# Trapezium problem

• Sep 12th 2005, 07:09 AM
totalnewbie
Trapezium problem
Look at the drawing and declare the x and y. The intersection is the point we get if we lenghten the c and d as far as they intersect.
The final task is to calculate the area of the formed triangle.
• Sep 12th 2005, 07:56 AM
hemza
Try
Hi,

Ok, there is an easy way : mathematically the area of the triangle CDE is the same in two figure (figure 1 and figure 2) because you see the two blue areas (figure 3) are the same so it does not change anything to the area of CDE if we replace DE without changing its length.

So I'll work with the second figure : we have two triangles ABC and CDE which have a common angle C and having DE parallel to AB, the angle CDE is the same as CAB so the two triangle have proportional sides. (the relation between the two triangles has a name but I just know it in french. sorry)

So all homologous sides have the same proportion k = m(AB)/m(DE) = m(CB)/m(CE) = m(CA)/m(CD) and we have one of them which is m(AB)/m(DE) = 1.8/1.2 = 1.5 so k=1.5= m(CB)/m(CE) = m(CA)/m(CD).

We replace what we know : k=1.5= 1.5/m(CE) = 1.2/m(CD). So m(CE) = 1 and m(CD) = 1.2/1.5=0.8 So we have all our sides.

Now you can use either the Heron formula or draw a height and calculate A(triangle CDE) = h*b/2

By Heron : the semiperimeter is p = (0.8+1+1.2)/2 = 1.5 and A(triangle CDE) = sqrt(1.5 (1.5-1) (1.5-0.8) (1.5-1.2)) = 0.397 units^2.

Done!
• Sep 12th 2005, 09:42 AM
totalnewbie
CD=2,4 CE=3 or contrary. You have a mistake on your solution.
• Sep 12th 2005, 11:12 AM
hemza
Sorry, I misunderstood your figure. I thought m(AC) = 1.2 when it was m(AD)=1.2.

ok,

"Ok, there is an easy way : mathematically the area of the triangle CDE is the same in two figure (figure 1 and figure 2) because you see the two blue areas (figure 3) are the same so it does not change anything to the area of CDE if we replace DE without changing its length.

So I'll work with the second figure : we have two triangles ABC and CDE which have a common angle C and having DE parallel to AB, the angle CDE is the same as CAB so the two triangle have proportional sides. (the relation between the two triangles has a name but I just know it in french. sorry)

So all homologous sides have the same proportion k = m(AB)/m(DE) = m(CB)/m(CE) = m(CA)/m(CD) and we have one of them which is m(AB)/m(DE) = 1.8/1.2 = 1.5 so k=1.5= m(CB)/m(CE) = m(CA)/m(CD).

We replace what we know : k=1.5= (1.5+m(CE))/m(CE) = (1.2+m(CD))/m(CD). So 1.5m(CE) = 1.5+m(CE) so m(CE) =3 and 1.5m(CD) = 1.2+m(CD) so m(CD) = 2.4. So we have all our sides.

Now you can use either the Heron formula or draw a height and calculate A(triangle CDE) = h*b/2

By Heron : the semiperimeter is p = (1.2 + 3 + 2.4)/2 = 3.3 and A(triangle CDE) = sqrt(3.3 (3.3-1.2) (3.3-2.4) (3.3-3)) = 1.378 units^2.

Done!"