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{8}[/tex]
LM=
MN=
NK=
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=
=
Also, let N be the mid point of DC, therefore,
N=
=
Now, length of the mid segment MN=
(D)Given: Kite PQRS, TS=6cm and TP=8cm
Then, From triangle TSP, we have
