Question 1. Draw a parallelogram READ where RE = 3.6cm, EA = 5.2cm, and ∠REA = 75°.

Solution:

I drew a parallelogram READ where RE = AD = 3.6cm, EA = RD = 5.2cm and ∠REA = 75°

Haryana Board Class 8 Maths Chapter 1 Solutions

Question 2. Draw a rectangle GOLD where GO = 5m and OL = 7.2cm

Solution:

I drew a parallelogram GOLD where GO = LD = 5cm, OL = GD = 7.2 cm and ∠GOL = 90°.

Class 8 Maths Chapter 1 Revision Notes Haryana Board

Question 3. Draw a rhombus BEST where BE = 4.8cm, ∠BES = 45°.

Solution:

The Diagonal Of a rhombus bisect each other perpendicularly.

I drew a rhombus BESt where BE = ES = BS = BT = ET = 4.8 cm

Haryana Board Class 8 Geometry Chapter 1 Important Questions

Question 4. Draw a square ROAD where RO = 6cm.

Solution:

Diagonals of a square bisect each other perpendicularly.

I drew a square where RA = OD = 6cm

Question 1. Find the measurement of Complementary angles of 38.5°, \(33 \frac{2}{3}^{\circ}\), 90°.

Solution:

The measurement of the Complementary angle of 38.5°, (90°-38.5°) or 51.5°.

The Complementary angle of \(33 \frac{2}{3}^{\circ}\) is (90° – \(33 \frac{2}{3}^{\circ}\)) or \(56 \frac{1}{3}^{\circ}\)

The Complementary angle of 90° is (90°-90°) or 0.

Question 2. Find the measurement of Supplementary angles of 120°, 0°, 128.6°, 90°.

Solution:

The measurement of the Supplementary angle of 120° is (180°-120°) or 60°.

The Supplementary angle of 0° is (180°-0°) or 180°

The Supplementary angle of 128.6° is (180°- 128.6°) or 51.4°

The Supplementary angle of 90° is (180°-90°) or 90°

Haryana Board Class 8 Maths Chapter 6 Solutions

Question 3. Which pair of angles are complementary?

1. 52°, 48°
2. x°, 90°-x°,
3. 60°, 120°.

Solution:

52°+48° = 100°

x°+90°-x° = 90° [complementary angles]

60°+128° = 180°

Angle-Side Relationship Theorem Class 8 Haryana Board

Question 4. Which pair of angles are Supplementary?

1. 42°, 139°
2. 70°, 110°,
3. 90°, 90°

Solution:

42°+1839° = 181°

70° +110° = 180° – Supplementary angles

90+90= 180° – Supplementary angles

Verification of the Relation Between Angles and Sides of a Triangle Class 8

Question 5. If measurements of two adjacent angles are 85.4° and 94.6%, then how external sides of those two angles are Situated?

Solution:

The Sum of two adjacent angles is 85.4°+ 94.6° = 180°

So, the external sides of the two angles are situated on the Same Straight line.

Question 6. A and B are Supplementary angles to each other. If ∠A = (x + 20)°, then find the value of ∠B.

Solution:

∠A and ∠B are Supplementary angles.

∠A + ∠B = 180

(x+20°) + ∠B = 180°

∠B = 180°- x° – 20°

∴ ∠B = (160-x)°

Haryana Board 8th Class Maths Chapter 6 Important Questions

Question 7. If one angle of the Complementary angle is u times the Other then find the measurement of the Smaller angle.

Solution:

Let the measurement of the Smaller angle be x.

∴ The other angle is 4x°

⇒ 4x°+2°-90°

⇒ 5x° = 90°

⇒ \(x^{\circ}=\frac{90^{\circ}}{5}\)

⇒ x°= 18°

The measurement of the Smaller angle is 18°.

Question 8. Find the Complementary and Supplementary angles of 72°13′24′′.

Solution:

The Complementary angle of 72° 13’24” is (90-72°13’24”)

⇒ \(\begin{aligned}
& 90^{\circ}=89^{\circ} 59^{\prime} 60^{\prime \prime} \\
&\quad \quad-72^{\circ} 13^{\prime} 24^{\prime \prime} \\
& \hline \quad \quad \quad 17^{\circ} 46^{\prime} 36^{\prime \prime} \\
& \hline
\end{aligned}\)

= 17°46’36”

The Supplementary angle of 72°13′24′′ is (180°-72°13′24′′)

⇒ \(\begin{aligned}
& 180^{\circ}=179^{\circ} 59^{\prime} 60^{\prime \prime} \\
&\quad \quad -72^{\circ} 13^{\prime} 24^{\prime \prime} \\
& \hline \quad \quad \quad 107^{\circ} 46^{\prime} 36^{\prime \prime} \\
& \hline
\end{aligned}\)

= 107°46’36”

Triangle Properties Class 8 Haryana Board Solutions

Question 9. In the adjacent figure, how are the line Segment OA and OE situated?

Solution:

∠AOF = ∠AOB + ∠BOC+ ∠COD + ∠DOE + ∠EOF

= 25° + 32° + 41°+ 40°+ 42° = 180° (one Straight line)

∴ OA and OF are on the Same straight line.

Question 10. Find the measurements of complementary and Supplementary angles of (2x-15)°

Solution:

The Complementary angles of (2x-15)° is (90-2x+15)° Or (105-2x)° and Supplementary angles is (180°-2x-+15)° or (195-2x)°.

Question 11. Choose the Correct answer:

1. The Complementary angle of 90° is

1. 90°
2. 45°
3. 0°
4. None of these.

Solution: (90°-90°) = 0°

The Correct answer is (3).

Class 8 Geometry Triangle Angle Sum Property Proof Haryana Board

2. The Supplementary angle of 45° is

1. 90°
2. 135°
3. 0°
4. 45°

Solution:

(180°-45°) = 135°

The Correct answer is (2)

3. Which pair of angles are not Supplementary?

1. 42°, 139°
2. 70°, 110°
3. 90°, 90°
4. x, 180°-x°

Solution:

42°+139° = 181°

70+110° = 180° – Supplementary angle

90°-+90° = 180° – Supplementary angle

X+180x = 180° – Supplementary angle

The Correct answer is (1)

Question 12. Write ‘True’ or ‘False’.

1. Any two adjacent angles are Complementary to each other.

Solution:

If the Sum of measurements of two angles is equal to 90°, then each angle is Called Complementary to the other angle.

The sum of any two adjacent angles maybe 90° or may not be 90°

So the statement is false.

Verification of Triangle Properties Class 8 Worksheet Haryana Board

2. The Supplementary angle of the right angle is a right angle.

Solution:

The Supplementary angle of the right angle is (180°-90°) or 90° which is a right angle.

So the statement is true.

Question 13. Fill in the blanks are not _________ to each other.

1. Two acute angles

Solution: Supplementary.

2. The Complementary angle of x° is

Solution: 90°-x°

Question 1. In the adjacent figure find the measurement of ∠AOE, ∠BOD, and ∠AOC.

Solution:

∠AOD = ∠BOC (vertically opposite angles)

∠AOD = 90°

i.e., ∠AOE + ∠DOE = 75°

⇒ ∠AOE = 75-30 = 45°

Again, ∠BOD + ∠BOC = 180° (AS BO stands on CD)

∠BOD + 75° = 180°

⇒ ∠BOD = 180°- 75° = 105°

∠AOC = ∠BOD (Vertically opposite angles) = 105°

Question 2. In the adjoining figure if ∠POR=2 ∠QOR, then find the value of ∠POS.

Solution:

∠PQR + ∠QOR = 180° (AS OR Stands on PQ]

2∠QOR + ∠QOR = 180°

⇒ 3∠QOR = 180°

⇒ \(\angle Q O R=\frac{180^{\circ}}{3}=60^{\circ}\)

∠POS = ∠QOR (vertically opposite angles) = 60°

Class 8 Maths Chapter 3 Vertically Opposite Angles Haryana Board

Question 3. Two Straight lines ∠PQ and ∠RS intersect at point O; OT is the bisector of ∠POS. If POR = 45°, then find ∠TOS.

Solution:

∠OP Stands On ∠RS.

∴ ∠POR + ∠POS = 180°

⇒ 45° + ∠POS = 180°

⇒∠POS= 180 °-45°= 135°

OT is the bisector of ∠POS

∴ ∠TOS = \(\frac{1}{2} \angle P O S=\frac{1}{2} \times 135^{\circ}=67 \frac{1}{2}^{\circ}\)

Haryana Board Class 8 Maths Vertically Opposite Angles Solutions

Question 4. If two straight lines intersect each other then four angles are formed. Find the Sum of the measurement of four angles.

Solution:

Let two straight lines AB and CD intersed at point O.

CO Stands on AB.

∴ ∠ADC + ∠COB = 180°

OD is stands on AB.

∴ ∠AOD + ∠BOD = 180°

∴ ∠AOC+ ∠COB + ∠AOD + ∠BOD = 130°+ 180°= 360°.

Question 5. In the adjacent figure find the value of x, y, and z?

Solution:

∠AOC = ∠BOD (vertically opposite angles)= 40°

∠AOP + ∠OOD + ∠BOD = 180°

60°+ y° + 48 = 180°

⇒ y° = 180°-100° = 80°

∠AOC + ∠COQ + ∠BOQ = 180°

40°+z°+30° = 180° ⇒ z° = 180°-70° = 110°

Haryana Board 8th Class Maths Vertically Opposite Angles Questions and Answers

Question 6. The straight lines AB and CD intersect at point O; ∠AOD + ∠BOC = 102°, If Op is the bisector of ∠BOD, then find the measurement of ∠BOP.

Solution:

∠AOD = ∠BOC [vertically opposite angles)

∠AOD + ∠BOC = 102°

∠AOD + ∠AOD = 120°

⇒ 2∠AOD = 102°

⇒ ∠AOD = \(\frac{102^{\circ}}{2}=57^{\circ}\)

OD stands on AB

∴ ∠AOD + ∠BOD = 180°

51°+ ∠BOD = 180°

⇒ ∠BOD = 180°-51° = 129°

Op is the bisector of ∠BOD

∴ \(\angle B O P=\frac{1}{2} \angle B O D=\frac{1}{2} \times 129^{\circ}=64 \frac{1}{2}^{\circ}\)

Question 7. Prove that internal and external bisectors of an angle are perpendicular to each other.

Solution:

Let op and OQ be the internal and external bisectors of ∠AOC respectively.

Required to prove: Op and OQ are perpendicular to each other.

Proof: OQ Is the external bisector of ∠AOC,

So OQ is the bisector of BOC.

∠POQ = ∠POC + ∠COQ

= \(\frac{1}{2}\) ∠AOC + \(\frac{1}{2}\) ∠COB

= \(\frac{1}{2}\) (∠AOC+ ∠COB) = \(\frac{1}{2}\) x ∠AOB

= \(\frac{1}{2}\) x 180° (one straight angle] = 90°

OP and OQ are perpendicular to each other.

Chapter 3 Vertically Opposite Angles Class 8 Solutions in Hindi Haryana Board

Question 8. PQ and RS are two straight lines intersecting at a point O. Prove that if the bisector of the LPOR is produced through O, it will bisect the ∠SOQ.

Solution:

Let AO is the bisect LPOR and let it be Produced to B.

Required to prove: OB bisects SOQ.

Proof: ∠SOB = ∠AOR [vertically opposite angles]

∠BOQ = ∠AOP [vertically opposite angles]

Again, ∠AOR = ∠AOP [AO is the bisector of POR]

∴ ∠SOB = ∠BOQ

∴OB bisects ∠SOQ (Proved).

Question 9. Choose the Correct answer:

1. In the adjacent figure if ∠1 = 35°, then find the value of ∠2 is

1. 35°
2. 145°
3. 70°
4. 55°

Solution:

∠1 + ∠2 = 180°

35 + ∠2 = 180°

⇒ ∠2 = 180°-35° = 145°

So the Correct answer is (1).

Haryana Board Class 8 Maths Exercise 3.1 Solutions

2. In the adjacent figure, if ∠TOS =20° and ∠ROQ = 60°, then the Value of ∠POT is

1. 66°
2. 120°
3. 40°
4. 80°

Solution:

∠POS = ∠ROQ (vertically opposite angles)

= 60°

i.e., ∠POT+∠TOS = 60°

⇒∠POT+20° = 60°

⇒∠POT = 60°-20° = 40°

So the Correct answer is (3)

Step-by-Step Solutions for Vertically Opposite Angles Class 8 Haryana Board

3. In the adjacent figure if ∠AOC + ∠BOP = 112°, the value of ∠BOC is

1. 112°
2. 56°
3. 68°
4. 124°

Solution:

∠AOC = ∠BOD

∠AOC+ ∠BOD = 112°

∠AOC + ∠AOC = 112°

⇒ 2∠A0C = 112°

⇒ ∠AOC = \(\frac{112^{\circ}}{2}=56^{\circ}\)

∴ ∠BOC + ∠AOC = 180°

⇒ ∠BOC = 180°-56° = 124°

So the Correct answer is (4)

Question 10. Write ‘True’ or ‘False”

1. The vertically opposite angle of 68° is 112°

Solution:

The vertically opposite angle of 68° is 68°

So the statement is false.

Important Questions for Class 8 Maths Chapter 3 Haryana Board

2. If op stands on line AB and ∠AOP = 100°, then the value of ∠BOP is 80°.

Solution:

OP is standing on AB

∴ ∠AOP + ∠BOP = 180°

∠AOP + 80° = 180°

⇒ ∠AOP = 180°-80° = 100°

So the Statement is true.

Question 11. Fill in the blanks:

1. If a ray Stands on a Straight line, then the Sum of measurement of two ________ angles so formed is 180°.

Solution: Adjacent.

2. The value of right angle is half of _________.

Solution: Straight angle.

Question 1. In the adjacent figure find the Value of x.

Solution:

AB||CD and EF is transversal

∠EGB = ∠GHD (corresponding angles] = 56°

∠AGE + ∠EGB = 180° [AS GE stands on AB)

x + 56°= 180°

⇒ x = 180 ° – 56°

⇒ x = 124°

Properties of Parallel Lines and Their Transversal Worksheet Class 8

Question 2. In the adjacent figure find the value of y.

Solution:

AB||CP and EF are transversal.

∠DHF = ∠BGH. (Corresponding angles] = 105°

∠CHF = ∠DHF = 180°

y + 105° = 180°

⇒ y = 180°-105°

⇒ y = 75°

Haryana Board Class 8 Maths Geometry Chapter 4 Solutions

Question 3. Examine the measurement of the angles given below Concludes logically that AB and CD are parallel.

Solution:

1. ∠BGH + ∠GHD = 140°+ 20° = 160° ≠ 180 °

∴ AB and CD are not parallel lines to each other.

2. ∠BGH = 180°- ∠AGH

= 180°- 130°

= 50°

Again, ∠DHF = 50°

∠BGH = ∠DHF and these angles are Corresponding angles.

∴ AB||CD

3. ∠BGH = 180° – ∠EGB = 180°-70°

= 110°

Again, ∠DHF = 110°

∴ ∠BGH = ∠DHF and these angles are Corresponding angles.

∴ AB||CD.

Question 4. In parallelogram PQRS, if ∠P=90°, then find the values of the other three angles.

Solution:

In parallelogram PQRS,

SP||RQ and PQ is transversal.

∴ ∠P + ∠Q = 180°

∴ 90° + ∠Q= 180°

⇒ ∠Q = 180°-90°

= 90°

Again, SR||PQ and RQ are transversal.

Properties of Parallel Lines and Their Transversal Class 8 Solutions

Question 5. If the adjacent figure is PQ||RS; if ∠BAQ = 3∠ABS, then find the value of ∠RBN.

Solution:

AB||RS and MN is transversal.

∴ ∠BAQ + ∠ABS = 180°

3∠ABS + ∠ABS = 180°

⇒ 4∠ABS = 180°

⇒ ∠ABS = \(\frac{180^{\circ}}{4}=45^{\circ}\)

∠RBN = ∠ABS = 45°

Question 6. Prove that Straight lines perpendicular to the Same straight line are parallel to one another.

Solution:

let PQ and Rs are both perpendicular to AB.

Required to prove: PQ||RS

Proof: PQ⊥AB

∴ ∠PQB = 90°

∴ RS⊥AB

∴ ∠RSB = 90°

∴ ∠PQB = ∠RSB and these are Corresponding angles.

∴ PQ||RS (Proved).

Parallel Lines and Transversals Class 8 Haryana Board Questions

Question 7. In the adjacend figure AB||CD, ∠RCD = 30°, ∠PAB = 50°, ∠PAC= 140°. Find the measurement of all the angles of ΔAQC.

Solution:

I drew Qs through Q which is parallel to AB.

As AB||CD and AB||QS

∴ AB||QS||CP

AB||QS and PQ is transversal.

∴ ∠PQS = ∠PAB [Conrresponding angles) = 50°

QS||CD and QR is transversal.

∴ ∠RQS = ∠RCD (Corresponding angles) = 30°

∠AQC = ∠PQS +∠ RQS = 50°+30° = 80°

∠QAC + ∠PAC = 180°

∠QAC+ 140° = 180°

⇒ ∠QAC = 180°-140° = 40°

In ∠AQC, ∠AQC = 80°, ∠QAC = 40°

∴ ∠ACQ = 180°- (80°+40°) [In ΔAQC, ∠QAC+ ∠AQC + ∠ACQ = 180°)

= 180°-120° = 60°

Question 8. In the adjacent figure, AB||CD and ∠EGB = 50% Find the Values of ∠AGE, ∠AGH, ∠CHF, and ∠DHF.

Solution.

∠AGE +∠ EGB = 180°

∠AGE +50° = 180°

∠AGE = 180°- 50° = 130°

∠AGH = ∠EGB [vertically opposite angles) = 50°

AB||CD and EF are transversal.

∴ ∠GHD = ∠AGH (Altemate angles] = 50°

∠CHF = ∠GHD = 50°

∠CHF + ∠OHF = 180°

50° + ∠DHF = 180°

∠DHF = 180°-50° = 130°

Class 8 Maths Chapter 4 Haryana Board Important Questions

Question 9. O is any point inside two parallel lines AB and CD. Op and OQ are two perpendiculars on AB and CD respectively. prove that P, O, and Q are Collinear.

Solution:

Through O the straight line RS is drawn parallel to AB.

AB||CD and AB||RS.

∴ AB||CD||RS

OP⊥AB ∴∠OPB = 90°

OQ⊥CD ∴ ∠OQD = 90°

AB||RS and op is transversal.

∴ ∠OPB + ∠POS = 180°

90°+ Pos = 180°

⇒ ∠POS = 180°-90° = 90°

CD||RS and OQ are transversal.

∠OQD + ∠QOS = 180°

90°+ ∠QOS = 180°

⇒ ∠QOS = 180°- 90° = 90°

∠POQ = ∠QOS + ∠POS = 90° + 90° = 180°

∴ OP and OQ lie on the Same Straight line.

∴ P, O, and Q are collinear.

Question 10. If the Sides of angles are respectively parallel to the Sides of another angle, then the angles are either equal or Supplementary.

Solution:

Given: Let in angles ABC and ∠DEF, AB||DE and BC||EF, BC and DE intersect at G.

Required to prove:

1. ∠ABC = ∠DCF

2. ∠ABC and ∠DEF Supplementary i.e., ∠ABC + ∠DEF = 180°

Proof:

From (1) AB||DE and BC are transversal.

∴ ∠ABC = ∠DGC (Corresponding angles)

Again, BC||EF and DE is transversal.

∴ ∠DGC = ∠DEF (Corresponding angles)

AS ∠ABC = ∠DGC and ∠DGC = ∠DEF

∴ ∠ABC = ∠DEF (Proved).

From (2) BC||EF and DE is transversal

∴ ∠DGB = ∠DEF (Corresponding angles)

Again, AB||DE and Bc is transversal.

∴ ∠ABC + ∠DGB = 180°

∴ ∠ABC + ∠DEF = 180°

∴ ∠ABC and ∠DEF are Supplementary angles.

Haryana Board 8th Class Maths Geometry Notes Chapter 4

Question 11. Choose the Correct answer:

1. In the adjacent figure. If AB||CD, then the value of x is,

1. 68°
2. 22°
3. 112°
4. 34°

Solution:

AB||CD and EF are transversal.

∴ ∠EGB = ∠GHD (Corresponding angles) = 68°

The ray GE is stands on line AB

∴ ∠AGE + ∠EGB = 180°

∠AGE + 63° = 180

⇒ ∠AGE = 180°- 68°-112° ⇒ x = 112°

So the Correct answer is (3).

Haryana Board Class 8 Chapter 4 Maths MCQ Questions

2. In the adjacent figure AB||CD, iF E∠GB = 50°, then the Value of x is

1. 130°
2. 40°
3. 50°
4. 60°

Solution:

∠GHD = ∠EGB = 50°

∠CHF = ∠GHD (vertically opposite angle) = 50°

So the Correct answer is (3).

Question 12. Write ‘True’ or ‘False”:

1. In the adjacent figure if 3 = 120° and 8 = 60°, then AB||CD.

Solution:

∠6 + ∠8 = 60°

∠3 + ∠8= 120° + 60° = 180°

∴ AB||CD

So the Statement is true.

Class 8 Geometry Parallel Lines Theorem Explanation Haryana Board

2. In the adjacent figure if ∠EGB = 75° and ∠PHF = 95°, then AB||CD

Solution:

∠AGH= ∠EGB (vertically opposite Angles) = 45°

∴ ∠GHC = ∠DHF = 95°

∠AGH + ∠GHC = 75°+95° = 170°

∴ AB and CD are not parallel to each other.

So the Statement is False.

Question 13. Fill in the blanks:

1. If a straight line intersects a pair of Straight lines and the measurement of one pair of Corresponding angles is equal, then the two straight lines are _______.

Solution: Parallel.

2. If the sides of an angle are respectively parallel to the sides of another angle, then the angles are either equal or _________.

Solution: Supplementary.

Question 1. If the measurement of the angle of an isosceles triangle is low, then find the measurement of the other two angles.

Solution:

The Sum of two acute angles is (180°-105°)

or 75°

Let ∠A = 105° and AB = AC

∴ \(\angle B=\angle C=\frac{180^{\circ}-105^{\circ}}{2}=\frac{75^{\circ}}{2}=37.5^{\circ}\)

Question 2. In an Isosceles triangle one angle of the base is 550, then find the measurement of the Vertical angle.

Solution:

In an isosceles triangle, one angle of the base is 55°.

The other angle of the base is 55°.

Then the vertical angle is 180°- (55°+ 55°) = 180°-110°= 70°

What is the relation between two sides of a triangle and their opposite angles?

Question 3. In the adjacent figure, in ΔABC, AB = AC, If ∠A+∠B = 115°, find the measurement of ∠A.

Solution:

In ΔABE, ∠A + ∠B + ∠C = 180°

115° + ∠C = 180°

⇒ ∠C = 180° = 115°

⇒ ∠C = 65°

As AB = AC

∴ ∠B = ∠C = 65°

∠A = 115°-65° = 50°

Question 4. Two line Segments AB and CD bisect each other at 0; If AC = 4cm, then find the length of BD.

Solution:

In ΔAOC and ΔBOD,

OA = OB, OC = OD

and ∠AOC = ∠BOD (vertically Opposite angle)

∴ ΔAOC ≅ ΔBOD (by AAS Congruency)

∴ Ac = BP

4cm = BD

∴ The length of BD is 40m.

How to prove the relation between sides and angles of a triangle in Class 8 Maths?

Question 5. In an isosceles triangle, the vertical angle is three times each angle of the base. Find the measurement of the Supplementary angle of the Vertical angle.

Solution:

Let the measurement of each angle of the base be x°

∴ The measurement of the vertical angle is 37°

The sum of the three angles of a triangle is 180°.

∴ 3x°+x°+x° = 180°

⇒ 5x°=180°

⇒ \(x^{\circ}=\frac{180^{\circ}}{5^{\circ}}=36^{\circ}\)

∴ The Vertical angle Is (3×36°) or 180°

The Supplementary angle of the Vertical angle is (180°-108°) Or 72°.

Question 6. In ΔABC, AB=AC, The bisector of ∠ABC intersects AC at D. If ∠A=56°, then find the Value of ∠ABD.

Solution:

In ΔABC,

AB = AC

∴ ∠ABC = ∠ACB

∠BAC=560

In ΔABC,

∠BAC + ∠ABC + ∠ACB = 180°

56° + ∠ABC + ∠ABC = 180°

⇒ 2 ∠ABC = 180°-56°=124°

⇒ \(\angle A B C=\frac{124^{\circ}}{2}=62^{\circ}\)

As BD is the bisector of ABC.

∴ \(\angle A B D=\frac{1}{2} \angle A B C=\frac{1}{2} \times 62^{\circ}=31^{\circ}\)

Question 7. In ΔABC, AB = AC; BC is extended to D Such that AC=CD; if ∠ABC=70°, then find the value of ∠BAD.

Solution:

In ABC, AB=AC

‍∴ ∠ACB = ∠ABC = 70°

Again, ∠ACB + ∠ACD = 180°

70° + ∠ACD = 180°

⇒ ∠ACD = 180°-70°

⇒ ∠ACD = 110°

In ΔACD, AC= CD ∴ ∠DAC = ∠ADC

∠ACD + ∠ADC + ∠DAC = 180°

110° + ∠ADC + ∠ADC = 180°

2∠ADC = 180°- 110° = 70°

⇒ \(\angle A D C=\frac{70^{\circ}}{2}=35^{\circ}\)

In ∠ABD + ∠ADB + ∠BAD = 180°

i.e., ∠ABC + ∠ADC+ ∠BAD = 180°

70°+35°+ BAD = 180°

⇒ ∠BAD = 180°-105° = 75°

What is the theorem on the relation between sides and angles of a triangle?

Question 8. AB is the hypotenuse of the isosceles right-angled triangle ABC AD is the bisector of ∠BAC and AD intersects BC at D. Prove that AC + CD = AB.

Solution:

In the right-angled Isosceles triangle ABC, AB is the hypotenuse.

AD is the bisector of ∠BAC and AD Intersects BC at D.

Required to prove: Ac+CD = AB.

Construction: Through D I draw DC which is perpendicular to AB.

Proof: In ΔACD and ΔADE,

∠CAD = ∠EAD [as AD is the bisector of ∠BAC]

∠ACD = ∠AED = 90° [∵ DE ⊥ AB]

and AD is Common side.

∴ ΔACD ≅ ΔADE [by AAS Congruency)

∴ AC = AE [Corresponding Sides of Congwent triangles]

and CD = DE [Corresponding Sides]

In ΔABC, ∠ACB = 90° and AC = BC.

∴ ∠BAC = \(\angle A B C=\frac{90^{\circ}}{2}=45^{\circ}\)

In ΔBDE, ∠BED = 90°, B=45°

∴ ∠BDE = 180°-90° = 45° = 45°

∴ ∠BDE = ∠B ∴ BE = DE

Again CD = DE DE = CD = BE

AC+ CD = AE + BE = AB (Proved)

Question 9. Choose the Correct Answer:

1. In the adjacent figure, in ΔABC, which relation is correct?

1. AB = BC
2. AB = AC
3. AC = BC
4. AC ≠ BC

Solution:

In ΔABC, ∠BAC = ∠ABC= 70°

∴ AC = BC (If the two angles of a triangle are equal then their opposite sides are equal)

So the Correct answer is (3).

What are the important theorems in Haryana Board Class 8 Maths Chapter 5?

2. In ΔABC, AB = AC; If ∠BAC = 70°, then the value of ∠ACB is

1. 70°
2. 110°
3. 35°
4. 55°

Solution:

In ΔABC, AB = AC ∴ ∠ACB = ∠ABC.

Again, ∠BAC+ ∠ABC + ∠ACB = 180°

70° + ACB + ACB = 180°

⇒ 2∠ACB = 180°-70°

⇒ 2∠ACB = 110°

⇒ \(\angle A C B=\frac{110^{\circ}}{2}=55^{\circ}\)

So the correct answer is (4).

3. In the adjacent figure, in ΔABC, AB = AC and DE||BC; If ∠AED = 50°, then the value of ∠ABC is

1. 50°
2. 80°
3. 100°
4. 70°

Solution:

DE||BC and AC is transversal.

∴ ∠ACB = ∠AED (Corresponding angles) = 50°

In ΔABC, AB=AC ∴ ∠ACB = ∠ABC = 50° = ∠ABC.

So the correct answer is (1).

Question 10. Write ‘True’ and ‘False”.

1. The external bisector of the Vertical angle of an isosceles triangle is parallel to the base.

Solution: The Statement is true.

2. The Corresponding angles of two Congruent triangles are equal.

Solution: The statement is true.

How to solve Haryana Board Class 8 Geometry Chapter 5 problems step by step?

Question 11. Fill in the blanks:

1. The lengths of the hypotenuse of two _______ right-angled triangles are equal.

Solution: Congruent.

2. In an isosceles obtuse angle triangle the unequal angle is _________.

Solution: Obtuse angle

Question 1. In the adjacent figure, find the value of x.

Solution:

I joined A, C and AC is produced to T.

In ΔABC,

Exterior ∠BCT = ∠BAC + ∠ABC

In ΔACO,

Exterior ∠BCT = ∠BAC+ ∠ABC

In ΔACD,

Exterior ∠DCT = ∠DAC + ∠ADC

∠BCT + ∠DCT = (∠BAC+ ∠DAC) + ∠ABC + ∠APC

i.e ∠BCD = ∠BAD + ∠ABC + ∠ADC

x = 72°+45+30° ⇒ x = 147°.

Haryana Board Class 8 Maths Chapter 6 Solutions

Question 2. In the adjacent figure find the Value of ∠A + ∠B + ∠C + ∠D + ∠E + ∠F.

Solution:

(∠A+∠B)+(∠C+∠D)+(∠E+∠F)

= ∠BOD + ∠DOF + ∠FOB = 360°

Question 3. In ABC, BC is produced to D. If ∠ACD = 126° and ∠B = \(\frac{3}{4}\) ∠A then find the Value of ∠A.

Solution:

In ΔABC,

∠A + ∠B = exterior ∠ACD

⇒ \(\angle A+\frac{3}{4} \angle A=126^{\circ}\)

⇒ \(\frac{7 \angle A}{4}=126^{\circ}\)

⇒ \(\angle A=\frac{4}{7} \times 126^{\circ}\)

⇒ ∠A = 4 x 126°

⇒ ∠A = 4 x 18°

⇒ ∠A = 72°

Verification of the Relation Between Angles and Sides of a Triangle Class 8

Question 4. If O is an interior point of ABC, then find the relation between ∠BOC and ∠BAC.

Solution:

I join A, O and AD is extended to T

In ΔAOB, the exterior ∠BOT = ∠BAO + ∠ABO

∴ ∠BOT > ∠BAO

Similarly, In ΔAOC, ∠COT > ∠CAO

∴ ∠BOT + ∠COT > ∠BAO + ∠CAO

i.e., ∠BOC > ∠BAC. This is the relation.

Question 5. Find the Sum of measurement of all angles of a quadrilateral.

Solution:

In quadrilateral ABCD, I join A,C.

In ΔABC,

∠BAC + ∠ABC + ∠ACB = 180°

In ΔADC, ∠DAC + ∠ADC + ∠ACD = 180°

(∠BAC + ∠DAC) + ∠ABC + ∠ADC + (∠ACB + ∠ACD) = 180° + 180°

∴ ∠BAD + ∠ABC+ ∠ADC + ∠BCD = 360°

Haryana Board Class 8 Maths Geometry Chapter 6

Question 6. In PQR, If ∠P = 80° and ∠Q= 70°, then find the relation between PQ and QR.

Solution:

In ΔPQR, ∠P = 80°, ∠Q = 70°

∴ ∠R = 180°- (80°+70°) = 30°

As ∠P > ∠R ∴ QR >PQ (This the relation)

Question 7. The hypotenuse of a right-angled triangle is greatest One ____ Explain.

Solution:

In ΔABC, ∠ABC = 90°

∴ Ac is the hypotenuse.

∠A and ∠C each are acute angles.

∴ ∠ABC > ∠A and ∠ABC > ∠C

As ∠ABC > ∠A ∴ ∠AC > ∠BC

As ∠ABC > ∠C ∴ AC > AB

∴ Ac is the largest side.

Triangle Angles and Sides Class 8 Haryana Board Questions

Question 8. If the ratio of measurement of angles of a triangle is 4:5:9; then write the nature of the triangle.

Solution:

Let the measurement of three angles are 4x°, 5x° and 9x°

4x°+5x°+9x° = 180°

⇒ 18x° = 180°

⇒ \(x^{\circ}=\frac{180^{\circ}}{18}=10^{\circ}\)

∴ The angles are 4×10° or 40°, 5×10° or 50° and 9×10° or 90°

∴ The triangle is a right-angled triangle.

Question 9. In ΔABC, the bisectors of ∠ABC and ∠ACB meet at 0. If AB > Ac then Prove that OB > OC.

Solution:

Given: In ΔABC, AB > AC, OB, and OC are the bisectors of ∠ABC and ∠ACB respectively.

RTP: OB > OC.

Proof: AB > AC

∴ ∠ACB > ∠ABC ⇒ ∠ACB > \(\frac{1}{2}\) ∠ABC

∴ ∠OCB > ∠OBC

∴ OB > OC (Proved).

Haryana Board Class 8 Maths Chapter 6 Important Questions

Question 10. In ΔPQR, the Internal bisector of ∠PQR and the external bisector of ∠PRQ intersect at T. If ∠QPR = 40° then find the Value of ∠QTR.

Solution:

In ΔPQR,

⇒ ∠QTR + ∠TQR = exterior ∠TRS

⇒ ∠QTR = \(\frac{1}{2}\) ∠PRS = \(\frac{1}{2}\) ∠PQR

⇒ \(\frac{1}{2}\) (∠QPR + ∠PQR) – \(\frac{1}{2}\) ∠PQR

⇒ \(\frac{1}{2}\) ∠QPR + \(\frac{1}{2}\) ∠PQR – \(\frac{1}{2}\) ∠PQR

⇒ \(\frac{1}{2}\) ∠QPR = \(\frac{1}{2}\) x 40°

⇒ 20°

Class 8 Geometry Triangle Theorem Explanation Haryana Board

Question 11. Choose the Correct answer:

1. In ΔABC, AB = AC, BC is produced to D. If ∠ACD = 112°, then the Value of ∠BAC is

1. 44°
2. 68°
3. 22°
4. 34°

Solution:

∠ACD + ∠ACB = 180°

112° + ∠ACB = 180°

Or, ∠ACB = 180°- 112° = 68°

In ΔABC, AB = AC, ∴ ΔABC = ∠ACB = 68°

In ΔABC, exterior ∠ACD = ∠BAC+ ∠ABC

112°= ∠BAC + 68°

⇒ ∠BAC = 112°-68° = 44°

So the Correct answer is (1).

2. In ΔABC, If ∠A =70° and ∠B=60°, then the relation between AB and BC is

1. AB = BC
2. AB > BC
3. AB < BC
4. None of these

Solution:

In ΔABC, ∠A + ∠B + ∠C = 180°

70° + 60° + ∠C = 180°

⇒ ∠C = 180° – 130° = 50°

As, ∠A > ∠C

∴ BC > AB ⇒ AB > BC

So the correct answer is (3).

Haryana Board Class 8 Maths Chapter 6 MCQ Questions

3. If the measurement of an angle of a triangle is equal to the Sum of the other two angles, then the triangle becomes.

1. Acute angled triangle
2. Obtuse angled triangle
3. Equilateral triangle
4. Right-angled triangle.

Solution:

In ΔABC, ∠A = ∠B + ∠C

∠A + ∠B + ∠C = 80°

∠A + ∠A = 180°

⇒ 2∠A = 180°

⇒ ∠A = 90°

The triangle is the right-angled triangle.

So the Correct answer is (4).

Question 12. Write ‘True’ or ‘False”.

1. If the ratio of measurements of the three angles of a triangle is 1:2:3, then the triangle becomes a right-angled triangle.

Solution:

Let the angles are x°, 2x°, and 3x°.

x+2x+3x = 180

⇒ 6x = 180

⇒ x = 30.

∴ The angles are 30°, 30°x2 or 6o° and 30°x3 or 90°

∴ The triangle is a right-angled triangle.

So the statement is true.

2. In the adjacent figure, if PQ || TS, then the Value of x is 80.

Solution:

∠QTS = ∠PQT [Alternate angle] = 55°

i.e., ∠RTS = 55°

In ΔRTS,

∠TRS + ∠RTS + ∠RST = 180°

x + 55° + 40° = 180°

⇒ x = 180° – 95° = 85°

So the statement is false.

Question 13. Fill in the blanks:

1. In an obtuse-angled triangle, the Opposite Side of __________ is the largest

Solution:

Let, In ΔABC, ∠B is an obtuse angle,

∴ ∠A and ∠C are both acute angles.

∴ ∠B > ∠A and ∠B > ∠C

As ∠B > A then AC > AB

∴ AC is largest

∴ The obtuse-angled triangle on the opposite side of the obtuse angle is the largest.

Relation Between Angles and Sides of a Triangle Class 8 Notes

2. In ΔABC, AB = AC; BC is produced to D. If ∠ACD = 105° then the value of ∠BAC is _________.

Solution:

In, ∠ABC, ∠ACD = 105°

∠ACB = 180°- ∠ACD

= 180°-105° = 75°

AB = AC

∴ ∠ABC = ∠ACB = 75°

∠BAC + ∠ABC = Exterior ∠ACD

∠BAC + 75°= 105°

⇒ ∠BAC = 105° – 75°

= 30°

Question 1. Multiply the following:

1. (4-5x) by (7x+6)

Solution:

Given (4-5x) by (7x+6)

(4-5x) by (7x+6) = (4-5x) (7x+6)

(4-5x) (7x+6) = 28x + 24 – 35x² = 30x

(4-5x) (7x+6) = -35x² = 2x+24

2. (a²-30+7) by (2a-5)

Solution:

Given (a²-3a+7) by (2a-5)

(a²-3a+7) by (2a-5) = (a²-3a+7) (2a-5)

(a²-3a+7) (2a-5) = \(2 a^3-6 a^2+14 a-5 a^2+15 a-35\)

(a²-3a+7) (2a-5) = \(2 a^3-11 a^2+29 a-35\)

Haryana Board Class 8 Maths Multiplication and Division of Polynomials Solutions

3. \(\left(x^2+x y+y^2\right) \text { by }\left(x^2-x y+y^2\right)\)

Solution:

Given \(\left(x^2+x y+y^2\right) \text { by }\left(x^2-x y+y^2\right)\)

= \(x^2\left(x^2+x y+y^2\right)-x y\left(x^2+x y+y^2\right)+y^2\left(x^2+x y+y^2\right)\)

= \(x^4+x^2 y+x^2 y^2-x x^y y-x^2 y^2-x y_3^3+x^2 y^2+x y_3^3+y^4\)

= \(x^4+x^2 y^2+y^4\)

4. \(\left(\frac{a^2}{b c}+\frac{b^2}{c a}\right) \text { by }\left(\frac{a}{b c}-\frac{b}{c a}\right)\)

Solution:

Given \(\left(\frac{a^2}{b c}+\frac{b^2}{c a}\right) \text { by }\left(\frac{a}{b c}-\frac{b}{c a}\right)\)

= \(\frac{a}{b c}\left(\frac{a^2}{b c}+\frac{b^2}{c a}\right)-\frac{b}{c a}\left(\frac{a^2}{b c}+\frac{b^2}{c a}\right)\)

= \(\frac{a^3}{b^2 c^2}+\frac{a b^2}{b c^2 a}-\frac{a^2 b}{a b c^2}-\frac{b^3}{a^2 c^2}\)

= \(\frac{a^3}{b^2 c^2}+\frac{b}{c^2}-\frac{a}{c^2}-\frac{b^3}{a^2 c^2}\)

Question 2. Find the Successive product of the following:

1. \((a+b),(a-b),\left(a^2-a b+b^2\right),\left(a^2+a b+b^2\right)\)

Solution:

Given \((a+b),(a-b),\left(a^2-a b+b^2\right),\left(a^2+a b+b^2\right)\)

= \(((a+b)(a-b))\left(\left(a^2-a b+b^2\right)\left(a^2+a b+b^2\right)\right)\)

= \(\left(a^2-a b+a b b-b^2\right)\left(a^4+a^3 b+a^2 b^2-a^3 b-a^2 b^2-a b^3+a^2 b^2+ab^3+b^4)\right.\)

= \(\left(a^2-b^2\right)\left(a^4+b^4\right)\)

= \(\left(a^2-b^2\right)\left(a^4-b^4\right)\)

= \(\left(a^6-b^6\right)\)

Class 8 Maths Chapter 2 Multiplication and Division of Polynomials Haryana Board

2. \(\left(a^2-b^2\right),\left(b^2-c^2\right),\left(c^2-a^2\right)\)

Solution:

Given \(\left(a^2-b^2\right),\left(b^2-c^2\right),\left(c^2-a^2\right)\)

= \(\left(\left(a^2-b^2\right)\left(b^2-c^2\right)\right)\left(c^2-a^2\right)\)

= \(\left(a^2 b^2-a^2 c^2-b^4+b^2 c^2\right)\left(c^2-a^2\right)\)

= \(c^2\left(a^2b^2-a^2 c^2-b^4+b^2 c^2\right)-a^2\left(a^2 b^2-a^2 c^2-b^4+b^2 c^2\right)\)

= \(\left(a^2 b^2 c^2-a^2 c^4-b^4 c^2+b^2 c^4-a^4 b^2+a^4 c^2+a^2 b^4-a^2 b^2 c^2\right)\)

= \(a^4 c^2-a^4 b^2-a^2 c^4-b^4 c^2+a^2 b^4+b^2 c^4\)

3. (x+1)(x-1)(x² + 1)(x⁴ +1)

Solution:

Given: (x+1), (x−1), (x² + 1), (x⁴+1)

= \(((x+1)(x-1))\left(\left(x^2+1\right)\left(x^4+1\right)\right)\)

= \(\left(x^2-x+x-1\right)\left(x^6+x^2+x^4+1\right)\)

= \(\left(x^2-1\right)\left(x^6+x^2+x^4+1\right)\)

= \(\left(x^2-1\right)\left(x^6+x^2+x^4+1\right)\)

= \(x^2\left(x^6+x^2+x^4+1\right)-1\left(x^6+x^2+x^4+1\right)\)

= \(x^8+x^4+x^8+x^4-x^6-x^2-x^4-1\)

= \(x^8-1\)

3. Simplify: (2x-3) (x+2)− (3x-5)(x-6) – (5+x)(7-x)

Solution:

Given: (2x-3)(x+2)-(3x-5) (x −6) − ( 5+x)(7-x)

(2x-3)(x+2)-(3x-5) (x −6) − ( 5+x)(7-x) = (2x²+4x-3x-6)-(3x²-18x-5x+30)-(35-5x+7x-x²)

(2x-3)(x+2)-(3x-5) (x −6) − ( 5+x)(7-x) = 2x²+x-6-3x²+23x-30-35-2x-x²

(2x-3)(x+2)-(3x-5) (x −6) − ( 5+x)(7-x) = 24x – 2x – 36 – 35 ⇒ 228-71

Question 4. If x= (a-b+c), y=(b-c+a) and 2 = (b+c-a) then find the value of (xy+yz+zx).

Solution:

Given x=(a+b+c), Y= (b-c+a), Z = (b+(-a)

xy+yz+Zx = ((a-b+c)(b-c+a)) + ((b-c+a) (b+c-a))+((b+c-a)(a-b+c))

xy+yz+Zx =(ab-ac + a²-b² + bc-ab+cb-c²+ac) + (b² + bc-ab-bc-c²+ac+ab+ac-a²)+(ab-b²+ bc+ac-bc+c²-a²+ab-ac)

xy+yz+Zx = a²– b² +bc+cb-c²+b²-c²+ac+ac-a²+ab-b²+c²-a²+ab

xy+yz+Zx = 2ab+2bc+2ac-a²-b²-c²

Haryana Board 8th Class Maths Multiplication and Division of Polynomials Questions and Answers

Question 5. The area of a rectangle is (84x²-xy-15y²) Sq.cm and the length is (12x+5y) cm. Find the breadth of the rectangle.

Solution:

Length of a rectangle = 12x+5y

Breadth of a rectangle = B

Area of rectangle = 84x²-xy-15y²

Breadth of a rectangle = 7x-3y

Question 6. The product of two numbers is (x³– 8) and one number is (x-2). Find the other number.

Solution:

Product of two numbers = x³-8

One number = (x-2)

Another number =?

The product of another number = x²+2x+4

Question 7. Divide:

1. (x³ +17x-8x²-10) by (x-5)

Solution:

Given (x³+17x-8x²-10) by (x-5)

∴ (x² – 3x+2)(x-5) = x³ +17x – 8x²-10

Chapter 2 Multiplication and Division of Polynomials Class 8 Solutions in Hindi Haryana Board

2. (x³-3x²y + 3xy²-y³) by (x-y)

Solution:

Given: (x³-3x²y + 3xy²-y³) by (x-y)

∴ (x²-2xy+y²)(x-y) = x³-3x²y+3xy²-y³

3. (a^2/3+a^1/3b^1/3+b^2/3) by (a-b)

Solution:

Given: (a^2/3+a^1/3b^1/3+b^2/3) by (a-b)

(a^2/3+a^1/3b^1/3+b^2/3) = (a-b)(a^1/3– b^1/3)

4. (5x – x² – 6) by (12x³ – 66 + 97x – x⁴– 52x²)

Solution:

Given: (5x – x² – 6) by (12x³ – 66 + 97x – x⁴– 52x²)

Question 8. Find the quotient and the remainder:

1. (x²-12x+30) by (x-5)

Solution:

Quotient = (x-7)

Remainder = -5

Haryana Board Class 8 Maths Exercise 2.1 Solutions

2. (x²-3x+8) by (x⁴+2x³-14x²+63x-57)

Solution:

Given: (x⁴+2x³-14x²+63x-57) by (x²-3x+8)

Quotient = (x²+5x-7)

Remainder = (2x-1)

3. (a³-2a²b+3ab²-5b³) by (a-2b)

Solution:

Given: (a³-2a²b+3ab²-5³) by (a-2b)

Quotient = (a²+3b²)

Remainder = -11³

4. (x⁴+x²y²+2y⁴) by (x²+xy+y²)

Solution:

Given: (x⁴+x²y²+2y⁴) by (x²+xy+y²)

Quotient = (x²-xy + y²)

Remainder = y⁴

Important Questions for Class 8 Maths Chapter 2 Haryana Board

Question 9. Choose the correct answer:

1. (a+3) (a+4) (a+5)= _____________

1. a³+12a²+47a+60
2. a²+47a²+12a+60
3. a³+12a² +60a+47
4. None of these

Solution: (a+3) (a+y) (a+5)

= (a²+4a+3a+12) (a+5) = (a²+7a+12)(a+5) = a³+35a+60+7a²+12a+5a²

= a³+12a²+47a+60

So, the Correct answer is (1)

2. (x⁴+2x²-x³+3+x) ÷ (x²+x+ 1) = _________

1. x²=2x+3
2. x²+2x-3
3. x²+2x+3
4. x²=2x-3

Solution:

So, the Correct answer is (1).

3. (8x⁸-8x²=-36x)÷ 4x = ____________

1. 2x⁸-2x²-9
2. 2x⁸-8x-36
3. 2x⁷-2x-9
4. 2x⁷-2x-18

Solution:

∴ \(\frac{8 x^8-8 x^2-36 x}{4 x}=2 x^7-2 x-9\)

The Correct answer is (3).

Step-by-Step Solutions for Multiplication and Division of Polynomials Class 8 Haryana Board

Question 10. write ‘True’ or ‘False”:

1. (x²+xy+y²) (x²-xy+y²) = x⁴+x²y²+y⁴

Solution:

(x²+xy+y²) (x²-xy+y²) = x⁴– x³y + x²y² + x3y – x²y² + xy³ + x²y² – xy² + y⁴

= x⁴ + x²y² + y⁴

So, the statement is true.

2. (x²+7x+7) ÷ (x+3) = (x+4) – \(\frac{5}{x+3}\)

Solution:

∴ \(\left(x^2+7 x+7\right) \div(x+3)=(x+4)-\frac{5}{x+3}\)

So the statement is true.

Question 11. Fill in the blanks:

1. (2x³-5x²-9x-8)=(x-4) = ______.

Solution:

∴ \(\left(2 x^3-5 x^2-9 x-8\right) \div(x-4)=2 x^2+3 x+3+\frac{4}{x-4}\)

2. -3r²t(4+r⁴⁵+2rt³) = __________

Solution:

∴ -3r²t(4+r⁴t⁵+2r³) = -12r⁶t⁶-6r³t⁴

3. (x²+xy+y²)(x-4)= __________

Solution:

(x²+xy+y²) (x-y)

(x²+xy+y²) (x-y) = x²(x-y)+xy(x-y) + y² (x-y)

(x²+xy+y²) (x-y) = x³-x²y+x²y-xy²+xy²-y³

(x²+xy+y²) (x-y) = x³-y³

Question 1. If P = 999, then find the volume of P (p²+3p+3).

Solution:

Given P = 999

= P(P²+3843)

= 999 (1999)²+3 (999)+3)

= 999(99800142997+3)

= 999 (1001001)

= 999,999,999

The volume of P (p²+3p+3) = 999,999,999

Question 2. Find the Cube of:

1. 1001

Solution:

Given: 1001

= (1001)³

– (1001)²(1001)

= 1002001 (1001)

= 1,003,003,001

(1001)³ = 1,003,003,001

2. 998

Solution:

Given: 998

= (998)³

= (998)²(992)

= 996004 (998)

= 994,001,1992

(998)³ = 994,001,1992

Haryana Board Class 8 Maths Cubes and Cube Roots Solutions

3. 5a-4b

Solution:

Given 5a-4b

= (5a-4b)³ [∵ (a-b)² = a³-3ab+3ab²+b³]

= (5a)³ – 3(5a)²4b + 3(5a)(4b)²-(4b)³

= 125a³ – 300a²b + 240ab² – 64b³

(5a-4b)³  = 125a³ – 300a²b + 240ab² – 64b³

4. 3a-b+4c

Solution:

= 3a-b+4c

= (3a-b+4c)³ [(∵(a+b+c)³ = a³ + b³ + c² + 3 (a+b) (b+c) (c+a)]

= (3a)³ + (-b)³ + (4c)³ + 3(3a-b)(-b+4c) (4c+3a)

= 27a³ – b³ + 64c³ + 3((3a-b) (-4bc – 3ab + 16c² + 12ac)

= 27a³-b³ + 64c³ +3[-12abc – 9a²b + 48ac² + 36²c + 4b²c + 3ab² – 16bc² = 12abc]

= 27a³-b³ + 64c³ – 36abc – 27a²b + 144ac² + 108a²C +12b²c + 9ab² – 48bc² – 36abc

= 27a³ – b³ + 64c³ – 27a²b + 9a² + 108a²c – 72abc + 12b²c +144ac² – 48bc²

(3a-b+4c)³= 27a³ – b³ + 64c³ – 27a²b + 9a² + 108a²c – 72abc + 12b²c +144ac² – 48bc²

Class 8 Maths Chapter 7 Cubes Haryana Board

Question 3. Simplify:

1. 3.2 x 3.2 x 3.2 – 3 x 3.2 x 3-2 x 1.2 + 3 x 3.2 × 1.2 x 1.2 −1·2 x 1.2 x 1.2

Solution:

Given:

= 3.2 x 32 x 3.2 – 3 x 3.2 x 3.2 x 1.2 + 3 x 3.2 x 1.2 x 1.2-1.2 x 1.2 x 1.2

= (3.2)³-3 (3-2)² x 1-2 + 3 x 3.2 x (1.2)² -(1.2)³

= 32.768 – 36.864 + 13.824 – 1.728

= 8

3.2 x 32 x 3.2 – 3 x 3.2 x 3.2 x 1.2 + 3 x 3.2 x 1.2 x 1.2-1.2 x 1.2 x 1.2 = 8

2. (a-b+c)³ + (a+b-c)³ + 6a [a²-(b-c)²]

Solution

Given:

= (a-b+c)³ + (a+b-c)³ + 6a [a²-(b-c)²]

= a³ – b³ + c³ + 3(a-b)(-b+ c)(c+a) + a³ + b² – c³ + 3(a+b)(b-c)(-c+a) + 6a [a²– b² + c² + 2bc]

= a³ – b³ + c³ + 3(a-b)(-bc-ba+c²+ca) + a³ + b³ – c³ + 3(a+b)(-bc+ba+c²-ca) + 6a [a²-b²+c²+2bc]

= a³ – b³ + c³ + 3 (-abc – a²b + ac² + a²c + b²c + b²a – bc² – abc) + a³ + b³ – c³ + 3(-abc + a²b + ac² – a²c -b²c + b²a + bc² – abc)+ 6a³ – 6ab² + 6ac² + 12abc

= a³– b³ + c³ – 3abc – 3a²b + 3ac²+ 3a²c + 3b²C + 3b²a – 3bc² – 3abc + a³ + b³ – c³ – 3abc + 3a²b + 3ac² – 3a²c – 3b²c + 3b²a + 3bc² – 3abc + 6ab³ – 6ab² + 6ac² + 12abc

= a³ – 3abc + 3ac² + 3b²a – 3abc – 3abc + 3ac² + 3b²a – 3abc + 6a³ – 6ab² + 6ac² + 12abc

= 8a³ – 12abc + 6ac² + 6ab² – 6ab² – 6ac² + 12abc

= 8a³

(a-b+c)³ + (a+b-c)³ + 6a [a²-(b-c)²] = 8a³

Haryana Board 8th Class Maths Cube and Cube Root Questions and Answers

Question 4. If a-b = 8 then find the value of (a³-b³-24ab)

Solution:

Given: a-b=8

= a³-b³-24ab

= (a)³ – (b)³ -24ab

= (a-b)³ + 3 x a x b(a-b)- 24ab

= (8)³ +3ab (8)-24ab

= 512 + 24ab – 24ab

= 512

The value of (a³-b³-24ab) = 512

Question 5. Find the value of 8x³-36x²+54x-30 lf x=-5

Solution:

Given: X=-5

= 8x³-36x²+54x – 30

= 8(-5)³ – 36(-5)² + 54(-5)-30

= 8(-125)-36(25)-270-30

= -1000-900-270-30

= 1900+300

= 2200

The value of 8x³-36x²+54x-30 = 2200

Chapter 7 Cubes Class 8 Solutions in Hindi Haryana Board

Question 6. Find the product of the following:

1. (x-3)(x²+3x+9)

Solution:

Given: (x-3)(x²+3x+9)

= x³ + 3x² + 9x – 3x² – 9x – 27

= x³-27

(x-3)(x²+3x+9) = x³-27

2. (a²+b²)(a⁴-a²b²+b⁴)

Solution:

Given (a²+b²)(a⁴-a²b²+b⁴)

= (a²+b²)(a⁴-a²b²+b⁴)

= a²(a⁴-a²b²+b⁴) + b²(a⁴-a²b²+b⁴)

= a⁶ – a⁴b² + a²b⁴ + b²á⁴ – a²b⁴+ b⁶

= a⁶+b⁶

(a²+b²)(a⁴-a²b²+b⁴) = a⁶+b⁶

Haryana Board Class 8 Maths Exercise 7.1 Solutions

3. (4a²-9b²) (4a²+6ab+9b²) (4a²-6ab+9b²)

Solution:

Given: (4a²-9b²) (4a²+6ab+9b²) (4a²-6ab+9b²)

= (4a²(4a²+6ab+9b²)-9b²(4a²+6ab+9b²))(4a²-6ab+9b²)

= (16a⁴ + 24a³b + 36a²b² – 36a²b² – 54ab³ – 81(b⁴)(4a²-6ab+9b²)

= 64a⁶ + 96a⁵b – 216a³b³ – 324a²b⁴ – 96a⁵b – 144a⁴b² + 324a²b⁴ + 486ab⁵ + 144a⁴b² + 216a³b³ – 486ab⁵ – 729b⁶

= 64a⁶ – 729b⁶

(4a²-9b²) (4a²+6ab+9b²) (4a²-6ab+9b²) = 64a⁶ – 729b⁶

Question 7. Resolve into factors:

1. \(a^3-9 b^3-3 a b(a-b)\)

Solution:

Given: \(a^3-9 b^3-3 a b(a-b)\)

= \(a^3-9 b^3-3 a b(a-b)\)

= \(a^3-b^3-3 a b(a-b)-8 b^3\)

= \((a-b)^3-(2 b)^3\)

= \((a-b-2 b)\left\{(a-b)^2+(a-b) \cdot 2 b+(2 b)^2\right\}\)

= \((a-3 b)\left(a^2-2 a b+b^2+2 a b-2 b^2+4 b^2\right)\)

= \((a-3 b)\left(a^2+3 b^2\right)\)

\(a^3-9 b^3-3 a b(a-b)\) = \((a-3 b)\left(a^2+3 b^2\right)\)

Important Questions for Class 8 Maths Chapter 7 Haryana Board

2. a¹² – b¹²

Solution:

Given a¹² – b¹²

= (a⁶)² – (b⁶)²

= \(\left(a^6+b^6\right)\left(a^6-b^6\right)\)

= \(\left\{\left(a^2\right)^3+\left(b^2\right)^3\right\}\left\{\left(a^3\right)^2-\left(b^3\right)^2\right\}\)

= \(\left(a^2+b^2\right)\left\{\left(a^2\right)^2-a^2 b^2+\left(b^2\right)^2\right\}\left(a^3+b^3\right)\left(a^3-b^3\right)\)

= \(\left(a^2+b^2\right)\left(a^4-a^2 b^2+b^4\right)(a+b)\left(a^2-a b+b^2\right)(a-b)\left(a^2+a b+b^2\right)\)

= \((a+b)(a-b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)\)

a¹² – b¹² = \((a+b)(a-b)\left(a^2+b^2\right)\left(a^2+a b+b^2\right)\left(a^2-a b+b^2\right)\left(a^4-a^2 b^2+b^4\right)\)

Question 8. Choose the Correct answer:

1. The Cube of 99 is

1. 972099
2. 970299
3. 979029
4. 972909

Solution:

=(99)³

= (99)² 99

= 9801 x 99

= 970299

(99)³ = 970299

The Correct answer is (2).

Step-by-Step Solutions for Cubes Class 8 Haryana Board

2. If a+b+c=0, then a³+b³+c³ = ?

1. 3abc
2. abc
3. c
4. none of these

Solution:

= a³ + b³ + c³

= (a+b)³ -3ab (a+b) + c³

= (-c)³-3ab (-c) +c³ [a+b+c=0)

= -c³+3abc+c³

= 3abc

a³ + b³ + c³ = 3abc

So the Correct answer is (1).

3. If a-b=1 and a³-b³=61, then the Value of ab is

1. 10
2. 20
3. 30
4. none of these

Solution:

= a³-b³=61

= (a-b)³ + 3ab (a-b) = 61

= (1)³ + 3ab(1) = 61

= 3ab = 61-1

= 3ab = 60

= ab = \(\frac{60}{3}\)

= ab = 20

So, the Correct answer is (2).

Question 9. write ‘True’ or ‘False”.

1. 216 is not a perfect Cube.

Solution:

216 = 36×6

216= 6x6x6

216=(6)³

The statement is False.

2. 1729 is a Hardy Ramanujam number.

Solution: The statement is true.

3. p³q³+1 = (pq-1)(p²q² + pq + 1)

Solution:

p³q³ + 1 = (pq)³ + (1)³

p³q³ + 1= (pq+1){(pq)² – pq.1 + (1)2}

p³q³ + 1= (pq+1) (p²q² – pq+1)

Statement is False.

Question 10. Fill in the blanks.

1. The Cube root of Single digit number may also be a _________ digit number.

Solution: Single.

2. a³+b³ = ___________ x (a²-ab+b²)

Solution: a³ + b³ = (a+b) (a²-ab+b²)

3. (a+b)³ = a² + b³ + __________.

Solution: (a+b)³ = a³ + b³ + 3ab(a+b)

Question 1. Resolve into factors:

1. x² – 22x + 120

Solution:

Given: x² – 22x + 120

= x²-10x-12x-120

= x(x-10)-12(x-10)

= (x-10) (x-12)

x² – 22x + 120 = (x-10) (x-12)

2. 40+3x-x²

Solution:

Given: 40+3x-x²

= 40+8x-5x-x²

= 8(5+x)-x(5+x)

= (8-x)(5+x)

40+3x-x² = (8-x)(5+x)

Haryana Board Class 8 Maths Factorisation Solutions

3. (a²-a)²-8(a²-a)+12

Solution:

Given: (a²-a)²-8(a²-a)+12

= (a²-a)² – 6(a²-a)-2(a²-a)+12

= (a²−a)((a²-a)−6)-2((a²-a)-6)

= (a²-a-2) (a²-a-6)

= (a²+a-2a-2)(a²3a+2a-6)

= (a(a+1)-2(a+1))(a(a-3)+2(9-3))

= (a+1) (a-2) (a+2) (a-3)

(a²-a)²-8(a²-a)+12 = (a+1) (a-2) (a+2) (a-3)

4. x²-√3x-6

Solution:

Given: x²-√3x-6

= x²-2√3x+√3x-6

= x(x-2√3)+ √3(x-2√3)

= (x+√3)(x-2√3)

x²-√3x-6 = (x+√3)(x-2√3)

Class 8 Maths Chapter 14 Factorisation Haryana Board

5. (x+1)(x+3)(x-4)(x-6)+13

Solution:

Given (x+1)(x+3)(x-4)(x-6)+ 13

= (x²+3x+x+3)(x²-6x-4x+24)+13

= (x²+4x+3)(x²-10x+24)+13

= x⁴-10x³+24x²+4x³40x²+96x+3x² = 30x+72+13

= x⁴-6x³-13x²-66+85

= x⁴-3x³-3x³-17x²+9x²– 5x²+51x+15x+85

= x⁴-3x³-17x²-3x³+9x²+51x-5x²+15x+85

= x²(x²-3x-17)-3x(x²-3x-17)-5(x²-3x-17)

= (x²-3x-5)(x²-3x-17)

(x+1)(x+3)(x-4)(x-6)+ 13 = (x²-3x-5)(x²-3x-17)

6. 21x² + 40xy – 21y²

Solution:

Given: 21x²+40xy-21y²

= 21x² – 9xy + 49xy – 21y²

= 3x(7x-3y) + 7y(7x-3y)

= (3x+7y)(7x-3y)

21x²+40xy-21y² = (3x+7y)(7x-3y)

Haryana Board 8th Class Maths Factorisation Questions and Answers

7. 4(2a-3)² -3(2a-3) (a-1)-7(a-1)²

Solution:

Given: 4(2a-3)²-3(2a-3) (a-1)-7(a-1)²

= 4(4a²-(2a+9)-3(2a²-2a-3a+3)-7(a²+1-2a)

= 16a² – 48a + 36 – 6a² + 6a + 9a – 9 – 7a² – 7 + 14

= 3a² – 19a + 20

= 3a² – 15a – 4a +20

= 3(a-5)-4(a-5)

4(2a-3)²-3(2a-3) (a-1)-7(a-1)² = (3a-4) (a-5)

8. (a+7) (a-10) + 16

Solution:

Given: (a+7)(a-10)+16

= a²-10a+7a-70+16

= a²-3a-54

= a-9a+6a-54

= a(a-9)+6(a-9)

=(a+6)(a-9)

(a+7)(a-10)+16 =(a+6)(a-9)

Chapter 14 Factorisation Class 8 Solutions in Hindi Haryana Board

9. (a²+4a)²+21(a²+4a)+98

Solution:

Given: (a²+4a)²+21(a²+4a)+98

= a⁴ + 8a³ + 16a² + 2(a²+4a)+ 98

= a⁴ + 8a³ + 37a² +84a +98

= a⁴+4a³+14a²+4a³+16a²+56a+7a²+28+98

= a² (a²+4a+14)+4a(a²+4a+14) +7(a²+4a+14)

= (a²+4a+7) (a² + 4a + 14)

(a²+4a)²+21(a²+4a)+98 = (a²+4a+7) (a² + 4a + 14)

10. (2a²+ 5a) (2a² + 5a-19) +84

Solution:

Given: (2a²+ 5a) (2a² + 5a-19) +84

= 4a⁴ + 10a³ – 38a² + 10a³ + 25a² – 95a + 84

= 4a⁴ + 20a³ – 13a² – 95a + 84

= 4a⁴ + 8a³ – 21a² + 12a³ + 24a² – 63a -16a² – 32a + 84

= a²(4a²+8a-21)+3a(4a²+8a-21)-4(4a²+8a-21)

= (a²+3a-4) (4a² +8a-21)

= (a²+4a-a-4) (4a² + 14a-6a-21)

= ((a+4)a-1(a+4)) (2a(2a+7)-3(2a+7))

= (a-1) (a+4) (2a-3)(2a+7)

(2a²+ 5a) (2a² + 5a-19) +84 = (a-1) (a+4) (2a-3)(2a+7)

11. 8x⁴+2x²-45

Solution:

Given: 8x⁴+2x²-45

= 8x² + 20x² – 18x² – 45

= 4x² (2x²+5)-9(2x²+5)

= (2x²+5)(4x²-9)

= (2x+5) (4x²-6x+6x-9)

= (2x²+5) (2x(2x-3)+3(2x-3))

= (2x²+5)(2x+3)(2x-3)

8x⁴+2x²-45 = (2x²+5)(2x+3)(2x-3)

Haryana Board Class 8 Maths Exercise 14.1 Solutions

12. xv-x-(a-3)(a-2)

Solution:

Given: x²-x-(a-3)(a-2)

= x²-x-(a²-2a-3a+6)

= x²-x-a²+5a-6

= x²-3x+2x-a²+ax-ax+3a+2a-6

= x²+ax-3x-ax-a²+3a+2x+2a-6

= x(x+a-3)-a(x+a-3)+2(x+a-3)

=(x-a+2)(x+a-3)

x²-x-(a-3)(a-2) =(x-a+2)(x+a-3)

13. 99x²-20xy+99y²

Solution:

Given: 99x²-20xy+99y²

= 99x²-81xy-121xy+99y²

= 9x (11x-9y)-11y (11x-94)

= (9x-114) (11x-94)

99x²-20xy+99y²  = (9x-114) (11x-94)

Question 2. Resolve the following expressions into factors by expressing them as the diffence of two Squares:

1. x²-5x-6

Solution:

Given: x²-5x-6

= x²-6x+x-6

= x(x-6)+1(x-6)

= (x+1)(x-6)

x²-5x-6 = (x+1)(x-6)

2. 3+x-10x²

Solution:

Given: 3+x-10x²

= 3+6x-5x-10x²

= 3(1+2x)-5x(1+2x)

= (3-5x) (1+2x)

3+x-10x² = (3-5x) (1+2x)

3. 8x-3-4x²

Solution:

Given: 8x-3-4x²

= 6x+2x-3-4x²

= 6x-3+2x-4x²

= 3(2x-1)-2x(2x-1)

= (3-2x) (2x-1)

8x-3-4x²= (3-2x) (2x-1)

4. 6(a+b)²+5(a²-b²)-6 (a-b)²

Solution:

Given: 6(a+b)²+5(a²-b²)-6(a-b)²

= 6(a²+b²+2ab)+5a²-5b²-6(a²+b²-2ab)

= 6a²+6b²+ 12ab+ 5a²-5b²-6a²-6b² +12ab

= 5a²+24ab-5b²

= 5a² +25ab-ab-5b²

= 5a(a+5b)-b(a+5b)

= (5a-b)(a+5b)

6(a+b)²+5(a²-b²)-6(a-b)² = (5a-b)(a+5b)

5. 6x²-13x+6

Solution:

Given: 6x²-13x+6

= 6x²-9x-4x+6

= 3x(2x-3)-2(2x-3)

=(3x-2)(2x-3)

6x²-13x+6 =(3x-2)(2x-3)

Question 3. Choose the Correct answer:

1. x²-3x-28 = ?

1. (x+4)(x+7)
2. (x+4)(x-7)
3. (x-4)(x+7)
4. (x-4)(x-7)

Solution:

x²-3x-28

= x²-7x+4x-28

= x(x-7)+4(x-7)

= (x+4) (x-7)

x²-3x-28 = (x+4) (x-7)

The Correct answer is (2)

Important Questions for Class 8 Maths Chapter 14 Haryana Board

2) If (5x²-4x-9) = (x+1)(5x+P), then the value of P is

1. 9
2. 5
3. -9
4. none of these

Solution:

= 5x²-4x-9

= 5x²-9x+5x-9 = x(5x-9)+1(5x-9)

= (5x-9)(x+1)

= (x+1)(5x+P)= (x+1) (5x-9)

= 5x + P = 5x-9

⇒ P = -9

The value of P = -9

The Correct answer is (3).

3. 2a²+b²-c²+3ab+ac = ?

1. (a+b+c)(2a+b+c)
2. (a+b+c)(2a+b-c)
3. (a+b+c) (2a-b-c)
4. none of these

Solution:

2a²+b²-c²+3ab+ac

= 2a²+ab-ac+2ab+b²-bc+2ac + bc-c²

= a(2a+b-c)+b(2a+b-c) + C(2a+b-c)

= (a+b+c) (2a+b-c)

2a²+b²-c²+3ab+ac = (a+b+c) (2a+b-c)

The Correct answer is (2).

Question 4. write ‘True’ or ‘False’:

1. The Factors of (x²-xy-30y²) is (x+5y)(x-64).

Solution:

x²-xy-30y² = x²-6xy+5xy-30y²

= x(x-6y) +5y(x-6y) = (x-6y) (x+5y)

The statement is true.

Step-by-Step Solutions for Factorisation Class 8 Haryana Board

2. a³-b³-a(a²-b²)+b(a-b)²=ab(a-b)

Solution:

a³-b³-a(a²-b²)+b(a-b)²=ab(a-b)

= a³-b³-a³+ab²+b(a²+b²-2ab)

= -b³+ab²+ba²+b³-2ab²

= a²b-ab² = ab(a-b)

The statement is true.

3.(x-1)(x+9)+21=(x+6)(x-2)

Solution:

(x-1)(x+9)+21

= x²+9x-x-9+21

= x²+8x+12

= x²+6x+2x+12

= x(x+6)+2(x+6)

= (x+2)(x+6)

The statement is false.

Question 5. Fill in the blanks:

1. (x+a)(x+b) = x²+(a+b)x + _______.

Solution:

(x+9)(x+6) = x(x+b)+a(x+b)

= x²+bx+ax+ab = x²+(a+b)x+ab

2. (a+b+c)³ = a³ + b² + c² + __________.

Solution:

3(a+b)(b+c)(c+a).