MP Board Class 11th Maths Solutions Chapter 1 समुच्चय Ex 1.1

MP Board Class 11th Maths Solutions Chapter 1 समुच्चय Ex 1.1

प्रश्न 1.
निम्नलिखित में कौन से समुच्चय हैं? अपने उत्तर का औचित्य बताइए।
(i) j अक्षर से प्रारम्भ होने वाले वर्ष के सभी महीनों का संग्रह।
(ii) भारत के दस सबसे अधिक प्रतिभाशाली लेखकों का संग्रह।
(iii) विश्व के सर्वश्रेष्ठ ग्यारह बल्लबाजों का संग्रह।
(iv) आप की कक्षा के सभी बालकों का संग्रह।।
(v) 100 से कम सभी प्राकृत संख्याओं का संग्रह।
(vi) लेखक प्रेमचन्द द्वारा लिखित उपन्यासों का संग्रह।
(vii) सभी सम पूर्णाकों का संग्रह।
(viii) इस अध्याय में आने वाले प्रश्नों का संग्रह।
(ix) विश्व में सबसे अधिक खतरनाक जानवरों का संग्रह।
हल:
(i) j से शुरु होने वाले महीनों के नाम : जनवरी, जून व जुलाई।
अतः यह एक समुच्चय है।
(ii) प्रतिभाशाली लेखक को परिभाषित नहीं किया जा सकता।
इसीलिए यह एक समुच्चय नहीं है।
(iii) सर्वश्रेष्ठ बल्लेबाज को परिभाषित नहीं कर सकते।
अतः यह एक समुच्चय नहीं है।
(iv) कक्षा के सभी विद्यार्थियों की संख्या निश्चित होती है।
अत: यह एक समुच्चय है।
(v) 100 से कम प्राकृत संख्याएँ 1, 2, 3, ……. 99 हैं।
अतः यह एक समुच्चय है।
(vi) लेखक प्रेमचन्द्र द्वारा लिखित उपन्यासों का संग्रह गबन, गोदान आदि द्वारा परिभाषित हैं।
अतः यह एक समुच्चय है।
(vii) समपूर्णांक {…. . – 6. – 4. 2, 4, 6, ….} हैं।
इसलिए यह एक समुच्चय है।
(viii) इस अध्याय के प्रश्न परिभाषित हैं।
अतः यह एक समुच्चय है।
(ix) संसार के सबसे अधिक खतरनाक पशुओं के संग्रह को परिभाषित नही किया जा सकता।
इसलिए यह एक समुच्चय नहीं है।

प्रश्न 2.
मान लीजिए A = {1, 2, 3, 4, 5, 6}. रिक्त स्थानों में उपयुक्त प्रतीक ϵ अथवा ∉ भरिए।
(i) 5…… A
(ii) 8 …….. A
(iii) 0 ……… A
(iv) 4……. A
(v) 2 …….. A
(vi) 10 ……. A
हल:
(i) 5 ϵ A
(ii) 8 ∉ A
(iii) 0 ∉ A
(iv) 4 ϵ A
(v) 2 ϵ A
(vi) 10 ∉ A

प्रश्न 3.
निम्नलिखित समुच्चयों को रोस्टर रूप में लिखिए :
(i) A = {x : x एक पूर्णांक है और -3 < x < 7}
(ii) B = {x : x संख्या 6 से कम एक प्राकृत संख्या है?
(iii) C = {x: x दो अंको की ऐसी प्राकृत संख्या है जिसके अंकों का योगफल 8 है।
(iv) D = {x : x एक अभाज्य संख्या है जो 60 की भाजक है?
(v) E = TRIGONOMETRY शब्द के सभी अक्षरों का समुच्चय
(vi) F = BETTER शब्द के सभी अक्षरों का समुच्च्य
हल:
(i) A = {-2, – 1, 0, 1, 2, 3, 4, 5, 6}
(ii) B = {1, 2, 3, 4, 5}
(iii) C = {17, 26, 35, 44, 53, 62, 71, 80}
(iv) D = {2, 3, 5}
(v) E = {T, R, I, G, O, N, M, E, Y}
(vi) F = {B, E, T, R}

MP Board Solutions

प्रश्न 4.
निम्नलिखित समुच्चयों को समुच्चय निर्माण रूप में व्यक्त कीजिए :
(i) {3, 6, 9, 12}
(ii) {2, 4, 8, 16, 32}
(ii) {5, 25, 125, 625}
(iv) {2, 4, 6, ….}
(v) {1, 4, 9, …… 100}
हल:
(i) {x : x = 3n और 1 ≤ n ≤ 4}
(ii) {x : x = 2n और 1 ≤ n ≤ 5}
(iii) {x : x = 5n और 1 ≤ n ≤ 4}
(iv) {x : x एक सम प्राकृत संख्या है?
(v) {x : x = n2, 1 ≤ n ≤ 10}

प्रश्न 5.
निम्नलिखित समुच्चयों के सभी अवयवों (सदस्यों) को सूचीबद्ध कीजिए।
(i) A = {x : x एक विषम प्राकृत संख्या है?
(ii) B = {x : x एक पूर्णांक है, \(-\frac{1}{2}\) < x < \(\frac{9}{2}\)}
(iii) C = {x : x एक पूर्णांक है, x2 ≤ 4}
(iv) D = {x : x, LOYAL शब्द का एक अक्षर है?
(v) F = {x : x वर्ष का एक ऐसा महीना है, जिसमें 31 दिन नहीं होते हैं।
(vi) F = {x : x अंग्रेजी वर्णमाला का एक व्यंजन है, जो k से पहले आता है।
हल:
(i) A = {1, 3, 5, 7, ………..}
(ii) B = { 0, 1, 2, 3, 4}
(iii) C = {-2, – 1, 0, 1, 2}
(iv) D = {L, O, Y, A}
(v) E = {फरवरी, अप्रैल, जून, सितम्बर, नवम्बर}
(vi) F = {b, c, d, f, g, h, j}

MP Board Solutions

प्रश्न 6.
बाई ओर रोस्टर रूप में लिखित और दाईं ओर समुच्चय निर्माण रूप में वर्णित समुच्चयों का सही मिलान कीजिए:
(i) {1, 2, 3, 6} – (a) {x : x एक अभाज्य संख्या है और 6 की भाजक है}
(ii) {2, 3} – (b) {x : x संख्या 10 से कम एक विषम प्राकृत संख्या है?
(iii) {M, A, T, H, E, I, C, S} – (c) {x : x एक प्राकृत संख्या है और 6 की भाजक है}
(iv) {1, 3, 5, 7, 9} – (d) {x : x MATHEMATICS शब्द का एक अक्षर है।
हल:
(i) ↔ (c)
(ii) ↔ (a)
(iii) ↔ (d)
(iv) ↔ (b)

MP Board Class 11th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.2

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.2

Question 1.
Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Solution:
(i) 322 = (30 + 2)2 = (30 + 2) (30 + 2)
= 30 (30 + 2) + 2 (30 + 2)
= 900 + 60 + 60 + 4 = 1024.
(ii) 352 = (30 + 5)2 = (30 + 5) (30 + 5)
= 30(30 + 5) + 5 (30 + 5)
= 900 + 150 + 150 + 25 = 1225.
(iii) 862 = (80 + 6)2 = (80 + 6) (80 + 6)
= 80(80 + 6) + 6(80 + 6)
= 6400 + 480 + 480 + 36 = 7396.
(iv) 932 = (90 + 3)2 = (90 + 3) (90 + 3)
= 90(90 + 3) + 3(90 + 3)
= 8100 + 270 + 270 + 9 = 8649.
(v) 712 = (70+ 1)2 = (70 + 1) (70 + 1)
= 70(70 + 1) + 1 (70 + 1)
= 4900 + 70 + 70 + 1 = 5041.
(vi) 462 = (40 + 6)2 = (40 + 6) (40 + 6)
= 40(40 + 6) + 6(40 + 6)
= 1600 + 240 + 240 + 36 = 2116.

MP Board Solutions

Question 2.
Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution:
We can get Pythagorean triplet by using general form 2m, m2 – 1, m2 + 1.
(i) Let us take 2m = 6 ⇒ m = 3
Thus, m2 – 1 = 32 – 1 = 9 – 1 = 8 and m2 + 1 = 32 + 1 = 9 + 1 = 10.
∴ The required triplet is 6, 8, 10

(ii) Let us take 2m = 14 ⇒ m = 7
Thus, m2 – 1 = 72 – 1 = 49 – 1 = 48
and m2 + 1 = 72 + 1 = 49 + 1 = 50
∴ The required triplet is 14, 48, 50.

(iii) Let us take 2m = 16 ⇒ m = 8
Thus, m2 – 1 = 82 – 1 = 64 – 1 = 63
and m2 + 1 = 82 + 1 = 64 + 1 = 65.
∴ The required triplet is 16, 63, 65.

(iv) Let us take 2m = 18 ⇒ m = 9
Thus, m2 – 1 = 92 – 1 = 81 – 1 = 80
and m2 + 1 = 92 + 1 = 81 + 1 = 82
∴ The required triplet is 18, 80, 82.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.1

MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.1

Question 1.
What will be the unit digit of the squares of the following numbers?
Solution:
(i) The unit digit of (81)2 is 1. Because when we multiply digit 1 by itself, we get 1.
(ii) The unit digit of (272)2 is 4. Because when we multiply digit 2 by itself, we get 4.
(iii) The unit digit of (799)2 is 1. Because when we multiply digit 9 by itself, we get 81.
(iv) The unit digit of (3853)2 is 9. Because when we multiply digit 3 by itself, we get 9.
(v) The unit digit of (1234)2 is 6. Because when we multiply digit 4 by itself, we get 16.
(vi) The unit digit of (26387)2 is 9. Because when we multiply digit 7 by itself, we get 49.
(vii) The unit digit of (52698)2 is 4. Because when we multiply digit 8 by itself, we get 64.
(viii) The unit digit of (99880)2 is 0. Because when we multiply digit 0 by itself, we get 0.
(ix) The unit digit of (12796)2 is 6. Because when we multiply the unit digit 6 by itself, we get 36.
(x) The unit digit of (55555)2 is 5. Because when we multiply the unit digit 5 by itself, we get 25.

Question 2.
The following numbers are obviously not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050.
Solution:
We know that number ending in 2, 3, 7 or 8 are not perfect squares.
∴ (i) 1057,
(ii) 23453,
(iii) 7928,
(iv) 222222
and
(vi) 89722 are not perfect squares.
Since, for perfect squares, there should be even number of zeroes at the end.
∴ (v) 64000,
(vii) 222000 and
(viii) 505050. Are not perfect squares.

Question 3.
The square of which of the following would be odd numbers ?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Solution:
(i) When we multiply the unit digit 1 by itself, we get 1 at the end, which shows that the square of 431 is an odd number.
(ii) When we multiply the unit digit 6 by itself we get 36, i.e., we get 6 at the end, which shows that the square of 2826 is an even number.
(iii) When we multiply the unit digit 9 by itself we get 81, i.e., we get 1 at the end, which shows that the square of 7779 is an odd number.
(iv) When we multiply the unit digit 4 by itself we get 16, i.e., we get 6 at the end, which shows that the square of 82004 is an even number.

Question 4.
Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 …… 2 …… 1 …….
100000012 = ……….
Solution:
1000012 = 10000200001
100000012 = 100000020000001.

Question 5.
Observe the following pattern and supply the missing numbers.
112 = 121
1012 = 10201
101012 = 102030201
10101012 = ……
…….2 = 10203040504030201
Solution:
10101012 = 1020304030201
1010101012 = 10203040504030201.

Question 6.
Using the given pattern, find the missing numbers.
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 +42 + 122 = 132
42 + 52 + _2 = 212
52 + _2 + 302 = 312
62 + 72 + _2 = _2
Solution:
In the pattern,
Third number = first number × second number and fourth number = third number + 1
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 + 422 = 432.

Question 7.
Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 +23.
Solution:
We have to find the sum of first 5 odd numbers,
1 + 3 + 5 + 7 + 9 = 52 = 25.
(ii) We have to find the sum of the first 10 odd numbers,
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 102 = 100.
(iii) We have to find the sum of first 12 odd numbers,
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 122 = 144.

Question 8.
(i) Express 49 as the sum of 7 odd numbers,
(ii) Express 121 as the sum of 11 odd numbers.
Solution:
(i) 49 = 72 = 1 + 3 + 5 + 7 + 9 + 11 + 13.
(ii) 121 = 112 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.

Question 9.
How many numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100.
Solution:
(i) We can find the number of terms between the squares of 12 and 13, by doubling the first term from 12 and 13. i.e., 2 × 12 = 24:
∴ Total number of terms = 24.
(ii) We can find the number of terms between the squares of 25 and 26, by doubling the first term from 25 and 26. i.e., 2 × 25 = 50.
∴ Total number of terms = 50.
(iii) We can find the number of terms between the squares of 99 and 100, by doubling the first term from 99 and 100. i.e., 2 × 99 = 198.
∴ Total number of terms = 198.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.5

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.5

Question 1.
Draw the following.
(i) The square READ with RE = 5.1 cm.
(ii) A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
(iii) A rectangle with adjacent sides of lengths 5 cm and 4 cm.
(iv) A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm.
Solution:
(i) Since, all 4 sides of a square are equal and each of 4 angles is equal to 90°.
Steps of Construction:
Step-1: Draw RE = 5.1 cm.
Step-2 : Draw ∠REX = 90°.
Step-3 : Cut off EA = 5.1 cm on \(\overrightarrow{E X}\)
Step-4: With R as centre and radius equal to 5.1 cm, draw an arc.
Step-5 : With A as centre and radius equal to 5.1 cm, cut off another arc on the arc drawn in step-4 at point D.
Step-6 : Join DA and DR.
Hence, READ is the required square.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.5 1

(ii) We know that diagonals of a rhombus bisect each other at right angles. Let AC = 5.2 cm and BD = 6.4 cm.
Steps of Construction:
Step-1: Draw AC = 5.2 cm.
Step-2 : Draw perpendicular bisector XY of AC which cut AC at point O.
Step-3 : Cut off OD = 3.2 cm on OX and OB = 3.2 cm on \(\overrightarrow{O Y}\).
Step-4 : Join AD, CD, AB and CB.
Hence, ABCD is the required rhombus.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.5 2

(iii) In a rectangle, opposite sides are equal and each of 4 angles is equal to 90°.
Let AB = 5 cm and BC = 4 cm
∴ AB = DC = 5 cm and BC = AD = 4 cm.
Also, ∠A = ∠B = ∠C = ∠D = 90°.
Steps of Construction:
Step-1: Draw AB = 5 cm.
Step-2 : Draw ∠ABX = 90°.
Step-3 : Cut off BC = 4 cm on BX .
Step-4: With A as centre and radius equal to 4 cm, cut off an arc.
Step-5 : With C as centre and radius equal to 5 cm cut off another arc on the arc drawn in step-4 at point D.
Step-6 : Join AD and CD.
Hence, ABCD is the required rectangle.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.5 3

(iv) Here, data given is incomplete. Since we know that to draw a quadrilateral at least five parts are necessary. In the present case, OK = 5.5 cm & KA = 4.2 cm is given.
We know that opposite sides of a parallelogram are equal.
∴ OK = YA = 5.5 cm and KA = OY = 4.2 cm
Here, only four parts are given. This means that one more part is necessary.
So, either one angle or diagonal of a parallelogram is required to construct it.
Hence, parallelogram OKAY cannot be drawn.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.4

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.4

Question 1.
Construct the following quadrilaterals.
(i) Quadrilateral
DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

MP Board Solutions

(ii) Quadrilateral
TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U = 120°
Solution:
(i) Steps of Construction:
Step-1: Draw EA = 5 cm.
Step-2: Make ∠AEX = 60° and ∠EAY = 90°.
Step-3 : Cut off arcs AR = 4.5 cm on \(\overrightarrow{A Y}\) and ED = 4 cm on \(\overrightarrow{E X}\).
Step-4: Join DR.
Hence, DEAR is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.4 1

MP Board Solutions

(ii) Steps of Construction :
Step-1: Draw RU = 3 cm.
Step-2 : Make ∠URX = 75° and ∠RUY = 120°
Step-3 : Cut off RT = 3.5 cm on \(\overrightarrow{R X}\) and UE = 4 cm on \(\overrightarrow{U Y}\).
Step-4: Join TE.
Hence, TRUE is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.4 2

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3

Question 1.
Construct the following quadrilaterals.
(i) Quadrilateral
MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 1050°
∠R = 105°

(ii) Quadrilateral
PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 6.5 cm
∠N = 85°

MP Board Solutions

(iii) Parallelogram
HEAR
HE =5 cm
EA = 6 cm
∠B = 85°

(iv) Rectangle
OKAY
OK = 7 cm
KA = 5 cm
Solution:
(i) Steps of Construction:
Step-1: Draw MO = 6 cm
Step-2 : Make ∠MOX = 105° and ∠OMY = 60°.
Step-3 : Cut off OR = 4.5 cm on OX.
Step-4 : At point R, draw ∠ORZ = 105° which cuts MY at E.
Hence, MORE is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3 1

(ii) Here, ∠P = 90°, ∠A = 110°, ∠N = 85°
∴ ∠L = 360° – (∠P + ∠A + ∠N)
= 360° – (90° + 110° + 85°)
= 360°- 285° = 75° ….. (A)
Steps of Construction:
Step-1: Draw PL = 4 cm.
Step-2 : Make ∠LPX = 90° and ∠PLY = 75°. [From (A)]
Step-3 : Cut off LA = 6.5 cm on \(\overrightarrow{L Y}\).
Step-4: Draw ∠LAZ = 110° which cut \(\overrightarrow{P X}\) at point N.
Hence, PLAN is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3 2

MP Board Solutions

(iii) Since, opposite sides and angles of a parallelogram are equal i.e., ∠R = ∠E = 85°, HE = RA = 5 cm and EA = HR = 6 cm.
Steps of Construction:
Step-1: Draw HE = 5 cm.
Step-2 : Draw ∠HEX = 85°.
Step-3 : Cut off EA = 6 cm on \(\overrightarrow{E X}\)
Step-4: With H as centre and radius equal to 6 cm, draw an arc.
Step-5: With A as centre and radius equal to 5 cm, cut another arc on the arc drawn in step-4 at point R.
Step-6 : Join HR and RA.
Hence HEAR is the required parallelogram.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3 3

(iv) We know that each of the four angles of a rectangle is equal to 90° and opposite sides are also equal.
OK = YA and KA = OY
Steps of Construction:
Step-1: Draw OK = 7 cm.
Step-2 : Make ∠OKX = 90°.
Step-3 : Cut off KA = 5 cm on \(\overrightarrow{K X}\).
Step-4: With O as centre and radius equal to 5 cm, cut an arc.
Step-5: With A as centre and radius equal to 7 cm cut another arc on the arc drawn in step-4 at point Y.
Step-6 : Join OY and YA.
Hence, OKAY is the required rectangle.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.3 4

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.2

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.2

Question 1.
Construct the following quadrilaterals
(i) Quadrilateral
LIFT
LI = 4 cm
IF = 3 cm
TL = 2.5 cm
LF= 4.5 cm
IT =4 cm

(ii) Quadrilateral
GOLD
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm

MP Board Solutions

(iii) Rhombus BEND
BN = 5.6 cm,
DE = 6.5 cm
Solution:
(i) Steps of Construction:
Step-1: Draw LI = 4 cm.
Step-2: With L as centre and radius equal to 2.5 cm, cut an arc.
Step-3: With I as centre and radius equal to 4 cm, cut another arc on the arc drawn in step-2 at point T.
Step-4 : With I as centre and radius equal to 3 cm, cut an arc.
Step-5: With L as centre and radius equal to 4.5 cm, cut another arc on the arc drawn in step-4 at point F.
Step-6 : Join LT, IT, LF, IF and TF.
Hence, LIFT is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.2 1

(ii) Steps of Construction:
Step-1: Draw OL = 7.5 cm.
Step-2: With L as centre and radius equal to 5 cm cut an arc.
Step-3 : With O as centre and radius equal to 10 cm, cut another arc on the arc drawn in step-2 at point D.
Step-4: With L as centre and radius equal to 6 cm, cut another arc.
Step-5: With D as centre and radius equal to 6 cm cut an arc on arc drawn in step-4 at point G.
Step-6 : Join LD, LG, OG, OD and DG. Hence, GOLD is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.2 2

MP Board Solutions

(iii) Here, BN = 5.6 cm and DE = 6.5 cm are given. These two sides are diagonals of a rhombus BEND. We know that diagonals of a rhombus bisect each other at right angles.

Steps of Construction:
Step-1: Draw BN = 5.6 cm.
Step-2 : Draw perpendicular bisector XY of BN which intersect BN at O.
Step-3 : Cut off OD = OF = 3.25 cm on \(\overrightarrow{\mathrm{OX}}\) and \(\overrightarrow{\mathrm{OY}}\) respectively.
Step-4 : Join BD, ND, BE and NE.
Hence, BEND is the required rhombus.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.2 3

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.1

MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.1

Question 1.
Construct the following quadrilaterals.
(i) Quadrilateral
ABCD
AB = 4.5 cm
BC= 5.5 cm
MP= 5 cm
AD = 6cm
AC = 7 cm

MP Board Solutions

(ii) Quadrilateral
JUMP
JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm

(iii) Parallelogram
MORE
OR = 6 cm
RE = 4.5 cm
EO = 7.5 cm

(iv) Rhombus
BEST
BE = 4.5 cm
ET = 6 cm
Solution:
(i) Steps of Construction :
Step-1: Draw AB = 4.5 cm.
Step-2: With B as centre and radius equal to 5.5 cm, cut an arc.
Step-3 : With A as centre and radius equal to 7 cm, cut another arc on the arc drawn in step-2 at point C.
Step-4 : With C as centre and radius equal to 4 cm, cut off an arc. Also, with A as centre and. radius equal to 6 cm cut off another arc on the arc just drawn at point D.
Step-5 : Join BC, AC, CD and AD.
Hence, ABCD is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.1 1

(ii) Steps of Construction :
Step-1: Draw JU = 3.5 cm.
Step-2: With J as centre and radius equal to 4.5 cm, cut an arc.
Step-3 : With U as centre and radius equal to 6.5 cm, cut another arc on the arc drawn in step-2 at point P.
Step-4: With P as centre and radius equal to 5 cm, cut off an arc.
Step-5: With U as centre and radius equal to 4 cm, cut another arc on the arc drawn in step-4 at point M.
Step-6 : Join JP, UP, UM and PM. Thus, JUMP is the required quadrilateral.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.1 2

MP Board Solutions

(iii) We know that opposite sides of a parallelogram are equal and parallel to each other.
∴ OR = ME and MO = ER.
Steps of Construction:
Step-1: Draw OR = 6 cm.
Step-2 : With R as centre and radius equal to 4.5 cm, cut an arc.
Step-3 : With O as centre and radius equal to 7.5 cm, cut another arc on the arc drawn in step-2 at point E.
Step-4: With E as centre and radius equal to 6 cm, cut an arc.
Step-5 : With O as centre and radius equal to 4.5 cm, cut an arc on the arc drawn in step-4 at point M.
Step-6 : Join RE, OE, OM and ME.
Hence, MORE is the required parallelogram.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.1 3

(iv) We know that all four sides of a rhombus are equal.
∴ BE = ES = ST = BT = 4.5 cm.
Steps of Construction:
Step-1: Draw BE = 4.5 cm.
Step-2: With B as centre and radius equal to 4.5 cm, draw an arc.
Step-3: With E as centre and radius equal to 6 cm, draw another arc, cutting the previous arc at point T.
Step-4: With E as centre and radius equal to 4.5 cm, cut an arc.
Step-5: With T as centre and radius equal to 4.5 cm, cut another arc on the previous arc at point S.
Step-6 : Join BT, ES, ET and ST.
Hence, BEST is the required rhombus.
MP Board Class 8th Maths Solutions Chapter 4 Practical Geometry Ex 4.1 4

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.4

MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.4

Question 1.
State whether True or False.
(a) All rectangles are squares.
(b) All rhombuses are parallelograms.
(c) All squares are rhombuses and also rectangles.
(d) All squares are not parallelograms.
(e) All kites are rhombuses.
(f) All rhombuses are kites.
(g) All parallelograms are trapeziums.
(h) All squares are trapeziums.
Solution:
(a) No, because in a square all sides are equal, but it is not true in case of rectangle.
(b) Yes, because both diagonals bisect each other and opposite sides are equal.
(c) Yes, because in square all sides are equal.
(d) No, all squares are parallelograms. Because all squares satisfies the conditions of a parallelogram.
(e) No, because only two pairs of consecutive sides are equal in kite whereas in rhombus all sides are of equal length.
(f) Yes, all rhombuses are kites.
(g) Yes, because in trapezium, there is only one pair of parallel opposite sides and in parallelograms, two pairs of opposite sides are parallel.
(h) Yes, because in trapezium, there is only one pair of parallel opposite sides and in square, two pairs of opposite sides are parallel.

MP Board Solutions

Question 2.
Identify all the quadrilaterals that have
(a) four sides of equal length
(b) four right angles
Solution:
(a) The quadrilaterals, those have four sides of equal length are square and rhombus.
(b) The quadrilaterals, those have four right angles, are square and rectangle.

Question 3.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) A square is four sided, so it is a quadrilateral.
(ii) Since a square has opposite sides parallel and diagonals bisect each other, so it is a parallelogram.
(iii) Since square is a parallelogram with all 4 sides equal, so it is a rhombus.
(iv) Since square is a parallelogram with each angle a right angle, so it is a rectangle.

MP Board Solutions

Question 4.
Name the quadrilaterals whose diagonals
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) The quadrilaterals in which diagonals bisect each other are rhombus, rectangle, square and parallelogram.
(ii) The quadrilaterals in which diagonals are perpendicular bisectors of each other are rhombus and square.
(iii) The quadrilaterals in which diagonals are equal are square and rectangle.

MP Board Solutions

Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution:
When we draw the diagonals joining the end points of a rectangle it lies in its interior. So, it is a convex quadrilateral.

Question 6.
ABC is a right-angled triangle and O is the mid point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.4 1
Solution:
Since ∆ABC is right angled at B. So ∠D = 90°, \(\overline{A D} \| \overline{B C}\) and \(\overline{A B} \| \overline{C D}\)
⇒ ABCD is a rectangle where AB = CD and AD = BC
AC and BD are the diagonals which bisects each other.
Thus AO = OC and BO = OD, also BO = OC which shows that O is equidistant from A, B and C.

MP Board Class 8th Maths Solutions

MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition or property used.
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 1
(i) AD = …..
(ii) ∠DCB = …….
(iii) OC = ……..
(iv) m∠DAB + m∠CDA =
Solution:
Given that ABCD is a parallelogram.
(i) AD = BC [∵ In a parallelogram opposite sides are equal]
(ii) ∠DCB = ∠DAB [∵ In a parallelogram opposite angles are equal]
(iii) OC = OA [∵ In a parallelogram diagonals bisect each other]
(iv) m∠DAB + m∠CDA = 180° [∵ Adjacent angles in a parallelogram are supplementary]

MP Board Solutions

Question 2.
Consider the following parallelograms. Find the values of the unknowns x, y, z.
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 2
Solution:
(i) ABCD is a parallelogram in which ∠B = 100° (given)
∴ ∠A + ∠B = 180° [∵ Sum of adjacent angles is 180°]
⇒ z +100° = 180°
⇒ z = 180° – 100° = 80°
Also ∠B = ∠D and ∠A = ∠C
[∵ Opposite angles are equal]
∴ ∠B = 100° = ∠D = y and ∠A = z = 80° = ∠C = x
∠ x = 80°, y = 100°, z = 80°.

(ii) y + 50° = 180° [∵ Sum of adjacent angles is 180°]
⇒ y = 180° – 50° = 130°
Also, y = x = 130° [∵ Opposite angles are equal in a parallelogram]
And 180°- z = 50° [Linear pair]
⇒ z = 180°- 50° ⇒ z = 130°

(iii) Clearly, x = 90°
[∵ Vertical opposite angles are equal] Also, x + y + 30° = 180°
[By using angle sum property of a triangle]
⇒ 90° + 30° + y = 180°
⇒ 120° + y = 180° ⇒ y = 180° – 120° = 60°
Since, alternate angles are equal in a parallelogram
∴ y = z = 60°
Thus x = 90°, y = 60° and z = 60°.

(iv) Since x + 80° = 180°
[Sum of adjacent angles is 180°] ⇒ x = 180° – 80° = 100°
Also 80° = y [∵ Opposite angles are equal in a parallelogram]
And x = 180° – z ⇒ 100° = 180° – 2° ⇒ 2 = 180° – 100° = 80°
Thus x = 100°, y = 80° and 2 = 80°.

(v) y = 112° [∵ Opposite angles are equal in a parallelogram]
y + x + 40° = 180°
[By angle sum property of a triangle]
⇒ 112°+ x+ 40° = 180°
⇒ x +152° = 180° ⇒ x = 180° – 152°
⇒ x = 28°
⇒ x = 2 = 28° [∵ Alternate angles are equal in a parallelogram]

MP Board Solutions

Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) ∠A = 70° and ∠C = 65°?
Solution:
(i) Yes, but need not be true.
(ii) No, because in a parallelogram opposite sides are equal but here AD ≠ BC.
(iii) No, because in a parallelogram opposite angles are equal but here ∠A ≠ ∠C.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 3
We can draw a figure of a kite in which exactly two opposite angles are equal.
Hence ∠D = ∠B.

MP Board Solutions

Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a given parallelogram.
Let ∠A = 3x and ∠B = 2x.
Since the sum of adjacent angles in a parallelogram is 180°
∴ m∠A + m∠B = 180°
⇒ 3x + 2x = 180°
⇒ 5x = 180° ⇒ x = 36°
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 4
∴ m∠A = 3 × 36° = 108° and m∠B = 2 × 36° = 72°
Also, m∠A = m∠C = 108°
[∵ In a parallelogram opposite angles are equal]
and m∠B = m∠D = 72°.

MP Board Solutions

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram, in which, let m∠A = m∠B = x
Since sum of two adjacent angles is 180°
∴ m∠A + m∠B = 180°
⇒ x + x = 180° ⇒ 2x = 180°
⇒ x = 90°
Also m∠A = m∠C = 90°
[∵ In a parallelogram opposite angles are equal]
and m∠B = m∠D = 90°.

Question 7.
The given figure HOPE is a parallelogram. Find the angle measures x, y and z. State the
properties you use to find them.
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 5
Solution:
y = 40°
[Since PO || HE ∴ alternate angles are equal]
70° = y + 2 [Exterior angle property of a triangle]
⇒ z = 70° – 40° = 30°
∠POH = 180° – 70° = 110° [Linear pair]
∠POH = x = 110° [Opposite angles are equal]
Thus x = 110°, y = 40°, 2 = 30°.

MP Board Solutions

Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 6
Solution:
(i) Since GUNS is a parallelogram.
∴ GS = NU and GU = SN
[∵ In a parallelogram opposite sides are equal]
⇒ 3x = 18 ⇒ x = 6
and 3y -1 = 26 ⇒ 3y = 1 + 26 = 27
⇒ y = 9
Thus x = 6 cm and y = 9 cm.

(ii) Since diagonals bisects each other in a parallelogram.
So, 20 = y + 7 ⇒ y = 13
Also, x + y= 16 ⇒ x = 16 – 13 = 3
Thus x = 3 cm and y = 13 cm.

Question 9.
In the given figure both RISK and CLUE are parallelograms. Find the value of x.
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 7
Solution:
Since RISK and CLUE are parallelograms.
∴ ∠SKR = ∠RIS = 120° [∵ Opposite angles are equal]
Also, ∠ULC = ∠UEC = 70° [Opposite angles are equal]
∠RIS + ∠ISK = 180° [Adjacent angles are supplementary]
⇒ ∠ISK = 180° – 120° = 60°
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 8
In ∆OES, we have
70° + x + 60° = 180° [Angle sum property]
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°.

MP Board Solutions

Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel?
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 9
Solution:
In a trapezium, only one pair of opposite sides are parallel, whereas other pair of opposite sides are non-parallel.
∴ KLMN is a trapezium because MN || KL.
[∵ Sum of two adjacent interior angles is 180° = ( 80° + 100°)].

Question 11.
Find m∠C in given figure if AB || DC.
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 10
Solution:
We are given, \(\overline{A B} \| \overline{D C}\)
∴ Sum of two adjacent interior angles is 180°
i. e., ∠B + ∠C = 180° ⇒ 120° + ∠C= 180°
⇒ ∠C = 180° -120° = 60°.
Thus ∠C = 60°.

MP Board Solutions

Question 12.
Find the measure of ∠P and ∠S if \(\overline{S P} \| \overline{R Q}\) in figure. (If you find m∠R, is there more than one method to find m∠P?)
MP Board Class 8th Maths Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 11
Solution:
Since \(\overline{S P} \| \overline{R Q}\)
Thus ∠R + ∠S = 180°
[∵ Sum of two adjacent interior angles is 180°]
⇒ 90° + ∠S = 180° ⇒ ∠S = 180° – 90° = 90°
∠P + ∠Q + ∠R + ∠S = 360° [∵ Sum of all angles of a quadrilateral is 360°]
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310°
⇒ ∠P = 50°
Thus ∠P = 50° and ∠S = 90°.
Also m∠P can be found as ∠P + ∠Q = 180°
[Adjacent angles are supplementary]
⇒ ∠P + 130° = 180°
⇒∠P = 180° -130° ⇒ ∠P = 50°

MP Board Class 8th Maths Solutions