ML AGGARWAL CLASS 8 Chapter 14 Constructions of Quadrilaterals Exercise 14.1

 Exercise 14.1

Question 1

Sol :










Steps.

1. Draw PQ = 4.5cm

2.From Q draw are of length 6cm

3.From p draw are of length 6.5cm, intersection point is 'R'

4.From R draw are of length 5.5cm

5.From P draw are of the length 5cm , Intersection point is 'S'

6.Join $\overline{\mathrm{Q} R }, \overline{\mathrm{SR}}_{ ,} \overline{\mathrm{PS}}$, required quadrilateral PQRS formed

Question 2

Sol :















Steps.

1. Draw AB=3.5 cm

2. From 'A' draw Arc AD=5 cm

3.From 'B'  draw Arc BD = 5.4cm , Intersection point is 'D'

4. From $B^{\prime}$ draw Arc BC=5 cm

5.From 'D' draw arc DC = 5.6 cm , Intersection Point is 'C'

6.Join $\overline{A D}, \overline{D C}, \overline{B C}$ The required quadrilateral ABCD is formed


Question 3

Sol :








Steps.

1. Draw PQ=3 cm line

2.From Q , draw arc QR 2.2cm

3.From P , draw arc PR = 4CM , Intersection Point is 'R'

4.From  Q , draw arc QS = 5CM

5. From P, Draw arc PS = 3.5 CM , Intersection point is 'S'

6.Join $\overline{Q R}$, $\overline{P S}_{,} \overline{S R}$, Now required quadrilateral is formed


Question 4

Sol :











Steps.

1.Draw AB=5 cm line

2.From 'A' draw AC =7cm

3.From 'B' draw arc BC = 5.5 cm , Intersection point 'C'

4.From 'A' draw Arc AD = 5.5 cm

5. From 'C' draw Arc CD = 4.5 cm , Intersection point 'D'

6. Join $\overline{B C}, \overline{D L}, \overline{A D}$, Now required quadrilateral ABCD formed.

Question 5

Sol :











Steps.

1. Draw BC=6 cm line

2. At 'C' Draw C= 90 line upto $\overline{DC}=4 \mathrm{~cm}$ name point 'D'

3. From 'B' Draw $\angle B=45^{\circ}$ line 

4. From 'D' draw $\angle D=120^{\circ}$ Line , Intersection point is 'C '

5. Join $\overline{B C}, \overline{C D}$ Now required quadrilateral ABCD formed.


Question 6

Sol :












Steps.

1.Draw PQ = 4cm line 

2. Draw $\angle$ Q = 90 line up $\overline QR$= 6cm and mark 'R'

3.Draw $\angle P 60^{\circ}$ line

4.Draw $\angle$ R=120 line , Intersection Point is 'S'

5. Join $\overline{\mathrm{PS}}, \overline{\mathrm{SR}}$, Now required quadrilateral PQRS formed
  

Question 7

Sol :










Steps.

1. Draw AB=5 cm line

2. Draw $\angle B=60^{\circ}$ line from $B^{\prime}$

3.Draw Cut $\angle B=60^{\circ}$ Line with Arc of BC = 4.2 cm point 'C'

4.At 'A' Draw $\angle A=90$ and at a distance of AD = 3.5cm mark 'D'

5.Join $\overline{D C}$ Now required quadrilateral ABCD is formed.

Question 8 

Sol :










Steps.

1. Draw PQ= 4 cm line

2.Draw $\angle Q=60^{\circ}$ line and at from Q, draw arc of QR = 5cm

3.Draw $\angle R=90^{\circ}$ line upto a distance $R S=$\angle$ 4.5 cm

4.Now Join $\overline PS$ Now required quadrilateral PQRS formed.

Question 9

Sol :











Steps.

1. Draw BE=3.8 cm line

2. Draw $\angle E=80^{\circ}$, and an arc ES=3.4 cm and both cut at point 's'

3. Draw ST= 4.5 cm arc from 'S'

4. Draw BT = 5cm Arc from 'B' and Intersection Point 'T'

5. Now join $\overline{B C}, \overline{T S}$ required quadrilateral BEST formed.

Question 10

Sol :











Steps.

1. Draw AB=4.5 cm line

2.Draw  $\angle B=60^{\circ}$ line From 'B' and draw an arc BC = 4cm from 'B' Intersection Point is 'C'

3. Draw arc Dc = 3.9 cm From 'C'

4. Draw arc AD =3.2 cm from 'A' Intersection Point 'D'

5.Join $\overline{C D}, \overline{A D}$ , Now required quadrilateral ABCD formed.

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more