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Svetradugi [14.3K]
3 years ago
11

The length of a certain rectangle exceeds its width by exactly 8 cm, and the perimeter of the rectangle is 66 cm. What is the wi

dth of the rectangle? Although you may be able to solve this problem using a method of your own, try the following approach, which starts by guessing the width of the rectangle. Study the first row of the table below, which is based on a 10-cm guess for the width. Then make your own guess and use it to fill in the next row of the table. If you have not guessed the correct width, use another row of the table and try again.
I attached a photo of the table that is shown as part of the question.

Mathematics
1 answer:
Svetradugi [14.3K]3 years ago
5 0
Well remember a perimeter is something that would add or mutiply in a problem but the area is where you have to double it like 2x2 would be 4 but with area and a square it would be 8
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I need to know the improper fractions answers for:
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Part 1) x=\frac{15}{2}\ units

Part 2) z=\frac{15\sqrt{3}}{2}\ units

Part 3) y= \frac{15\sqrt{3}}{4}\ units

Part 4) b= \frac{45}{4}\ units

Step-by-step explanation:

step 1

Find the value of x

In the large right triangle

cos(60^o)=\frac{x}{15} ----> by CAH (adjacent side divided by the hypotenuse)

Remember that

cos(60^o)=\frac{1}{2}

substitute

\frac{1}{2}=\frac{x}{15}

solve for x

x=\frac{15}{2}\ units ---> improper fraction

step 2

Find the value of z

In the large right triangle

Applying the Pythagorean Theorem

15^2=x^2+z^2

substitute the value of x

15^2=(\frac{15}{2})^2+z^2

solve for z

z^2=15^2-(\frac{15}{2})^2

z^2=225-\frac{225}{4}

z^2=\frac{675}{4}

z=\frac{\sqrt{675}}{2}\ units

simplify

z=\frac{15\sqrt{3}}{2}\ units

step 3

Find the value of y

In the right triangle of the right

sin(30^o)=\frac{y}{z} ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values of y and z

Remember that

sin(30^o)=\frac{1}{2}

so

\frac{1}{2}=y:\frac{15\sqrt{3}}{2}

solve for y

\frac{1}{2}= \frac{2y}{15\sqrt{3}}

y= \frac{15\sqrt{3}}{4}\ units

step 4

Find the value of b

In the right triangle of the right

cos(30^o)=\frac{b}{z} ---> by CAH (adjacent side divided by the hypotenuse)

substitute the given values of y and z

Remember that

cos(30^o)=\frac{\sqrt{3}}{2}

so

\frac{\sqrt{3}}{2}=b:\frac{15\sqrt{3}}{2}

solve for y

\frac{\sqrt{3}}{2}= \frac{2b}{15\sqrt{3}}

b= \frac{45}{4}\ units

7 0
3 years ago
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