Here we will solve some numerical problems on the properties of isosceles triangles.
1. Find x° from the below figures.
Solution:
In ∆XYZ, XY = XZ.
Therefore, ∠XYZ = ∠XZY = x°.
Now, ∠YXZ + ∠XYZ + XZY = 180°
⟹ 84° + x° + x° = 180°
⟹ 2x° = 180° - 84°
⟹ 2x° = 96°
⟹ x° = 48°
2. Find x° from the given figures.
Solution:
LMN, LM = MN.
Therefore, ∠MLN = ∠MNL
Thus, ∠MLN = ∠MNL = 55°, [since ∠MLN = 55°]
Now, ∠MLN + ∠LMN + ∠MNL = 180°
⟹ 55° + x° + 55° = 180°
⟹ x° + 110° = 180°
⟹ x° = 180° - 110°
⟹ x° = 70°
3. Find x° and y° from the given figure.
Solution:
In ∆XYP,
∠YXP = 180° - ∠QXY, as they form a linear pair.
Therefore, ∠YXP = 180° - 130°
⟹ ∠YXP = 50°
Now, XP = YP
⟹ ∠YXP = ∠XYP = 50°.
Therefore, ∠XPY = 180° - (∠YXP + ∠XYP), as the sum of three angles of a triangle is 180°
⟹ ∠XPY = 180° - (50° + 50°)
⟹ ∠XPY = 180° - 100°
⟹ ∠XPY = 80°
Now, x° = ∠XPZ = 180° - ∠XPY (linear pair).
⟹ x° = 180° - 80°
⟹ x° = 100°
Also, in ∆XPZ we have,
XP = ZP
Therefore, ∠PXZ = ∠XZP = z°
Therefore, in ∆XPZ we have,
∠XPZ + ∠PXZ + ∠XZP = 180°
⟹ x° + z° + z° = 180°
⟹ 100° + z° + z° = 180°
⟹ 100° + 2z° = 180°
⟹ 2z° = 180° - 100°
⟹ 2z° = 80°
⟹ z° = \(\frac{80°}{2}\)
⟹ z° = 40°
Therefore, y° = ∠XZR = 180° - ∠XZP
⟹ y° = 180° - 40°
⟹ y° = 140°.
4. In the adjoining figure, it is given that XY = 3y, XZ = 7x, XP = 9x and XQ = 13 + 2y. Find the values of x and y.
Solution:
It is given that XY = XZ
Therefore, 3y = 7x
⟹ 7x - 3y = 0 ............................ (I)
Also, we have XP = XQ
Therefore, 9x = 13 + 2y
⟹ 9x – 2y – 13 = 0 ............................ (II)
Multiplying (I) by (II), we get:
14x - 6y = 0 ............................ (III)
Multiplying (II) by (III), we get:
27x – 6y – 39 = 0 ............................ (IV)
Subtracting (III) from (IV) we get,
13x - 39 = 0
⟹ 13x = 39
⟹ x = \(\frac{39}{13}\)
⟹ x = 3
Substituting x = 3 in (I) we get,
7 × 3 – 3y = 0
⟹ 21 – 3y =0
⟹ 21 = 3y
⟹ 3y = 21
⟹ y = \(\frac{21}{3}\)
⟹ y = 7.
Therefore, x = 3 and y = 7.
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