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Sagot :
Given
The base and height of a parallelogram is (x-7)meters and (x+9)meters respectively.
[tex]\begin{gathered} \text{Base(b) = (x-7)meters} \\ Height=\text{ (x+9)meters} \end{gathered}[/tex]Formula
[tex]\begin{gathered} \text{The area of a Paralelogram = Base}\times Height \\ \end{gathered}[/tex][tex]\text{The area is 192m}^2[/tex]We now substitute into the formula
[tex]\begin{gathered} 192=\mleft(x-7\mright)\mleft(x+9\mright) \\ 192=x^2+9x-7x-63 \\ 192=x^2+2x-63 \\ \text{REARRANGE} \\ x^2+2x-63-192=0 \\ x^2+2x-255=0 \\ \end{gathered}[/tex]It is now Quadratic Equation
[tex]\begin{gathered} a=1,\text{ b=2 and c=-255} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ We\text{ can now substitute into the quadratic formula} \\ x=\frac{-2\pm\sqrt[]{2^2^{}-4\times1\times-255}}{2\times1} \\ \\ x=\frac{-2\pm\sqrt[]{4^{}--1020}}{2} \\ \\ \end{gathered}[/tex][tex]\begin{gathered} x=\frac{-2\pm\sqrt[]{1024}}{2} \\ \\ x=\frac{-2\pm32}{2} \\ \\ x=\frac{-2+32}{2}=\frac{30}{2}=15 \\ \\ or \\ x=\frac{-2-32}{2}=-\frac{34}{2}=-17 \end{gathered}[/tex]For distance or dimension it can't be negative, so we choose the positive
x=15
Recall from the question
Base=(x-7)meters
Height= (x+9)meters
We can now replace x with 15
[tex]\begin{gathered} \text{Base}=\text{ 15-7} \\ \text{Base}=8m \\ \\ \text{Height =15+9} \\ \text{Height}=24m \end{gathered}[/tex]The final answer
Base is 8m
Height is 24m
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