Mathematics Solutions Solutions for Class 8 Math Chapter 1 Rational And Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Rational And Irrational Numbers are extremely popular among Class 8 students for Math Rational And Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Page No 43:

#### Question 1:

Construct the following quadrilaterals of given measures.

(1) In ☐ MORE, *l*(MO) = 5.8 cm, *l*(OR) = 4.4 cm, *m*∠M = 58$\xb0$, *m*∠O = 105$\xb0$, *m*∠R = 90$\xb0$.

(2) Construct ☐ DEFG such that *l*(DE) = 4.5 cm, *l*(EF) = 6.5 cm, *l*(DG) = 5.5 cm, *l*(DF) = 7.2 cm, *l*(EG) = 7.8 cm.

(3) In ☐ ABCD *l*(AB) = 6.4 cm, *l*(BC) = 4.8 cm, *m*∠A = 70$\xb0$, *m*∠B = 50$\xb0$, *m*∠C = 140$\xb0$.

(4) Construct ☐ LMNO such that *l*(LM) = *l*(LO) = 6 cm, *l*(ON) = *l*(NM) = 4.5 cm, *l*(OM) = 7.5 cm.

#### Answer:

(1)

**Steps of Construction:
Step 1: **Draw MO = 5.8 cm.

**Step 2:**Draw ∠MOX = 105$\xb0$.

**Step 3:**With O as centre and radius 4.4 cm, draw an arc cutting ray OX at R.

**Step 4:**Draw ∠ORY = 90$\xb0$.

**Step 5:**Draw ∠OMZ = 58$\xb0$ such that ray MZ and ray RY intersect each other at E.

Here, MORE is the required quadrilateral.

(2)

**Steps of Construction:**

Step 1:Draw DE = 4.5 cm.

Step 1:

**Step 2:**With D as centre and radius 7.2 cm, draw an arc.

**Step 3:**With E as centre and radius 6.5 cm, draw an arc cutting the previous arc at F.

**Step 4:**Join EF.

**Step 5:**With D as centre and radius 5.5 cm, draw an arc.

**Step 6:**With E as centre and radius 7.8 cm, draw an arc cutting the previous arc at G.

**Step 7:**Join DG and GF.

Here, DEFG is the required quadrilateral.

(3)

**Steps of Construction:**

Step 1:Draw AB = 6.4 cm.

Step 1:

**Step 2:**Draw ∠ABX = 50$\xb0$.

**Step 3:**With B as centre and radius 4.8 cm, draw an arc cutting ray BX at C.

**Step 4:**Draw ∠BCY = 140$\xb0$.

**Step 5:**Draw ∠BAZ = 70$\xb0$ such that ray AZ and ray CY intersect each other at D.

Here, ABCD is the required quadrilateral.

(4)

**Steps of Construction:**

Step 1:Draw LM = 6 cm.

Step 1:

**Step 2:**With L as centre and radius 6 cm, draw an arc.

**Step 3:**With M as centre and radius 7.5 cm, draw an arc cutting the previous arc at O.

**Step 4:**Join OL.

**Step 5:**With O as centre and radius 4.5 cm, draw an arc.

**Step 6:**With M as centre and radius 4.5 cm, draw an arc cutting the previous arc at N.

**Step 7:**Join ON and MN.

Here, LMNO is the required quadrilateral.

#### Page No 46:

#### Question 1:

Draw a rectangle ABCD such that *l*(AB) = 6.0 cm and *l* (BC) = 4.5 cm.

#### Answer:

**Steps of Construction:
Step 1:** Draw AB = 6 cm.

**Step 2:**Draw ∠ABX = 90º.

**Step 3:**With B as centre and radius 4.5 cm, draw an arc cutting the ray BX at C.

**Step 4:**With C as centre and radius 6 cm, draw an arc.

**Step 5:**With A as centre and radius 4.5 cm, draw an arc cutting the previous arc at D.

**Step 6:**Join AD and CD.

Here, ABCD is the required rectangle.

#### Page No 46:

#### Question 2:

Draw a square WXYZ with side 5.2 cm.

#### Answer:

**Steps of Construction:
Step 1:** Draw WX = 5.2 cm.

**Step 2:**Draw ∠WXP = 90º.

**Step 3:**With X as centre and radius 5.2 cm, draw an arc cutting the ray XP at Y.

**Step 4:**With Y as centre and radius 5.2 cm, draw an arc.

**Step 5:**With W as centre and radius 5.2 cm, draw an arc cutting the previous arc at Z.

**Step 6:**Join YZ and WZ.

Here, WXYZ is the required square.

#### Page No 46:

#### Question 3:

Draw a rhombus KLMN such that its side is 4 cm and *m*∠K = 75$\xb0$.

#### Answer:

**Steps of Construction:
Step 1:** Draw KL = 4 cm.

**Step 2:**Draw ∠LKX = 75º.

**Step 3:**With K as centre and radius 4 cm, draw an arc cutting the ray KX at N.

**Step 4:**With N as centre and radius 4 cm, draw an arc.

**Step 5:**With L as centre and radius 4 cm, draw an arc cutting the previous arc at M.

**Step 6:**Join LM and NM.

Here, KLMN is the required rhombus.

#### Page No 46:

#### Question 4:

If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.

#### Answer:

Suppose ABCD is a rectangle.

Here, segment AC is a diagonal and segment AD is one side of the rectangle ABCD.

*l*(AC) = 26 cm and *l*(AD) = 24 cm.

In right ∆ACD,

*l*(AC)^{2} = *l*(AD)^{2 }+ *l*(CD)^{2} (Pythagoras theorem)

⇒ *l*(CD)^{2 }= *l*(AC)^{2} − *l*(AD)^{2}

⇒ *l*(CD)^{2 }= (26)^{2} − (24)^{2}

⇒ *l*(CD)^{2 }= 676 − 576 = 100

⇒ *l*(CD)^{ }= $\sqrt{100}$ = 10 cm

Thus, the other side of the rectangle is 10 cm.

#### Page No 47:

#### Question 5:

Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.

#### Answer:

ABCD is a rhombus.

Here, segment AC and segment BD are the diagonals of the rhombus ABCD.

*l*(AC) = 16 cm and *l*(BD) = 12 cm.

Diagonals of a rhombus are perpendicular bisectors of each other.

∴ *m*∠AOD = 90º

Also, *l*(OA) = $\frac{1}{2}$*l*(AC) = $\frac{1}{2}$ × 16 = 8 cm

*l*(OD) = $\frac{1}{2}$*l*(BD) = $\frac{1}{2}$ × 12 = 6 cm

In right ∆AOD,

*l*(AD)^{2} = *l*(OA)^{2 }+ *l*(OD)^{2} (Pythagoras theorem)

⇒ *l*(AD)^{2 }= (8)^{2} + (6)^{2}

⇒ *l*(AD)^{2 }= 64 + 36 = 100

⇒ *l*(AD)^{ }= $\sqrt{100}$ = 10 cm

All sides of a rhombus are equal.

∴ Perimeter of the rhombus ABCD = 4 × Side of a rhombus = 4 × 10 = 40 cm

Thus, the side and perimeter of the rhombus are 10 cm and 40 cm, respectively.

#### Page No 47:

#### Question 6:

Find the length of diagonal of a square with side 8 cm.

#### Answer:

Suppose ABCD is a square of side 8 cm.

Here, segment AC is a diagonal of square ABCD.

In ∆ABC,

*l*(AC)^{2} = *l*(AB)^{2 }+ *l*(BC)^{2} (Pythagoras theorem)

⇒ *l*(AC)^{2 }= (8)^{2} + (8)^{2}

⇒ *l*(AC)^{2 }= 64 + 64 = 128

⇒ *l*(AC)^{ }= $\sqrt{128}$ cm

Thus, the length of diagonal of square is $\sqrt{128}$ cm.

#### Page No 47:

#### Question 7:

Measure of one angle of a rhombus is 50$\xb0$. find the measures of remaining three angles.

#### Answer:

Suppose ABCD is a rhombus.

Let *m*∠A = 50º.

Opposite angles of a rhombus are congruent.

∴ *m*∠C = *m*∠A = 50º and *m*∠B = *m*∠D

Now,

*m*∠A + *m*∠B + *m*∠C + *m*∠D = 360º

∴ 50º + *m*∠B + 50º + *m*∠D = 360º

⇒ *m*∠B + *m*∠D = 360º − 100º = 260º

⇒ 2 *m*∠B = 260º (*m*∠B = *m*∠D)

⇒ *m*∠B = $\frac{260\xb0}{2}$ = 130º

∴ *m*∠D = *m*∠B = 130º

Thus, the measures of the remaining angles of the rhombus are 130º, 50º and 130º.

#### Page No 49:

#### Question 1:

Measures of opposite angles of a parallelogram are (3*x *− 2)$\xb0$ and (50 − *x*)$\xb0$. Find the measure of its each angle.

#### Answer:

Let ABCD be the parallelogram.

Suppose ∠A = (3*x *− 2)º and ∠C = (50 − *x*)º.

We know that the opposite angles of a parallelogram are congruent.

∴ *m*∠A = *m*∠C

⇒ 3*x* − 2 = 50 − *x*

⇒ 3*x* + *x* = 50 + 2

⇒ 4*x* = 52

⇒ *x* = 13

∴ *m*∠A = (3*x *− 2)º = (3 × 13* *− 2)º= (39* *− 2)º= 37º

So, *m*∠C = *m*∠A = 37º

Also, the adjacent angles of a parallelogram are supplementary.

∴ *m*∠A + *m*∠D = 180º

⇒ 37º + *m*∠D = 180º

⇒ *m*∠D = 180º − 37º = 143º

Now,

*m*∠B = *m*∠D = 143º (Opposite angles of a parallelogram are congruent)

Thus, the measure of the angles of the parallelogram are 37º, 143º, 37º and 143º.

#### Page No 50:

#### Question 2:

Referring the adjacent figure of a parallelogram, write the answer of questions given below.

(1) If *l*(WZ) = 4.5 cm then *l*(XY) = ?

(2) If *l*(YZ) = 8.2 cm then *l*(XW) = ?

(3) If *l*(OX) = 2.5 cm then *l*(OZ) = ?

(4) If *l*(WO) = 3.3 cm then *l*(WY) = ?

(5) If *m*∠WZY = 120$\xb0$ then *m*∠WXY = ? and *m*∠XWZ = ?

#### Answer:

WXYZ is a parallelogram.

(1)

*l*(XY) = *l*(WZ) = 4.5 cm (Opposite sides of a parallelogram are congruent)

(2)

*l*(XW) = *l*(YZ) = 8.2 cm (Opposite sides of a parallelogram are congruent)

(3)

*l*(OZ) = *l*(OX) = 2.5 cm (Diagonals of parallelogram bisect each other)

(4)

*l*(WY) = 2 × *l*(WO) = 2 × 3.3 = 6.6 cm (Diagonals of parallelogram bisect each other)

(5)

*m*∠WXY = *m*∠WZY = 120º (Opposite angles of a parallelogram are congruent)

Now,

*m*∠WZY + *m*∠XWZ = 180º (Adjacent angles of a parallelogram are supplementary)

⇒ 120º + *m*∠XWZ = 180º

⇒ *m*∠XWZ = 180º − 120º = 60º

#### Page No 50:

#### Question 3:

Construct a parallelogram ABCD such that *l*(BC) = 7 cm, *m*∠ABC = 40$\xb0$, *l*(AB) = 3 cm.

#### Answer:

**Steps of Construction:**

**Step 1: **Draw AB = 3 cm.

**Step 2: **Draw ∠ABX = 40$\xb0$.

**Step 3: **With B as centre and radius 7 cm, draw an arc cutting the ray BX at C.

**Step 4: **With C as centre and radius 3 cm, draw an arc.

**Step 5: **With A as centre and radius 7 cm, draw an arc cutting the previous arc at D.

**Step 6: **Join AD and CD.

Here, ABCD is the required parallelogram.

#### Page No 50:

#### Question 4:

Ratio of consecutive angles of a quadrilateral is 1:2:3:4. Find the measure of its each angle. Write, with reason, what type of a quadrilateral it is.

#### Answer:

Suppose PQRS is a quadrilateral.

Let *m*∠P : *m*∠Q : *m*∠R : *m*∠S = 1 : 2 : 3 : 4

So, *m*∠P = *k*, *m*∠Q = 2*k*, *m*∠R = 3*k* and *m*∠S = 4*k*, where *k* is some constant

Now,

*m*∠P + *m*∠Q + *m*∠R + *m*∠S = 360º

∴ *k *+ 2*k* + 3*k* + 4*k* = 360º

⇒ 10*k* = 360º

⇒ *k* = 36º

∴ *m*∠P = 36º

*m*∠Q = 2*k *=* *2 × 36º = 72º

m∠R = 3*k* = 3 × 36º = 108º

*m*∠S = 4*k *= 4 × 36º = 144º

Now, *m*∠P + *m*∠S = 36º + 144º = 180º

We know if two lines are intersected by a transversal such that the sum of interior angles on the same of the transversal are supplementary, then the two lines are parallel.

∴ Side PQ || Side SR

Also, *m*∠P + *m*∠Q = 36º + 72º = 108º ≠ 180º

So, side PS is not parallel to side QR.

In quadrilateral PQRS, only one pair of opposite sides is parallel. Therefore, quadrilateral PQRS is a trapezium.

#### Page No 50:

#### Question 5:

Construct ☐ BARC such that *l*(BA) = *l*(BC) = 4.2 cm, *l*(AC) = 6.0 cm, *l*(AR) = *l*(CR) = 5.6 cm

#### Answer:

**Steps of Construction:
Step 1: **Draw BA = 4.2 cm.

**Step 2:**With B as centre and radius 4.2 cm, draw an arc.

**Step 3:**With A as centre and radius 6 cm, draw an arc cutting the previous arc at C.

**Step 4:**Join BC.

**Step 5:**With A as centre and radius 5.6 cm, draw an arc.

**Step 6:**With C as centre and radius 5.6 cm, draw an arc cutting the previous arc at R.

**Step 7:**Join AR and CR.

Here, BARC is the required quadrilateral.

#### Page No 50:

#### Question 6:

Construct ☐ PQRS, such that *l*(PQ) = 3.5 cm, *l*(QR) = 5.6 cm, *l*(RS) = 3.5 cm, *m*∠Q = 110$\xb0$, *m*∠R = 70$\xb0$.

If it is given that ☐ PQRS is a parallelogram, which of the given information is unnecessary?

#### Answer:

**Steps of Construction:
Step 1: **Draw PQ = 3.5 cm.

**Step 2:**Draw ∠PQX = 110$\xb0$.

**Step 3:**With Q as centre and radius 5.6 cm, draw an arc cutting ray QX at R.

**Step 4:**Draw ∠QRY = 70$\xb0$.

**Step 5:**With R as centre and radius 3.5 cm, draw an arc cutting ray RY at S.

**Step 6:**Join PS.

Here, PQRS is the required quadrilateral.

If it is given that quadrilateral PQRS is a parallelogram, then the information

*l*(RS) = 3.5 cm and

*m*∠R = 70$\xb0$ is unnecessary.

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