Distributive Property in Geometry - Pre Algebra

1. A rhombus has side length $(2p + 1)$ and height $(p - 3)$. If we use the formula $A = \text{base} \times \text{height}$, which shows the correct first step of applying the distributive property?
   Answer: $A = 2p(p - 3) + 1(p - 3)$ When you see a problem asking for the "first step" of applying the distributive property, you need to set up the multiplication without actually completing it. The distributive property breaks down multiplication of expressions like $(a + b)(c + d)$ into separate parts. Here, you're finding the area using $A = \text{base} \times \text{height} = (2p + 1)(p - 3)$. The first step of the distributive property is to distribute each term in the first expression to the entire second expression. This means taking $2p$ and multiplying it by $(p - 3)$, then taking $1$ and multiplying it by $(p - 3)$. So the correct setup is $A = 2p(p - 3) + 1(p - 3)$, which is choice C. Choice A shows what happens after you've already distributed and partially simplified—this skips the "first step" the question asks for. Choice B goes even further by completely combining like terms, making it a final answer rather than a first step. Choice D attempts to distribute but incorrectly breaks apart the second expression $(p - 3)$ into separate multiplications with $p$ and $3$, which isn't how the distributive property works. Remember that when a question asks for the "first step" of the distributive property, you're looking for the setup that shows each term from the first expression being distributed to the complete second expression. Don't be tempted by answers that show partial or complete simplification—stick to what the question specifically requests.
2. A parallelogram has a base of $(4y + 3)$ units and a height of $(y - 2)$ units. If the area formula for a parallelogram is $A = bh$, which step correctly applies the distributive property to find the area?
   Answer: $A = 4y(y - 2) + 3(y - 2)$ The distributive property can be applied by factoring out each term in the first expression: $(4y + 3)(y - 2) = 4y(y - 2) + 3(y - 2)$. This correctly shows the first step of distribution. Choice A shows an incorrect intermediate step with wrong signs. Choice B shows the complete distribution but isn't the clearest first step. Choice C shows terms after distribution but with incorrect arrangement.
3. A trapezoid has an area formula $A = \frac{1}{2}(b_1 + b_2)h$. If $b_1 = (2n + 1)$, $b_2 = (n + 5)$, and $h = (3n - 2)$, which expression shows the correct application of the distributive property before simplification?
   Answer: $\frac{1}{2}[(2n + 1)(3n - 2) + (n + 5)(3n - 2)]$ The area formula $A = \frac{1}{2}(b_1 + b_2)h$ becomes $A = \frac{1}{2}[(2n + 1) + (n + 5)](3n - 2)$. Using the distributive property, this equals $\frac{1}{2}[(2n + 1)(3n - 2) + (n + 5)(3n - 2)]$. Choice A correctly shows this distribution step. Choice B incorrectly simplifies the bases first. Choice C doesn't apply the distributive property yet. Choice D shows the result after full expansion but doesn't clearly demonstrate the distributive property application.
4. The surface area of one face of a rectangular prism is $(3z + 2)(z + 4)$. Using the distributive property, which expression correctly shows the intermediate step before combining like terms?
   Answer: $3z^2 + 12z + 2z + 8$ Using the distributive property: $(3z + 2)(z + 4) = 3z(z) + 3z(4) + 2(z) + 2(4) = 3z^2 + 12z + 2z + 8$. Choice A correctly shows this intermediate step with terms in the proper order before combining like terms. Choice B shows the final simplified form but skips the distributive property demonstration. Choice C shows an alternative distributive form but doesn't complete the distribution. Choice D has the same terms but in incorrect order, which could indicate computational errors.
5. The area of a triangle is given by $A = \frac{1}{2}bh$. If the base is $(6m + 4)$ and the height is $(m + 3)$, which expression represents $\frac{1}{2}$ times the distributed product $bh$?
   Answer: $\frac{1}{2}(6m^2 + 18m + 4m + 12)$ First distribute $(6m + 4)(m + 3) = 6m(m) + 6m(3) + 4(m) + 4(3) = 6m^2 + 18m + 4m + 12$. Then multiply by $\frac{1}{2}$: $A = \frac{1}{2}(6m^2 + 18m + 4m + 12)$. Choice B correctly shows this step. Choice A shows the final simplified result but skips showing the distributive property application. Choice C shows the result in fraction form but doesn't show the distribution step. Choice D incorrectly distributes the $\frac{1}{2}$ only to some terms.
6. A pentagon can be divided into a rectangle with dimensions $(t + 3)$ by $(2t - 1)$ and a triangle with base $(t + 3)$ and height $t$. Using the distributive property in the area formulas, which expression represents the total area?
   Answer: $(t + 3)\left[(2t - 1) + \frac{t}{2}\right]$ When working with composite shapes, you need to find the total area by adding the areas of individual components. The key insight here is recognizing when you can use the distributive property to factor expressions efficiently. The pentagon consists of a rectangle with area $(t + 3)(2t - 1)$ and a triangle with area $\frac{1}{2}(t + 3)(t)$. The total area is their sum: $(t + 3)(2t - 1) + \frac{1}{2}(t + 3)(t)$. Notice that both terms share the common factor $(t + 3)$. Using the distributive property, you can factor this out: $(t + 3)\left[(2t - 1) + \frac{t}{2}\right]$. This is answer choice C. Looking at the wrong answers: Choice A gives the correct sum but doesn't apply the distributive property as the question requires. Choice B multiplies everything out completely, which again doesn't demonstrate using the distributive property in the area formulas. Choice D looks similar to C but is missing the crucial brackets around the entire second factor $\left[(2t - 1) + \frac{t}{2}\right]$. Without proper grouping, this expression would be interpreted as $(t + 3)(2t - 1) + \frac{t}{2}$, which incorrectly adds $\frac{t}{2}$ instead of $\frac{1}{2}(t + 3)(t)$. Study tip: When asked to use the distributive property with area problems, look for common factors in your area expressions. Factor out shared terms rather than expanding everything – this often leads to cleaner, more manageable expressions.

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