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The perimeter is \(4\sqrt{7}\) and \(\sqrt{7}\) is irrational. A non-zero rational multiplier keeps irrationality.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय संख्या / Irrational number. The perimeter is \(4\sqrt{7}\) and \(\sqrt{7}\) is irrational. A non-zero rational multiplier keeps irrationality.
Step 3
Exam Tip
परिमाप \(4\sqrt{7}\) होगा और \(\sqrt{7}\) अपरिमेय है। शून्येतर परिमेय गुणक से अपरिमेयता बनी रहती है।
B. असांत अनावर्ती दशमलव/Non-terminating non-repeating decimal
Step 1
Concept
An irrational number has a non-terminating and non-repeating decimal. In exams check the repeating part carefully.
Step 2
Why this answer is correct
The correct answer is B. असांत अनावर्ती दशमलव / Non-terminating non-repeating decimal. An irrational number has a non-terminating and non-repeating decimal. In exams check the repeating part carefully.
Step 3
Exam Tip
अपरिमेय संख्या का दशमलव असांत और अनावर्ती होता है। परीक्षा में दोहराने वाले भाग को ध्यान से देखें।
The side will be \(\sqrt{11}\) and (11) is not a perfect square. So \(\sqrt{11}\) is irrational.
Step 2
Why this answer is correct
The correct answer is D. अपरिमेय संख्या / Irrational number. The side will be \(\sqrt{11}\) and (11) is not a perfect square. So \(\sqrt{11}\) is irrational.
Step 3
Exam Tip
भुजा \(\sqrt{11}\) होगी और (11) पूर्ण वर्ग नहीं है। इसलिए \(\sqrt{11}\) अपरिमेय है।
It has no fixed repetition so it is irrational. Do not mistake a non-repeating pattern for a repeating one.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. It has no fixed repetition so it is irrational. Do not mistake a non-repeating pattern for a repeating one.
Step 3
Exam Tip
इसमें निश्चित दोहराव नहीं है इसलिए यह अपरिमेय है। अनावर्ती पैटर्न को आवर्ती मानने की गलती न करें।
(5) is not a perfect cube so \(\sqrt[3]{5}\) is irrational. In cube roots check perfect cubes.
Step 2
Why this answer is correct
The correct answer is C. यह अपरिमेय है / It is irrational. (5) is not a perfect cube so \(\sqrt[3]{5}\) is irrational. In cube roots check perfect cubes.
Step 3
Exam Tip
(5) पूर्ण घन नहीं है इसलिए \(\sqrt[3]{5}\) अपरिमेय है। घनमूल में पूर्ण घन देखना चाहिए।
Subtracting irrational \(\sqrt{2}\) from rational (6) gives an irrational number. The irrational part does not disappear.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय / Irrational. Subtracting irrational \(\sqrt{2}\) from rational (6) gives an irrational number. The irrational part does not disappear.
Step 3
Exam Tip
परिमेय (6) में से अपरिमेय \(\sqrt{2}\) घटाने पर अपरिमेय संख्या मिलती है। अपरिमेय भाग खत्म नहीं होता।
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so the sum is \(7\sqrt{3}\). Simplify both radicals first.
Step 2
Why this answer is correct
The correct answer is A. \(7\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{75}=5\sqrt{3}\), so the sum is \(7\sqrt{3}\). Simplify both radicals first.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{75}=5\sqrt{3}\), इसलिए योग \(7\sqrt{3}\) है। पहले दोनों मूल सरल करें।
Multiplying \(\frac{1}{\sqrt{3}}\) by \(\frac{\sqrt{3}}{\sqrt{3}}\) gives \(\frac{\sqrt{3}}{3}\). This is called rationalising the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{3}}{3}\). Multiplying \(\frac{1}{\sqrt{3}}\) by \(\frac{\sqrt{3}}{\sqrt{3}}\) gives \(\frac{\sqrt{3}}{3}\). This is called rationalising the denominator.
Step 3
Exam Tip
\(\frac{1}{\sqrt{3}}\) को \(\frac{\sqrt{3}}{\sqrt{3}}\) से गुणा करने पर \(\frac{\sqrt{3}}{3}\) मिलता है। इसे हर का परिमेयकरण कहते हैं।
\(\pi\) is irrational and (2) is rational, so the sum is irrational. Do not treat \(\pi\) as equal to (22/7).
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. \(\pi\) is irrational and (2) is rational, so the sum is irrational. Do not treat \(\pi\) as equal to (22/7).
Step 3
Exam Tip
\(\pi\) अपरिमेय है और (2) परिमेय है, इसलिए योग अपरिमेय है। \(\pi\) को (22/7) के बराबर न मानें।
A. यह \(\pi\) का निकट मान है/It is an approximation of \(\pi\)
Step 1
Concept
(22/7) is rational and an approximation of \(\pi\). Do not treat an approximation as exact equality.
Step 2
Why this answer is correct
The correct answer is A. यह \(\pi\) का निकट मान है / It is an approximation of \(\pi\). (22/7) is rational and an approximation of \(\pi\). Do not treat an approximation as exact equality.
Step 3
Exam Tip
(22/7) परिमेय है और \(\pi\) का निकट मान है। निकट मान को वास्तविक बराबरी न समझें।
This decimal is non-terminating and non-repeating so it is irrational. There is no fixed repeating block.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. This decimal is non-terminating and non-repeating so it is irrational. There is no fixed repeating block.
Step 3
Exam Tip
यह दशमलव असांत और अनावर्ती है इसलिए अपरिमेय है। निश्चित दोहराने वाला ब्लॉक नहीं है।
\(\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}\) and \(\sqrt{7}\) is irrational. Check perfect squares even in fractions.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. \(\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}\) and \(\sqrt{7}\) is irrational. Check perfect squares even in fractions.
Step 3
Exam Tip
\(\sqrt{\frac{7}{4}}=\frac{\sqrt{7}}{2}\) है और \(\sqrt{7}\) अपरिमेय है। भिन्न में भी पूर्ण वर्ग जाँचें।
The \(\sqrt{2}\) terms cancel and the value is (1). Sometimes irrational terms can cancel to give a rational result.
Step 2
Why this answer is correct
The correct answer is B. परिमेय (1) / Rational (1). The \(\sqrt{2}\) terms cancel and the value is (1). Sometimes irrational terms can cancel to give a rational result.
Step 3
Exam Tip
\(\sqrt{2}\) के पद कट जाते हैं और मान (1) है। कभी-कभी अपरिमेय पद हटकर परिमेय परिणाम दे सकते हैं।
A. समकोण त्रिभुज जिसकी भुजाएँ (1) और (1) हों/Right triangle with legs (1) and (1)
Step 1
Concept
In a right triangle, the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). Understand construction using Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is A. समकोण त्रिभुज जिसकी भुजाएँ (1) और (1) हों / Right triangle with legs (1) and (1). In a right triangle, the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). Understand construction using Pythagoras theorem.
Step 3
Exam Tip
समकोण त्रिभुज में कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। पाइथागोरस प्रमेय से निर्माण समझें।
C. परिमेय और अपरिमेय दोनों/Both rational and irrational numbers
Step 1
Concept
The number line represents real numbers, including both rational and irrational numbers. Irrationals can be constructed geometrically.
Step 2
Why this answer is correct
The correct answer is C. परिमेय और अपरिमेय दोनों / Both rational and irrational numbers. The number line represents real numbers, including both rational and irrational numbers. Irrationals can be constructed geometrically.
Step 3
Exam Tip
संख्या रेखा पर वास्तविक संख्याएँ दिखाई जाती हैं जिनमें परिमेय और अपरिमेय दोनों शामिल हैं। अपरिमेयों का निर्माण ज्यामिति से किया जा सकता है।
The positive number is \(\sqrt{19}\) because (\(\sqrt{19}\)2=19). Since (19) is not a perfect square, it is irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{19}\). The positive number is \(\sqrt{19}\) because (\(\sqrt{19}\)2=19). Since (19) is not a perfect square, it is irrational.
Step 3
Exam Tip
धनात्मक संख्या \(\sqrt{19}\) है क्योंकि (\(\sqrt{19}\)2=19)। (19) पूर्ण वर्ग नहीं है इसलिए यह अपरिमेय है।
\(\sqrt{2}\) is irrational so its decimal is non-terminating and non-repeating. Do not conclude from only a few decimal digits.
Step 2
Why this answer is correct
The correct answer is C. असांत अनावर्ती / Non-terminating non-repeating. \(\sqrt{2}\) is irrational so its decimal is non-terminating and non-repeating. Do not conclude from only a few decimal digits.
Step 3
Exam Tip
\(\sqrt{2}\) अपरिमेय है इसलिए इसका दशमलव असांत और अनावर्ती होता है। दशमलव के कुछ अंक देखकर निष्कर्ष न लगाएं।
D. हर वास्तविक संख्या परिमेय होती है/Every real number is rational
Step 1
Concept
Every real number is not rational because irrational numbers are also real. The number system includes both types.
Step 2
Why this answer is correct
The correct answer is D. हर वास्तविक संख्या परिमेय होती है / Every real number is rational. Every real number is not rational because irrational numbers are also real. The number system includes both types.
Step 3
Exam Tip
हर वास्तविक संख्या परिमेय नहीं होती क्योंकि अपरिमेय संख्याएँ भी वास्तविक हैं। संख्या तंत्र में दोनों प्रकार शामिल हैं।
B. \(\sqrt{2}\) और \(\sqrt{5}\)/\(\sqrt{2}\) and \(\sqrt{5}\)
Step 1
Concept
\(\sqrt{2}\) and \(\sqrt{5}\) are both irrational because (2) and (5) are not perfect squares. Check each number separately.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\) और \(\sqrt{5}\) / \(\sqrt{2}\) and \(\sqrt{5}\). \(\sqrt{2}\) and \(\sqrt{5}\) are both irrational because (2) and (5) are not perfect squares. Check each number separately.
Step 3
Exam Tip
\(\sqrt{2}\) और \(\sqrt{5}\) दोनों अपरिमेय हैं क्योंकि (2) और (5) पूर्ण वर्ग नहीं हैं। प्रत्येक संख्या को अलग जाँचें।
\(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the difference is \(4\sqrt{2}\). Simplify like radicals first.
Step 2
Why this answer is correct
The correct answer is D. \(4\sqrt{2}\). \(\sqrt{72}=6\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the difference is \(4\sqrt{2}\). Simplify like radicals first.
Step 3
Exam Tip
\(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\), इसलिए अंतर \(4\sqrt{2}\) है। पहले समान मूलों को सरल करें।
The positive number is \(\sqrt{37}\) because (\(\sqrt{37}\)2=37). Since (37) is not a perfect square, it is irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{37}\). The positive number is \(\sqrt{37}\) because (\(\sqrt{37}\)2=37). Since (37) is not a perfect square, it is irrational.
Step 3
Exam Tip
धनात्मक संख्या \(\sqrt{37}\) है क्योंकि (\(\sqrt{37}\)2=37)। (37) पूर्ण वर्ग नहीं है इसलिए यह अपरिमेय है।
This decimal has no fixed repeating block. So it is a non-terminating non-repeating decimal and an irrational number.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय / Irrational. This decimal has no fixed repeating block. So it is a non-terminating non-repeating decimal and an irrational number.
Step 3
Exam Tip
इस दशमलव में निश्चित दोहराने वाला खंड नहीं है। इसलिए यह असांत अनावर्ती दशमलव और अपरिमेय संख्या है।
\(\sqrt{44}=2\sqrt{11}\) and \(\sqrt{99}=3\sqrt{11}\), so the sum is \(5\sqrt{11}\). Simplify before adding like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(5\sqrt{11}\). \(\sqrt{44}=2\sqrt{11}\) and \(\sqrt{99}=3\sqrt{11}\), so the sum is \(5\sqrt{11}\). Simplify before adding like radicals.
Step 3
Exam Tip
\(\sqrt{44}=2\sqrt{11}\) और \(\sqrt{99}=3\sqrt{11}\), इसलिए योग \(5\sqrt{11}\) है। समान मूलों को जोड़ने से पहले सरल करें।
