Then, we calculate the area of rectangle ABCD, which is 48. For triangle ABC, the base is 4 and the height is 8, so we multiply 4 and 8, then divide by 2 to get 24. For triangle DCM, the base is 4 and the height is 6, so we multiply 4 and 6, then divide by 2 to get 12. We can subtract the total areas of triangles DCM and ABC from the rectangle ABCD. Since the rectangle's area is 48, 1/4 of 48 would be 12. And it is also true that triangle MAC is half of triangle DAC, so triangle MAC would just be 1/4 of the whole rectangle. So since M is the midpoint of DA, we can see that triangle DAC is half of the whole rectangle. We can find the area of the entire rectangle, DCBA to be and find DCM area to be and BCA to be. This means that a triangle with half of the base of the rectangle and also the same height means its area is one quarter of the rectangle's area. Ī triangle with the same height and base as a rectangle is half of the rectangle's area. Use the triangle area formula for triangles: where is the area, is the base, and is the height. 3 Video Solution (THINKING CREATIVELY!!!).Simplifying this equation we get the equation. With this information, we can write the equation Doing some more angle chasing, we can find that, as they both share and they both have a right angle. Let point be the point on line so that lines and are perpendicular, and we get that and. Through simplifying this equation, we get that. Using this information, we can write the equation. We find that through some angle chasing (they both have a right angle, and they both share angle. We find that (through Pythagorean Theorem),, and. Solution 7 (system of equations through angle similarity)įirst, using given information, we can find the values of some line segments in the figure. Plugging this into our equation for line gives us, so Solving for the intersection by letting, we get. First, we know that its slope is, and we know that it passes through, so line has equation. We can use point-slope form to find the equation for line. Let be the origin of our coordinate system. The y-coordinate of this intersection point is indeed our answer. We solve the system of equations with these three lines. Our goal is to find the y-coordinate of that intersection point. Three lines, line, line, and line, intersect at. Since there are only lines, you can resort to coordinate bashing. Since and we have Thus, Suppose and Thus, we have Additionally, now note that which is pretty obvious from insight, but can be proven by AA with extending to meet From this new pair of similar triangles, we have Therefore, we have by combining those two equations, Solving, we have and therefore Solution 5 Thus,, from which we have, thus Solution 4 and are the altitudes of and, respectively. From the similarity of triangles and, we have the ratio (as, and ). Since is a rectangle, it follows that, therefore. Since can't have 2 different lengths, both expressions for must be equal. Similarly, must equal 3.īecause and are similar, the ratio of and, must also hold true for and. In rectangle, we have, , is on with, is on with, line intersects line at, and is on line with. 2.7 Solution 7 (system of equations through angle similarity).
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