Question

In isosceles △ABC (AC = BC) with base angle 30° CD is a median. How long is the leg of △ABC, if sum of the perimeters of △ACD and △BCD is 20 cm more than the perimeter of △ABC?

Answer 1

Note necessary facts about isosceles triangle ABC:

• The median CD drawn to the base AB is also an altitude to tha base in isosceles triangle (CD⊥AB). This gives you that triangles ACD and BCD are congruent right triangles with hypotenuses AC and BC, respectively.
• The legs AB and BC of isosceles triangle ABC are congruent, AC=BC.
• Angles at the base AB are congruent, m∠A=m∠B=30°.

1. Consider right triangle ACD. The adjacent angle to the leg AD is 30°, so the hypotenuse AC is twice the opposite leg CD to the angle A.

AC=2CD.

2. Consider right triangle BCD. The adjacent angle to the leg BD is 30°, so the hypotenuse BC is twice the opposite leg CD to the angle B.

BC=2CD.

3. Find the perimeters of triangles ACD, BCD and ABC:

$P_{ACD}=AC+CD+AD=2CD+CD+AD=3CD+AD;$

$P_{BCD}=BC+CD+BD=2CD+CD+AD=3CD+AD;$

$P_{ABC}=AC+BC+AB=2CD+2CD+AD+BD=4CD+2AD.$

4.  If sum of the perimeters of △ACD and △BCD is 20 cm more than the perimeter of △ABC, then

$P_{ACD}+P_{BCD}=P_{ABC}+20,\\ \\3CD+AD+3CD+AD=4CD+2AD+20,\\ \\6CD+2AD=4CD+2AD+20,\\ \\2CD=20.$

5. Since AC=BC=2CD, then the legs AC and BC of isosceles triangles have length 20 cm.

Answer: 20 cm.

Answer 2

We have been given that AC = BC.

Let AC = BC = x.

Now, perimeter of Δ ABC is AC + BC + AB = 2x + AB.

Sum of the perimeters of Δ ACD and Δ BCD = [AC + (AB/2) + CD] + [BC + (AB/2) + CD]

                                                                          = 2 AC + AB + 2 CD

                                                                          = 2x + AB + 2 CD

But, it is given that the above sum is 20 cm more than the perimeter of Δ ABC.

Therefore, 2x + AB + 2 CD = 2x + AB + 20

2x + AB + 2 CD - 2x - AB = 20

2 CD = 20

CD = 10 cm

So, leg of Δ ABC = 10 cm.