The Question has Parts A and B and i cant put it all up here.

Question

Triangle JKL has the dimensions shown.

Part A. What is the measure of KM rounded to the nearest hundredth of a centimeter?

Part B. What is the perimeter of triangle JKL rounded to the nearest hundredth of a centimeter?

Answer 1

Part (A) The drawing shows a right triangle KML where you know two of the three sides, so can use Pythagorean Theorem: a² + b² = h². In this case KM² + 35² = 37² → KM² + 1225 = 1369 → KM² = 1369 - 1225 = 144

So KM = √144 = 12 cm. (Why do they need it rounded to nearest hundredth?)

Part (B) Now we have right triangle KMJ with sides 12 and 12 opposite the hypotenuse.

12² + 12² = JK² → 144 + 144 = 288 = JK²

So JK = √288 = 16.97056..., rounds to 16.97 cm.

Now the perimeter of ΔJKL is 12 + 35 + 37 + 16.97 = 100.97 cm.

Answer 2

KM is the mean proportioanl between the two segments it creates

12 / KM = KM / 35

Use KM and Pythagoras to determin JK