Posted by **olav** on Sunday, December 12, 2010 at 1:58pm.

I don't see a problem but I will rewrite the structure.

Make two rectangles attached together.

Call the first rectangle ABHG (AG top, BH bottom).

Call the second rectangle GHCD (HC top, GD bottom).

Draw a horizontal line from line AB to line GH and name the points E on line AB and F on line GH.

example:

A----G-------------D

|.........|..................|

|.........|..................|

E----F..................|

|.........|..................|

|.........|..................|

|.........|..................|

|.........|..................|

B----H-------------C

Rectangles AEFG, EBHF and GHCD are similar.

The length of AE is 4cm.

The length of EB is 9cm.

What is the area of rectangle GHCD ?

What are the lengths of HC and EF ?

- Gr. 9 Math (continued) -
**MathMate**, Sunday, December 12, 2010 at 3:16pm
Given AE=4, EB=9,

If AE≠EB, the only way that AEFG and EBHF are similar is that AE<EF<EB, which means really that

AEFG is similar to FEBH to ensure the vertices correspond.

If that's the case,

Let EF=x, then

AE/x = x/EB, or

x² = AE*EB = 36

EF = x = 6 (negative root rejected)

Now for the rectangle GHCD, there are two ways to be similar to rectangle AEFG, i.e.

HC=(2/3)GH, or

HC=(3/2)GH

I will let you complete the problem.

- Gr. 9 Math (continued) -
**olav**, Sunday, December 12, 2010 at 4:12pm
I looked at it another way.

I used rectangle FEBH; area = 9 x EF.

I used rectangle AEFG; area = 4 x EF.

Ratio of area FEBH to AEFG = 2.25

The square root of area is ratio of the perimeters = 1.5

Perimeter of AEFG = 2 x 9 + 2 x EF

Perimeter of FEBH = 2 x 4 + 2 x EF.

Since FEBH is 1.5 larger than AEFG then

1.5 x (8 + 2EF) = 18 + 2EF,

12 + 3EF = 18 + 2EF,

1EF = 6.

Now I know ratio of AE / EF = 2/3.

It works out the same FE / HF = 2/3.

Therefore GH / HC = 2/3,

13 x 3 / 2 = HC = 19.5.

Therefore the area of GHCD =

13 x 19.5 = 253.5

Is this correct ?

- Gr. 9 Math (continued) -
**MathMate**, Sunday, December 12, 2010 at 4:22pm
That is correct, but HC=19.5 is only one of the two possible solutions.

If HC=13*(2/3)=26/3, the rectangles are still similar (but turned 90°).

So the other solution is

Area = 13*(26/3) = 112.67

- Gr. 9 Math (continued) -
**olav**, Sunday, December 12, 2010 at 10:33pm
Thank you for all your help.

- Gr. 9 Math (continued) -
**Alex**, Wednesday, March 2, 2011 at 8:19pm
O_O

## Answer This Question

## Related Questions

- math - There is on large rectangle ABCD, top line is AD and bottom is BC. A line...
- Grade 9 Math - There is one large rectangle ABCD, top line is AD and bottom is ...
- Gr. 9 math - There is one large rectangle ABCD, top line is AD and bottom is BC...
- Algebra - a rectangle with perimeter of 198 cm is divided into 5 congruent ...
- Optimization (Math) - The mayor of a village wants to build a library of which ...
- geometry - Two rectangles are similar. If the height of the first rectangle is 3...
- Calculus - You are given a 60 inch by 30 inch piece of cardboad and asked to ...
- MATH Middle School - The area of two rectangles is 360 square centimeters. The ...
- math - You have several 5x11 rectangles, using only these rectangles, what can ...
- Geometry - I have a rectangle MNOP ~ rectangle STUV. What is the length of line ...

More Related Questions