Hello, anshulbshah!
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 . . .