Let ABCD be a convex quadrilateral, and let P, Q, R, S, T, U, V, and W be the trisection points of the sides of ABCD, as shown. If the area of quadrilateral ABCD is 180, then find the area of hexagon AQRCUV.
+26388 Let ABCD be a convex quadrilateral,
and let P, Q, R, S, T, U, V, and W be the trisection points of the sides of ABCD, as shown.
If the area of quadrilateral ABCD is 180, then find the area of hexagon AQRCUV.
Let AB = a
Let BC = b
Let CD = c
Let DA = d
Let \(\angle{ABC}\) = B
Let \(\angle{UDV}\) = D
\(\begin{array}{|rcll|} \hline Area_{\text{ABC}} &=& \frac{ab\sin(B)}{2} \\ Area_{\text{CDA}} &=& \frac{cd\sin(D)}{2} \\ Area_{\text{ABCD}} &=& Area_{\text{ABC}} + Area_{\text{CDA}} \\ &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline Area_{\text{QBR}} &=& \frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} \\ Area_{\text{UDV}} &=& \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} - Area_{\text{QBR}} - Area_{\text{UDV}} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\left(1-\frac19 \right) + \frac{cd\sin(D)}{2}\left(1-\frac19 \right) \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\cdot \frac89 + \frac{cd\sin(D)}{2}\cdot \frac89 \\ Area_{\text{hexagon}} &=& \frac89 \cdot \left( \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \right) \\ Area_{\text{hexagon}} &=& \frac89 \cdot Area_{\text{ABCD}} \\ Area_{\text{hexagon}} &=& \frac89 \cdot 180 \\ \mathbf{ Area_{\text{hexagon}} } & \mathbf{=} & \mathbf{160} \\ \hline \end{array}\)
The area of hexagon AQRCUV is 160
+129840 Draw AC
Now, because BQ =(1/3) BA and BR =(1/3)BC, Q and R divide BA and BC proportionally...
Thus.....BQ/BA =BQ/BC...and by Euclid, QR is parallel to AC
And <BQR = < BAC and <QBR = <ABC
So, by AA congruency ΔQBR ~ ΔABC
And since AB=3QB.....then the area of ΔABC will be 3^2 = 9 times that of ΔQBR
By similar reasoning, we can show that the area of ΔADC = 9 times that of ΔVDU
And area of ΔABC + area of ΔADC = 180....so....
9 * area of ΔQBR + 9 * area of ΔVDU = 180
9 * [ area of ΔQBR + area of ΔVDU] = 180 ....so...
[ area of ΔQBR + area of ΔVDU] = 180/9 = 20
So......the area of hexagon AQRCUV = 180 - 20 = 160
+26388 Let ABCD be a convex quadrilateral,
and let P, Q, R, S, T, U, V, and W be the trisection points of the sides of ABCD, as shown.
If the area of quadrilateral ABCD is 180, then find the area of hexagon AQRCUV.
Let AB = a
Let BC = b
Let CD = c
Let DA = d
Let \(\angle{ABC}\) = B
Let \(\angle{UDV}\) = D
\(\begin{array}{|rcll|} \hline Area_{\text{ABC}} &=& \frac{ab\sin(B)}{2} \\ Area_{\text{CDA}} &=& \frac{cd\sin(D)}{2} \\ Area_{\text{ABCD}} &=& Area_{\text{ABC}} + Area_{\text{CDA}} \\ &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline Area_{\text{QBR}} &=& \frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} \\ Area_{\text{UDV}} &=& \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} - Area_{\text{QBR}} - Area_{\text{UDV}} \\ Area_{\text{hexagon}} &=& Area_{\text{ABCD}} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} -\frac{\frac{a}{3}\frac{b}{3}\sin(B)}{2} - \frac{\frac{c}{3}\frac{d}{3}\sin(D)}{2} \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\left(1-\frac19 \right) + \frac{cd\sin(D)}{2}\left(1-\frac19 \right) \\ Area_{\text{hexagon}} &=& \frac{ab\sin(B)}{2}\cdot \frac89 + \frac{cd\sin(D)}{2}\cdot \frac89 \\ Area_{\text{hexagon}} &=& \frac89 \cdot \left( \frac{ab\sin(B)}{2} + \frac{cd\sin(D)}{2} \right) \\ Area_{\text{hexagon}} &=& \frac89 \cdot Area_{\text{ABCD}} \\ Area_{\text{hexagon}} &=& \frac89 \cdot 180 \\ \mathbf{ Area_{\text{hexagon}} } & \mathbf{=} & \mathbf{160} \\ \hline \end{array}\)
The area of hexagon AQRCUV is 160
