= b x h
= ab sinÎ¸
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 Î”PQR

Given Î”PQR

Area of Î”PQR =

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

â‡’ $$\normalsize \frac {1}{2} \scriptsize \: \times \: 7 \: \times \: 4 = \normalsize \frac {1}{2} \scriptsize \: \times \: 8 \: \times \: PM$$

$$\scriptsize 14 = 4PM$$

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

|PM| = 3.5cm

Example 2:

The figure below shows PQRS is a parallelogram, HSR is a straight line, and HPQ = 90o. 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