MATH SOLVE

5 months ago

Q:
# If the original square had a side length ofx = 2 inches, then what is the area of thesecond rectangle? Show how you arrived atyour answer.

Accepted Solution

A:

Answer:Part a) The new rectangle labeled in the attached figure N 2Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is [tex]x^{2} +11x+28[/tex]Part c) The area of the second rectangle is 54 in^2Part d) see the explanationStep-by-step explanation:The complete question in the attached figure N 1Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described abovewe know thatThe dimensions of the new rectangle will be[tex]Length=(x+4)\ in[/tex][tex]width=(x+7)\ in[/tex]The diagram of the new rectangle in the attached figure N 2Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomialThe diagram of the new rectangle with their areas in the attached figure N 3we have thatTo find out the area of each portion, multiply its length by its width[tex]A1=(x)(x)=x^{2}\ in^2[/tex][tex]A2=(4)(x)=4x\ in^2[/tex][tex]A3=(x)(7)=7x\ in^2[/tex][tex]A4=(4)(7)=28\ in^2[/tex]The total area of the second rectangle is the sum of the four areas[tex]A=A1+A2+A3+A4[/tex]State the product of (x+4) and (x+7) as a trinomial[tex](x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28[/tex]Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?we know thatThe area of the second rectangle is equal to [tex]A=A1+A2+A3+A4[/tex]For x=2 insubstitute the value of x in the area of each portion[tex]A1=(2)(2)=4\ in^2[/tex][tex]A2=(4)(2)=8\ in^2[/tex][tex]A3=(2)(7)=14\ in^2[/tex][tex]A4=(4)(7)=28\ in^2[/tex][tex]A=4+8+14+28[/tex][tex]A=54\ in^2[/tex]Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 inWe have thatThe trinomial is [tex]A(x)=x^{2} +11x+28[/tex]For x=2 insubstitute and solve for A(x)[tex]A(2)=2^{2} +11(2)+28[/tex][tex]A(2)=4 +22+28[/tex][tex]A(2)=54\ in^2[/tex] ----> verifiedthereforeThe trinomial represent the total area of the second rectangle