Question

In the process of construction of ${45}^{\circ }$, what is the sum of all the other angles you would be constructing during the course?

• A. ${270}^{\circ }$
• B. ${45}^{\circ }$
• C. ${180}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A

${270}^{\circ }$

Following are steps of construction:

Draw a line. At any point, say O as a centre and any radius, draw an arc which cuts this line in A and B. We now have an angle of ${180}^{\circ }$ (∠AOB). With A and B as centres and any radius more than ½(AB), draw arcs which cut each other in C. Join OC, which would be the angular bisector and now $\angle COB={90}^{\circ }$ If we bisect this angle, we will get ${45}^{\circ }$.

So, the sum of the intermediate angles = ${180}^{\circ }$ + ${90}^{\circ }$ = ${270}^{\circ }$