In this exercise we are asked to find the lengths of six segments. Let's take a look at the given picture. If we call be the length of KP as x, then the length of PM will be 2x.
Let's notice that △ KML and △ KNM are similar by the Angle-Angle Similarity Theorem as they are both right triangles and they share ∠K. Therefore, corresponding sides of these figures are proportional.
KM/KN=KL/KM
⇓
KM/9=9+ 16/KM
We can solve the above proportion using cross multiplication.
The length of KP is 5. This means that the length of MP is 10. Let's add this information to our picture.
Next, as â–³ MNK is a right triangle, we can use the Pythagorean Theorem to evaluate the length of MN. According to this theorem, the sum of the squared legs of a triangle is equal to its squared hypotenuse.
The length of MN is 12. Using this information, we will evaluate the length of ML. To do this, let's again use the Pythagorean Theorem, as â–³ MNL is also a right triangle.
The length of ML is 20. Finally, we still need to evaluate lengths of MR and PR. Let's look at the picture for the last time.
Since we are given that PR is parallel to KL, angles ∠MRP and ∠MLK are congruent as well as ∠MPR and ∠MKL. This means that △ MPR and △ MKL are similar by Angle-Angle Similarity Theorem.
We can write that corresponding sides are proportional.
PR/KL=MP/MK=MR/ML
Let's solve the above proportion by substituting appropriate side lengths. We will start with the left and the middle ratio.
