Question

PQRS is a trapezium with \[PQ\parallel SR\] and \[\angle P=\angle Q={{50}^{\circ }}\]
(a) Prove that \[PS=QR\]
(b) Find \[\angle S=\angle R=?\]

Answer 1

Hint: We solve this problem by constructing the height from vertex S to PQ and R to PQ as follows

We use the condition that altitudes between two parallel lines are always equal to prove the two triangles we have as congruent to prove that \[PS=QR\]
We use the condition that if two triangles are congruent to each other then all corresponding sides and angles are equal to each other.
Then we use the condition that the sum of all angles in a quadrilateral is equal to \[{{360}^{\circ }}\] so that we can find the required angles.

Complete step by step answer:
We are given that \[PQ\parallel SR\] and \[\angle P=\angle Q={{50}^{\circ }}\]
Now, let us construct perpendicular from S to PQ and R to PQ as shown in the figure.
We know that the condition that altitudes between two parallel lines are always equal
By using the above condition we can say that
\[\Rightarrow SM=RN.......equation(i)\]
We know that \[SM\bot PQ\] and \[RN\bot PQ\] so that the angle at M and N are equal that is
\[\Rightarrow \angle SMP=\angle RNQ......equation(ii)\]
We are given that the angle at P and Q are equal that is
\[\Rightarrow \angle SPM=\angle RQN......equation(iii)\]
Now, let us consider the triangles \[\Delta SPM\] and \[\Delta RQN\] then we have three condition that from equation (i), equation (ii) and equation (ii) as
\[\Rightarrow SM=RN\]
\[\Rightarrow \angle SMP=\angle RNQ\]
\[\Rightarrow \angle SPM=\angle RQN\]
Here we can see that two angles and one side are equal
So, by using the A.A.S congruence we can say that \[\Delta SPM\] and \[\Delta RQN\] are congruent to each other.
We know that if two triangles are congruent then all corresponding sides and angles are equal
By using the above condition we can conclude that
\[\Rightarrow SP=RQ\]
Hence the required result has been proved.
We know that the sum of all angles in a quadrilateral is equal to \[{{360}^{\circ }}\]
By using the above condition to PQRS we get
\[\Rightarrow \angle P+\angle Q+\angle R+\angle S={{360}^{\circ }}.....equation(iv)\]
We are given that
\[\angle P=\angle Q={{50}^{\circ }}\]
We are asked to find the value \[\angle S=\angle R=?\]
Here we can see that the angles \[\angle S,\angle R\] are equal to each other.
By substituting the required values in equation (iv) we get
\[\begin{align}
  & \Rightarrow {{50}^{\circ }}+{{50}^{\circ }}+\angle S+\angle S={{360}^{\circ }} \\
 & \Rightarrow 2\angle S={{360}^{\circ }}-{{100}^{\circ }} \\
 & \Rightarrow \angle S=\dfrac{{{260}^{\circ }}}{2}={{130}^{\circ }} \\
\end{align}\]
Therefore we can conclude that
\[\Rightarrow \angle S=\angle R={{130}^{\circ }}\]

Note: We can find the angles \[\angle S,\angle R\] in other methods also.
We have the standard condition that sum of opposite angles in a quadrilateral is equal to \[{{180}^{\circ }}\]
By using the above condition to \[\angle S,\angle P\] then we get
\[\Rightarrow \angle S+\angle P={{180}^{\circ }}\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow \angle S+{{50}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow \angle S={{130}^{\circ }} \\
\end{align}\]
Similarly by applying the above condition to \[\angle R,\angle Q\] then we get
\[\Rightarrow \angle R+\angle Q={{180}^{\circ }}\]
By substituting the required values in above equation we get
\[\begin{align}
  & \Rightarrow \angle R+{{50}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow \angle R={{130}^{\circ }} \\
\end{align}\]
Therefore we can conclude that
\[\Rightarrow \angle S=\angle R={{130}^{\circ }}\]