Question 1208996
.
ABCD is a trapezoid in which AD: BC=3:5. If the area of
triangle AMD is 315 cm2, find the area of the trapezoid, in cm2.
https://ibb.co/HTrxtP9
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I will give another solution, to get you totally different view on this problem.



<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:


&nbsp;&nbsp;&nbsp;&nbsp;if you are given a trapezoid, &nbsp;divided by its diagonals in four triangles, &nbsp;then

         &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(a) &nbsp;&nbsp;the triangles, adjacent to the bases, always are similar;

         &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(b) &nbsp;&nbsp;the triangles adjacent to lateral sides always have equal areas.


&nbsp;&nbsp;&nbsp;&nbsp;If, in addition, you are given the area "a" of a triangle adjacent to a base and the base-to-base ratio  "k", &nbsp;then

         &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(c) &nbsp;&nbsp;the areas of all triangles of the division can be easily found using the similarity coefficient:


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;- the area of the opposite triangle is &nbsp;{{{k^2*a}}};

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;- the areas of the triangles adjacent to lateral sides are &nbsp;k*a;


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;- the whole area of the trapezoid, &nbsp;in terms of &nbsp;" a " &nbsp;and &nbsp;" k " &nbsp;is &nbsp;&nbsp;{{{a + 2ka + k^2*a}}} = {{{(1+k)^2*a}}}.