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The perimeter of a rectangle is 56 meters. The ratio of its length to its width is 4:3. What is
the length in meters of a diagonal of the rectangle?

Apr 7, 2020

#1
+932
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If the perimeter of the rectangle is 56, that means that the length + the width equals 28. Since the ratio of the sides is 4:3, we can find that the length is 16 and the width is 12. From there we use Pythagorean thereom to find the length of the diagonal. 12^2+16^2= 400= 20^2. So the length of tge diagonal of the rectangle is 20 meters.

Hope it helps!

Apr 7, 2020
#2
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Length is   4/7     width is     3 / 7

32                        24

EACH side is                   16                        12

Diagonal (via Pythagorean theorem)     sqrt (16^2  + 12^2)

Apr 7, 2020
#3
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Perimeter  = 2 ( W +L)

56  =  2 (W + L)

And

L / W  = 4/3

L = (4/3)W

So we have  that

56  = 2  ( W + (4/3)W )     divide both sides  by 2

28  =  W ( 1 + 4/3)

28  = W( 7/3)     multiply both sides by  3/7

28 ( 3/7)  = W

(28/7) * 3  =  W

4 * 3  = W

12  = W

And  L = (4/3)(12)  = (12/3) * 4 = 4 * 4  =16

And  the  length  of  the diagonal  =√ [ W^2 + L^2 ] =  √[ 12^2 + 16^2 ] = √[144 + 256]  = √400 = 20 m

Apr 7, 2020
#4
+50
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Let's say that the length is x and the width is y, then the perimeter is 2x + 2y, and making that into an equation we get...

2x + 2y = 56

Simplifiying we get...

x + y = 28

The ratio of the length and width is 4 to 3, so we can make the equation...

4x = 3y

3(x + y) = 28

3x + 3y = 84

Plugging in 3y as 4x we get

3x + 4x = 84

7x = 84

x = 12

Now we plug x = 12, into the original equation:

x + y = 28

x = 12

y = 28 - 12 = 16

The question asks us to find the diagonal, and that is going to be using the Pythagorean theorem.

In this case, the hypotenuse is the diagonal of the triangle if we split the rectangle along the diagonal.

a^2 + b^2 = c^2

12^2 + 16^2 = 400

c^2 = 400

c = 20

Since we know that the hypotenuse (c) is diagonal, and that is what the question is asking for it is our answer.