असमानताओं \(x\ge 1\), \(y\ge 0\), \(x+y\le 5\), \(x\le 4\) से बने क्षेत्र का क्षेत्रफल क्या है?
What is the area of the region formed by \(x\ge 1\), \(y\ge 0\), \(x+y\le 5\), and \(x\le 4\)?
Explanation opens after your attempt
Correct Answer
C. \(\frac{15}{2}\) वर्ग इकाई\(\frac{15}{2}\) square units
Concept
The vertices are ((1,0)), ((4,0)), ((4,1)), and ((1,4)). With parallel sides (4) and (1) and distance (3), the area is (\frac{1}{2}(4+1)3=\frac{15}{2}).
Why this answer is correct
The correct answer is C. \(\frac{15}{2}\) वर्ग इकाई / \(\frac{15}{2}\) square units. The vertices are ((1,0)), ((4,0)), ((4,1)), and ((1,4)). With parallel sides (4) and (1) and distance (3), the area is (\frac{1}{2}(4+1)3=\frac{15}{2}).
Exam Tip
क्षेत्र के शीर्ष ((1,0)), ((4,0)), ((4,1)), ((1,4)) हैं। समांतर भुजाओं (4) और (1) तथा दूरी (3) से क्षेत्रफल (\frac{1}{2}(4+1)3=\frac{15}{2}) है।
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