(5,2) & (-3, 6)
1 answer:
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Refer to the diagram shown below.
The dimensions of the outside of the picture rame are:
Width = x + 1 + 1 = x + 2 inches
Length = (x+2) + 1 + 1 = x + 4 inches.
The perimeter of the outside of the picture frame is 28 inches.
Therefore
2(x + 2) + 2(x+4) = 28
Solve for x.
2x + 4 + 2x + 8 = 28
4x + 12 = 28
4x = 16
x = 4 inches
Answer: 4 inches
The dimensions of the outside of the picture rame are:
Width = x + 1 + 1 = x + 2 inches
Length = (x+2) + 1 + 1 = x + 4 inches.
The perimeter of the outside of the picture frame is 28 inches.
Therefore
2(x + 2) + 2(x+4) = 28
Solve for x.
2x + 4 + 2x + 8 = 28
4x + 12 = 28
4x = 16
x = 4 inches
Answer: 4 inches
Answer:
9 units²
Step-by-step explanation:
calculate the area of Δ ABC as
area of rectangle - area of 3 outer right triangles
area of rectangle = 5 × 4 = 20 units²
area of left Δ = × 3 × 3 = 4
units²
area of Δ on right = × 2 × 4 = 4 units²
area of top Δ = × 5 × 1 = 2
units²
Then
area of Δ ABC = 20 - 4 - 4 - 2
= 20 - 11 = 9 units²