H: 5 square yards is the correct answer, break the figure into squares and then multiply length by width to get area
Answer:
DE = 4
Step-by-step explanation:
The bisector divides the triangle into proportional segments, so ...
FK/DF = EK/DE
5/10 = 2/DE . . . . substitute given values
DE = 4 . . . . . multiply by 2DE
Check the picture below, so, those red ones are the midpoints you found.

![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) \stackrel{midpoint}{JK}&({{ -1}}\quad ,&{{ 2}})\quad % (c,d) \stackrel{midpoint}{ML}&({{ 2}}\quad ,&{{ -2}}) \end{array} \\\\\\ d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\implies d_2=\sqrt{[2-(-1)]^2+[-2-2]^2} \\\\\\ d_2=\sqrt{(2+1)^2+(-4)^2}\implies d_2=\sqrt{9+16}\implies d_2=5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%5Cstackrel%7Bmidpoint%7D%7BJK%7D%26%28%7B%7B%20-1%7D%7D%5Cquad%20%2C%26%7B%7B%202%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%5Cstackrel%7Bmidpoint%7D%7BML%7D%26%28%7B%7B%202%7D%7D%5Cquad%20%2C%26%7B%7B%20-2%7D%7D%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%5Cimplies%20d_2%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%5B-2-2%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad_2%3D%5Csqrt%7B%282%2B1%29%5E2%2B%28-4%29%5E2%7D%5Cimplies%20d_2%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20d_2%3D5)
now, that many units are each distance for those "medians", hmmm now, recall that each unit is meant to be 10 meters, so, just multiply each by 10, to get how many meters long each one is.
Answer:
61 bikes are not mountain bikes.
Step-by-step explanation:
To figure this out you have to simply subtract the amount of mountain bikes from the total amount of bikes.
98 - 37 = 61
Now you know that 61 of the bikes are not mountain bikes.
I hope this helped! :)
~<em>CaityConcerto</em>