The ratio of the lengths of the sides of a triangle abc is 3: 4: 6. m, n, and k are the
Question:
Answers
Given that in a trianlgle the sides AB, BC, CA are in the ratio 3:4:6.
Let AB = 3k, BC = 4k and CA = 6k.
Then perimeter =3k+4k+6k = 13k
M, N, K are mid points of the sides.
By mid point theorem MN = 3k/2, NK = 4k/2 and KM = 6k/2
Hence perimeter of MNK = 13k/2 =5.2 (given)
Solve for k
k=2(5.2)/13 = 0.8
Hence sides are
AB = 3k = 3(0.8) = 2.4 in
BC = 4k= 3.2 in
CA = 4.8 in
Each midsegment is half the length of its corresponding parallel segment, so the perimeter of ∆MNK is half the perimeter of ∆ABC. Then the perimeter of ∆ABC is ...
... 2 × 5.2 in = 10.4 in
The side length ratios total 3+4+6 = 13 "ratio units" (corresponding to the total of side lengths, the perimeter), so the sides are 3/13, 4/13, and 6/13 of the perimeter, respectively. That is, the side lengths are ...
... 3/13 × 10.4 in = 2.4 in
... 4/13 × 10.4 in = 3.2 in
... 6/13 × 10.4 in = 4.8 in
(1/x + 1)^2 + 6 (1/x + 1) +5
6.5
step-by-step explanation:
you said it was in the question
~courtney