Question

9. If MN, NO, and MO are midsegments, find the perimeter of AJKL.K13x - 1049NI 6x-1 Zo42

Answer 1

Due MN, NO and MO are midsegments, the four triangles in the picture are congruents, so:

And also, we know that the triangle MNO is similar to JKL, and the sides of JKL are the double of the respective sides of MNO, so we can found the value of x with the following equation:

`\begin{gathered} KL=2\cdot MO \\ 13x-10=2\cdot(6x-1) \\ 13x-10=12x-2 \\ 13x-12x=10-2 \\ x=8 \end{gathered}`

And the Perimeter of KLJ is the double of the perimeter of MNO, so:

`\begin{gathered} \text{Per}_{\text{KLJ}}=2\cdot\text{Per}_{\text{MNO}} \\ \text{Per}_{\text{KLJ}}=2\cdot(49+42+(6\cdot8-1),\text{ Note we replace x=8} \\ \text{Per}_{\text{KLJ}}=2\cdot(91+48-1_{}) \\ \text{Per}_{\text{KLJ}}=276 \end{gathered}`

The perimeter of triangle KLJ is 276.