1. Trapezoid KLMN has vertices K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2) .Is this trapezoid an isosceles trapezoid?2. In kite WXYZ, m∠XWY=47° and m∠ZYW=18°.What is m∠WZY?Enter your answer in the box.3. What is the length of the midsegment of this trapezoid?Enter your answer in the box. In kite PQRS, TS=6 cm and TP=8 cm.What is SP ?Enter your answer in the box.

Answer:
Step-by-step explanation:(A) The vertices of the trapezoid KLMN are K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2)Now, KL= [tex]\sqrt{(3-1)^{2}+(1-3)^{2}}[/tex]=\sqrt{8}[/tex]LM=[tex]\sqrt{(3-3)^{2}+(0-1)^{2}}=1[/tex]MN=[tex]\sqrt{(1-3)^{2}+(-2-0)^{2}}=\sqrt{8}[/tex]NK=[tex]\sqrt{(1-1)^{2}+(-2-3)^{2}}=5[/tex]Now, as KL and MN  are equal, therefore, KLMN is an isosceles trapezoid.(B) Since  m∠XWY=47°, therefore ∠XWZ=47+47=94° and ∠ZYW=18°, therefore ∠XYZ=36°( as diagonals bisect the angles)In ΔXWO,∠XWO+∠WXO+∠WOX=180°(Angle sum property)∠WXO=43°Also, from ΔXOY,∠OXY=72° using the angle sum property.Therefore, ∠WXY=43+72=115°Now, sum of all the angles of a quadrilateral is equal to 360°, therefore∠WXY+∠XYZ+∠YZW+∠ZWX=360°115°+36°+∠YZW+94°=360°∠YZW=115°Therefore,∠WZY=115°(C) Since, AB and CD are the two parallel lines as the slope of both the sides are equal.LetM be the mid point of AB, therefore M=[tex](\frac{-2+4}{2}, \frac{4+3}{2})[/tex]=[tex](1,\frac{7}{2})[/tex]Also, let N be the mid point of DC, therefore,N=[tex](\frac{4-2}{2},\frac{-2-5}{2})[/tex]=[tex](1,\frac{-7}{2})[/tex]Now, length of the mid segment MN= [tex]\sqrt{(1-1)^{2}+(\frac{7}{2}+\frac{7}{2})^{2}}=\sqrt{0+49}=7 cm[/tex](D)Given: Kite PQRS,  TS=6cm and TP=8cmThen, From triangle TSP, we have [tex](SP)^{2}=(TS)^{2}+(TP)^{2}[/tex][tex](SP)^{2}=36+64[/tex][tex]SP=10cm[/tex]