Expert Mathematics Quadratic Equations Class 10 Level 29

एक समकोण त्रिभुज का आधार (x+4), ऊंचाई (x+8) और क्षेत्रफल (72) है। सही समीकरण कौन-सा है?

A right triangle has base (x+4), height (x+8), and area (72). Which equation is correct?

Explanation opens after your attempt
Correct Answer

A. \(x^2+12x-112=0\)

Step 1

Concept

The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).

Step 2

Why this answer is correct

The correct answer is A. \(x^2+12x-112=0\). The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).

Step 3

Exam Tip

क्षेत्रफल (\frac{1}{2}(x+4)(x+8)=72) होगा। इसलिए ((x+4)(x+8)=144) और \(x^2+12x-112=0\)।

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एक समकोण त्रिभुज का आधार (x+4), ऊंचाई (x+8) और क्षेत्रफल (72) है। सही समीकरण कौन-सा है? / A right triangle has base (x+4), height (x+8), and area (72). Which equation is correct?

Correct Answer: A. \(x^2+12x-112=0\). Explanation: क्षेत्रफल (\frac{1}{2}(x+4)(x+8)=72) होगा। इसलिए ((x+4)(x+8)=144) और \(x^2+12x-112=0\)। / The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).

Which concept should I revise for this Mathematics MCQ?

The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).

What exam hint can help solve this Mathematics question?

क्षेत्रफल (\frac{1}{2}(x+4)(x+8)=72) होगा। इसलिए ((x+4)(x+8)=144) और \(x^2+12x-112=0\)।

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