A square park has the same area as a rectangular park. The side of the square is 30 m, while the breadth of the rectangle is (x) m and its length is (x+15) m. What is the positive value of the rectangle’s breadth (x)?
Answer and explanation
Correct answer: m
Equality of the areas gives \(x(x+15)=30^2=900\), so \(x^2+15x-900=0\). Solving it, \(x=\frac{-15\pm\sqrt{3825}}{2}=\frac{15(-1\pm\sqrt{17})}{2}\). Since a breadth must be positive, the valid value is \(x=\frac{15(\sqrt{17}-1)}{2}\approx23.42\) m; the other root is negative and therefore invalid. Exam tip: In area word problems, form the equation first and retain only the positive root for a physical length.
Frequently asked questions
What is the correct answer to this question?
m
Why is this the correct answer?
Equality of the areas gives \(x(x+15)=30^2=900\), so \(x^2+15x-900=0\). Solving it, \(x=\frac{-15\pm\sqrt{3825}}{2}=\frac{15(-1\pm\sqrt{17})}{2}\). Since a breadth must be positive, the valid value is \(x=\frac{15(\sqrt{17}-1)}{2}\approx23.42\) m; the other root is negative and therefore invalid. Exam tip: In area word problems, form the equation first and retain only the positive root for a physical length.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Word Problems and Applications.
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