Question

Angles of a quadrilateral are (4x)⁰, 5(x+2)⁰, (7x-20)⁰ and 6(x+3)⁰. Find each angle of the quadrilateral.

Answer 1

Hint: A quadrilateral, you know is a shape with 4 sides and 4 angles. The interior angles add up to 360⁰. For any quadrilateral, we can draw a diagonal line to divide it into two triangles. Each triangle has an angle sum of 180 degrees. Therefore, the total angle sum comes out to be 360 degrees. Students only need to apply this point in solving the above question. We solve for x and then place it in each angle value to obtain our answer.

Complete step by step solution:
Step 1: As we know the sum of the interior angles of a quadrilateral is equal to 360⁰. So, when we add up the given angles, they must be equal to 360⁰.
Step 2: So we get an equation as:
$$$$$\begin{array}{rl}& 4x{}^{\circ }+5(x+2){}^{\circ }+(7x-20){}^{\circ }+6(x+3){}^{\circ }=360{}^{\circ }\\ & \Rightarrow 4x+5x+10+7x-20+6x+18=360\\ & \Rightarrow 22x+8=360\\ & \Rightarrow 22x=352\\ & \Rightarrow x=16\\ & \therefore x=16{}^{\circ }\end{array}$
Step 3: Now, placing the value of x in the given angles, we get the value of each angle.
$\begin{array}{rl}& 4x{}^{\circ }=4\times 16{}^{\circ }=64{}^{\circ }\\ & 5(x+2){}^{\circ }=5(16+2){}^{\circ }=90{}^{\circ }\\ & (7x-20){}^{\circ }=(7\times 16-20){}^{\circ }=92{}^{\circ }\\ & 6(x+3){}^{\circ }=6(16+3){}^{\circ }=114{}^{\circ }\end{array}$
Thus, the four angles of the quadrilateral are 64⁰, 90⁰, 92⁰ and 114⁰.

Note: Students can verify their answer by adding up the angles and if the outcome is 360 degrees, then their solution is correct. So, 64⁰+90⁰+92⁰+112⁰=360⁰.
Thus, the angles are properly calculated. These kinds of questions are easy to solve and are seen frequently in different exams. Students should know that the sum of the angles of a quadrilateral is 360⁰. You can also look up the sum of angles of other geometrical shapes too; such as for a triangle, it is 180⁰, for pentagon, it is 540⁰, for hexagon it is 720⁰. This knowledge will always come in handy for all the students.