Answer:
a) Parallelogram
b) 18 units
c) 12 + 2√13 units²
d) see attached and below
e) perimeter = 72 units
area = 24 + 4√13 units²
Step-by-step explanation:
Given points:
- M = (-2, 1)
- A = (0, 4)
- T = (6, 4)
- H = (4, 1)
Part (a)
Shape of MATH: Parallelogram
Part (b)
[tex]\begin{aligned}\textsf{Area of a parallelogram} & = \sf base \times height\\& = (x_H-x_M) \times (y_A-y_M)\\& = (4-(-2)) \times (4-1)\\& = 6 \times 3\\& = 18\: \sf units^2\end{aligned}[/tex]
Part (c)
[tex]\begin{aligned}\textsf{Perimeter of MATH} & =2 \times \sf base+2 \times side\\& = 2 \textsf{MH} + 2 \textsf{AM}\\& = 2(4-(-2))+2(\sqrt{2^2+3^2})\\& = 2(6)+2(\sqrt{13})\\& = 12+2\sqrt{13}\: \sf units\end{aligned}[/tex]
Part (d)
To dilate MATH with a dilation center at (0,0) and a dilation factor of 2, multiply the x and y coordinates of MATH by sf 2:
- M' = (-4, 2)
- A '= (0, 8)
- T' = (12, 8)
- H' = (8, 2)
Part (e)
As M'A'T'H' is an enlargement of MATH by a scale factor of 2, the perimeter of M'A'T'H' is twice that of MATH:
[tex]\begin{aligned}\textsf{Perimeter of M'A'T'H'} & =2 \times \textsf{perimeter of MATH}\\& = 2(12+2\sqrt{13})\\& = 24+4\sqrt{13}\: \sf units\end{aligned}[/tex]
As M'A'T'H' is an enlargement of MATH by a scale factor of 2, the area of M'A'T'H' is 2² that of MATH (as area is in 2 dimensions):
[tex]\begin{aligned}\textsf{Area of M'A'T'H'} & =2^2 \times \textsf{Area of MATH}\\& = 4(18)\\& = 72\: \sf units^2 \end{aligned}[/tex]
