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construction MCQ Questions for Class 10
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6 MCQs
Class 10 • construction
Practice Questions
6 questions tagged with construction.
संख्या रेखा पर \(\sqrt{3}\) बनाने में समकोण त्रिभुज की कौन-सी भुजाएँ उपयोगी हो सकती हैं?
Which legs of a right triangle can be useful to construct \(\sqrt{3}\) on the number line?
#polynomials
#number-line
#construction
#pythagoras
A \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
B (2) और (1) / (2) and (1)
C (3) और (1) / (3) and (1)
D \(\sqrt{3}\) और (1) / \(\sqrt{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
Step 1
Concept
Because (\(\sqrt{2}\)2 +12 =3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1). Because (\(\sqrt{2}\)2 +12 =3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 3
Exam Tip
क्योंकि (\(\sqrt{2}\)2 +12 =3), अतः कर्ण \(\sqrt{3}\) होगा। क्रमिक वर्गमूल बनाने में पाइथागोरस का प्रयोग होता है।
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संख्या रेखा पर \(\sqrt{29}\) बनाने के लिए समकोण त्रिभुज की कौन सी दो लंब भुजाएं सही हो सकती हैं?
To construct \(\sqrt{29}\) on the number line, which two perpendicular sides of a right triangle can be correct?
#number-line
#construction
#pythagoras
#square-root
A (5) और (2) / (5) and (2)
B (4) और (4) / (4) and (4)
C (3) और (5) / (3) and (5)
D (6) और (1) / (6) and (1)
Explanation opens after your attempt
Correct Answer
A. (5) और (2) / (5) and (2)
Step 1
Concept
\(5^2+2^2=29\), so the hypotenuse will be \(\sqrt{29}\). Check the sum of squares of the sides.
Step 2
Why this answer is correct
The correct answer is A. (5) और (2) / (5) and (2). \(5^2+2^2=29\), so the hypotenuse will be \(\sqrt{29}\). Check the sum of squares of the sides.
Step 3
Exam Tip
\(5^2+2^2=29\), इसलिए कर्ण \(\sqrt{29}\) होगा। भुजाओं के वर्गों का योग जांचें।
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समकोण त्रिभुज में भुजाएं (3) और (4) लेकर संख्या रेखा पर कौन सा मान बनाया जा सकता है?
Using sides (3) and (4) in a right triangle, which value can be constructed on the number line?
#number-line
#construction
#pythagoras
#real-numbers
A (5)
B (6)
C (7)
D (12)
Explanation opens after your attempt
Step 1
Concept
The hypotenuse is \(\sqrt{3^2+4^2}=5\). This is a direct use of a Pythagorean triple.
Step 2
Why this answer is correct
The correct answer is A. (5). The hypotenuse is \(\sqrt{3^2+4^2}=5\). This is a direct use of a Pythagorean triple.
Step 3
Exam Tip
कर्ण \(\sqrt{3^2+4^2}=5\) होता है। यह पाइथागोरस त्रिक का सीधा उपयोग है।
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संख्या रेखा पर \(\sqrt{17}\) बनाने के लिए समकोण त्रिभुज की लंब भुजाएं (4) और (1) ली गई हैं। कर्ण की लंबाई क्या होगी?
To construct \(\sqrt{17}\) on the number line, the perpendicular sides of a right triangle are taken as (4) and (1). What will be the length of the hypotenuse?
#number-line
#square-root
#construction
#pythagoras
A \(\sqrt{15}\)
B \(\sqrt{17}\)
C (5)
D (17)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{17}\)
Step 1
Concept
The hypotenuse is \(\sqrt{4^2+1^2}=\sqrt{17}\). Use Pythagoras in this type of construction.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{17}\). The hypotenuse is \(\sqrt{4^2+1^2}=\sqrt{17}\). Use Pythagoras in this type of construction.
Step 3
Exam Tip
कर्ण \(=\sqrt{4^2+1^2}=\sqrt{17}\) होगा। ऐसी रचना में पाइथागोरस प्रमेय लगाएं।
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संख्या रेखा पर \(\sqrt{13}\) को बनाने के लिए (3) और (2) लंब भुजाओं वाला समकोण त्रिभुज बनाया जाए तो कर्ण क्या होगा?
To construct \(\sqrt{13}\) on the number line, if a right triangle has perpendicular sides (3) and (2), what is the hypotenuse?
#number-line
#construction
#pythagoras
#square-root
A \(\sqrt{5}\)
B \(\sqrt{11}\)
C \(\sqrt{13}\)
D (5)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{13}\)
Step 1
Concept
The hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Apply Pythagoras in a right triangle.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{13}\). The hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Apply Pythagoras in a right triangle.
Step 3
Exam Tip
कर्ण \(=\sqrt{3^2+2^2}=\sqrt{13}\) होता है। समकोण त्रिभुज में पाइथागोरस प्रमेय लागू करें।
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संख्या रेखा पर \(\sqrt{2}\) को बनाने के लिए समकोण त्रिभुज की दो लंब भुजाएं (1) और (1) ली गई हैं। कर्ण की लंबाई क्या होगी?
To construct \(\sqrt{2}\) on the number line, two perpendicular sides of a right triangle are taken as (1) and (1). What will be the length of the hypotenuse?
#number-line
#real-numbers
#square-root
#construction
A (1)
B \(\sqrt{2}\)
C (2)
D \(\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{2}\)
Step 1
Concept
By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 3
Exam Tip
पाइथागोरस से कर्ण \(=\sqrt{1^2+1^2}=\sqrt{2}\) होगा। ऐसी रचनाओं में वर्गों का योग जरूर देखें।
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