Question

Some wire is used to make 3 rectangles: A, B, and C. Rectangle B’s dimensions are 3/5 cm larger than Rectangle A’s dimensions, and Rectangle C’s dimensions are 3/5 cm larger than Rectangle B’s dimensions. Rectangle A is 2 cm by 3 1/5 cm.
a. What is the total area of all three rectangles?
b. If a 40-cm coil of wire was used to form the rectangles, how much wire is left?

Answer 1

Answer:

a. $30.36\ cm^2$

b. $1.6\ cm$

Step-by-step explanation:

a. You know that the dimensions of Rectangle A are $2\ cm* 3\frac{1}{5}\ cm=2\ cm* 3.2 cm$

Since Rectangle B’s dimensions are $\frac{3}{5}\ cm$ (which is 0.6 cm) larger than Rectangle A’s dimensions, then the dimensions of Rectangle B are:

$(2\ cm+0.6\ cm)( 3.2\ cm+0.6\ cm)=2.6\ cm*3.8\ cm$

Since Rectangle C’s dimensions are $\frac{3}{5}\ cm$ (which is 0.6 cm) larger than Rectangle B's dimensions, then the dimensions of Rectangle C are:

$(2.6\ cm+0.6\ cm)( 3.8\ cm+0.6\ cm)=3.2\ cm*4.4\ cm$

The find the total area of all three rectangles you must add the products obtained when you multiply their dimensions. Then:

$A_t=(2\ cm* 3.2 cm)+(2.6\ cm*3.8\ cm)+(3.2\ cm*4.4\ cm)\\\\A_t=30.36\ cm^2$

b. The perimeter of a rectangle can be calculated with this formula:

$P=2l+2w$

Where "l" is the lenght and "w" is the width.

Knowing the dimensions of each rectangleg, you can calculate the total perimeter as follows:

$P_t=(2)[(2\ cm+ 3.2 cm)+(2.6\ cm+3.8\ cm)+(3.2\ cm+4.4\ cm)]\\\\P_t=38.4\ cm$

Then, if a 40-cm coil of wire was used to form the rectangles, the amount of wire that is left is:

$40\ cm-38.4\ cm=1.6\ cm$