असमानताओं \(x\geq 2\), \(y\geq 0\), \(x+y\leq 9\), \(2x+y\leq 12\) से बने हल-क्षेत्र का क्षेत्रफल कितना है?
What is the area of the solution region formed by \(x\geq 2\), \(y\geq 0\), \(x+y\leq 9\), and \(2x+y\leq 12\)?
Explanation opens after your attempt
Correct Answer
A. (20) वर्ग इकाई(20) square units
Concept
The vertices are ((2,0)), ((6,0)), ((3,6)), and ((2,7)). Using the shoelace method or splitting into parts gives area (20).
Why this answer is correct
The correct answer is A. (20) वर्ग इकाई / (20) square units. The vertices are ((2,0)), ((6,0)), ((3,6)), and ((2,7)). Using the shoelace method or splitting into parts gives area (20).
Exam Tip
शीर्ष ((2,0)), ((6,0)), ((3,6)), ((2,7)) हैं। शूलेस विधि या भागों में बांटकर क्षेत्रफल (20) मिलता है।
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