Question 11 Mark
Complete the following statements by means of one of those given in brackets against each:
If one pair of opposite sides are equal and parallel, then the figure is _____________. (parallelogram, rectangle, trapezium).
If one pair of opposite sides are equal and parallel, then the figure is _____________. (parallelogram, rectangle, trapezium).
Answer
View full question & answer→If one pair of opposite sides are equal and parallel, then the figure is parallelogram.Explanation:
In $\triangle\text{ABC}$ and $\triangle\text{CDA},$ AB = DC (Given) AC = AC (Common)$\angle\text{BAC}=\angle\text{DCA}$ (Because AB || CD, Alternate interior angle are equal)
So, by SAS Congruence rule, we have $\triangle\text{ABC}\cong\triangle\text{CDA}$ Also,$\angle\text{BCA}=\angle\text{DAC}$ (Corresponding parts of congruent triangles are equal)
But, these are alternate interior angles, which are equal. AD || BC Thus, AB || CD and AD || BC. Hence, quadrilateral ABCD is parallelogram.
In $\triangle\text{ABC}$ and $\triangle\text{CDA},$ AB = DC (Given) AC = AC (Common)$\angle\text{BAC}=\angle\text{DCA}$ (Because AB || CD, Alternate interior angle are equal)So, by SAS Congruence rule, we have $\triangle\text{ABC}\cong\triangle\text{CDA}$ Also,$\angle\text{BCA}=\angle\text{DAC}$ (Corresponding parts of congruent triangles are equal)
But, these are alternate interior angles, which are equal. AD || BC Thus, AB || CD and AD || BC. Hence, quadrilateral ABCD is parallelogram.
ABCD is a quadrilateral in which AB = CD and BC = DA. We need to show that ABCD is a parallelogram. In $\triangle\text{ACB}$ and $\triangle\text{CAD},$ we have AC = CA (Common) CB = AD (Given) AB = CD (Given) So, by SSS criterion of congruence, we have$\triangle\text{ACB}\cong\triangle\text{CAD}$
We have, $\angle\text{A}=90^\circ$ In a parallelogram, opposite angles are equal. Therefore,$\angle\text{C}=90^\circ$
ABCD is a quadrilateral in which $\angle\text{A}=\angle\text{C}$ and $\angle\text{B}=\angle\text{D}.$ We need to show that ABCD is a parallelogram. In quadrilateral ABCD, we have$\angle\text{A}=\angle\text{C}$
