Lesson 2, Topic 2
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# Isosceles & Equilateral Triangles

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In the study of isosceles and equilateral triangles and parallelograms, we shall recall some of the concepts discussed under congruent triangles.

### Isosceles Triangle:

Definition: An Isosceles triangle has two equal sides. The third side is called the base. The equal sides meet at the vertex. The angle at the vertex is called the vertical angle.

The Base angles of an Isosceles Triangle are Equal.

### Properties of Isosceles Triangle:

(i). Two sides equal |BA| = |CA|

(ii). Two angles equal $$\scriptsize \hat{ABD} = \hat{ACD}$$

(iii). The bisector of the vertical angle meets the base at a right angle

(iv). The bisector of the vertical angle bisects the base, (i.e. |BD| = |DC| ).

### Equilateral Triangle:

This is a triangle with three equal sides or it is a triangle with three angles equal (i.e. 60Âº).

Note that an equilateral triangle is a special Isosceles triangle because all the properties of an Isosceles triangle are true for an equilateral triangle.

### Properties of Equilateral Triangle:

(i). All the three sides are equal (i.e. all angles equal)

(ii). Angles are 60o each.

(iii). It is a special Isosceles triangle

### Evaluation 1

For each of the diagrams below, name the triangle which is congruent to âˆ†XYZ giving the letters in the correct order. In each case state the condition of Congruency using the abbreviations RHS, SSS, SAS, ASA, or AAS.

(i)

Solution: $$\scriptsize \Delta ZDX \equiv \Delta XYZ \: (AAS)$$

(ii)

Solution: $$\scriptsize \Delta XLK \equiv \Delta XYZ \: (SAS)$$

(iii)

Solution: $$\scriptsize \Delta TYS \equiv \Delta XYZ \: (SAS)$$

(iii)

Solution: $$\scriptsize \Delta NMZ \equiv \Delta XYZ \: (SAS)$$

### Evaluation 2

In the diagram below, DEF is a triangle with EDF = 2XÂº.
$$\scriptsize \bar{DE} \: and \: \bar{DF}$$are produced to G and H respectively, so that |EF| = |EG| = |FH|
$$\scriptsize \bar{EH} \: and \: \bar{FG}$$intersect at K. Show that EKG = 90Âº â€“ xÂº

Solution

âˆ EFG = âˆ EGF = a (equal angles of Isosceles triangle)

âˆ FEH = âˆ FHE = b (equal angles of Isosceles triangle)

âˆ DFE = 2bÂº (ext. Angle of âˆ† EFH)

âˆ DEF = 2aÂº (ext. Angle of âˆ† EGF)

Consider âˆ†EKG

2a + b = âˆ EKG + a (ext. Angle of âˆ† EKG)

âˆ EKG + a = 2a + b

âˆ EKG = 2a – a + b

âˆ EKG = (a + b)Âº

Consider âˆ†DEF

2xÂº + 2aÂº + 2bÂº = 180Âº (sum of angles in a triangle)

i.e. 2(x + a + b) = 180Âº

x + a + b = $$\frac{180^o}{2}$$

â‡’  x + a + b = 90Âº

â‡’ a + b = 90Âº – xÂº

But we know âˆ EKG = (a + b)

âˆ´ âˆ EKG = 90Âº – xÂº

### Evaluation 3

In the diagram below, |AD| = |DB| = |BC|
and $$\scriptsize \hat{ADB} = \hat{BDC} = 62^o$$
Calculate $$\scriptsize \hat{ABC}$$

Solution:

$$\scriptsize \hat{ABC} =\hat{ABD} \: + \: \hat{DBC}$$

$$\scriptsize \hat{DAB} = \hat{ABD}$$ (Equal angles of Isosceles triangle)

If $$\scriptsize \hat{DAB} \: and \: \hat{ABD} = x \\ \scriptsize \: then \: \hat{ABD} \: + \: \hat{DAB} = 2x$$

$$\scriptsize 180 = 2x \: + \: 62$$ (Sum of angles in an Isosceles triangle)

$$\scriptsize x = \normalsize \frac{180 \: – \: 62}{2} = \scriptsize 59^o$$

$$\scriptsize \therefore \scriptsize \hat{ABD} = 59^o$$

Consider $$\scriptsize \Delta DBC$$ which is an iscoceles triangle

Therefore, $$\scriptsize \hat{BCD} = \hat{CDB} = 62^o$$

$$\scriptsize 180 = \hat{DBC}\: + \: \hat{BCD}\: + \: \hat{CDB}$$

$$\scriptsize \therefore \hat{DBC} = 180 \: â€“ \: (62 \: + \:62)$$ (Sum of angles in a triangle)

$$\scriptsize \therefore \hat{DBC} = 180 \: – \: 124 = 56$$

$$\scriptsize \hat{ABC} =\hat{ABD} \: + \: \hat{DBC}$$

$$\scriptsize \hat{ABC} = 59 \: + \: 56$$

â‡’ $$\scriptsize \hat{ABC} = 115^o$$

### Evaluation 4

From the diagram below, â§¸PRâ§¸ = â§¸PQâ§¸, â§¸QSâ§¸ = â§¸QRâ§¸ and RPQ = 40Âº. Calculate PQS. (WAEC)

Solution:

$$\scriptsize \hat{PRQ} = \hat{PQR}$$ (Base angles of Isosceles triangle)

If $$\scriptsize \hat{PRQ} \: and \: \hat{PQR} = x \\ \scriptsize \: then \: \hat{PRQ} \: + \: \hat{PQR} = 2x$$

$$\scriptsize 180 = 2x \: + \: 40$$ (Sum of angles in an Isosceles triangle)

$$\scriptsize x = \normalsize \frac{180 \: – \: 40}{2} = \scriptsize 70^o$$

$$\scriptsize \therefore \scriptsize \hat{PRQ} = 70^o$$

$$\scriptsize \scriptsize \hat{PRQ} = \hat{RSQ}$$ (Equal angles of Isosceles triangle)

â‡’ $$\scriptsize \hat{RSQ}$$ = 70Âº

but $$\scriptsize \hat{RSQ} = \hat{SPQ} \: + \: \hat{PQS}$$ (Exterior angle of a triangle)

â‡’ $$\scriptsize 70^o = 40^o \: + \: \hat{PQS}$$

$$\scriptsize \hat{PQS} = 70^o \: – \: 40^o$$

$$\scriptsize \hat{PQS} = 30^o$$

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