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Lesson 3, Topic 1
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### 1. Parallelogram:

This is a quadrilateral which has both pairs of opposite sides parallel.

### Properties of Parallelogram:

• The opposite sides are parallel
• The opposite sides are equal |AD| = |BC|, |AB| = |DC|
• The opposite angles are equal Y2 = Y1, X2 = X1
• The diagonals bisect one another

### Properties of Rhombus:

• All four sides are equal
• The opposite sides are parallel
• The opposite angles are equal (∠S = ∠Q, ∠P = ∠R)
• The diagonals bisect the angles
• Adjacent angles are supplementary (For eg., ∠S + ∠R = 180°).

### Properties of Rectangle:

• All of the properties of a parallelogram are found
• All four angles are right angles
• The diagonals are equal

### Properties of a Kite:

• Diagonals bisect at right angles
• Only one pair of opposite angles are equal
• Each pair of adjacent sides equal

### Properties of a Square:

• All of the properties of a rhombus are found
• All four angles are right angles
• The diagonals are equal

### Example 1:

RSTUV is a regular pentagon. Find the angle of RUV.

Solution:

Join R to U

Sum of interior angles of RSTUV (Pentagon) = $$\scriptsize (n \: – \: 2) \: \times \: 180$$

Where “n” = 5

i.e. Sum of angles = $$\scriptsize (5 \: – \: 2) \: \times \: 180$$

Sum of angles = $$\scriptsize 3 \: \times \: 180 \\ \scriptsize = 540^o$$

Recall ∆ RUV is an Isosceles triangle

Also, each interior angle is given as $$\frac{540}{5} \\ \scriptsize = 108^o$$

V = 108°

|RV| = |VU| (equal sides of regular pentagon)

i.e. $$\scriptsize \angle R = \angle U$$  (equal angles of Isosceles triangle)

Let R and U = y, therefore y = y

If V = 108°

Then 108° + y + y = 180 (sum of angles in a triangle)

Or 2y = 180° – 108°

2y = 72o

y = $$\frac{72}{2}$$

y = 36°

The angles of ∆ RUV are V = 108°, R = 36° and U = 36°

### Example 2:

In the diagonal below, find the value of the angle marked x (WAEC).

Solution:

The concept of the exterior angle of a triangle being equal to the sum of two opposite interior angles will be employed here.

Extend a line from point P to meet |QR| at point A.

RAD = AQP + QPA (The exterior angle of a triangle is equal to the sum of the opposite interior angles)

i.e. RAD = 90° + 33°

=123°

Also,

x = PDR

PDR = RAD + ARD (ext. angle of ΔDAR)

PDR = 123° + 22°

PDR = 145°

∴ x = 145°

### Example 3:

In the diagram below |MP| is the bisector of LMN and |NP| is the bisector of LNM, If L = 68°, find the size of the reflex MPN.

Solution:

Let the equal angles be x and y respectively.

i.e.

⇒  M + L + N = 180° (Sum of angles in a triangle).

⇒  y + y + 68 + x + x = 180°

2y + 2x + 68 = 180°

⇒  2(x + y) = 180° – 68° = 112°

⇒ x + y = $$\frac{112}{2} = \scriptsize 56^o$$

MPN (obtuse) = 180 – (x + y)

= 180° – 56°

= 124°

Recall MPN (Obtuse) + MPN (reflex) = 360° (Sum of angles at a point)

⇒ 124° + MPN (reflex) = 360°

MPN (reflex) = 360° – 124°

⇒ reflex MPN = 236o

### Example 4:

A regular polygon has an angle of size 160° each. How many sides does the polygon have?

Solution:

Consider the line BC in the sketch below and extend it to point D.

BCE + ECD = 180°(Sum of angles on a straight line)

i.e. exterior ECD = 20°

recall, the sum of exterior angles of a polygon is equal to 360°

the number of sides “n” = $$\frac {360}{20} \scriptsize = 18 \: sides$$

### Example 5:

If the exterior angles of a quadrilateral are y°, (y + 5°), (y + 10°) and (y + 25°). Find y. (WAEC).

Solution:

Recall the sum of exterior angles of a polygon (quadrilateral) is 360°

y + y + 5 + y + 10 + y + 25 = 360°

4y + 40 = 360°

4y = 320°

y = 80°

### Example 6:

A regular polygon of n sides has each exterior angle to be 45°, find the value of n. (WAEC)

Solution:

Recall from example 5,

Sum of exterior angles = 360°

Since each exterior = 45°

Number of sides = $$\frac{360}{45} \scriptsize = 8 \: sides$$

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