Question

1. Trapezoid KLMN has vertices K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2) .

Answer 1

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= $\sqrt{(3-1)^{2}+(1-3)^{2}}$=\sqrt{8}[/tex]

LM=$\sqrt{(3-3)^{2}+(0-1)^{2}}=1$

MN=$\sqrt{(1-3)^{2}+(-2-0)^{2}}=\sqrt{8}$

NK=$\sqrt{(1-1)^{2}+(-2-3)^{2}}=5$

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=$(\frac{-2+4}{2}, \frac{4+3}{2})$

=$(1,\frac{7}{2})$

Also, let N be the mid point of DC, therefore,

N=$(\frac{4-2}{2},\frac{-2-5}{2})$

=$(1,\frac{-7}{2})$

Now, length of the mid segment MN= $\sqrt{(1-1)^{2}+(\frac{7}{2}+\frac{7}{2})^{2}}=\sqrt{0+49}=7 cm$

(D)Given: Kite PQRS,  TS=6cm and TP=8cm

Then, From triangle TSP, we have

$(SP)^{2}=(TS)^{2}+(TP)^{2}$

$(SP)^{2}=36+64$

$SP=10cm$