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A right triangle has base \((x+5)\), height \((x+9)\), and area 95 square units. Which quadratic equation represents this situation?

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Answer and explanation

Correct answer: \(x^2+14x-145=0\)

Area of the triangle is given by \(\frac{1}{2}\times\text{base}\times\text{height}\). So \(\frac{1}{2}(x+5)(x+9)=95\). Multiplying both sides by 2 gives \((x+5)(x+9)=190\). Expanding yields \(x^2+14x+45=190\), which simplifies to \(x^2+14x-145=0\); hence option A is correct. Option B is the common mistake of setting \((x+5)(x+9)=95\) (forgetting the factor 1/2). Options C and D have incorrect constant terms. Exam tip: write the area formula first and multiply by 2 immediately to avoid sign/constant errors when forming the quadratic.

Related tags

Quadratic-EquationsWord-ProblemTriangle-AreaAlgebraQuadratic-Formation

Frequently asked questions

What is the correct answer to this question?

\(x^2+14x-145=0\)

Why is this the correct answer?

Area of the triangle is given by \(\frac{1}{2}\times\text{base}\times\text{height}\). So \(\frac{1}{2}(x+5)(x+9)=95\). Multiplying both sides by 2 gives \((x+5)(x+9)=190\). Expanding yields \(x^2+14x+45=190\), which simplifies to \(x^2+14x-145=0\); hence option A is correct. Option B is the common mistake of setting \((x+5)(x+9)=95\) (forgetting the factor 1/2). Options C and D have incorrect constant terms. Exam tip: write the area formula first and multiply by 2 immediately to avoid sign/constant errors when forming the quadratic.

Which subject and chapter does this question cover?

This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.

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