Question

$△$ABC is constructed such that AB = 5 cm, AC = 5 cm and $\angle$B = 50${}^{\circ }$.

Statement A: $\angle$C = $\angle$A

Statement B: $\angle$C + $\angle$B = 100${}^{\circ }$

• A. Statement A is true and Statement B is false.
• B. Statement A is false and Statement B is true.
• C. Both Statement A and Statement B are true.
• D. Both Statement A and Statement B are false.

Answer 1

The correct option is B

Statement A is false and Statement B is true.

Step 1:
Draw the base line AB of the length 5 cm.

Step 2:
Measure the angle B with your protractor. Make a mark at 50 degrees and draw a line from B passing through the mark.

Now we must locate C. It should be 5 cm from A and also should be on the upper line, in order to form the required triangle. All points at a distance of 5 cm from A are on the circle centered at B of radius 5 cm. Thus, we have the next step.

Step 3:
Draw a circle with centre at A and radius 5 cm. The point at which this circle cuts the line through B is the required point C.

Thus $△$ABC is formed.

From the above construction, we see that the measure of angle A is 80${}^{\circ }$ and the measure of angle C is 50${}^{\circ }$.

Thus we have, $\angle$C $\ne$ $\angle$A and hence statement A is false.

Also, $\angle$C + $\angle$B = 50${}^{\circ }$ + 50${}^{\circ }$ = 100${}^{\circ }$. Hence statement B is true.