Answer:
The dimensions of the original rectangle are 8 cm x 6 cm
Step-by-step explanation:
Let
x----> the length of the original rectangle
y----> the width of the original rectangle
we know that
The area of the original rectangle is
[tex]A=xy[/tex]
[tex]A=48\ cm^{2}[/tex]
so
[tex]48=xy[/tex]
[tex]x=48/y[/tex] -----> equation A
If the length and width are each increased by 4 centimeters, the area of the larger rectangle is 120 square centimeters
so
[tex]120=(x+4)(y+4)\\120=xy+4x+4y+16[/tex]
[tex]xy+4x+4y-104=0[/tex] -----> equation B
Substitute equation A in equation B
[tex]48+4(48/y)+4y-104=0[/tex]
Multiply by y both sides
[tex]48y+192+4y^{2}-104y=0\\ \\4y^{2}-56y+192=0[/tex]
using a graphing tool------> solve the quadratic equation
I assume that the length is greater than the width
The solution is [tex]y=6\ cm[/tex]
see the attached figure
Find the value of x
[tex]x=48/6=8\ cm[/tex]
The dimensions of the original rectangle are 8 cm x 6 cm
The dimensions of the original rectangle is 6cm and 8cm respectively
The area of the rectangle is
[tex]xy = 48 ...equation(i)[/tex]
When the dimension of the rectangle changed,
[tex]l = x+4\\w=y+4[/tex]
The new area of the rectangle becomes
[tex]A=(x+4)(y+4)=120\\xy+4x+4y+16=120[/tex]
substitute the value of xy, x and y into the equation
[tex]48+4x+4(48/x)+16=120\\4x+192/x+64=120\\x+48/x+16=30\\x^2+48+16x=30x\\x^2-14x+48=0[/tex]
solve the quadratic equation by factorization method
[tex]x^2-14x+48=0\\x^2-8x-6x+48=0\\(x-8)(x-6)=0\\x=8 \\or\\ x = 6[/tex]
From the calculations above, the dimensions of the original rectangle is 6cm and 8cm respectively.
Learn more on area of a rectangle here;
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