Hard Mathematics Quadratic Equations Class 10 Level 29

एक समकोण त्रिभुज का आधार (x+1), ऊंचाई (x+3) और क्षेत्रफल (30) है। सही समीकरण कौन-सा है?

A right triangle has base (x+1), height (x+3), and area (30). Which equation is correct?

Explanation opens after your attempt
Correct Answer

A. \(x^2+4x-57=0\)

Step 1

Concept

The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).

Step 2

Why this answer is correct

The correct answer is A. \(x^2+4x-57=0\). The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).

Step 3

Exam Tip

क्षेत्रफल (\frac{1}{2}(x+1)(x+3)=30) होगा। इसलिए ((x+1)(x+3)=60) और \(x^2+4x-57=0\)।

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Mathematics Answer, Explanation and Revision Hints

एक समकोण त्रिभुज का आधार (x+1), ऊंचाई (x+3) और क्षेत्रफल (30) है। सही समीकरण कौन-सा है? / A right triangle has base (x+1), height (x+3), and area (30). Which equation is correct?

Correct Answer: A. \(x^2+4x-57=0\). Explanation: क्षेत्रफल (\frac{1}{2}(x+1)(x+3)=30) होगा। इसलिए ((x+1)(x+3)=60) और \(x^2+4x-57=0\)। / The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).

Which concept should I revise for this Mathematics MCQ?

The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).

What exam hint can help solve this Mathematics question?

क्षेत्रफल (\frac{1}{2}(x+1)(x+3)=30) होगा। इसलिए ((x+1)(x+3)=60) और \(x^2+4x-57=0\)।

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