Maharashtra Board Class 8 Maths Chapter 1 Rational and Irrational Numbers Set 1.4 Solutions

Question 1. The number \( \sqrt{2} \) is shown on a number line. Steps are given to show \( \sqrt{3} \) on the number line using \( \sqrt{2} \). Fill in the boxes properly and complete the activity.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक संख्या रेखा पर \( \sqrt{2} \) और \( \sqrt{3} \) के निर्माण को दर्शाता है। संख्या रेखा पर -1, 0, 1 अंक हैं। मूलबिंदु O से Q बिंदु \( \sqrt{2} \) पर स्थित है। Q से संख्या रेखा पर लंबवत एक इकाई लंबाई का खंड QR खींचा गया है, जिससे एक समकोण त्रिभुज OQR बनता है। इस त्रिभुज का कर्ण OR \( \sqrt{3} \) को निरूपित करता है।
Answer:
The point Q on the number line shows the number \( \sqrt{2} \)
A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
Right angled \( \triangle OQR \) is obtained by drawing seg OR.
I(OQ) = \( \sqrt{2} \), I(QR) = 1
..By Pythagoras theorem,
\[ \text{[I(OR)]}^2 = \text{[I(OQ)]}^2 + \text{[I(QR)]}^2 \]
\[ = (\boxed{\sqrt{2}})^2 + \boxed{1}^2 \]
\[ = \boxed{2} + \boxed{1} \]
\[ = \boxed{3} \]
\( \implies \)
\( \text{I(OR)} = \boxed{\sqrt{3}} \)
...[Taking square root of both sides]
Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number \( \sqrt{3} \)
In simple words: This activity demonstrates how to geometrically construct irrational numbers like √3 on a number line by repeatedly applying the Pythagorean theorem, using a perpendicular unit length from a previously constructed irrational point.

🎯 Exam Tip: Constructing irrational numbers like √3, √5, √7 on the number line using the Pythagorean theorem is a common question. Ensure precise measurements for the unit length and perpendicular lines.

Question 2. Show the number \( \sqrt{5} \) on the number line.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र संख्या रेखा पर \( \sqrt{5} \) के निर्माण को दर्शाता है। मूलबिंदु O से, बिंदु Q को 2 इकाई पर चिह्नित किया गया है। Q पर संख्या रेखा के लंबवत 1 इकाई लंबाई का एक रेखाखंड QR खींचा गया है। समकोण त्रिभुज OQR का कर्ण OR \( \sqrt{5} \) को निरूपित करता है। O को केंद्र मानकर OR त्रिज्या से खींचा गया एक चाप संख्या रेखा को C पर काटता है, जो \( \sqrt{5} \) को दर्शाता है।
Answer:
Solution:
Draw a number line and take a point Q at 2
such that I(OQ) = 2 units.
Draw a line QR perpendicular to the number line through the point Q such that I(QR)
= 1 unit.
Draw seg OR.
\( \triangle OQR \) formed is a right angled triangle.
By Pythagoras theorem,
\[ \text{[I(OR)]}^2 = \text{[I(OQ)]}^2 + \text{[I(QR)]}^2 \]
\[ = 2^2 + 1^2 \]
\[ = 4 + 1 \]
\[ = 5 \]
\( \implies \)
\[ \text{I(OR)} = \sqrt{5} \text{ units} \]
...[Taking square root of both sides]
Draw an arc with centre O and radius OR. Mark the point of intersection of the
number line and arc as C. The point C shows the number \( \sqrt{5} \).
In simple words: To show √5, we use a right-angled triangle where one leg is 2 units and the other is 1 unit. The hypotenuse will be √5, which is then transferred to the number line using an arc from the origin.

🎯 Exam Tip: Remember that for √5, you can use a base of 2 units and a perpendicular height of 1 unit. Practice drawing the arcs accurately from the origin to mark the point on the number line.

Question 3. Show the number \( \sqrt{7} \) on the number line.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र संख्या रेखा पर \( \sqrt{5} \), \( \sqrt{6} \) और \( \sqrt{7} \) के सर्पिल निर्माण को दर्शाता है। यह \( \sqrt{5} \) के निर्माण से शुरू होता है। फिर, \( \sqrt{5} \) (C) को दर्शाने वाले बिंदु से, 1 इकाई का एक लंबवत रेखाखंड CD खींचा जाता है। कर्ण OD \( \sqrt{6} \) को दर्शाता है, जिसे संख्या रेखा पर E के रूप में चिह्नित किया गया है। E से, 1 इकाई का एक और लंबवत रेखाखंड EP खींचा जाता है। कर्ण OP तब \( \sqrt{7} \) को दर्शाता है, जिसे संख्या रेखा पर F के रूप में चिह्नित किया गया है।
Answer:
Solution:
Draw a number line and take a point Q at 2 such that I(OQ) = 2 units.
Draw a line QR perpendicular to the number line through the point Q such that I(QR)
= 1 unit.
Draw seg OR.
\( \triangle OQR \) formed is a right angled triangle.
By Pythagoras theorem,
\[ \text{[I(OR)]}^2 = \text{[I(OQ)]}^2 + \text{[I(QR)]}^2 \]
\[ = 2^2 + 1^2 \]
\[ = 4 + 1 \]
\[ = 5 \]
\( \implies \)
\[ \text{I(OR)} = \sqrt{5} \text{ units} \]
[Taking square root of both sides]
Draw an arc with centre O and radius OR. Mark the point of intersection of the number line and arc as C. The point C shows the number \( \sqrt{5} \).
Similarly, draw a line CD perpendicular to the number line through the point C such that I(CD) = 1 unit.
By Pythagoras theorem,
\[ \text{I(OD)} = \sqrt{6} \text{ units} \]
The point E shows the number \( \sqrt{6} \).
Similarly, draw a line EP perpendicular to the number line through the point E such that I(EP) = 1 unit.
By Pythagoras theorem,
\[ \text{I(OP)} = \sqrt{7} \text{ units} \]
The point F shows the number \( \sqrt{7} \).
In simple words: To show √7, you first construct √5 by making a right triangle with legs 2 and 1. Then, from the point representing √5, you construct √6 by adding another perpendicular unit length. Finally, from the point representing √6, you construct √7 similarly.

🎯 Exam Tip: Constructing √7 requires a sequential construction, building upon previously constructed irrational numbers (like √5 and √6). Ensure each step is precise, especially drawing perpendiculars and arcs.