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?

1 answer:

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

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, 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

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

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$