6 results found for "triangle-area" in Class 10.
Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक समकोण त्रिभुज का आधार (x+5), ऊंचाई (x+9) और क्षेत्रफल (95) है। सही समीकरण कौन-सा है?
A right triangle has base (x+5), height (x+9), and area (95). Which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+14x-145=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+5)(x+9)=95). Thus ((x+5)(x+9)=190) and \(x^2+14x-145=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+14x-145=0\). The area is (\frac{1}{2}(x+5)(x+9)=95). Thus ((x+5)(x+9)=190) and \(x^2+14x-145=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+5)(x+9)=95) होगा। इसलिए ((x+5)(x+9)=190) और \(x^2+14x-145=0\)।
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Question
Expert Mathematics
Quadratic Equations Introduction to 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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Question
Expert Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 28
एक समकोण त्रिभुज का आधार (x+3), ऊंचाई (x+7) और क्षेत्रफल (55) है। सही समीकरण कौन-सा है?
A right triangle has base (x+3), height (x+7), and area (55). Which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+10x-89=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+3)(x+7)=55). Thus ((x+3)(x+7)=110) and \(x^2+10x-89=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+10x-89=0\). The area is (\frac{1}{2}(x+3)(x+7)=55). Thus ((x+3)(x+7)=110) and \(x^2+10x-89=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+3)(x+7)=55) होगा। इसलिए ((x+3)(x+7)=110) और \(x^2+10x-89=0\)।
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Question
Hard Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक समकोण त्रिभुज का आधार (x+2), ऊंचाई (x+6) और क्षेत्रफल (40) है। सही समीकरण कौन-सा है?
A right triangle has base (x+2), height (x+6), and area (40). Which equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+8x-68=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+2)(x+6)=40). Thus ((x+2)(x+6)=80) and \(x^2+8x-68=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+8x-68=0\). The area is (\frac{1}{2}(x+2)(x+6)=40). Thus ((x+2)(x+6)=80) and \(x^2+8x-68=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+2)(x+6)=40) होगा। इसलिए ((x+2)(x+6)=80) और \(x^2+8x-68=0\)।
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Question
Hard Mathematics
Quadratic Equations Introduction to 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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Question
Medium Mathematics
Quadratic Equations Introduction to Quadratic Equations Class 10 Level 30
एक समकोण त्रिभुज का आधार (x), ऊँचाई (x+2) और क्षेत्रफल (24) है। सही द्विघात समीकरण कौन-सा है?
A right triangle has base (x), height (x+2), and area (24). Which quadratic equation is correct?
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x-48=0\)
Step 1
Concept
The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x-48=0\). The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}x(x+2)=24) होगा। इसलिए (x(x+2)=48) और \(x^2+2x-48=0\) मिलता है।
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