Question

question 22 of 27
what is the measure of ∠bdc, given that figure adcb is a rectangle?
image of rectangle adcb with diagonals intersecting, ∠dac (or similar) marked 50°
a. 90°
b. 40°
c. 60°
d. 50°

Explanation:

Step1: Recall rectangle properties

In a rectangle, all angles are \(90^\circ\), and the diagonals are equal and bisect each other. Also, \(AD\parallel BC\) and \(AB\parallel DC\), and \(AD = BC\), \(AB=DC\). In rectangle \(ADCB\), \(\angle ADC = 90^\circ\) (since it's a rectangle, all angles are right angles). We know \(\angle ADB=50^\circ\).

Step2: Calculate \(\angle BDC\)

Since \(\angle ADC=\angle ADB + \angle BDC\) and \(\angle ADC = 90^\circ\), \(\angle ADB = 50^\circ\), we can find \(\angle BDC\) by subtracting \(\angle ADB\) from \(\angle ADC\). So, \(\angle BDC=90^\circ - 50^\circ=40^\circ\).

Answer:

B. 40°