Hard Mathematics Quadratic Equations Class 10 Level 30

एक आयत की लंबाई (x+7) और चौड़ाई (x-3) है। क्षेत्रफल (90) हो तो सही समीकरण कौन-सा है?

A rectangle has length (x+7) and breadth (x-3). If its area is (90), which equation is correct?

Explanation opens after your attempt
Correct Answer

A. \(x^2+4x-111=0\)

Step 1

Concept

The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).

Step 2

Why this answer is correct

The correct answer is A. \(x^2+4x-111=0\). The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).

Step 3

Exam Tip

क्षेत्रफल ((x+7)(x-3)=90) होगा। विस्तार से \(x^2+4x-21=90\) और \(x^2+4x-111=0\) मिलता है।

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एक आयत की लंबाई (x+7) और चौड़ाई (x-3) है। क्षेत्रफल (90) हो तो सही समीकरण कौन-सा है? / A rectangle has length (x+7) and breadth (x-3). If its area is (90), which equation is correct?

Correct Answer: A. \(x^2+4x-111=0\). Explanation: क्षेत्रफल ((x+7)(x-3)=90) होगा। विस्तार से \(x^2+4x-21=90\) और \(x^2+4x-111=0\) मिलता है। / The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).

Which concept should I revise for this Mathematics MCQ?

The area is ((x+7)(x-3)=90). Expanding gives \(x^2+4x-21=90\) and \(x^2+4x-111=0\).

What exam hint can help solve this Mathematics question?

क्षेत्रफल ((x+7)(x-3)=90) होगा। विस्तार से \(x^2+4x-21=90\) और \(x^2+4x-111=0\) मिलता है।

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