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20 gr1.3 solves mathematical or real - world problems by plotting order…

Question

20 gr1.3 solves mathematical or real - world problems by plotting ordered pairs of rational numbers on a coordinate plane, including finding the perimeter or area of a rectangle: contained in one quadrant (l3)
a coordinate plane is showing a rectangle. the rectangle is formed by the points: c(-7, 6) a(-2, 6) r(-2, 4) s(-7, 4) what is the area, in square units, of the rectangle shown?
a coordinate plane is showing a rectangle. the rectangle is formed by the points: p(1, -3) l(12, -3) a(12, -5) y(1, -5) what is the area, in square units, of the rectangle shown?

Explanation:

First Rectangle (Points C, A, R, S)

Step1: Find length and width

For the first rectangle with points \( C(-7, 6) \), \( A(-2, 6) \), \( R(-2, 4) \), \( S(-7, 4) \):

  • Length (horizontal side): Distance between \( C \) and \( A \). Since \( y \)-coordinates are same (\( 6 \)), length \( = |-2 - (-7)| = 5 \).
  • Width (vertical side): Distance between \( A \) and \( R \). Since \( x \)-coordinates are same (\( -2 \)), width \( = |6 - 4| = 2 \).

Step2: Calculate area

Area of rectangle \( = \text{length} \times \text{width} = 5 \times 2 = 10 \) square units.

Second Rectangle (Points P, L, A, Y)

Step1: Find length and width

For the second rectangle with points \( P(1, -3) \), \( L(12, -3) \), \( A(12, -5) \), \( Y(1, -5) \):

  • Length (horizontal side): Distance between \( P \) and \( L \). Since \( y \)-coordinates are same (\( -3 \)), length \( = |12 - 1| = 11 \).
  • Width (vertical side): Distance between \( L \) and \( A \). Since \( x \)-coordinates are same (\( 12 \)), width \( = |-3 - (-5)| = 2 \).

Step2: Calculate area

Area of rectangle \( = \text{length} \times \text{width} = 11 \times 2 = 22 \) square units.

Answer:

s:

  • Area of first rectangle: \(\boldsymbol{10}\) square units.
  • Area of second rectangle: \(\boldsymbol{22}\) square units.