Lesson A trapezoid divided in four triangles by its diagonals
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<H2>A trapezoid divided in four triangles by its diagonals<H2> <H3>Problem 1</H3>ABCD is a trapezoid in which AD:BC = 3:5. If the area of triangle AMD is 315 cm^2, find the area of the trapezoid. <B>Solution</B> <pre> {{{drawing(400,400,-4,4.14,-4,3, line(-2,2,2,2),line(2,2,3.14,-2.98),line(3.14,-2.98,-3,-3),line(-3,-3,-2,2),line(-2,2,3.14,-2.98),line(2,2,-3,-3), locate(-2.5,2,"A"),locate(-3,-3,"B"),locate(3.14,-2.98,"C"),locate(2.3,2,"D"),locate(0.0316205534-0.5,0.0316205534,"M"), line(-2,2,0.64,2),line(0.64,2,0.4114,1.868),line(0.64,2,0.4114,2.132),line(-3,-3,0.3800292838,-2.9889901326),line(0.3800292838,-2.9889901326,0.0878292838,-3.1589901326),line(0.3800292838,-2.9889901326,0.0867292838,-2.8209901326), locate(-3.5,-3.5,matrix(1,4,"Diagram","not","to","scale")) ) }}} Triangles AMD and BMC are similar (since their three angles are congruent, in pairs). The similarity coefficient is k = 5/3, from the greater triangle to the smaller. Lat a = AM, b = BM, c = CM and D = DM. Then b = {{{(5/3)*d}}}; c = {{{(5/3)*a}}}, due to similarity. For any triangle with the side p and q and the concluded angle {{{theta}}}, the area of this triangle is {{{(1/2)*p*q*sin(theta)}}}. So, for the area of triangle AMD we have {{{area[AMD]}}} = {{{(1/2)*a*d*sin(M)}}} = 315 cm^2. Here M is the angle between sides a = AM and d = DM . Next, {{{area[BMC]}}} = {{{(1/2)*b*c*sin(N)}}} = {{{(1/2)*(5/3)d*(5/3)a*sin(N))}}}, where N is the supplementary angle to M. Since sin(N) = sin(M), we can re-group the formula above and to get {{{area[BMC]}}} = {{{(5/3)^2}}}.{{{(1/2)*a*d*sin(M)}}} = {{{(5/3)^2}}}*area(AMD) = {{{(5/3)^2*315}}} = {{{25/9)*315}}} = 25*35 = 875. This calculation re-tells us a well known fact that the areas of similar triangles do relate as the square of the similarity coefficient. Now let's consider the most interesting things - the areas of triangles adjacent to lateral sides. {{{area[AMB]}}} = {{{(1/2)*a*b*sin(M)}}} = {{{(1/2)*a*(5/3)*d*sin(M)}}} = {{{(5/3)*area[AMD]}}} = {{{(5/3)*315}}} = 5*105 = 525, {{{area[DMC]}}} = {{{(1/2)*d*c*sin(M)}}} = {{{(1/2)*d*(5/3)*a*sin(M)}}} = {{{(5/3)*area[AMD]}}} = {{{(5/3)*315}}} = 5*105 = 525. Thus the total area of the trapezoid ABCD is 315 + 875 + 525 + 525 = 2240 cm^2. </pre> At this point, the problem is solved completely. The lesson to learn from this my solution: if you are given a trapezoid, divided by its diagonals in four triangles, then (a) the triangles, adjacent to the bases, always are similar; (b) the triangles adjacent to lateral sides always have equal areas. If, in addition, you are given the area "a" of a triangle adjacent to a base and the base-to-base ratio "k", then (c) the areas of all triangles of the division can be found easily using the similarity coefficient: - the area of the opposite triangle is {{{k^2*a}}}; - the areas of the triangles adjacent to lateral sides are k*a. 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