Question

If $ABCD$ is an isosceles trapezium, $\angle C$ is equal to
A.$\angle B$
B.$\angle A$
C.$\angle D$
D.${{90}^{\circ }}$

Answer 1

Hint: We recall the shape of trapezium and isosceles trapezium. We recall that the base angle of isosceles trapezium are always of equal measure. We find which pair of angles in ABCD trapezium are base angles and choose the correct option.

Complete answer:
We know that trapezium is a convex quadrilateral with at least one pair of sides that is parallel to each other. If the quadrilateral AB and CD are parallel to each other then we write the name of the quadrilateral is ABCD.
The sides which are not parallel to each other are called legs of the trapezium. The pair of adjacent angles at the two consecutive vertices of parallel sides are called base angle.
We know that isosceles trapezium is a trapezium where the legs are of equal length and base angles are of equal measure.
We are given the question that ABCD is an isosceles trapezium and are asked to which angle the measure of $\angle C$ is equal to. Since ABCD is trapezium by conventional naming the parallel sides are AB and CD which means
\[AB||CD\]
The sides that are not parallel are AD and BC and hence they will be equal length that is
\[AD=BC\]
 So the adjacent angles on the parallel side AB are $\angle A,\angle B$ and the adjacent angles on the parallel side CD are $\angle C,\angle D$. So these are base angle and hence they are of equal measure which means’
\[\begin{align}
  & \angle A=\angle B \\
 & \angle C=\angle D \\
\end{align}\]

So the correct option is C.

Note:
We note that diagonals of trapezium are also of equal length which means $AC=BD$. We should always follow conventional notation of a trapezium to find the parallel sides. We should also note that the opposite angles are supplementary which means in trapezium ABCD, we have $\angle A+\angle C=\angle B+\angle D={{180}^{\circ }}$.