Question 1208996
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Answer: <font color=red>2240 square cm</font>


Explanation


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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"))
)
}}}


AD is parallel to BC due to the arrow markers.


These parallel segments allow us to determine these two facts:
angle ADB = angle DBC
angle DAC = angle BCA
The reason for each is the Alternate Interior Angle Theorem.


We can then use the Angle Angle Theorem to prove triangle AMD is similar to triangle CMB.


Segments AD and BC are in ratio 3:5
That scales up to 3x:5x
AD = 3x
BC = 5x
where x is some positive real number.


Rephrased another way: AD = (3/5)*BC = 0.6BC


Let h be the height of triangle CMB
(3/5)h = 0.6h is the height of triangle AMD because the triangles are similar.


area of triangle = 0.5*base*height
areaAMD = 0.5*AD*0.6h
areaAMD = 0.5*3x*0.6h
areaAMD = 0.9x*h
0.9x*h = 315
x*h = 315/0.9
<font color=blue>x*h = 350</font>
Let's call this <font color=blue>equation (1)</font> to use later.


Since h and 0.6h are the heights of triangles CMB and AMD respectively, this means the height of the trapezoid is h+0.6h = 1.6h


area of trapezoid = 0.5*height*(base1+base2)
areaABCD = 0.5*1.6h*(AD+BC)
areaABCD = 0.5*1.6h*(3x+5x)
areaABCD = 0.5*1.6h*(8x)
areaABCD = 0.5*1.6*8*(<font color=blue>x*h</font>)
areaABCD = 0.5*1.6*8*(<font color=blue>350</font>) ........ applying <font color=blue>equation (1)</font>
areaABCD = <font color=red>2240 square cm</font>


I used <a href="https://www.geogebra.org/calculator">GeoGebra</a> to confirm the answer is correct.


A similar question is found <a href="https://www.algebra.com/algebra/homework/Surface-area/Surface-area.faq.question.1148964.html">here</a>
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