Question

the graph shows quadrilaterals efgh and lmno. is efgh similar to lmno? justify your answer.

Explanation:

Step1: Calculate the side - length ratios

• For quadrilaterals \(EFGH\) and \(LMNO\), use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) (or count the grid units for horizontal/vertical sides).
• \(EF\): \(E(-6,2)\), \(F(-8,2)\), \(EF=\vert-6-(-8)\vert = 2\); \(LM\): \(L(-3,1)\), \(M(-4,1)\), \(LM=\vert-3-(-4)\vert = 1\).
• \(FG\): \(F(-8,2)\), \(G(-8,-8)\), \(FG=\vert2-(-8)\vert = 10\); \(MN\): \(M(-4,1)\), \(N(-4,-4)\), \(MN=\vert1-(-4)\vert = 5\).
• The ratio of \(EF\) to \(LM\) is \(\frac{EF}{LM}=\frac{2}{1}\), and the ratio of \(FG\) to \(MN\) is \(\frac{FG}{MN}=\frac{10}{5}=\frac{2}{1}\).
• Let's check another pair of sides. \(EH\): \(E(-6,2)\), \(H(-5,-4)\), \(EH=\sqrt{(-5 + 6)^2+(-4 - 2)^2}=\sqrt{1 + 36}=\sqrt{37}\); \(LO\): \(L(-3,1)\), \(O(-1,-2)\), \(LO=\sqrt{(-1+3)^2+(-2 - 1)^2}=\sqrt{4 + 9}=\sqrt{13}\). The ratio \(\frac{EH}{LO}=\frac{\sqrt{37}}{\sqrt{13}}

eq2\).

Answer:

No, \(EFGH\) is not similar to \(LMNO\). Although the ratios of \(EF\) to \(LM\) and \(FG\) to \(MN\) are \(2:1\), the ratios of other corresponding sides (e.g., \(EH\) to \(LO\)) are not equal to \(2:1\). For two polygons to be similar, all corresponding side - length ratios must be equal.