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nekit [7.7K]
3 years ago
15

The length of a rectangle is 4 more than its width. The area of the rectangle is 21m^2 . What is the measure of the width?

Mathematics
1 answer:
saw5 [17]3 years ago
5 0
Make the length x+4 and the width x  and then make a formula x(x+4)=21
then u ll get x^2+4x-21=0  and then solve for x ..... the width is 3 

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Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a
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Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

Area_{A}=0.25*Area_{B} (Equation\ 1)\\

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}

put this value into equation 1

Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\  Area_{B}=80\ cm^{2}

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

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Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm

Now, put the value of length(2B) into equation 3 to know length (1B)

length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm

I really hope this helps you, have a great day.

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Step-by-step explanation:

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Answer:

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