Question

1. consider a 36×36 grid that is used to make one large rectangle that is joined together. all of the grid is used except for the shaded section of 14×14. use the formula for the area of a rectangle to write an equation for the area of the three combined rectangles as a product of a sum and a difference in terms of 36 and 14. choose the correct answer below

○ a. (36−14)^2=36^2−36(14)+14^2
○ b. 36^2−14^2=(36+14)(36−14)
○ c. 36^2−14^2=(36−14)(36−14)
○ d. (36−14)^2=(36+14)(36−14)

Explanation:

Step1: Recall the formula for the area of a rectangle

The area of a rectangle is \(A = l\times w\). Also, recall the formula \((a - b)^2=a^{2}-2ab + b^{2}\) and \(a^{2}-b^{2}=(a + b)(a - b)\)

Step2: Analyze option A

For \((36-14)^{2}\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), \((36-14)^{2}=36^{2}-2\times36\times14 + 14^{2}
eq36^{2}-36\times14+14^{2}\)

Step3: Analyze option B

For \(36^{2}-14^{2}\), using the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\), where \(a = 36\) and \(b = 14\), \(36^{2}-14^{2}=(36 + 14)(36-14)\)

Step4: Analyze option C

\(36^{2}-14^{2}=(36 + 14)(36-14)
eq(36-14)(36-14)\)

Step5: Analyze option D

\((36 - 14)^{2}=36^{2}-2\times36\times14+14^{2}
eq(36 + 14)(36-14)\)

Answer:

B. \(36^{2}-14^{2}=(36 + 14)(36-14)\)