Rs Aggarwal 2018 for Class 8 Math Chapter 1 - Rational Numbers

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Rs Aggarwal 2018 Solutions for Class 8 Math Chapter 1 Rational Numbers are provided here with simple step-by-step explanations. These solutions for Rational Numbers are extremely popular among Class 8 students for Math Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 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 Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 198:

Question 1:

Construct a quadrilateral ABCD in which AB = 4.2 cm, BC = 6 cm, CD = 5.2 cm, DA = 5 cm and AC = 8 cm.

Answer:

Steps of construction:
Step 1: Draw $AB = 4.2 cm$ .
Step 2: With A as the centre and radius equal to $8 cm$ , draw an arc.
Step 3: With B as the centre and radius equal to $6 cm$ , draw another arc, cutting the previous arc at C.
Step 4: Join BC.
Step 5: With A as the centre and radius equal to $5 cm,$ draw an arc.
Step 6: With C as the centre and radius equal to $5.2 cm$ , draw another arc, cutting the previous arc at D.
Step 7: Join AD and CD.

Thus, ABCD is the required quadrilateral.

Page No 198:

Question 2:

Construct a quadrilateral PQRS in which PQ = 5.4 cm, QR = 4.6 cm, RS = 4.3 cm, SP = 3.5 cm and diagonal PR = 4 cm.

Answer:

Steps of construction:
Step 1: Draw $PQ = 5.4 cm$ .
Step 2: With P as the centre and radius equal to $4 cm$ , draw an arc.
Step 3: With Q as the centre and radius equal to $4.6 cm$ , draw another arc, cutting the previous arc at R.
Step 4: Join QR.
Step 5: With P as the centre and radius equal to $3.5 cm,$ draw an arc.
Step 6: With R as the centre and radius equal to $4.3 cm$ , draw another arc, cutting the previous arc at S.
Step 7: Join PS and RS.

Thus, PQRS is the required quadrilateral.

Page No 198:

Question 3:

Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 3.8 cm, CD = DA = 4.5 cm and diagonal BD = 5.6 cm.

Answer:

Steps of construction:
Step 1: Draw $AB = 3.5 cm$ .
Step 2: With B as the centre and radius equal to $5.6 cm$ , draw an arc.
Step 3: With A as the centre and radius equal to $4.5 cm$ , draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With D as the centre and radius equal to $4.5 cm,$ draw an arc.
Step 6: With B as the centre and radius equal to $3.8 cm$ , draw another arc, cutting the previous arc at C.
Step 7: Join BC and CD.

Thus, ABCD is the required quadrilateral.

Page No 198:

Question 4:

Construct a quadrilateral ABCD in which AB = 3.6 cm, BC = 3.3 cm, AD = 2.7 cm, diagonal AC = 4.6 cm and diagonal BD = 4 cm.

Answer:

Steps of construction:
Step 1: Draw $AB = 3.6 cm$ .
Step 2: With B as the centre and radius equal to $4 cm$ , draw an arc.
Step 3: With A as the centre and radius equal to $2.7 cm$ , draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With A as the centre and radius equal to $4.6 cm,$ draw an arc.
Step 6: With B as the centre and radius equal to $3.3 cm$ , draw another arc, cutting the previous arc at C.
Step 7: Join AC, BC and CD.

Thus, ABCD is the required quadrilateral.

Page No 198:

Question 5:

Construct a quadrilateral PQRS in which QR = 7.5 cm, PR = PS = 6 cm, RS = 5 cm and QS = 10 cm. Measure the fourth side.

Answer:

Steps of construction:
Step 1: Draw $Q R = 7.5 cm .$
Step 2: With Q as the centre and radius equal to $10 cm$ , draw an arc.
Step 3: With R as the centre and radius equal to $5 cm$ , draw another arc, cutting the previous arc at S.
Step 4: Join QS and RS.
Step 5: With S as the centre and radius equal to $6 cm,$ draw an arc.
Step 6: With R as the centre and radius equal to $6 cm$ , draw another arc, cutting the previous arc at P.
Step 7: Join PS and PR.
Step 8: PQ = 4.9 cm
Thus, PQRS is the required quadrilateral.

Page No 198:

Question 6:

construct a quadrilateral ABCD in which AB =3.4 cm, CD = 3 cm, DA = 5.7 cm, AC = 8 cm and BD = 4 cm.

Answer:

Steps of construction:
Step 1: Draw $A B = 3.4 cm .$
Step 2: With B as the centre and radius equal to $4 cm$ , draw an arc.
Step 3: With A as the centre and radius equal to $5.7 cm$ , draw another arc, cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: With A as the centre and radius equal to $8 cm,$ draw an arc.
Step 6: With D as the centre and radius equal to $3 cm$ , draw another arc, cutting the previous arc at C.
Step 7: Join AC, CD and BC.

Thus, ABCD is the required quadrilateral.

Page No 198:

Question 7:

Construct a quadrilateral ABCD in which AB = BC = 3.5 cm, AD = CD = 5.2 cm and ∠ABC = 120°.

Answer:

Steps of construction:
Step 1: Draw AB= $3.5 cm$ .
Step 2: Make $∠ A B C = 120^{\circ}$ .
Step 3: With B as the centre, draw an arc $3.5 cm$ and name that point C.
Step 4: With C as the centre, draw an arc $5.2 cm$ .
Step 5: With A as the centre, draw another arc $5.2 cm$ , cutting the previous arc at D.
Step 6: Join CD and AD.
Thus, $A B C D$ is the required quadrilateral.

Page No 198:

Question 8:

Construct a quadrilateral ABCD in which AB = 2.9 cm, BC = 3.2 cm, CD = 2.7 cm, DA = 3.4 cm and ∠A = 70°.

Answer:

Steps of construction:
Step 1: Draw AB= $2.9 c m$
Step 2: Make $∠ A = 70^{\circ}$
Step 3: With A as the centre, draw an arc of $3.4 c m$ . Name that point as D.
Step 4: With D as the centre, draw an arc of $2.7 c m$ .
Step 5: With B as the centre, draw an arc of 3.2 cm, cutting the previous arc at C.
Step 6: Join CD and BC.
Then, $A B C D$ is the required quadrilateral.

Page No 198:

Question 9:

Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 5 cm, CD = 4.6 cm, ∠B = 125° and ∠C = 60°.

Answer:

Steps of construction:
Step 1: Draw BC= $5 c m$
Step 2: Make $∠ B = 125^{\circ} a n d ∠ C = 60^{\circ}$
Step 3: With B as the centre, draw an arc of $3.5 c m$ . Name that point as A.
Step 4: With C as the centre, draw an arc of $4.6 c m$ . Name that point as D.
Step 5: Join A and D.
Then, $A B C D$ is the required quadrilateral.

Page No 198:

Question 10:

Construct a quadrilateral PQRS in which PQ = 6 cm, QR = 5.6 cm, RS = 2.7 cm, ∠Q = 45° and ∠R = 90°.

Answer:

Steps of construction:
Step 1: Draw QR= $5.6 c m$
Step 2: Make $∠ Q = 45^{\circ} and ∠ R = 90^{\circ}$
Step 3: With Q as the centre, draw an arc of $6 c m$ . Name that point as P.
Step 4: With R as the centre, draw an arc of $2.7 c m$ . Name that point as S.
Step 6: Join P and S.
Then, $P Q R S$ is the required quadrilateral.

Page No 198:

Question 11:

Construct a quadrilateral ABCD in which AB = 5.6 cm, BC = 4 cm, ∠A = 50°, ∠B = 105° and ∠D = 80°.

Answer:

Steps of construction:
Step 1: Draw AB= $5.6 c m$
Step 2: Make $∠ A = 50^{\circ} a n d ∠ B = 105^{\circ}$
Step 3: With B as the centre, draw an arc of $4 c m$ .
Step 3: Sum of all the angles of the quadrilateral is $360^{\circ}$ .
$∠ A + ∠ B + ∠ C + ∠ D = 360^{\circ} 50^{\circ} + 105^{\circ} + ∠ C + 80^{\circ} = 360^{\circ} 235^{\circ} + ∠ C = 360^{\circ} ∠ C = 360^{\circ} - 235^{\circ} ∠ C = 125^{\circ}$
Step 5: With C as the centre, make $∠ C equal to ∠ 125^{\circ}$ .
Step 6: Join C and D.
Step 7: Measure $∠ D = 80^{\circ}$
Then, $A B C D$ is the required quadrilateral.

Page No 199:

Question 12:

Construct a quadrilateral PQRS in which PQ = 5 cm, QR = 6.5 cm, ∠P = ∠R = 100° and ∠S = 75°.

Answer:

Steps of construction:
Step 1: Draw PQ= $5 c m$
Step 2:
$∠ P + ∠ Q + ∠ R + ∠ S = 360^{\circ} 100^{\circ} + ∠ Q + 100^{\circ} + 75^{\circ} = 360^{\circ} 275^{\circ} + ∠ Q = 360^{\circ} ∠ Q = 360^{\circ} - 275^{\circ} ∠ Q = 85^{\circ}$
Step 3: Make $∠ P = 100^{\circ} a n d ∠ Q = 85^{\circ}$
Step 3: With Q as the centre, draw an arc of $6.5 c m$ .
Step 4: Make $∠ R = 100^{\circ}$
Step 6: Join R and S.
Step 7: Measure $∠ S = 75^{\circ}$
Then, $P Q R S$ is the required quadrilateral.

Page No 199:

Question 13:

Construct a quadrilateral ABCD in which AB = 4 cm, AC = 5 cm, AD = 5.5 cm and ∠ABC = ∠ACD = 90°.

Answer:

Steps of construction:
Step 1: Draw $A B = 4 c m$
Step 2: $M a k e ∠ B = 90^{\circ}$
Step 3: $A C^{2} = A B^{2} + B C^{2} 5^{2} = 4^{2} + B C^{2} 25 - 16 = B C^{2} B C = 3 c m$
With B as the centre, draw an arc equal to 3 cm.
Step 4: Make $∠ C = 90^{\circ}$
Step 5: With A as the centre and radius equal to $5.5 c m$ , draw an arc and name that point as D.
Then, $A B C D$ is the required quadrilateral.

Page No 201:

Question 1:

Construct a parallelogram ABCD in which AB = 5.2 cm, BC = 4.7 cm and AC = 7.6 cm.

Answer:

Steps of construction:
Step 1: Draw AB = $5.2 c m$
Step 2: With B as the centre, draw an arc of $4.7 c m$ .
Step 3: With A as the centre, draw another arc of $7.6 c m$ , cutting the previous arc at C.
Step 4: Join A and C.
Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with C as the centre, draw an arc of $5.2 c m$ .
Step 6: With A as the centre, draw another arc of $4.7 c m$ , cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.

Page No 201:

Question 2:

Construct a parallelogram ABCD in which AB = 4.3 cm, AD = 4 cm and BD = 6.8 cm.

Answer:

Steps of construction:
Step 1: Draw AB= $4.3 c m$
Step 2: With B as the centre, draw an arc of $6.8 c m$ .
Step 3: With A as the centre, draw another arc of $4 c m$ , cutting the previous arc at D.
Step 4: Join BD and AD.
Step 5: We know that the opposite sides of a parallelogram are equal.
Thus, with D as the centre, draw an arc of $4.3 c m$ .
Step 6: With B as the centre, draw another arc of $4 c m$ , cutting the previous arc at C.
Step 7: Join CD and BC.
then, ABCD is the required parallelogram.

Page No 201:

Question 3:

Construct a parallelogram PQRS in which QR = 6 cm, PQ = 4 cm and ∠PQR = 60° cm.

Answer:

Steps of construction:
Step 1: Draw PQ= 4 cm
Step 2: Make $∠ P Q R = 60^{\circ}$
Step 2: With Q as the centre, draw an arc of 6 cm and name that point as R.
Step 3: With R as the centre, draw an arc of 4 cm and name that point as S.
Step 4: Join SR and PS.
Then, PQRS is the required parallelogram.

Page No 201:

Question 4:

Construct a parallelogram ABCD in which BC = 5 cm, ∠BCD = 120° and CD = 4.8 cm.

Answer:

Steps of construction:
Step 1: Draw BC= $5 c m$
Step 2: Make an $∠ B C D = 120^{\circ}$
Step 2: With C as centre draw an arc of $4.8 cm$ , name that point as D
Step 3: With D as centre draw an arc $5 cm$ , name that point as A
Step 4: With B as centre draw another arc $4.8 cm$ cutting the previous arc at A.
Step 5: Join AD and AB
then, ABCD is a required parallelogram.

Page No 201:

Question 5:

Construct a parallelogram, one of whose sides is 4.4 cm and whose diagonals are 5.6 cm and 7 cm. Measure the other side.

Answer:

We know that the diagonals of a parallelogram bisect each other.
Steps of construction:
Step 1: Draw AB= $4.4 c m$
Step 2: With A as the centre and radius $2.8 c m$ , draw an arc.
Step 3: With B as the centre and radius $3.5 c m$ , draw another arc, cutting the previous arc at point O.
Step 4: Join OA and OB.
Step 5: Produce OA to C, such that OC= AO. Produce OB to D, such that OB=OD.
Step 5: Join AD, BC, and CD.
Thus, ABCD is the required parallelogram. The other side is 4.5 cm in length.

Page No 201:

Question 6:

Construct a parallelogram ABCD in which AB = 6.5 cm, AC = 3.4 cm and the altitude AL from A is 2.5 cm. Draw the altitude from C and measure it.

Answer:

Steps of construction:
Step 1: Draw AB= 6.5cm
Step 2: Draw a perpendicular at point A. Name that ray as AX. From point A, draw an arc of length 2.5 cm on the ray AX and name that point as L.
Step 3: On point L, make a perpendicular. Draw a straight line YZ passing through L, which is perpendicular to the ray AX.
Step 4: Cut an arc of length 3.4 cm on the line YZ and name it as C.
Step 5: From point C, cut an arc of length 6.5 cm on the line YZ. Name that point as D.
Step 6: Join BC and AD.

Therefore, quadrilateral ABCD is a parallelogram.

The altitude from C measures 2.5 cm in length.

Page No 201:

Question 7:

Construct a parallelogram ABCD, in which diagonal AC = 3.8 cm, diagonal BD = 4.6 cm and the angle between AC and BD is 60°.

Answer:

We know that the diagonals of a parallelogram bisect each other.

Steps of construction:
Step 1: Draw AC= $3.8 cm$
Step 2: Bisect AC at O.
Step 3: Make $∠ C O X = 60^{\circ}$
Produce XO to Y.
Step 4:
$O B = \frac{1}{2} (4.6) cm O B = 2.3 cm and O D = \frac{1}{2} (4.6) cm O D = 2.3 cm$
Step 5: Join AB, BC, CD and AD.
Thus, ABCD is the required parallelogram.

Page No 201:

Question 8:

Construct a rectangle ABCD whose adjacent sides are 11 cm and 8.5 cm.

Answer:

Steps of construction:
Step 1: Draw AB = $11 c m$
Step 2: Make $∠ A = 90^{\circ} ∠ B = 90^{\circ}$
Step 3: Draw an arc of 8.5 cm from point A and name that point as D.
Step 4: Draw an arc of 8.5 cm from point B and name that point as C.
Step 5: Join C and D.
Thus, ABCD is the required rectangle.

Page No 201:

Question 9:

Construct a square, each of whose sides measures 6.4 cm.

Answer:

All the sides of a square are equal.
Steps of construction:
Step 1: Draw AB = $6.4 c m$
Step 2: Make $∠ A = 90^{\circ} ∠ B = 90^{\circ}$
Step 3: Draw an arc of length 6.4 cm from point A and name that point as D.
Step 4: Draw an arc of length 6.4 cm from point B and name that point as C.
Step 5: Join C and D.
Thus, ABCD is a required square.

Page No 201:

Question 10:

Construct a square, each of whose diagonals measures 5.8 cm.

Answer:

We know that the diagonals of a square bisect each other at right angles.
Steps of construction:
Step 1: Draw AC= $5.8 cm$
Step 2: Draw the perpendicular bisector XY of AC, meeting it at O.
Step 3:
: $F r o m O : O B = \frac{1}{2} (5.8) cm = 2.9 cm O D = \frac{1}{2} (5.8) cm = 2.9 cm$
Step 4: Join AB, BC, CD and DA.
ABCD is the required square.

Page No 201:

Question 11:

Construct a rectangle PQRS in which QR = 3.6 cm and diagonal PR = 6 cm. Measure the other side of the rectangle.

Answer:

Steps of construction:
Step 1: Draw QR = $3.6 c m$
Step 2: Make $∠ Q = 90^{\circ} ∠ R = 90^{\circ}$
Step 3:
$P R^{2} = P Q^{2} + Q R^{2} 6^{2} = P Q^{2} + 3 . 6^{2} P Q^{2} = 36 - 12.96 P Q^{2} = 23.04 P Q = 4.8 cm$

Step 3: Draw an arc of length 4.8 cm from point Q and name that point as P.
Step 4: Draw an arc of length 6 cm from point R, cutting the previous arc at P.
Step 5: Join PQ
Step 6: Draw an arc of length 4.8 cm from point R.
From point P, draw an arc of length 3.6 cm, cutting the previous arc. Name that point as S.
Step 7: Join P and S.
Thus, PQRS is the required rectangle. The other side is 4.8 cm in length.

Page No 201:

Question 12:

Construct a rhombus the lengths of whose diagonals are 6 cm and 8 cm.

Answer:

We know that the diagonals of a rhombus bisect each other.
.Steps of construction:
Step 1: Draw AC= $6 c m$
Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.
Step 3:
$O B = \frac{1}{2} (8) cm O B = 4 cm and O D = \frac{1}{2} (8) cm O D = 4 cm$
Draw an arc of length 4 cm on OX and name that point as B.
Draw an arc of length 4 cm on OY and name that point as D.
Step 4 : Join AB, BC, CD and AD.
Thus, ABCD is the required rhombus, as shown in the figure.

Page No 201:

Question 13:

Construct a rhombus ABCD in which AB = 4 cm and diagonal AC is 6.5 cm.

Answer:

Steps of construction:
Step 1: Draw AB = $4 c m$
Step 2: With B as the centre, draw an arc of $4 cm$ .
Step 3: With A as the centre, draw another arc of $6.5 cm$ , cutting the previous arc at C.
Step 4: Join AC and BC.
Step 5: With C as the centre, draw an arc of 4 cm.
Step 6: With A as the centre, draw another arc of $4 cm$ , cutting the previous arc at D.
Step 7: Join AD and CD.
ABCD is the required rhombus.

Page No 201:

Question 14:

Draw a rhombus whose side is 7.2 cm and one angle is 60°.

Answer:

Steps of construction:
Step1: Draw AB = $7.2 cm$
Step2: Draw $∠ A B Y = 60 ° ∠ B A X = 120 °$
Sum of the adjacent angles is 180°.
$∠ B A X + ∠ A B Y = 180 ° = > ∠ B A X = 180 ° - 60 ° = 120 °$
Step 3:
$Set off A D (7.2 cm) along A X and B C (7.2 cm) along B Y .$
Step 4: Join C and D.
Then, ABCD is the required rhombus.

Page No 201:

Question 15:

Construct a trapezium ABCD in which AB = 6 cm, BC = 4 cm, CD = 3.2 cm, ∠B = 75° and DC||AB.

Answer:

Steps of construction:
Step 1: Draw AB= $6 cm$
Step 2: Make $∠ A B X = 75^{\circ}$
Step 3: With B as the centre, draw an arc at $4 c m$ . Name that point as C.
Step 4: $A B ∥ C D$

$∴ ∠ A B X + ∠ B C Y = 180 ° \Rightarrow ∠ B C Y = 180 ° - 75 ° = 105 °$
Make $∠ B C Y = 105 °$
At C, draw an arc of length $3.2 cm$ .
Step 5: Join A and D.
Thus, ABCD is the required trapezium.

Page No 201:

Question 16:

Draw a trapezium ABCD in which AB||DC, AB = 7 cm, BC = 5 cm, AD = 6.5 cm and ∠B = 60°.

Answer:

Steps of construction :
Step1: Draw AB equal to 7 cm.
Step2: Make an angle, $∠ A B X, equal to 60 ° .$
Step3: With B as the centre, draw an arc of $5 cm$ . Name that point as C. Join B and C.
Step4:
$A B ∥ D C ∴ ∠ A B X + ∠ B C Y = 180 ° \Rightarrow ∠ B C Y = 180 ° - 60 ° = 120 °$

Draw an angle, $∠ B C Y, equal to 120 ° .$

Step4: With A as the centre, draw an arc of length $6.5 cm$ , which cuts CY. Mark that point as D.

Step5: Join A and D.

Thus, ABCD is the required trapezium.

Page No 202:

Question 1:

Define the terms:
(i) Open curve
(ii) Closed curve
(iii) Simple closed curve

Answer:

( i) Open curve: An open curve is a curve where the beginning and end points are different.
Example: Parabola

(ii) Closed Curve: A curve that joins up so there are no end points.
Example: Ellipse

(iii) Simple closed curve: A closed curve that does not intersect itself.

Page No 202:

Question 2:

The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4. Find the measure of each angle.

Answer:

Let the angles be $(x) °, (2 x) °, (3 x) ° and (4 x) ° .$
Sum of the angles of a quadrilateral is $360^{\circ}$ .
$x + 2 x + 3 x + 4 x = 360 10 x = 360 x = \frac{360}{10} x = 36$

$(2 x) ° = (2 \times 36)^{\circ} = 72^{\circ} (3 x) ° = (3 \times 36)^{\circ} = 108^{\circ} (4 x) ° = (4 \times 36)^{\circ} = 144^{\circ}$

The angles of the quadrilateral are $36^{\circ}, 72^{\circ}, 108^{\circ} and 144^{\circ} .$

Page No 202:

Question 3:

Two adjacent angles of a parallelogram are in the ratio 2 : 3. Find the measure of each of its angles.

Answer:

$Let the two adjacent angles of the parallelogram b e (2 x) ° and (3 x) ° .$
Sum of any two adjacent angles of a parallelogram is $180^{\circ}$ .

$∴ 2 x + 3 x = 180 \Rightarrow 5 x = 180 \Rightarrow x = 36$

$(2 x) ° = (2 \times 36) ° = 72^{\circ} (3 x) ° = (3 \times 36) ° = 108^{\circ}$

Measures of the angles are $72^{\circ} a n d 108^{\circ}$ .

Page No 202:

Question 4:

The sides of a rectangle are in the ratio 4 : 5 and its perimeter is 180 cm. Find its sides.

Answer:

Let the length be $4 x$ cm and the breadth be $5 x$ cm.
Perimeter of the rectangle =180 $c m$
Perimeter of the rectangle= $2 (l + b)$

$2 (l + b) = 180 \Rightarrow 2 (4 x + 5 x) = 180 \Rightarrow 2 (9 x) = 180 \Rightarrow 18 x = 180 \Rightarrow x = 10$

$∴ Length = 4 x cm = 4 \times 10 = 40 cm Breadth = 5 x cm = 5 \times 10 = 50 cm$

Page No 202:

Question 5:

Prove that the diagonals of a rhombus bisect each other at right angles.

Answer:

Rhombus is a parallelogram.

Consider:

∆ A O B and ∆ C O D ∠ O A B = ∠ O C D (alternate angle) ∠ O D C = ∠ O B A (alternate angle) ∠ D O C = ∠ A O B (vertically opposite angles) ∆ A O B ≅ C O B ∴ A O = C O O B = O D

Therefore, the diagonals bisects at O.

Now, let us prove that the diagonals intersect each other at right angles.

Consider

∆ C O D a n d ∆ C O B

C D = C B (all sides of a rhombus are equal) C O = C O (common side) O D = O B (p oint O bisects B D)

∴

∆ C O D ≅ ∆ C O B

∴

∠ C O D = ∠ C O B

(corresponding parts of congruent triangles)

Further,

∠ C O D + ∠ C O B = 180 ° (l inear pair)

∴

∠ C O D = ∠ C O B = 90 °

It is proved that the diagonals of a rhombus are perpendicular bisectors of each other.

Page No 202:

Question 6:

The diagonals of a rhombus are 16 cm and 12 cm. Find the length of each side of the rhombus.

Answer:

All the sides of a rhombus are equal in length.
The diagonals of a rhombus intersect at $90^{\circ}$ .
The diagonal and the side of a rhombus form right triangles.

In $△ A O B$ :
$A B^{2} = A O^{2} + O B^{2} = 8^{2} + 6^{2} = 64 + 36 = 100 A B = 10 cm$
Therefore, the length of each side of the rhombus is 10 cm.

Page No 202:

Question 7:

Mark (✓) against the correct answer:
Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. The measures of all its angles are
(a) 97°, 83°, 97°, 83°
(b) 37°, 143°, 37°, 143°
(c) 76°, 104°, 76°, 104°
(d) none of these

Answer:

(b) 37^o, 143^o, 37^o 143^o

Opposite angles of a parallelogram are equal.
$∴ 3 x - 2 = 50 - x \Rightarrow 3 x + x = 50 + 2 \Rightarrow 4 x = 52 \Rightarrow x = 13$

Therefore, the first and the second angles are:
${(3 x - 2)}^{\circ} = {(2 \times 13 - 2)}^{\circ} = 37^{\circ} {(50 - x)}^{\circ} = {(50 - 13)}^{\circ} = 37^{\circ}$
Sum of adjacent angles in a parallelogram is $180^{\circ}$ .
Adjacent angles = $180^{\circ} - 37^{\circ} = 143^{\circ}$

Page No 202:

Question 8:

Mark (✓) against the correct answer:
The angles of quadrilateral are in the ratio 1 : 3 : 7 : 9. The measure of the largest angle is
(a) 63°
(b) 72°
(c) 81°
(d) none of these

Answer:

(d) none of the these

Let the angles be $(x) °, (3 x) °, (7 x) ° and (9 x) °$ .

Sum of the angles of the quadrilateral is $360^{\circ}$ .

$x + 3 x + 7 x + 9 x = 360 20 x = 360 x = 18$

$Angles : (3 x) ° = (3 \times 18) = 54^{\circ} (7 x) ° = (7 \times 18) ° = 126^{\circ} (9 x) ° = (9 \times 18) ° = 162^{\circ}$

Page No 202:

Question 9:

Mark (✓) against the correct answer:
The length of a rectangle is 8 cm and each of its diagonals measures 10 cm. The breadth of the rectangle is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 9 cm

Answer:

(b) 6 cm
Let the breadth of the rectangle be x cm.
Diagonal =10 cm
Length= 8 cm
The rectangle is divided into two right triangles.
$D i a g o n a l^{2} = L e n g t h^{2} + B r e a d t h^{2} 10^{2} = 8^{2} + x^{2} 100 - 64 = x^{2} x^{2} = 36 x = 6 c m$

Breadth of the rectangle = 6 cm

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Question 10:

Mark (✓) against the correct answer:
In a square PQRS, if PQ = (2x + 3) cm and QR = (3x − 5) cm then
(a) x = 4
(b) x = 5
(c) x = 6
(d) x = 8

Answer:

(d) x = 8
All sides of a square are equal.

$P Q = Q R (2 x + 3) = (3 x - 5) = > 2 x - 3 x = - 5 - 3 = > x = 8 cm$

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Question 11:

Mark (✓) against the correct answer:
The bisectors of two adjacent angles of a parallelogram intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(d) 90°

We know that the opposite sides and the angles in a parallelogram are equal.
Also, its adjacent sides are supplementary, i.e. sum of the sides is equal to 180.
Now, the bisectors of these angles form a triangle, whose two angles are:
$\frac{A}{2} and \frac{B}{2} or \frac{A}{2} = (90 - \frac{A}{2}) Sum of the angles of a triangle is 180 ° . \frac{∠ A}{2} + 90 - \frac{∠ A}{2} + ∠ O = 180^{\circ} ∠ O = 180 - 90 ∠ O = 90^{\circ}$

Hence, the two bisectors intersect at right angles.

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Question 12:

Mark (✓) against the correct answer:
How many diagonals are there in a hexagon?
(a) 6
(b) 8
(c) 9
(d) 10

Answer:

(c) 9
Hexagon has six sides.
$Number of diagonals = \frac{n (n - 3)}{2} (where n is the number of side s) = \frac{6 (6 - 3)}{2} = 9$

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Question 13:

Mark (✓) against the correct answer:
Each interior angle of a polygon is 135. How many sides does it have?
(a) 10
(b) 8
(c) 6
(d) 5

Answer:

(b) 8
$Interior angle = \frac{180 (n - 2)}{n} \Rightarrow 135 = \frac{180 (n - 2)}{n} \Rightarrow 135 n = 180 n - 360 \Rightarrow 360 = 180 n - 135 n \Rightarrow n = 8$

It has 8 sides.

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Question 14:

Fill in the blanks.
For a convex polygon of n sides, we have:
(i) Sum of all exterior angles = .........
(ii) Sum of all interior angles = .........
(iii) Number of diagonals = .........

Answer:

(i) Sum of all exterior angles = $360^{\circ}$

(ii) Sum of all interior angles = $(n - 2) \times 180 °$

(iii) Number of diagonals = $\frac{n (n - 3)}{2}$

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Question 15:

Fill in the blanks.
For a regular polygon of n sides, we have:
(i) Sum of all exterior angles = .........
(ii) Sum of all interior angles = .........

Answer:

(i) Sum of all exterior angles of a regular polygon is $360^{\circ}$ .

(ii) Sum of all interior angles of a polygon is $(n - 2) \times 180 °, where n is the number of sides .$

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Question 16:

Fill in the blanks.
(i) Each interior angle of a regular octagon is (.........)°.
(ii) The sum of all interior angles of a regular hexagon is (.........)°.
(iii) Each exterior angle of a regular polygon is 60°. This polygon is a .........
(iv) Each interior angle of a regular polygon is 108°. This polygon is a .........
(v) A pentagon has ......... diagonals.

Answer:

(i) Octagon has 8 sides.
$∴ Interior angle = \frac{180 ° n - 360 °}{n} Int erior angle = \frac{(180 ° \times 8) - 360 °}{8} = 135 °$
(ii) Sum of the interior angles of a regular hexagon = $(6 - 2) \times 180^{\circ} = 720^{\circ}$

(iii) Each exterior angle of a regular polygon is $60^{\circ}$ .
$∴ \frac{360}{60} = 6$
Therefore, the given polygon is a hexagon.

(iv) If the interior angle is $108^{\circ}$ , then the exterior angle will be $72^{\circ}$ . (interior and exterior angles are supplementary)
Sum of the exterior angles of a polygon is 360°.

Let there be n sides of a polygon.

$72 n = 360 n = \frac{360}{72} n = 5$

Since it has 5 sides, the polygon is a pentagon.

(v) A pentagon has 5 diagonals.

$If n is the number of s ides, the number of diagonals = \frac{n (n - 3)}{2} \frac{5 (5 - 3)}{2} = 5$

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Question 17:

Write 'T' for true and 'F' for false for each of the following:
(i) The diagonals of a parallelogram are equal.
(ii) The diagonals of a rectangle are perpendicular to each other.
(iii) The diagonals of a rhombus bisect each other at right angles.
(iv) Every rhombus is a kite.

Answer:

(i) F
The diagonals of a parallelogram need not be equal in length.

(ii) F
The diagonals of a rectangle are not perpendicular to each other.

(iii) T

(iv) T

Adjacent sides of a kite are equal and this is also true for a rhombus. Additionally, all the sides of a rhombus are equal to each other.

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Question 18:

Construct a quadrilateral PQRS in which PQ = 4.2 cm, ∠PQR = 60°, ∠QPS = 120°, QR = 5 cm and PS = 6 cm.

Answer:

Steps of construction:
Step 1: Take PQ = 4.2 cm
Step 2: $Make ∠ X P Q = 120^{\circ}, ∠ Y Q P = 60^{\circ}$
Step 3: Cut an arc of length 5 cm from point Q. Name that point as R.
Step 4: From P, make an arc of length 6 cm. Name that point as S.
Step 5: Join P and S.
Thus, PQRS is a quadrilateral.

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