Grades 9-10 Video Solutions 2014-Levels 9&10 Video Solutions 2014 problem29

Levels 9&10 Video Solutions 2014 problem29

Video Transcription

Question number nine. Let ABC be a triangle such that its leg here AB measures 6 centimeters, its leg AC measures 8 centimeters, its leg BC  measures 10 centimeters, but is bisected by the point M, so BM and MC are both 5 centimeters each for a total of 10. Now in the picture A M D E here is a square  and the side MD of the square intersects the leg AC of the triangle at the point F. So together A F D and E those four points make a quadrilateral which is  shaded here and our job is to compute its area. So let's make the following plan. If we know the area of the square we can compute the area of the shaded  region by subtracting from the area of the square the area of this triangle here. And since AM and MD are on the square the angle here I will mark in red  is a right angle so to compute the area of that right triangle we would have to find the lengths of its legs AM and MF. So that's the plan. So let's begin by  computing the length here of the segment AM. I'll just call that alpha and I'll use the law of cosines twice. Once on this angle here and then the second time  on this angle here. So first for the angle A and B we have that 6 squared is equal to alpha squared plus 5 squared minus 2 times alpha times 5 times the  cosine of that angle A and B. So that's 36 is equal to alpha squared plus 25 minus 10 alpha cosine of the angle A and B. And then we note that pi minus the  measure of the angle A and B is exactly equal to the supplementary angle which is the angle A and C. And so the cosine of pi minus the measure of the angle  here A and B is going to be equal to exactly the measure of the cosine rather of the angle A and C and that we can compute in several ways is going to be  the negative of the cosine of the angle A and B. So using the law of cosines here a second time on the second triangle we have now that 8 squared is equal to 5  squared plus alpha squared minus 2 times alpha times 5 cosine of the angle here A and C. And then simplifying we have 64 is equal to alpha squared plus 25 minus  10 alpha negative cosine of the angle A and B. So we can cancel the minuses here and get a plus so we can erase here this minus and then what we will do is add  these equations together. So add these together and what we obtain is that 100 is equal to 2 alpha squared plus 50 so that 25 is equal to alpha squared, 5 is  equal to alpha. So now we know that the length of that segment from A to M is equal to 5, which tells us that the square area is equal to 5 squared. So  that's AM squared or 25 centimeters squared. And from here what we do is the following. I will draw the median of the triangle here AMC so roughly connect  the midpoint of the side AC with the point M and now what I have obtained is the following triangle which I will try to copy here so we don't have too much  clutter. We have the points M over here, we have this new point that I will call N over here, we have the point C over here, and somewhere we have the point F  in between. This measures 5 from N to C, that side measures 4 so MN has to be 3. That's a right triangle then and we know the area here of the triangle AMN. So  triangle AMN has area equal to 1 half its base times its height so 6. And then what we need is the triangle MNF. So for that calculation I will draw here in  orange the line connecting M to F like that and we see that's a right triangle but we don't know the area here of that segment let me call it X from N to F. So  we'll use the Pythagorean theorem twice. We have 3 squared plus X squared and that's going to be equal to MF squared so let me just call that Y here so  that's Y squared. And now looking at the red triangle here in the big picture MF is Y so we have Y squared plus alpha squared which is 5 squared is equal to  its hypotenuse so that would be from AN to NF which is 4 plus X quantity squared.  And when we do that when we compute here and solve we see that substituting that gives us 3 squared plus X squared that's equal to Y squared so plus 5 squared is equal to  4 plus X quantity squared we have to solve that for X and we can do that so we have  9 plus X squared plus 25 is 16 plus 8X plus X squared the X squares cancel we have 34 here is equal to 16 plus 8X subtracting we have 18 is equal to 8X and that finally  gives us that X is equal to 18 divided by 8 or 9 divided by 4.  And then finally with those computations we have the area of the shaded region so let me see if I  can write that over here that would be the area of the square which is 25 so 25 minus the area of the triangle over here which I will shade in purple and that was calculated to be  6 and then minus the area of the other triangle which I will shade in purple its dimensions are  bases 9 fourths as we calculated and then that's NF and MN is the median and that measures  uh 3 so after simplifying here we obtain that this is equal to 125 divided by 8 and after all  this work we have our answer and that is choice B 125 over 8 for the area of the shaded region.

Video Summary

The problem involves finding the area of a shaded quadrilateral within a triangle ABC and a square AMDE. The triangle has sides AB = 6 cm, AC = 8 cm, and BC = 10 cm, bisected at M. The quadrilateral includes points A, F, D, and E. The plan involves calculating the area of the square by finding the side AM using the law of cosines. The area of the shaded region is determined by subtracting the areas of two triangles (AMF and MNF) from the area of the square, resulting in an area of \(125/8\) cm².