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In a trapezoid $ABCD$ with $AB$ parallel to $CD$, the diagonals $AC$ and $BD$ intersect at $E$. If the area of triangle $ABE$ is 50 square units, and the area of triangle$ADE$ is 20 square units, what is the area of trapezoid $ABCD$?

Guest Nov 19, 2018
#1
+20680
+8

In a trapezoid $ABCD$ with $AB$ parallel to $CD$, the diagonals $AC$ and $BD$ intersect at $E$.

If the area of triangle $ABE$ is 50 square units, and the area of triangle$ADE$ is 20 square units,

what is the area of trapezoid $ABCD$?

$$\text{Let Area A_1=[ADE] = 20 }\\ \text{Let Area A_2=[ABE] = 50 }\\ \text{Let Area A_3=[DEC] }\\ \text{Let Area A_4=[BEC] }\\ \text{Let Area A_5=[ADC] }\\ \text{Let Area A_6=[BDC] }$$

$$\begin{array}{|rcll|} \hline \dfrac{A_2}{A_1} &=& \dfrac{50}{20} \\ &=& \dfrac52 \\\\ \hline A_1 &=& \dfrac{DE\cdot AH}{2} \\\\ A_2 &=& \dfrac{BE\cdot AH}{2} \\\\ A_3 &=& \dfrac{DE\cdot CK}{2} \\\\ A_4 &=& \dfrac{BE\cdot CK}{2} \\ \hline \\ \dfrac{A_2}{A_1} &=& \dfrac{\dfrac{BE\cdot AH}{2}}{\dfrac{DE\cdot AH}{2}} \\\\ &=& \dfrac{BE\cdot AH}{DE\cdot AH} \\\\ \dfrac{A_2}{A_1} &=& \dfrac{BE}{DE} \quad & | \quad \dfrac{A_2}{A_1} = \dfrac52 \\\\ \mathbf{\dfrac52} & \mathbf{=}& \mathbf{\dfrac{BE}{DE}} \\\\ \hline \dfrac{A_4}{A_3} &=& \dfrac{\dfrac{BE\cdot CK}{2}}{\dfrac{DE\cdot CK}{2}} \\\\ &=& \dfrac{BE\cdot CK}{DE\cdot CK} \\\\ \dfrac{A_4}{A_3} &=& \dfrac{BE}{DE} \quad & | \quad \dfrac{BE}{DE} = \dfrac52 \\\\ \mathbf{\dfrac{A_4}{A_3}} &\mathbf{=}& \mathbf{\dfrac52} \qquad \text{or} \qquad \mathbf{A_3 = \dfrac{2}{5}\cdot A_4 } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline AM &=& BN \\ \hline \\ A_5 &=& \dfrac{DC\cdot AM}{2} \\\\ A_6 &=& \dfrac{DC\cdot BN}{2} \\ \hline \\ \dfrac{A_5}{A_6} &=& \dfrac{\dfrac{DC\cdot AM}{2}}{\dfrac{DC\cdot BN}{2}} \\\\ &=& \dfrac{DC\cdot AM}{DC\cdot BN} \quad & | \quad AM=BN \\\\ &=& \dfrac{DC\cdot BN}{DC\cdot BN} \\\\ \dfrac{A_5}{A_6} &=& 1 \\\\ \mathbf{ A_5 } & \mathbf{=} & \mathbf{A_6} \quad & | \quad A_5 = A_1+A_3,\quad A_6 =A_4+A_3 \\ A_1+A_3 &=& A_4+A_3 \quad & | \ -A_3\\ A_1 &=& A_4 \quad & | \quad A_1 = 20\\ \mathbf{20} & \mathbf{=}& \mathbf{ A_4 } \\ \hline \\ \mathbf{A_3} & \mathbf{=} & \mathbf{ \dfrac{2}{5}\cdot A_4 } \quad | \quad A_4 = 20 \\ A_3 &=& \dfrac{2}{5}\cdot 20 \\ \mathbf{A_3} & \mathbf{=} & \mathbf{ 8 } \\ \hline \end{array}$$

$$\text{The area of trapezoid ABCD = A_1+A_2+A_3+A_4 = 20+50+8+20=\mathbf{98} }$$

heureka  Nov 20, 2018
#2
+92808
+2

Very nice, heureka  !!!!

CPhill  Nov 20, 2018
#3
+20680
+9

Thank you, CPhill

heureka  Nov 21, 2018