Question

In the above figure both RISK and CLUE are parallelograms. Find the value of \[x\].

Answer 1

Hint: First we will take the parallelogram RISK from the above diagram. Then we will use the property that adjacent angles are supplementary in a parallelogram, so the sum of angles \[\angle {\text{RKS}}\] and \[\angle {\text{ISK}}\] is equal \[180^\circ \] to find the value of\[\angle {\text{ISK}}\]. Then we will take the parallelogram CLUE from the above diagram and use the property that the opposite angles are equal in a parallelogram to find the value of \[\angle {\text{UEC}}\]. Then we will consider the triangle EOS, we will use the property that the sum of angles of a triangle is equal \[180^\circ \] to find the value of \[x\].

Complete step by step solution: We are given that RISK and CLUE are parallelograms in the diagram.

Let us assume that the point at which the line segment EC and SI intersect is O.

First, we will take the parallelogram RISK from the above diagram.

We know that the adjacent angles are supplementary in a parallelogram, so the sum of angles \[\angle {\text{RKS}}\] and \[\angle {\text{ISK}}\] is equal to \[180^\circ \].

Then, we have
\[ \Rightarrow \angle {\text{RKS}} + \angle {\text{ISK = 180}}^\circ \]

Substituting the value of the angle \[\angle {\text{RKS}}\] from the diagram in the above equation, we get

\[ \Rightarrow 120^\circ + \angle {\text{ISK = 180}}^\circ \]

Subtracting the above equation by \[120^\circ \] on both sides, we get

\[
   \Rightarrow 120^\circ + \angle {\text{ISK}} - 120^\circ {\text{ = 180}}^\circ - 120^\circ \\
   \Rightarrow \angle {\text{ISK = 60}}^\circ \\
 \]

We also know that the opposite angles in a parallelogram, so we have that \[\angle {\text{ISK = }}\angle {\text{IRK}}\] and \[\angle {\text{EKR = }}\angle {\text{SIR}}\].

Now, we will take the parallelogram CLUE from the above diagram.

We also know that the opposite angles in a parallelogram, so we have that \[\angle {\text{ULC = }}\angle {\text{UEC}}\].
Then, we have \[\angle {\text{UEC}} = 70^\circ \].

Consider the triangle EOS from the above diagram, we will use the property that the sum of angles of a triangle is equal to \[180^\circ \], we have

\[ \Rightarrow \angle {\text{UEC + }}\angle {\text{ISK + }}\angle {\text{SOE = 180}}^\circ \]

Replacing \[70^\circ \] for \[\angle {\text{UEC}}\], \[60^\circ \] for \[\angle {\text{ISK}}\] and \[x\] for \[\angle {\text{SOE}}\] in the above equation, we get
\[
   \Rightarrow 70^\circ + 60^\circ + x = 180^\circ \\
   \Rightarrow 130^\circ + x = 180^\circ \\
 \]

Subtracting the above equation by \[130^\circ \] on both sides, we get

\[
   \Rightarrow 130^\circ + x - 130^\circ = 180^\circ - 130^\circ \\
   \Rightarrow x = 50^\circ \\
 \]

Note: In solving these types of questions, you should remember that the sum of angles of a parallelogram. And there are four sides in a parallelogram out of which opposite sides are equal and adjacent sides are different. Also, students should always remember that when all the angles are less than 90 degrees, than it is an acute angle and if one of the angles is 90 degrees then the triangle is right-angled. The possibility of error in this question can be that you assume the sum is equal to 90 degrees, which is wrong. Also, avoid calculation mistakes.