Question

if abcd is a rectangle and m∠cbd = 47°, what is the value of x? a. 47 b. 133 c. 94 d. 86 e. 43 f. cannot be determined

Explanation:

Step1: Recall rectangle properties

In rectangle \(ABCD\), the diagonals \(AC\) and \(BD\) are equal and bisect each other. \(\triangle BCD\) is a right - triangle with \(\angle BCD = 90^{\circ}\).

Step2: Find \(\angle BDC\)

In \(\triangle BCD\), since \(\angle BCD=90^{\circ}\) and \(\angle CBD = 47^{\circ}\), using the angle - sum property of a triangle (\(\angle CBD+\angle BDC+\angle BCD = 180^{\circ}\)), we have \(\angle BDC=180^{\circ}-\angle BCD - \angle CBD=180^{\circ}-90^{\circ}-47^{\circ}=43^{\circ}\).

Step3: Use vertical angles

The angle \(x\) is a vertical angle to the non - right angle of the isosceles triangle formed by the diagonals. In rectangle, the diagonals are equal and bisect each other. Let the intersection of the diagonals be \(O\). \(\triangle BOC\) and \(\triangle AOD\) are isosceles triangles. The angle \(x\) is the exterior angle of \(\triangle DOC\). Since \(\angle ODC=\angle OCD = 43^{\circ}\) (because \(OC = OD\) in rectangle), then \(x = 180^{\circ}-2\times43^{\circ}=94^{\circ}\) (using the exterior - angle property of a triangle: \(x=\angle ODC+\angle OCD\)).

Answer:

C. 94