Respuesta :

Answer:

1. (A)

2 (H)

3 (B)

4 None

Step-by-step explanation:

Note : after multiplying with dilution constant a it's all side become a times original one.

    1. Side of hexagon = 6*t

          Thus perimeter of this hexagon =6*6t =C

       Let dilution parameter be a such that it s new perimeter will be 9*t .

      Now new perimeter = a*36*t =9*t

                [tex]\alpha =\frac{1}{4}[/tex]

         So option A is correct

    2.Let initial radius be R

      Dilated circle is a times the original radius

     So , after dilated by factor of 0.5 radius = 0.5R

    Dilating image mean change in area of image. if area of circle is decreased by 0.25 times then radius is decreased by 0.5 times of original one.

     Now, two times dilated radius = 0.5*0.25*R

                                                         = [tex]\frac{1}{8} *R[/tex]

 Perimeter =[tex]2*\pi *\frac{R}{8} = 0.25R[/tex]

    3. On plotting rectangle on graph you will get a rectangle of dimension 4*5

  Required area is = 5

  Let dilation factor be [tex]\alpha[/tex]  

 New area = [tex]5*4*\alpha ^{2}[/tex]=5

  ⇒[tex]\alpha = \frac{1}{2}[/tex]

  Measure of length of length = 5*0.5 =2.5

   4. Dilation factor for triangle given = [tex]\frac{final side length}{initial side length}[/tex]

   [tex]\alpha =\frac{4}{6} =\frac{8}{12}  = \frac{14}{3*7}[/tex] =[tex]\frac{2}{3}[/tex]

  Now let side of square be a then area = [tex]a^{2}[/tex]

After dilation area of square[tex](\frac{2}{3}) ^{2} *a^{2}[/tex]

 Fraction of new squares which could be fitted would be = [tex]a^{2}[/tex]/[tex](\frac{2}{3}) ^{2} *a^{2}[/tex]

 = [tex]\frac{9}{4}[/tex]