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

Exercise #1

Calculate the area of the rectangle

y+2y+2y+2x+5x+5x+5

Video Solution

Step-by-Step Solution

Let's begin by reminding ourselves of the formula to calculate the area of a rectangle: width X length

S=wh S=w⋅h

Where:

S = area

w = width

h = height

We extract the data from the sides of the rectangle in the figure.

w=x+5 w=x+5 h=y+2 h=y+2

We then substitute the above data into the formula in order to calculate the area of the rectangle:

S=wh=(x+5)(y+2) S=w⋅h=(x+5)(y+2)

We use the formula of the extended distributive property:

(a+b)(c+d)=ac+ad+bc+bd (a+b)(c+d)=ac+ad+bc+bd

We once again substitute and solve the problem as follows:

S=(x+5)(y+2)=(x)(y)+(x)(2)+(5)(y)+(5)(2) S=(x+5)(y+2)=(x)(y)+(x)(2)+(5)(y)+(5)(2)

(x)(y)+(x)(2)+(5)(y)+(5)(2)=xy+2x+5y+10 (x)(y)+(x)(2)+(5)(y)+(5)(2)=xy+2x+5y+10

Therefore, the correct answer is option C: xy+2x+5y+10.

Answer

xy+2x+5y+10 xy+2x+5y+10

Exercise #2

Express the area of the rectangle below in terms of y and z.

3y3y3yy+3z

Video Solution

Step-by-Step Solution

Let us begin by reminding ourselves of the formula to calculate the area of a rectangle: width X height

S=wh S=w⋅h

Where:

S = area

w = width

h = height

We must first extract the data from the sides of the rectangle shown in the figure.

w=3y w=3y h=y+3z h=y+3z

We then insert the known data into the formula in order to calculate the area of the rectangle:

S=wh=(y+3z)(3y) S=w⋅h=(y+3z)(3y)

We use the distributive property formula:

a(b+c)=ab+ac a\left(b+c\right)=ab+ac

We substitute all known data and solve as follows:

S=(y+3z)(3y)=(3y)(y+3z) S=(y+3z)(3y)=(3y)(y+3z)

(3y)(y+3z)=(3y)(y)+(3y)(3z) (3y)(y+3z)=(3y)(y)+(3y)(3z)

(3y)(y)+(3y)(3z)=3y2+9yz (3y)(y)+(3y)(3z)=3y^2+9yz

Keep in mind that because there is a multiplication operation, the order of the terms in the expression can be changed, hence:

(y+3z)(3y)=(3y)(y+3z) (y+3z)(3y)=(3y)(y+3z)

Therefore, the correct answer is option D: 3y2+9yz 3y^2+9yz

Answer

3y2+9yz 3y^2+9yz

Exercise #3

Calculate the area of the rectangle below in terms of a and b.

a+3a+3a+3b+8b+8b+8

Video Solution

Step-by-Step Solution

Let us begin by reminding ourselves of the formula to calculate the area of a rectangle: width X length

S=wh S=w⋅h

When:

S = area

w = width

h = height

We take data from the sides of the rectangle in the figure.w=b+8 w=b+8 h=a+3 h=a+3

We then substitute the above data into the formula in order to calculate the area of the rectangle:

S=wh=(b+8)(a+3) S=w⋅h = (b+8)(a+3)

We use the formula of the extended distributive property:

(a+b)(c+d)=ac+ad+bc+bd (a+b)(c+d)=ac+ad+bc+bd

We substitute once more and solve the problem as follows:

S=(b+8)(a+3)=(b)(a)+(b)(3)+(8)(a)+(8)(3) S=(b+8)(a+3)=(b)(a)+(b)(3)+(8)(a)+(8)(3)

(b)(a)+(b)(3)+(8)(a)+(8)(3)=ab+3b+8a+24 (b)(a)+(b)(3)+(8)(a)+(8)(3)=ab+3b+8a+24

Therefore, the correct answer is option B: ab+8a+3b+24.

Keep in mind that, since there are only addition operations, the order of the terms in the expression can be changed and, therefore,

ab+3b+8a+24=ab+8a+3b+24 ab+3b+8a+24=ab+8a+3b+24

Answer

ab + 8a + 3b + 24

Exercise #4

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 #5

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

Whilst the width of the house x+2y x+2y

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

Express the area of the house shape in the drawing :

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 obtain the total height.

Let's insert 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 as follows:

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

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