Solving for x on a triangle.

Question

Figure image: Triangle has B on top left side, A on bottom corner left side, D on bottom left side, and C on bottom corner right side.

In ABC, suppose AB=15 cm, BC=15 cm, AD=2x-8 cm, and DC=4x - 20 cm. Solve for x.

Answer 1

If I understood the problem right, then the main triangle is Δ AEC. In this triangle vertex A is from the left, vertex C is from the right, vertex E is on top of the Δ AEC. Points D and B are on the left side AE of the main Δ AEC. It is given to us, that

AB = BC = 15 cm

AD = 2x – 8 cm

DC = 4x – 20 cm

Now, when the picture is clear, let’s solve the problem. Consider the Δ DBC. I will use the fact that in any triangle the sum of lengths of any two sides is greater than the length of the third side.

I can write: 

DC + DB > BC                   (1)

DB + BC  > DC                 (2)

DC + BC > DB                  (3)

At first, we need to find the length of DB. Look at the side AE of the main triangle. The length of the segment AB is 15 cm. Knowing that AB = AD + DB and AD = 2x – 8, we get

                                                                15 = 2x – 8 + DB

Hence DB = 23 – 2x cm

Putting values of length of all segments in (1), we get

                                            4x – 20 + 23 – 2x > 15                 or x > 6

Similarly, from (2)

                                             23 – 2x + 15 > 4x – 20                 or x < 9.66

And finally, from (3)             4x – 20 + 15 > 23 – 2x                or x > 4.66

Now it is obvious that                     6 < x < 9.66

                                              Answer: 6 < x < 9.66 or in interval notation: (6, 9.66)