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3 years ago

The area of four walls of a rectangular room is 96m2 and its height is 4m.

Mathematics

1 answer:

Delicious77 [7]3 years ago

7 0

Answer:

<h2>The answer is 56 m</h2>

Step-by-step explanation:

Perimeter of a rectangle = 2l + 2w

where

l is the length

w is the width

To find the area we must first find the length of the rectangle using it's area.

That's

Area = length × height

From the question

Area = 96 m²

height = 4 m

So we have

96 = 4l

Divide both sides by 4

length = 24 m

Now we have

Perimeter = 2(24) + 2(4)

= 48 + 8

We have the final answer as

Hope this helps you

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3 years ago

(47 Points)

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case 1)

Line 1

Let A (2,5) B (-3,-5)

Line 2

Let C (3,0) D (0,-3)

Find the equation of the line 1

Line 1

Let A (2,5) B (-3,-5)

slope m=(y2-y1) /(x2-x1)------> m=(-5-5) /(-3-2)-----> m=2

with m=2 and the point A (2,5) find the equation of the line 1

y-y1=m*(x-x1)------->y-5=2*(x-2)------>y=2x-4+5------> y=2x+1

Find the equation of the line 2

Line 2

Let C (3,0) D (0,-3)

slope m=(y2-y1) /(x2-x1)------> m=(-3-0) /(0-3)-----> m=1

with m=1 and the point C (3,0) find the equation of the line 2

y-y1=m*(x-x1)------->y-0=1*(x-3)------>y=x-3

using a graph tool

see the attached figure N 1

the solution is the point (-4,-7)

case 2)

Line 1

Let A (1,1) B (2,3)

Line 2

Let C (0,3) D (2,5)

Find the equation of the line 1

Line 1

Let A (1,1) B (2,3)

slope m=(y2-y1) /(x2-x1)------> m=(3-1) /(2-1)-----> m=2

with m=2 and the point A (1,1) find the equation of the line 1

y-y1=m*(x-x1)------->y-1=2*(x-1)------>y=2x--2+1------> y=2x-1

Find the equation of the line 2

Line 2

Let C (0,3) D (2,5)

slope m=(y2-y1) /(x2-x1)------> m=(5-3) /(2-0)-----> m=1

with m=1 and the point C (0,3) find the equation of the line 2

y-y1=m*(x-x1)------->y-3=1*(x-0)------>y=x+3

using a graph tool

see the attached figure N 2

the solution is the point (4,7)

case 3)

Line 1

Let A (1,0) B (0,-1)

Line 2

Let C (0,3) D (-2,-1)

Find the equation of the line 1

Line 1

Let A (1,0) B (0,-1)

slope m=(y2-y1) /(x2-x1)------> m=(-1-0) /(0-1)-----> m=1

with m=1 and the point A (1,0) find the equation of the line 1

y-y1=m*(x-x1)------->y-0=1*(x-1)------>y=x-1

Find the equation of the line 2

Line 2

Let C (0,3) D (-2,-1)

slope m=(y2-y1) /(x2-x1)------> m=(-1-3) /(-2-0)-----> m=2

with m=2 and the point C (0,3) find the equation of the line 2

y-y1=m*(x-x1)------->y-3=2*(x-0)------>y=2x+3

using a graph tool

see the attached figure N 3

the solution is the point (-4,-5)

case 4)

Line 1

Let A (2,0) B (0,-2)

Line 2

Let C (4,5) D (3,3)

Find the equation of the line 1

Line 1

Let A (2,0) B (0,-2)

slope m=(y2-y1) /(x2-x1)------> m=(-2-0) /(0-2)-----> m=1

with m=1 and the point A (2,0) find the equation of the line 1

y-y1=m*(x-x1)------->y-0=1*(x-2)------>y=x-2

Find the equation of the line 2

Line 2

Let C (4,5) D (3,3)

slope m=(y2-y1) /(x2-x1)------> m=(3-5) /(3-4)-----> m=2

with m=1 and the point C (4,5) find the equation of the line 2

y-y1=m*(x-x1)------->y-5=2*(x-4)------>y=2x-8+5-----> y=2x-3

using a graph tool

see the attached figure N 4

the solution is the point (1,-1)

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4 years ago