Question

In the adjoining figure, $ABCD$ is a quadrilateral in which $AB$ is the longest side and $CD$ is the shortest side, then

• A. $\angle C>\angle A$ and $\angle B>\angle D$
• B. $\angle C<\angle A$ and $\angle D<\angle B$
• C. $\angle C<\angle A$ and $\angle D=\angle B$
• D. $\angle C>\angle A$ and $\angle D>\angle B$

Answer 1

The correct option is B $\angle C<\angle A$ and $\angle D<\angle B$

Given:

In quadrilateral $ABCD,$ $AB$ smallest & $CD$ is longest sides.

Construction: Join $AC$

.

Mark the angles as shown in the figure..

In $△ABC$ , $AB$ is the shortest side.

$BC>AB$

$\angle 2>\angle 4-\left(i\right)$

[Angle opposite to longer side is greater]

In $△ADC$ , $CD$ is the longest side

$CD>AD$

$\angle 1>\angle 3-\left(ii\right)$

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

$\angle 2+\angle 1>\angle 4+\angle 3$

$⟹\angle A>\angle C$

Similarly, by joining $BD$, we can prove that

$\angle B>\angle D$