The length of a rectangle is 6 inches greater than the width. If each dimension is increased by 4 inches, the new area will be 104 square inches larger, Find the area of the original rectangle. es A) 81 in 2 B) 96 in? 108 in? D) 112 in

Question

03-11-2022
Mathematics

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The length of a rectangle is 6 inches greater than the width. If each dimension is increased by 4 inches, the new area will be 104 square inches larger, Find the area of the original rectangle. es A) 81 in 2 B) 96 in? 108 in? D) 112 in

Sagot :

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Solve 4/9x+7=1/2. Solve 1/8x−12=3/4.

KhyirR435014 KhyirR435014 · Answer 1 · 2022-11-03T03:05:57-04:00

Step 1: Problem

Area of a rectangle.

Step 2: Concept

Area of a rectangle = Length x width

Draw the diagram with a label.

Step 3: Method

Area of original rectangle = Length x width

= x(x + 6)

[tex]=x^2\text{ + 6x}[/tex]

Find the area of the new ractangle.

Area = Length x width

= (x + 10)(x + 4)

[tex]\begin{gathered} x^2\text{ + 4x + 10x + 40} \\ x^2\text{ + 14x + 40 } \\ x^2\text{ + 14x + 40 } \\ x^2\text{ + 14x }+\text{ 40} \end{gathered}[/tex]

New area - Original area = 104

[tex]\begin{gathered} x^2+14x+40-(x^2\text{ + 6x) = 104} \\ x^2+14x+40-x^2\text{ - 6x = 104} \\ 8x\text{ + 40 = 104} \\ 8x\text{ = 104 - 40} \\ 8x\text{ = 64} \\ x\text{ = }\frac{64}{8} \\ \text{x = 8} \end{gathered}[/tex]

Step 4: Final answer

Area of the original figure = x(x + 6)

= 8(8 + 6)

= 8 x 14

= 112 in square

Sagot :

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