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Given that ABCD is a rhombus, what is the value of x?

finding the internal angles of the rhombus

Solution:

Given rhombus ABCD

We know that the diagonal of a rhombus bisect the corner angles into equal halves

So, ∠B is double of 4x - 25

∠B = 2(4x - 25) = 8x - 50 -------> [a]

∠A is double of x

∠A = 2x ------> [b]

Let us consider BC and AD as 2 parallel lines and AC as the transversal

We have a property, when 2 parallel lines are intersected by a transversal then the opposite angles of the transversal are the same.

So, ∠CAD = ∠ACB, hence ∠C = 2x

Similarly, assume BD is the transversal to the same parallel lines, then

∠BDA = 4x-25

So, ∠D = 8x - 50

We know that total sum of all angles of a polygon is equal to 180°

∠A + ∠B + ∠C + ∠D = 180°

2x + 8x - 50 + 2x + 8x - 50 = 180°

20x -100 = 180°

20x = 180 +100

20x = 280

x = 14°

Summary:

Given that ABCD is a rhombus, the value of x is 14°.

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