Notes: Marisol is painting on a piece of canvas that has an area of 180 square inches. The length of the painting is 1 1/4 times the width. What are the...

Wednesday, June 24, 2015

Marisol is painting on a piece of canvas that has an area of 180 square inches. The length of the painting is 1 1/4 times the width. What are the...

We are given a rectangular canvas with area 180 square inches with a length 5/4 as long as the width and we are asked to find the dimensions:

Area = length times width, so we let w=width in inches and l=5/4w=length in inches.

Then $\displaystyle{w}{\left(\frac{{5}}{{4}}{w}\right)}={180}$

$\displaystyle\frac{{5}}{{4}}{{w}}^{{2}}={180}$

$\displaystyle{{w}}^{{2}}={180}\cdot\frac{{4}}{{5}}$

$\displaystyle{{w}}^{{2}}={144}$

Since the width must be positive, w=12.

Then $\displaystyle{l}=\frac{{5}}{{4}}{w}=\frac{{5}}{{4}}\cdot{12}={15}$

The dimensions are 12" by 15"

...

We are given a rectangular canvas with area 180 square inches with a length 5/4 as long as the width and we are asked to find the dimensions:

Area = length times width, so we let w=width in inches and l=5/4w=length in inches.

Then $\displaystyle{w}{\left(\frac{{5}}{{4}}{w}\right)}={180}$

$\displaystyle\frac{{5}}{{4}}{{w}}^{{2}}={180}$

$\displaystyle{{w}}^{{2}}={180}\cdot\frac{{4}}{{5}}$

$\displaystyle{{w}}^{{2}}={144}$

Since the width must be positive, w=12.

Then $\displaystyle{l}=\frac{{5}}{{4}}{w}=\frac{{5}}{{4}}\cdot{12}={15}$

The dimensions are 12" by 15"

Check: 12x15=180 as required.

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