Question

Help please! 10 pts!
if mCED=62 what is x?

Answer 1

Answer: x = 59 degrees

Explanation:

The given quadrilateral ABCD is a rectangle. In a triangle, the opposite sides are equal. Also, the diagonals are equal and bisect each other at the midpoint.

If CE and DE are equal, it means that angle ADC and angle BCD are equal.

The sum of the angles in a triangle is 180 degrees. Therefore

ADC + BCD + 62 = 180

ADC + BCD = 180 - 62 = 118

ADC = BCD = 118/2 = 59 degrees

Angle x and angle ADC are alternate angles. Since alternate angles are equal, then

x = 59 degrees

Answer 2

The value of
`x^(0)` is
`59^(0)`.

Explanation:

The figure provided to is a rectangle, named ABDC.

All the angles, m∠CAB = m∠ABD= m∠BDC = m∠DCA =
`90^(0)`.

The lines AD and BC are diagonals of the rectangle ABDC.

According to the diagonal property of rectangles, they bisect each other.

Then,

• m∠CED = m∠AEB and m∠BED = m∠AEC
• AE = BE, CE = ED
• The opposite angle at the points where the diagonals meet are congruent, i.e. m∠DAB = m∠ADC and m∠DCB = m∠ABC.

Now, consider the triangle CED.

Since the triangle CED has two equal sides, i.e. CE = ED, it is an isosceles triangle. And hence the angles m∠DCE = m∠EDC =
`a^(0)` (say).

Compute the value of m∠DCE and m∠EDC using the sum of angles property of a triangle i.e. the sum of all three angles of a triangle is
`180^(0)`.

Solve for
`a^(0)` as follows:

m∠CED + m∠ECD + m∠EDC =
`180^(0)`

`62^(0)` +
`a^(0)` +
`a^(0)` =
`180^(0)`

`62^(0)+2a^(0)=180^(0)\\2a^(0)=180^(0)-62^(0) \\a^(0)=(118^(0) )/(2)\\=59^(0)`

So, m∠EDC =
`59^(0)` = m∠ADC.

As the opposite angles at the points where the diagonals meet are congruent, then,

m∠DAB = m∠ADC =
`59^(0)`.

Thus, the value of
`x^(0)` is
`59^(0)`.