Answer:
1)
Given the triangle RST with Coordinates R(2,1), S(2, -2) and T(-1 , -2).
A dilation is a transformation which produces an image that is the same shape as original one, but is different size.
Since, the scale factor [tex]\frac{5}{3}[/tex] is greater than 1, the image is enlargement or a stretch.
Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor [tex]\frac{5}{3}[/tex]
Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image [tex]S{}'[/tex] are same.
Now, the distances from the center of the dilation at point S to the other points R and T.
The dilation image will be[tex]\frac{5}{3}[/tex] of each of these distances,
[tex]SR=3[/tex], so [tex]S{}'R{}'[/tex]=5 ;
[tex]ST=3[/tex], so [tex]S{}'T{}'=5[/tex]
Now, draw the image of RST i.e R'S'T'
Since, [tex]RT=3\sqrt{2}[/tex] [By using hypotenuse of right angle triangle] and [tex]R{}'T{}'=5\sqrt{2}[/tex].
2)
(a)
Disagree with the given statement.
Side Angle Side postulate (SAS) states that:
If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.
Given: B is the midpoint of [tex]\overline{AC}[/tex] i.e [tex]\overline{AB}\cong \overline{BC}[/tex]
In the triangle ABD and triangle CBD, we have
[tex]\overline{AB}\cong \overline{BC}[/tex] (SIDE) [Given]
[tex]\overline{BD}\cong \overline{BD}[/tex] (SIDE) [Reflexive post]
Since, there is no included angle in these triangles.
∴ [tex]\Delta ABD[/tex] is not congruent to [tex]\Delta CBD[/tex] .
Therefore, these triangles does not follow the SAS congruence postulates.
(b)
SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.
Since it is also given that [tex]\overline{AD}\cong \overline{CD}[/tex].
therefore, in the triangle ABD and triangle CBD, we have
[tex]\overline{AB}\cong \overline{BC}[/tex] (SIDE) [Given]
[tex]\overline{AD}\cong \overline{CD}[/tex] (SIDE) [Given]
[tex]\overline{BD}\cong \overline{BD}[/tex] (SIDE) [Reflexive post]
therefore by, SSS postulates [tex]\Delta ABD\cong \Delta CBD[/tex].
3)
Given that: [tex]\angle1=\angle 3[/tex] are vertical angles, as they are formed by intersecting lines.
Therefore
, by the definition of linear pairs
[tex]\angle 1[/tex] and [tex]\angle 2[/tex] and [tex]\angle 3[/tex] and [tex]\angle 2[/tex] are linear pair.
By linear pair theorem, [tex]\angle 1[/tex] and [tex]\angle 2[/tex] are supplementary, [tex]\angle 2[/tex] and [tex]\angle 3[/tex] are supplementary.
[tex]m\angle1+m\angle 2=180^{\circ}[/tex]
[tex]m\angle2+m\angle 3=180^{\circ}[/tex]
Equate the above expressions:
[tex]m\angle 1+m\angle 2=m\angle 2+m\angle 3[/tex]
Subtract the angle 2 from both sides in the above expressions
∴[tex]m\angle 1=m\angle 3[/tex]
By Congruent Supplement theorem: If two angles are supplements of the same angle, then the two angles are congruent.
therefore, [tex]\angle 1\cong \angle 3[/tex].
