Question

A right triangle ABC right angled at A with lengths of perpendicular sides 4 cm and 3 cm is constructed. Keeping the third side as the base, another triangle BCD is constructed such that $\angle$DBC = 40${}^{\circ }$ and $\angle$BCD = 30${}^{\circ }$. Then the perimeter of triangle BCD is

• A. 8.4 cm
• B. 7.7 cm
• C. 11.1 cm
• D. 6.1 cm

Answer 1

The correct option is C

11.1 cm

Step 1:

Draw a pair of perpendicular lines and name the point where they meet as A.

Step 2:

Mark B on one of the lines, 3 cm from A and mark C on the other line, 4 cm from A.

Step 3:

Join the points B and C to get the required right angles triangle.

From the above construction, we see that the length of side BC is 5 cm.

Keeping side BC as the base, triangle BCD is constructed, as follows:

Step 4:
Measure the $\angle$BCD with your protractor. Make a mark at 30 degrees and draw a line from C passing through the mark.

Step 5:
Measure the $\angle$CBD with your protractor. Make a mark at 40 degrees and draw a line from B passing through the mark.

Step 6:
Name the point of intersection of these two lines as D and $△$BCD is complete.

Now, from the above construction, we see that the length of the side CD is 3.4 cm and the length of the side BD is 2.7 cm.

Now we shall find the perimeter of triangle BCD.

Perimeter of triangle BCD = BC + CD + BD = 5 cm + 3.4 cm + 2.7 cm = 11.1 cm