In the figure given below, ABCD is a rhombus. Find the value of x and

1.	In the figure given below, ABCD is a rhombus. Find the value of x and y.
Answer» We know that all the sides are equal in a rhombus Consider △ ABD We get AB = AD and ∠ ABD = ∠ ADB It can be written as x = y ……. (1) Consider △ ABC We get AB = BC and ∠ CAB = ∠ ACB We know that ∠ ACB = 40^o By using the sum property of a triangle ∠ B + ∠ CAB + ∠ ACB = 180^o By substituting values in the above equation ∠ B + 40^o + 40^o = 180^o On further calculation ∠ B = 180^o – 40^o – 40^o By subtraction ∠ B = 180^o – 80^o So we get ∠ B = 100^o ∠ DBC can be written as ∠ DBC = ∠ B – x^o By substituting the values ∠ DBC = 100^o – x^o From the figure we know that ∠ DBC and ∠ ADB are alternate angles ∠ DBC = ∠ ADB = y^o By substituting the value of ∠ DBC 100^o – x^o = y^o Consider the equation (1) we know that x = y 100^o – x^o = x^o On further calculation 2x^o = 100^o By division x^o = 50^o Therefore, x = y = 50^o.

In the figure given below, ABCD is a rhombus. Find the value of x and y.

Answer»

We know that all the sides are equal in a rhombus

Consider △ ABD

We get

AB = AD and

∠ ABD = ∠ ADB

It can be written as

x = y ……. (1)

Consider △ ABC

We get

AB = BC and

∠ CAB = ∠ ACB

We know that

∠ ACB = 40^o

By using the sum property of a triangle

∠ B + ∠ CAB + ∠ ACB = 180^o

By substituting values in the above equation

∠ B + 40^o + 40^o = 180^o

On further calculation

∠ B = 180^o – 40^o – 40^o

By subtraction

∠ B = 180^o – 80^o

So we get

∠ B = 100^o

∠ DBC can be written as

∠ DBC = ∠ B – x^o

By substituting the values

∠ DBC = 100^o – x^o

From the figure we know that ∠ DBC and ∠ ADB are alternate angles

∠ DBC = ∠ ADB = y^o

By substituting the value of ∠ DBC

100^o – x^o = y^o

Consider the equation (1) we know that x = y

100^o – x^o = x^o

On further calculation

2x^o = 100^o

By division

x^o = 50^o

Therefore, x = y = 50^o.