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 an

Important details about isosceles triangle ABC:

• The median CD, which is drawn to the base AB, also acts as an altitude to that base in the isosceles triangle (CD⊥AB). This indicates that triangles ACD and BCD are congruent right triangles, each with hypotenuses AC and BC.
• In isosceles triangle ABC, the sides AB and BC are equal, meaning AC=BC.
• The base angles at AB are equal, m∠A=m∠B=30°.

1. Consider the right triangle ACD. The angle adjacent to side AD is 30°, which dictates that the hypotenuse AC is double the length of the opposite side CD relating to angle A.

AC=2CD.

2. Now, for right triangle BCD, the angle next to side BD is also 30°, so hypotenuse BC is twice the opposite leg CD linked to angle B.

BC=2CD.

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

4. If the total of the perimeters of triangles ACD and BCD is 20 cm greater than the perimeter of triangle ABC, then

5. Given that AC=BC=2CD, the lengths of legs AC and BC of the isosceles triangles are 20 cm.

Answer: 20 cm.

We know that AC = BC.

Let's denote AC = BC = x.

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

The combined perimeter of triangles Δ ACD and Δ BCD = [AC + (AB/2) + CD] + [BC + (AB/2) + CD]

                                                                          = 2 AC + AB + 2 CD

                                                                          = 2x + AB + 2 CD

According to the given information, this total exceeds the perimeter of Δ ABC by 20 cm.

Thus, 2x + AB + 2 CD = 2x + AB + 20.

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

2 CD = 20.

Therefore, CD = 10 cm.

Thus, the length of the leg of Δ ABC equals 10 cm.