Kindly answer this one. Thank you.
1 answer:
Answer:
See below.
Step-by-step explanation:
I'm assuming these questions are about the Midline Theorem (segment AL joins the midpoints of the non-parallel sides.
♦ The midline's length is the average of the lengths of the top and bottom parallel sides.
Use this equation and substitute values given in each problem, then solve for the missing information.
1. AL = x, CE = 9, OR = 5
2. AL = <em>m</em> - 4, CE = 15, OR = 17
3. OR = y + 5, AL = 15, CE = 18
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Check the picture below, so, that'd be the square inscribed in the circle.
so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2} \\\\\\ d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10](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%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%205%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%20-3%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%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%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-8-%28-2%29%5D%5E2%2B%5B-3-5%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-8%2B2%29%5E2%2B%28-3-5%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-6%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B36%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B100%7D%5Cimplies%20d%3D10)
that means the radius r = 5.
now, what's the center? well, the Midpoint of the diagonals, is really the center of the circle, let's check,
so, now we know the center coordinates and the radius, let's plug them in,
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{ k}})\qquad radius=&{{ r}}\\ &-5&1&5 \end{array} \\\\\\\ [x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0Acenter%5C%20%28%26%7B%7B%20h%7D%7D%2C%26%7B%7B%20k%7D%7D%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-5%261%265%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-5%29%5D%5E2-%5By-1%5D%5E2%3D5%5E2%5Cimplies%20%28x%2B5%29%5E2-%28y-1%29%5E2%3D25)
so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.
that means the radius r = 5.
now, what's the center? well, the Midpoint of the diagonals, is really the center of the circle, let's check,
so, now we know the center coordinates and the radius, let's plug them in,
Answer:
Step-by-step explanation:
Given
See attachment
Required
The value of x that makes l and m parallel
Let the angles be ABC (See attachment 2)
is corresponding to 50 degrees. So:
is corresponding to 55 degrees. So:
This means that:
Solve for x
