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ShapePerimeter Area
Rectangle
rect e1622467379500
Perimeter
= 2(l + b)
Area
= l x b
= lb
Parallelogram
parallelgram e1622468673751
Perimeter
= 2(a + b)
Area
= b x h
= ab sinθ
Trapezium
trap ss1 e1622469923306
Perimeter
= a + b + c + l
Area
= \( \frac{1}{2} \scriptsize (a +b) \: \times \: h\)
Triangle
TRIANGU e1622469972594
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
square ss1 e1622470022703
Perimeter
= 4d
Area
= \( \scriptsize d^2 \)
Circle
circl ss1 e1622470070794
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)

mensuration Ex
Figure 1.

Solution:

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

tri mensu 1 e1622480792490
Figure 2.
tri mensu 2 e1622480851344
Figure 3.

Area of Triangle in figure 2 is equal to the area of triangle in figure 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| = 3.5cm

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)

example 2 mensuration 1 e1622482714277

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

triang mens e1622483143150

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

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