Question

select the correct answer from each drop - down menu.
wxyz is a quadrilateral graphed in the coordinate plane with vertices w(0,5), x(-3,2), y(0,-4), and z(3,2). what are the lengths of the sides of the quadrilateral, and what is the correct name for the figure?
the length of \\(\overline{wx}\\) is about \\(\boldsymbol{\square}\\). the length of \\(\overline{xy}\\) is about \\(\boldsymbol{\square}\\)
the length of \\(\overline{yz}\\) is about \\(\boldsymbol{\square}\\). the length of \\(\overline{zw}\\) is about \\(\boldsymbol{\square}\\)
the best name for this quadrilateral is \\(\boldsymbol{\square}\\)

Explanation:

Step1: Recall Distance Formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).

Step2: Calculate \( \overline{WX} \) length

For \(W(0,5)\) and \(X(-3,2)\):
\(d_{WX}=\sqrt{(-3 - 0)^2+(2 - 5)^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9 + 9}=\sqrt{18}\approx4.24\)

Step3: Calculate \( \overline{XY} \) length

For \(X(-3,2)\) and \(Y(0,-4)\):
\(d_{XY}=\sqrt{(0 - (-3))^2+(-4 - 2)^2}=\sqrt{(3)^2+(-6)^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.71\)

Step4: Calculate \( \overline{YZ} \) length

For \(Y(0,-4)\) and \(Z(3,2)\):
\(d_{YZ}=\sqrt{(3 - 0)^2+(2 - (-4))^2}=\sqrt{(3)^2+(6)^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.71\)

Step5: Calculate \( \overline{ZW} \) length

For \(Z(3,2)\) and \(W(0,5)\):
\(d_{ZW}=\sqrt{(0 - 3)^2+(5 - 2)^2}=\sqrt{(-3)^2+(3)^2}=\sqrt{9 + 9}=\sqrt{18}\approx4.24\)

Step6: Determine Quadrilateral Type

Since \(WX = ZW\approx4.24\) and \(XY = YZ\approx6.71\), and also check symmetry (axis of symmetry along y - axis as x - coordinates of W and Y are 0, X is - 3, Z is 3). Also, check slopes: Slope of WX: \(\frac{2 - 5}{-3 - 0}=\frac{-3}{-3}=1\), Slope of XY: \(\frac{-4 - 2}{0 - (-3)}=\frac{-6}{3}=-2\), Slope of YZ: \(\frac{2 - (-4)}{3 - 0}=\frac{6}{3}=2\), Slope of ZW: \(\frac{5 - 2}{0 - 3}=\frac{3}{-3}=-1\). Wait, actually, since \(WX = YZ\) and \(XY = ZW\) (wait no, earlier calculation: \(WX\approx4.24\), \(YZ\approx6.71\)? Wait no, wait Step2: \(WX\): \(W(0,5)\), \(X(-3,2)\): \(\Delta x=-3\), \(\Delta y=-3\), so distance \(\sqrt{9 + 9}=\sqrt{18}\approx4.24\). Step4: \(ZW\): \(Z(3,2)\), \(W(0,5)\): \(\Delta x=-3\), \(\Delta y = 3\), so distance \(\sqrt{9+9}=\sqrt{18}\approx4.24\). Step3: \(XY\): \(X(-3,2)\), \(Y(0,-4)\): \(\Delta x = 3\), \(\Delta y=-6\), distance \(\sqrt{9 + 36}=\sqrt{45}\approx6.71\). Step4: \(YZ\): \(Y(0,-4)\), \(Z(3,2)\): \(\Delta x = 3\), \(\Delta y=6\), distance \(\sqrt{9 + 36}=\sqrt{45}\approx6.71\). So \(WX = ZW\approx4.24\) and \(XY = YZ\approx6.71\). Also, check if it's a kite? No, because two pairs of adjacent sides equal? Wait \(WX\) and \(ZW\) are adjacent? \(WX\) is from W to X, \(XY\) from X to Y, \(YZ\) from Y to Z, \(ZW\) from Z to W. So \(WX\) and \(YZ\) are opposite, \(XY\) and \(ZW\) are opposite? Wait no, \(WX\): W(0,5)-X(-3,2); \(XY\): X(-3,2)-Y(0,-4); \(YZ\): Y(0,-4)-Z(3,2); \(ZW\): Z(3,2)-W(0,5). So \(WX\) and \(YZ\): \(WX\) length \(\approx4.24\), \(YZ\approx6.71\)? No, wait no, I made a mistake. Wait \(WX\): \(W(0,5)\), \(X(-3,2)\): distance \(\sqrt{(-3 - 0)^2+(2 - 5)^2}=\sqrt{9 + 9}=\sqrt{18}\approx4.24\). \(XY\): \(X(-3,2)\), \(Y(0,-4)\): \(\sqrt{(0 + 3)^2+(-4 - 2)^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.71\). \(YZ\): \(Y(0,-4)\), \(Z(3,2)\): \(\sqrt{(3 - 0)^2+(2 + 4)^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.71\). \(ZW\): \(Z(3,2)\), \(W(0,5)\): \(\sqrt{(0 - 3)^2+(5 - 2)^2}=\sqrt{9 + 9}=\sqrt{18}\approx4.24\). So \(WX = ZW\) and \(XY = YZ\)? No, \(WX\) and \(ZW\) are adjacent? \(WX\) is W - X, \(ZW\) is Z - W. So sides: \(WX\approx4.24\), \(XY\approx6.71\), \(YZ\approx6.71\), \(ZW\approx4.24\). So opposite sides are equal: \(WX = ZW\)? No, \(WX\) and \(YZ\): \(WX\approx4.24\), \(YZ\approx6.71\) no. Wait \(WX\) and \(YZ\): no, \(WX\) is between W(0,5) and X(-3,2), \(YZ\) is between Y(0,-4) and Z(3,2). Wait \(WX\) length: \(\sqrt{(-3)^2+(-3)^2}=\sqrt{18}\), \(YZ\) length: \(\sqrt{(3)^2+(6)^2}=\sqrt{45}\). Wait I think I messed up the sides. Wait the quadrilateral is W(0,5), X(-3,2), Y(0,-4), Z(3,2). So the order is W - X - Y - Z - W. So sides: WX, XY, YZ, ZW. So WX: W to X, XY: X to Y, YZ: Y to Z, ZW: Z to W. So WX: distance between (0,5) and (-3,2): \(\sqrt{(-3 - 0)^2+(2 - 5)^2}=\sqrt{9 + 9}=\sqrt{18}\approx4.24\). XY: distance between (-3,2) and (0,-4): \(\sqrt{(0 + 3)^2+(-4 - 2)^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.71\). YZ: distance between (0,-4) and (3,2): \(\sqrt{(3 - 0)^2+(2 + 4)^2}=\sqrt{9 + 36}=\sqrt{45}\approx6.71\). ZW: distance between (3,2) and (0,5): \(\sqrt{(0 - 3)^2+(5 - 2)^2}=\sqrt{9 + 9}=\sqrt{18}\approx4.24\). Ah! So WX = ZW (\(\approx4.24\)) and XY = YZ (\(\approx6.71\))? No, WX and ZW are adjacent? No, WX is W - X, ZW is Z - W. Wait, WX: length ~4.24, ZW: length ~4.24. XY: ~6.71, YZ: ~6.71. So actually, WX = ZW and XY = YZ, but also, check if it's a kite? No, a kite has two pairs of adjacent sides equal. But here, WX and ZW are not adjacent (WX is W - X, ZW is Z - W; adjacent sides are WX - XY, XY - YZ, YZ - ZW, ZW - WX). Wait WX and XY: WX ~4.24, XY ~6.71. XY and YZ: ~6.71 and ~6.71 (equal). YZ and ZW: ~6.71 and ~4.24. ZW and WX: ~4.24 and ~4.24 (equal). So actually, two pairs of adjacent sides: ZW and WX are adjacent (ZW - WX) with length ~4.24, XY and YZ are adjacent (XY - YZ) with length ~6.71. So it's a kite? Wait no, a kite has two distinct pairs of adjacent sides equal. But also, check the diagonals: diagonal WY: from W(0,5) to Y(0,-4), length is \(|5 - (-4)| = 9\). Diagonal XZ: from X(-3,2) to Z(3,2), length is \(| - 3 - 3|=6\). In a kite, one diagonal is the perpendicular bisector of the other. Let's check if WY (vertical line x = 0) bisects XZ (horizontal line y = 2). Midpoint of XZ: \((\frac{-3 + 3}{2},\frac{2+2}{2})=(0,2)\), which lies on WY (x = 0). And WY is vertical, XZ is horizontal, so they are perpendicular. So it is a kite? Wait but also, the lengths: WX = ZW (\(\sqrt{18}\)) and XY = YZ (\(\sqrt{45}\)), so two pairs of adjacent sides equal, diagonals perpendicular, so it's a kite. But also, let's check if it's an isosceles trapezoid? No, because in trapezoid, only one pair of sides parallel. Slope of WX: \(\frac{2 - 5}{-3 - 0}=1\), slope of YZ: \(\frac{2 - (-4)}{3 - 0}=2\), not parallel. Slope of XY: \(\frac{-4 - 2}{0 - (-3)}=-2\), slope of ZW: \(\frac{5 - 2}{0 - 3}=-1\), not parallel. Wait but earlier, we saw that WX = ZW and XY = YZ, and diagonals perpendicular, so it's a kite. But also, the sides: WX ~4.24, XY ~6.71, YZ ~6.71, ZW ~4.24. So the lengths: WX ≈ 4.24, XY ≈ 6.71, YZ ≈ 6.71, ZW ≈ 4.24. And the figure is a kite (or maybe a rhombus? No, because sides are not all equal. \(\sqrt{18}\approx4.24\), \(\sqrt{45}\approx6.71\), not equal). Wait, maybe I made a mistake in the side order. Let's plot the points: W(0,5) is on y - axis, X(-3,2) is left of y - axis, Y(0,-4) is on y - axis below origin, Z(3,2) is right of y - axis. So the figure is symmetric about the y - axis. So W and Y are on y - axis, X and Z are symmetric about y - axis. So WX and ZW: WX is from (0,5) to (-3,2), ZW is from (0,5) to (3,2)? No, Z is (3,2), so ZW is from (3,2) to (0,5). So WX: (-3,2) to (0,5), ZW: (3,2) to (0,5). So these two sides are symmetric about y - axis, so equal in length. XY: (-3,2) to (0,-4), YZ: (0,-4) to (3,2), also symmetric about y - axis, equal in length. So the quadrilateral has two pairs of adjacent sides equal (WX and ZW? No, WX is adjacent to XY, ZW is adjacent to WX? Wait no, the order is W - X - Y - Z - W, so adjacent sides are WX & XY, XY & YZ, YZ & ZW, ZW & WX. So WX and ZW are not adjacent, they are opposite? Wait no, in the order W - X - Y - Z - W, the sides are WX (W to X), XY (X to Y), YZ (Y to Z), ZW (Z to W). So WX and YZ are opposite, XY and ZW are opposite. Wait, WX length: \(\sqrt{(-3)^2+(-3)^2}=\sqrt{18}\), YZ length: \(\sqrt{(3)^2+(6)^2}=\sqrt{45}\) – no, that's not equal. Wait I think I messed up the coordinates of Y. Wait Y is (0,-4), Z is (3,2). So from Y(0,-4) to Z(3,2): \(\Delta x = 3\), \(\Delta y=6\), so distance \(\sqrt{9 + 36}=\sqrt{45}\). From X(-3,2) to Y(0,-4): \(\Delta x = 3\), \(\Delta y=-6\), distance \(\sqrt{9 + 36}=\sqrt{45}\). From Z(3,2) to W(0,5): \(\Delta x=-3\), \(\Delta y = 3\), distance \(\sqrt{9 + 9}=\sqrt{18}\). From W(0,5) to X(-3,2): \(\Delta x=-3\), \(\Delta y=-3\), distance \(\sqrt{9 + 9}=\sqrt{18}\). Ah! So WX and ZW: WX is \(\sqrt{18}\), ZW is \(\sqrt{18}\) (because \(\Delta x=-3\), \(\Delta y = 3\), squared is 9 and 9, sum 18). XY and YZ: XY is \(\sqrt{45}\), YZ is \(\sqrt{45}\) (because \(\Delta x = 3\), \(\Delta y=-6\) for XY, squared 9 and 36; \(\Delta x = 3\), \(\Delta y = 6\) for YZ, squared 9 and 36, sum 45). So actually, WX = ZW (\(\sqrt{18}\approx4.24\)) and XY = YZ (\(\sqrt{45}\approx6.71\))? No, WX is between W and X, ZW is between Z and W – they are adjacent? No, in the order W - X - Y - Z - W, ZW is between Z and W, which is adjacent to WX (between W and X). So ZW and WX are adjacent, both length \(\sqrt{18}\). XY and YZ are adjacent, both length \(\sqrt{45}\). So two pairs of adjacent sides equal, so it's a kite. Also, the diagonals: WY is from (0,5) to (0,-4), length 9 (vertical line). XZ is from (-3,2) to (3,2), length 6 (horizontal line). They intersect at (0,2), which is the midpoint of XZ (since X(-3,2), Z(3,2), midpoint (0,2)) and a point on WY. Also, WY is perpendicular to XZ (vertical and horizontal lines are perpendicular). So it's a kite.

Answer:

• Length of \(\overline{WX}\): \(\sqrt{18}\approx4.24\)
• Length of \(\overline{XY}\): \(\sqrt{45}\approx6.71\)
• Length of \(\overline{YZ}\): \(\sqrt{45}\approx6.71\)
• Length of \(\overline{ZW}\): \(\sqrt{18}\approx4.24\)
• Best name: Kite (or Isosceles Kite, or considering symmetry, also could be a quadrilateral with two pairs of adjacent sides equal and diagonals perpendicular, so kite)

(Note: If drop - down options have numerical values like 4.24, 6.71 etc., the lengths would be approximately