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3 years ago

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The area of a square is 144 square centimeters. Find the length of the diagonal. Write your answer in simplest radical form.

1 answer:

maks197457 [2]3 years ago

3 0

Answer:

$12\sqrt{2}\ cm$

Step-by-step explanation:

step 1

Find the length side of the square

we know that

The area of a square is equal to

$A=b^{2}$

where

b is the length side

we have

$A=144\ cm^2$

substitute in the formula of area

$144=b^{2}$

solve for b

square root both sides

$b=12\ cm$

step 2

Find the length of the diagonal

Applying the Pythagorean Theorem

$d^2=b^2+b^2$

see the attached figure to better understand the problem

substitute the given values

$d^2=12^2+12^2$

$d^2=288$

square root both sides

$d=\sqrt{288}\ cm$

simplify

$d=12\sqrt{2}\ cm$

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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

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The volume of the box is = area of the base × height

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The cost to construct of the bottom of the box is

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The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

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The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

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Differentiating with respect to x

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Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

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