Question

$△$ABC is constructed such that AB = 6 cm, BC = 4 cm and AC = 4 cm. Taking side AB as base, another triangle ABD is constructed such that the length of side BD is 3 cm and length of side AD is 5 cm, as shown below. Then $\angle$ACB + $\angle$ADB equals

• A. 172${}^{\circ }$
• B. 191${}^{\circ }$
• C. 180${}^{\circ }$
• D. 184${}^{\circ }$

Answer 1

The correct option is B

191${}^{\circ }$

Step 1:
Draw the base line AB of the length 6 cm.

Step 2:
Use your compass to make a circle with centre at point A and the radius 4 cm.

Step 3:
Use your compass to make a circle with centre at point B and the radius 4 cm.

Step 4:
In the point of intersection between the two circles, you'll find the point C.
Complete the triangle by drawing the lines between all three points.

Now, we need to construct another triangle, taking AB as the base.

Step 5:
Use your compass to make a circle with centre at point A and the radius 5 cm.

Step 6:
Use your compass to make a circle with centre at point B and the radius 3 cm.

Step 7:
In the point of intersection between the two circles, you'll find the point D.
Complete the triangle by drawing the lines between all three points.

Now, by above construction, we see that the $\angle$ACB = 97${}^{\circ }$ and $\angle$ADB = 94${}^{\circ }$.

Now, $\angle$ACB + $\angle$ADB = 97${}^{\circ }$ + 94${}^{\circ }$ = 191${}^{\circ }$