Examples with solutions for Area of a Rectangle: Extended distributive law

Exercise #1

Given the rectangle ABCD

Given BC=X and the side AB is 4 timis larger than the side BC.

The area of the rectangle is 64 cm².

Calculate the side BC

S=64S=64S=644X4X4XXXXAAABBBCCCDDD

Video Solution

Step-by-Step Solution

Let's begin by calculating the area of the rectangle using the given data:

64=4x×x 64=4x\times x

64=4x2 64=4x^2

Next we will divide both sides by 4:

16=x2 16=x^2

And we will remove the square root:

4=x 4=x

Therefore, BC equals 4.

Answer

4

Exercise #2

The perimeter of a rectangle is 14 cm.

The area of the rectangle is 12 cm².

What are the lengths of its sides?

Video Solution

Step-by-Step Solution

Since in a rectangle each pair of opposite sides are equal to each other, let's call each pair of sides X and Y

Now let's set up a formula to calculate the perimeter of the rectangle:

2x+2y=14 2x+2y=14

Let's divide both sides by 2:

x+y=7 x+y=7

From this formula, we'll calculate X:

x=7y x=7-y

We know that the area of the rectangle equals length times width:

x×y=12 x\times y=12

We know that X equals 7 minus Y, let's substitute this in the formula:

(7y)×y=12 (7-y)\times y=12

7yy2=12 7y-y^2=12

y27y+12=0 y^2-7y+12=0

(y3)×(y4)=0 (y-3)\times(y-4)=0

From this we can claim that:

y=3,y=4 y=3,y=4

Let's go back to the formula we found earlier:

x=7y x=7-y

Let's substitute y equals 3 and we get:

x=73=4 x=7-3=4

Now let's substitute y equals 4 and we get:

x=74=3 x=7-4=3

Therefore, the lengths of the rectangle's sides are 4 and 3

Answer

3, 4

Exercise #3

The height of the house in the drawing is 12x+9 12x+9

its width x+2y x+2y

Given the ceiling height is half the height of the square section.

Express the area of the house shape in the drawing band x and and.

Video Solution

Step-by-Step Solution

Let's draw a line in the middle of the drawing that divides the house into 2

Meaning it divides the triangle and the rectangular part.

The 2 lines represent the heights in both shapes.

If we connect the height of the roof with the height of the rectangular part, we get the total height

Let's put the known data in the formula:

12hsquare+hsquare=12x+9 \frac{1}{2}h_{\text{square}}+h_{square}=12x+9

32hsquare=12x+9 \frac{3}{2}h_{\text{square}}=12x+9

We'll multiply by two thirds and get:

hsquare=2(12x+9)3=2(4x+3) h_{\text{square}}=\frac{2(12x+9)}{3}=2(4x+3)

hsquare=8x+6 h_{\text{square}}=8x+6

If the height of the triangle equals half the height of the rectangular part, we can calculate it using the following formula:

htriangle=12(8x+6)=4x+3 h_{\text{triangle}}=\frac{1}{2}(8x+6)=4x+3

Now we can calculate the area of the triangular part:

(x+2y)×(4x+3)2=4x2+3x+8xy+6y2=2x2+1.5x+4xy+3y \frac{(x+2y)\times(4x+3)}{2}=\frac{4x^2+3x+8xy+6y}{2}=2x^2+1.5x+4xy+3y

Now we can calculate the rectangular part:

(x+2y)×(8x+6)=8x2+6x+16xy+12y (x+2y)\times(8x+6)=8x^2+6x+16xy+12y

Now let's combine the triangular area with the rectangular area to express the total area of the shape:

S=2x2+1.5x+4xy+3y+8x2+6x+16xy+12y S=2x^2+1.5x+4xy+3y+8x^2+6x+16xy+12y

S=10x2+20xy+7.5x+15y S=10x^2+20xy+7.5x+15y

Answer

3x2+8xy+112x+4y2+3y 3x^2+8xy+1\frac{1}{2}x+4y^2+3y