A right triangle has a base of \((x+3)\) units, a height of \((x+7)\) units, and an area of 55 square units. Which quadratic equation represents this situation?
Answer and explanation
Correct answer: \(x^2+10x-89=0\)
The area of a triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Hence, \(\frac{1}{2}(x+3)(x+7)=55\), so \((x+3)(x+7)=110\). Expanding gives \(x^2+10x+21=110\), and therefore \(x^2+10x-89=0\). Option B results from mishandling the factor \(\frac{1}{2}\) in the area formula. Exam tip: In triangle-area problems, first multiply both sides by 2 and then expand the product.
Frequently asked questions
What is the correct answer to this question?
\(x^2+10x-89=0\)
Why is this the correct answer?
The area of a triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Hence, \(\frac{1}{2}(x+3)(x+7)=55\), so \((x+3)(x+7)=110\). Expanding gives \(x^2+10x+21=110\), and therefore \(x^2+10x-89=0\). Option B results from mishandling the factor \(\frac{1}{2}\) in the area formula. Exam tip: In triangle-area problems, first multiply both sides by 2 and then expand the product.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
