# finding the perimeter

• Oct 1st 2009, 09:07 PM
anshulbshah
finding the perimeter
Attachment 13161
Seven congruent rectangles form a bigger rectangle as shown in the attachment.If the area of the bigger attachment is 336 square units,what is the perimeter of the bigger rectangle.(Sorry fig. not accurate)
• Oct 1st 2009, 10:28 PM
Matt Westwood
Quote:

Originally Posted by anshulbshah
Attachment 13161
Seven congruent rectangles form a bigger rectangle as shown in the attachment.If the area of the bigger attachment is 336 square units,what is the perimeter of the bigger rectangle.(Sorry fig. not accurate)

You can get an equation relating the lengths of the sides of one of the smaller rectangles straight away, which will help.

Let the sides of the small rects be a and b.

Then $\displaystyle 3a = 4b$.

Then the perimeter is $\displaystyle 4b + 2 (a+b) + 3a$.

The area is given by $\displaystyle (a+b)(3a) = 336$.

Then I expect you just need to eliminate either a or b from this system of simultaneous equations and the solution should drop into your lap.
• Oct 1st 2009, 10:30 PM
jashansinghal
area of 7 rectangles=336 square unit
area of 1 rectangle=48 square units
lets take the breadth of smaller rectangle as B and length as L
[B will be the smaller side as per the figure]

As per the figure
4. B= 3. L
B= 3L/4

area of smaller rectangle=48square units
so,
L.B=48
now we substitute the value of B in this equation
L. 3L/4=48
so
L=8
B=6
(Speechless)(Speechless)
as per the figure
area of bigger rectangle=
5L+6B
=> 5X8+6X6
=>40 + 36
= 76 units
• Oct 1st 2009, 10:33 PM
Soroban
Hello, anshulbshah!

Quote:

Seven congruent rectangles form a bigger rectangle as shown in the diagram.
If the area of the bigger rectangle is 336 square units,
what is the perimeter of the bigger rectangle?

Code:

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Let: .$\displaystyle \begin{array}{ccc}x &=& \text{length of a small rectangle} \\ y &=& \text{width of a small rectangle} \end{array}$

From the top and bottom edges of the rectangle,
. . we see that: .$\displaystyle 4y \:=\:3x \quad\Rightarrow\quad y \:=\:\tfrac{3}{4}x$ .[1]

The length of the big rectangle is: $\displaystyle L \,=\,3x$
The width of the big rectanble is: $\displaystyle W \,=\,x + y$
. . Hence, its area is: .$\displaystyle L\cdot W \:=\:(3x)(x+y) \:=\:3x^2 + 3xy$

We are told that the area is 336 unitsē.
. . So we have: .$\displaystyle 3x^2 + 3xy \:=\:336$ .[2]

Substitute [1] into [2]: .$\displaystyle 3x^2 + 3x\left(\tfrac{3}{4}x\right) \:=\:336 \quad\Rightarrow\quad 3x^2 + \tfrac{9}{4}x^2 \:=\:336$

. . $\displaystyle \tfrac{21}{4}x^2 \:=\:336 \quad\Rightarrow\quad x^2 \:=\:64 \quad\Rightarrow\quad \boxed{x \:=\:8}$

Substitute into [1]: .$\displaystyle y \:=\:\tfrac{3}{4}(8) \quad\Rightarrow\quad \boxed{y \:=\:6}$

The length of the big rectangle is: .$\displaystyle L \:=\:3x \:=\:3(8) \quad\Rightarrow\quad L\:=\:24$
The width of the big rectangle is: .$\displaystyle W \:=\:x+y \:=\:8+6 \quad\Rightarrow\quad W\:=\:14$

Therefore, the perimeter is: .$\displaystyle 2L + 2W \;=\;2(24) + 2(14) \;=\;76$ units.

Edit: Too slow again . . .
• Oct 1st 2009, 10:36 PM
Matt Westwood
Quote:

Originally Posted by Soroban
Edit: Too slow again . . .

... too complete as well ...