Question

select the correct answers from each drop - down menu. complete the steps in the proof that show quadrilateral kite with vertices k(0, - 2), i(1, 2), t(7, 5), and e(4, - 1) is a kite. using the distance formula, $ki = \sqrt{(2 - (-2))^{2}+(1 - 0)^{2}} = \sqrt{17}$, $ke = $, $it = $, and $te = $. therefore, kite is a kite because .

Explanation:

Step1: Calculate KE

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(K(0,-2)\) and \(E(4,-1)\), we have \(x_1 = 0,y_1=-2,x_2 = 4,y_2=-1\). So \(KE=\sqrt{(4 - 0)^2+(-1-(-2))^2}=\sqrt{16 + 1}=\sqrt{17}\).

Step2: Calculate IT

For points \(I(1,2)\) and \(T(7,5)\), using the distance formula: \(x_1 = 1,y_1 = 2,x_2=7,y_2 = 5\). Then \(IT=\sqrt{(7 - 1)^2+(5 - 2)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}\)? Wait, no, wait, let's recalculate. Wait, \(I(1,2)\) and \(T(7,5)\): \(7 - 1=6\), \(5 - 2 = 3\), so \((6)^2+(3)^2=36 + 9=45\), so \(IT=\sqrt{45}=3\sqrt{5}\)? But wait, maybe I made a mistake. Wait, no, the other side: let's calculate \(TE\). For \(T(7,5)\) and \(E(4,-1)\): \(x_1 = 7,y_1 = 5,x_2=4,y_2=-1\). So \(TE=\sqrt{(4 - 7)^2+(-1 - 5)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5}\). Wait, but \(KI=\sqrt{17}\), \(KE=\sqrt{17}\), \(IT = \sqrt{45}\), \(TE=\sqrt{45}\). So a kite has two distinct pairs of adjacent sides equal. So \(KI = KE=\sqrt{17}\) and \(IT=TE=\sqrt{45}\), so two pairs of adjacent sides are equal.

Wait, let's recalculate \(KE\) again. \(K(0,-2)\), \(E(4,-1)\): \(x\) difference is \(4-0 = 4\), \(y\) difference is \(-1-(-2)=1\). So \((4)^2+(1)^2=16 + 1=17\), so \(KE=\sqrt{17}\), correct. \(IT\): \(I(1,2)\), \(T(7,5)\): \(x\) difference \(7 - 1=6\), \(y\) difference \(5 - 2=3\), so \(6^2+3^2=36 + 9=45\), so \(IT=\sqrt{45}\). \(TE\): \(T(7,5)\), \(E(4,-1)\): \(x\) difference \(4 - 7=-3\), \(y\) difference \(-1 - 5=-6\), so \((-3)^2+(-6)^2=9 + 36=45\), so \(TE=\sqrt{45}\). Then, since \(KI=\sqrt{17}\), \(KE=\sqrt{17}\) (so \(KI = KE\)) and \(IT=\sqrt{45}\), \(TE=\sqrt{45}\) (so \(IT = TE\)), and in a kite, two distinct pairs of adjacent sides are equal. So the sides: \(KI\) and \(KE\) are adjacent (from \(K\) to \(I\) and \(K\) to \(E\)), and \(IT\) and \(TE\) are adjacent (from \(I\) to \(T\) and \(T\) to \(E\))? Wait, no, the quadrilateral is KITE, so the vertices are K, I, T, E. So the sides are KI, IT, TE, EK. So KI and EK (KE) are adjacent? Wait, no, K to I, I to T, T to E, E to K. So the sides are KI, IT, TE, EK. So KI and EK: KI is from K to I, EK is from E to K. So adjacent sides: KI and IT (at I), IT and TE (at T), TE and EK (at E), EK and KI (at K). So for a kite, two pairs of adjacent sides are equal. So KI and EK (KE) should be equal? Wait, KI is \(\sqrt{17}\), KE is \(\sqrt{17}\), so KI = KE. Then IT and TE: IT is \(\sqrt{45}\), TE is \(\sqrt{45}\), so IT = TE. So two pairs of adjacent sides (KI & KE, IT & TE) are equal, so it's a kite.

So to fill in:

\(KE=\sqrt{(4 - 0)^2+(-1-(-2))^2}=\sqrt{16 + 1}=\sqrt{17}\)

\(IT=\sqrt{(7 - 1)^2+(5 - 2)^2}=\sqrt{36 + 9}=\sqrt{45}\) (or \(3\sqrt{5}\))

\(TE=\sqrt{(4 - 7)^2+(-1 - 5)^2}=\sqrt{9 + 36}=\sqrt{45}\) (or \(3\sqrt{5}\))

And KITE is a kite because two distinct pairs of adjacent sides are equal (KI = KE and IT = TE).

Answer:

\(KE=\boldsymbol{\sqrt{17}}\), \(IT=\boldsymbol{\sqrt{45}}\) (or \(3\sqrt{5}\)), \(TE=\boldsymbol{\sqrt{45}}\) (or \(3\sqrt{5}\)), and KITE is a kite because two distinct pairs of adjacent sides are equal (KI = KE and IT = TE).