Examples with solutions for Extended Distributive Property: Distributive property in geometry

Exercise #1

Look at the rectangles in the diagram below.

Which has a larger area and by how much?

x+4x+4x+4x+5x+5x+5ABx+3x+2

Video Solution

Step-by-Step Solution

Let's calculate the area of each rectangle step by step:

Step 1: Calculate the area of Rectangle A.
- Dimensions are (x+4)(x+4) and (x+3)(x+3).
- The area is calculated as (x+4)×(x+3) (x+4) \times (x+3) .

Expanding this expression using the distributive property, we get:
(x+4)(x+3)=x⋅x+x⋅3+4⋅x+4⋅3(x+4)(x+3) = x \cdot x + x \cdot 3 + 4 \cdot x + 4 \cdot 3
⇒x2+3x+4x+12\Rightarrow x^2 + 3x + 4x + 12
⇒x2+7x+12\Rightarrow x^2 + 7x + 12

Step 2: Calculate the area of Rectangle B.
- Dimensions are (x+5)(x+5) and (x+2)(x+2).
- The area is calculated as (x+5)×(x+2) (x+5) \times (x+2) .

Using the distributive property to expand:
(x+5)(x+2)=x⋅x+x⋅2+5⋅x+5⋅2(x+5)(x+2) = x \cdot x + x \cdot 2 + 5 \cdot x + 5 \cdot 2
⇒x2+2x+5x+10\Rightarrow x^2 + 2x + 5x + 10
⇒x2+7x+10\Rightarrow x^2 + 7x + 10

Step 3: Compare the areas of Rectangle A and B.
- Area of Rectangle A: x2+7x+12 x^2 + 7x + 12
- Area of Rectangle B: x2+7x+10 x^2 + 7x + 10

Subtract the area of Rectangle B from the area of Rectangle A:
(x2+7x+12)−(x2+7x+10)=(x2+7x+12)−x2−7x−10 (x^2 + 7x + 12) - (x^2 + 7x + 10) = (x^2 + 7x + 12) - x^2 - 7x - 10
⇒x2−x2+7x−7x+12−10 \Rightarrow x^2 - x^2 + 7x - 7x + 12 - 10
⇒2 \Rightarrow 2

Thus, the area of Rectangle A is larger by 2 2 area units.

The correct answer, therefore, is: The area of rectangle A is larger by 2 area units.

Answer

The area of rectangle A is larger by 2 area units.

Exercise #2

What is the area of the trapezoid in the figure?

h=(12x−8) h=(12x-8)

x+5x+5x+513x-213x-213x-2hhh

Video Solution

Step-by-Step Solution

To solve this problem, we apply the formula for the area of a trapezoid:

  • Step 1: Identify and substitute the expressions for the bases and height:
    b1=x+5 b_1 = x + 5 , b2=13x−2 b_2 = 13x - 2 , h=12x−8 h = 12x - 8 .
  • Step 2: Calculate b1+b2 b_1 + b_2 :
    b1+b2=(x+5)+(13x−2)=14x+3 b_1 + b_2 = (x + 5) + (13x - 2) = 14x + 3 .
  • Step 3: Substitute into the area formula:
    A=12×(14x+3)×(12x−8) A = \frac{1}{2} \times (14x + 3) \times (12x - 8) .
  • Step 4: Distribute and simplify:
    A=12×((14x+3)⋅(12x−8)) A = \frac{1}{2} \times ((14x + 3) \cdot (12x - 8)) .
  • Step 5: Expand the product:
    (14x+3)(12x−8)=14x⋅12x+14x⋅(−8)+3⋅12x+3⋅(−8)=168x2−112x+36x−24(14x + 3)(12x - 8) = 14x \cdot 12x + 14x \cdot (-8) + 3 \cdot 12x + 3 \cdot (-8) = 168x^2 - 112x + 36x - 24.
  • Step 6: Combine like terms:
    168x2−76x−24 168x^2 - 76x - 24 .
  • Step 7: Divide by 2 to find the area:
    A=12×(168x2−76x−24)=84x2−38x−12 A = \frac{1}{2} \times (168x^2 - 76x - 24) = 84x^2 - 38x - 12 .

Therefore, the area of the trapezoid is 84x2−38x−12 84x^2 - 38x - 12 .

Answer

84x2−38x−12 84x^2-38x-12

Exercise #3

Look at the rectangle in the figure below. What is its area?

What do a and x need to be for the rectangle to exist?

-a+3x-a+3x-a+3x-5+4x

Video Solution

Step-by-Step Solution

To solve this problem, we'll follow these steps:

  • Step 1: Calculate the area of the rectangle using the area formula for a rectangle.

  • Step 2: Identify the conditions required for a valid rectangle by ensuring positive dimensions.

  • Step 3: Analyze the provided choices to identify the correct answer.

Now, let's work through each step:

Step 1: The width of the rectangle is given as −a+3x-a + 3x, and the height is −5+4x-5 + 4x. The area of a rectangle is calculated by multiplying these two dimensions:

Area=(−a+3x)(−5+4x) \text{Area} = (-a + 3x)(-5 + 4x)

Step 2: We'll expand the expression for the area:

=(−a+3x)(−5+4x)=(−a)(−5)+(−a)(4x)+(3x)(−5)+(3x)(4x) = (-a + 3x)(-5 + 4x) = (-a)(-5) + (-a)(4x) + (3x)(-5) + (3x)(4x)

Step 3: Simplifying each term, we get:

=5a−4ax−15x+12x2 = 5a - 4ax - 15x + 12x^2

Step 4: Reorganize the terms:

=12x2−15x−4ax+5a = 12x^2 - 15x - 4ax + 5a

Next, let's determine the conditions for the rectangle to exist, which means both dimensions must be positive:

  • Width: −a+3x>0  ⟹  3x>a -a + 3x > 0 \implies 3x > a

  • Height: −5+4x>0  ⟹  4x>5  ⟹  x>54=114 -5 + 4x > 0 \implies 4x > 5 \implies x > \frac{5}{4} = 1\frac{1}{4}

Therefore, the conditions for the rectangle to exist are 3x>a 3x > a and x>114 x > 1\frac{1}{4} .

By evaluating the provided choices, we can see the correct choice is:

Area: 12x2−15x−4ax+5a 12x^2-15x-4ax+5a

Conditions: x>114 x > 1\frac{1}{4} and 3x>a 3x > a .

Thus, the correct choice is option 4. Confirming with the given correct answer, our solution matches perfectly.

Answer

Area:

12x2−15x−4ax+5a 12x^2-15x-4ax+5a

Conditions:

x>114 x > 1\frac{1}{4}

3x>a 3x>a

Exercise #4

Look at the circle in the figure.

r=7−5a r=7-5a

What is the area of the circle?

rrr

Video Solution

Step-by-Step Solution

To solve this problem, we'll calculate the area of the circle using the given radius expression. The process involves substituting and simplifying expressions:

  • Step 1: Recognize that the area of a circle is given by the formula A=πr2 A = \pi r^2 , where r r is the radius.
  • Step 2: Substitute the given expression for the radius: r=7−5a r = 7 - 5a .
  • Step 3: Calculate r2 r^2 by expanding (7−5a)2 (7 - 5a)^2 using the identity for squaring a binomial.

Let's apply these steps:

First, substitute the expression for the radius into the area formula:
A=π(7−5a)2 A = \pi (7 - 5a)^2 .

Next, expand (7−5a)2 (7 - 5a)^2 using the distributive property or binomial expansion:
(7−5a)2=72−2⋅7⋅5a+(5a)2=49−70a+25a2 (7 - 5a)^2 = 7^2 - 2 \cdot 7 \cdot 5a + (5a)^2 = 49 - 70a + 25a^2 .

Substituting back, we find:
A=π(49−70a+25a2) A = \pi (49 - 70a + 25a^2) .

The area of the circle, simplified, is:
A=25πa2−70πa+49π A = 25\pi a^2 - 70\pi a + 49\pi .

Therefore, the area of the circle in terms of a a is 25πa2−70πa+49π 25\pi a^2 - 70\pi a + 49\pi .

Answer

25πa2−70πa+49π 25\pi a^2-70\pi a+49\pi

Exercise #5

Look at the triangle in the figure.

h=x−5 h=x-5

What is the area of the triangle in terms of x?hhh

Video Solution

Answer

112x2−3.5x−20 1\frac{1}{2}x^2-3.5x-20