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3 years ago

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the length of a rectangle is 5 minutes longer than its with if the perimeter of the rectangle must be at least 34 units what are

the possible lamps of the rectangle right in any quality to represent the problem then solve the inequality

1 answer:

stealth61 [152]3 years ago

7 0

Answer:

w ≥ 6

l ≥ 11

Step-by-step explanation:

The perimeter of a rectangle is equal to P = 2l+2w where l=length and w=width. Here the length is 5 feet longer than the width or 5+w. This means the width is w. Substitute P = 34, l=5+w and w into the perimeter equation. Then solve for w.

34 ≤ 2*(5+w) + 2w

34 ≤ 10+2w+2w

34 ≤ 10+4w

24 ≤ 4w

6 ≤ w

This means the width must be at least 6 so the solution is w ≥ 6.

To find the length substitute 6 into l ≤ 5+w.

l ≤ 5 + 6

l ≤ 11

The length is l ≥ 11.

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Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

$(x,y)\to (k(x-a)+a,k(y-b)+b)$

According to the given information, the scale factor is $\dfrac{1}{3}$ and the center of dilation is at (4,2).

$(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)$ ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

$C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)$

$C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)$

$C(-2,11)\to C'(-2+4,3+2)$

$C(-2,11)\to C'(2,5)$

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

$A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)$

From the given figure it is clear that the image of point A is (8,4).

$A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)$

On comparing both sides, we get

$\dfrac{1}{3}(m-4)+4=8$

$\dfrac{1}{3}(m-4)=8-4$

$(m-4)=3(4)$

$m=12+4$

$m=16$

And,

$\dfrac{1}{3}(n-2)+2=4$

$\dfrac{1}{3}(n-2)=4-2$

$(n-2)=3(2)$

$n=6+2$

$n=8$

Therefore, the coordinates of point A are (16,8).

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