A rectangular cloth must be 14 1/2 cm longer than ots width. What are the largest possible demensions of teh rectangle if teh perimeter is at most 275 cm?

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Posted on Jan 02, 2017

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Solve the resulting quadratic equation

2X^2+3X-90=0

Discriminant: (3)^2-4(2)(-90)=729=(27)^2

Two roots

X_1=(1/4)*(-3+27)=6

X-2=(1/4)*(-3-27) =-(15/2), negative

Since the width must be positive, reject the negative root and keep X_1=6

**Width =6**

Length=2(6)+3=15

Check 6*(15)=90. Checks OK

2X^2+3X-90=0

Discriminant: (3)^2-4(2)(-90)=729=(27)^2

Two roots

X_1=(1/4)*(-3+27)=6

X-2=(1/4)*(-3-27) =-(15/2), negative

Since the width must be positive, reject the negative root and keep X_1=6

Length=2(6)+3=15

Check 6*(15)=90. Checks OK

Apr 23, 2014 | Office Equipment & Supplies

Right Rectangular prism, Length L, width W, Height H

Volume=2058 cm^3=L*W*H

However

L=3W, and H=2W

(3W)*W*(2W)=**6 W^3=2058**

W^3=(2058/6)=343

**W**=Cube root of 343=**(343)^(1/3)=7 cm**

**L=3*7=21 cm**

H=2*7=14 cm

Volume=2058 cm^3=L*W*H

However

L=3W, and H=2W

(3W)*W*(2W)=

W^3=(2058/6)=343

H=2*7=14 cm

Oct 29, 2013 | Computers & Internet

The rectangle is 11cm by 19cm.

The perimeter is 60, so the width and length must add to 30. The length is 8 more than the width, or width+width+8=30. Solve that for width, then you can calculate the length based on the width.

If this is homework, be sure to show your work.

The perimeter is 60, so the width and length must add to 30. The length is 8 more than the width, or width+width+8=30. Solve that for width, then you can calculate the length based on the width.

If this is homework, be sure to show your work.

Oct 11, 2013 | Lands Phones

Since the 'base' and 'top' of any prism have the same shape, the surface area can be found by

- surface area of prism = 2 * area of base + perimeter of base * H

- surface area = 2LW + 2(L+W)H
- = 2 * 7 * 4 + 2*(7+4)*5
- = 56 + 2 * 11 * 5
- = 56 +110
- = 166 sq cm

Sep 05, 2011 | Encore Math Advantage Algebra II and...

Area = length X width, and you know the area and the width. So,

4122 sq cm = length X 3 cm

length = 4122 sq cm / 3 cm

length = 1374 cm

(That is one long, skinny rectangle, isn't it?)

4122 sq cm = length X 3 cm

length = 4122 sq cm / 3 cm

length = 1374 cm

(That is one long, skinny rectangle, isn't it?)

Mar 02, 2011 | Office Equipment & Supplies

1. Well this one is pretty simple. Since you are not supposed to trim the length trim the width instead.

to make the perimeter 0.4 m shorter cut the cloth by 0.2 m width the picture will make it clear.

2. let the length of rectangle be x cm

and breadth = 20 cm

now original perimeter = 2(x+20) cm

new length = x-30 cm

breadth remains same = 20 cm

new perimeter = 2(x-30+20) cm = 2(x-10) cm

According to question

new perimeter = old perimeter /2

2(x-10) = 2(x+20) /2

2x -20 = x + 20

x = 40 cm

hence length of longer side 40 cm.

If you want to directly comm ankitsharma.220@gmail.com

to make the perimeter 0.4 m shorter cut the cloth by 0.2 m width the picture will make it clear.

2. let the length of rectangle be x cm

and breadth = 20 cm

now original perimeter = 2(x+20) cm

new length = x-30 cm

breadth remains same = 20 cm

new perimeter = 2(x-30+20) cm = 2(x-10) cm

According to question

new perimeter = old perimeter /2

2(x-10) = 2(x+20) /2

2x -20 = x + 20

x = 40 cm

hence length of longer side 40 cm.

If you want to directly comm ankitsharma.220@gmail.com

Feb 13, 2011 | Vivendi Excel@ Mathematics Study Skills...

The diameter of the circle is the diagonal of the rectangle use Pythagoras this rectangle is a made from two 3,4,5 triangles, so the diameter is 5cm, the radius will be 2.5cm don't forget the BODMAS ordering rule (Brackets of Division Multiplication Addition and Subtraction) Also rounding to 2 decimal places. This solution is based on all four corners of the rectangle touching the circumference of the circle.

Area of a circle is P (22/7) i x Radius (squared ^2) ====> 22/7 x 2.5cm ^ 2 ====> 22/7 x 6.25cm

TOTAL AREA of CIRCLE = 19.64 cm^2

Area of the rectangle is length x width ====> 4 cm x 3 cm

TOTAL AREA of RECTANGLE = 12 cm ^2

The remaining portion is the circle area minus the rectangle 19.64-12 = 7.64 cm^2

SIMPLES brought to you by SpideRaY @ http://www.windowstipsclub.com

Area of a circle is P (22/7) i x Radius (squared ^2) ====> 22/7 x 2.5cm ^ 2 ====> 22/7 x 6.25cm

TOTAL AREA of CIRCLE = 19.64 cm^2

Area of the rectangle is length x width ====> 4 cm x 3 cm

TOTAL AREA of RECTANGLE = 12 cm ^2

The remaining portion is the circle area minus the rectangle 19.64-12 = 7.64 cm^2

SIMPLES brought to you by SpideRaY @ http://www.windowstipsclub.com

Jan 31, 2011 | Computers & Internet

Let x = the short side. Then the long side is the short side plus 6 meters

now we have 2 sides that are x meters long, add those together gives us 2x for the length

of both short sides

The length of the long side is x + 6 for one side and x + 6 for the other side. Added together we have x + 6 + x + 6 = 2x + 12

Adding the 4 sides together we get 2x ( the 2 short sides) + 2x + 12 (the long sides) we get 4x + 12 = the perimeter or 628. 4x + 12 = 628. Subtract 12 from both sides of the equation leaves 4x = 616. divide both sides by 4 leaves x = 154 So the width is 154 meters and the length is 160 meters.

Hope this helps

Good luck Loringh

now we have 2 sides that are x meters long, add those together gives us 2x for the length

of both short sides

The length of the long side is x + 6 for one side and x + 6 for the other side. Added together we have x + 6 + x + 6 = 2x + 12

Adding the 4 sides together we get 2x ( the 2 short sides) + 2x + 12 (the long sides) we get 4x + 12 = the perimeter or 628. 4x + 12 = 628. Subtract 12 from both sides of the equation leaves 4x = 616. divide both sides by 4 leaves x = 154 So the width is 154 meters and the length is 160 meters.

Hope this helps

Good luck Loringh

Oct 06, 2008 | SoftMath Algebrator - Algebra Homework...

The length of a rectangle is twice its width w. a second rectangle, which is 8 cm longer and 3 cm narrower than the first rectangle, has perimeter 154 cm. Make a sketch of the rectangles expressing all dimensions of w. Then find the dimensions of each rectangle.

Oct 03, 2008 | Texas Instruments BA Real Estate...

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