Rs Aggarwal 2019 2020 Solutions for Class 8 Math Chapter 17 Construction Of Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Construction Of Quadrilaterals are extremely popular among Class 8 students for Math Construction Of Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 8 Math Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 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 QR=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 AB=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 AB3.5 cm.
Step 2: Make ABC=120.
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, ABCD 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 AB2.9cm
Step 2: Make A=70
Step 3: With A as the centre, draw an arc of 3.4cm. Name that point as D.
Step 4: With D as the centre, draw an arc of 2.7cm.
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, ABCD 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 BC5cm
Step 2: Make B=125 and C=60
Step 3: With B as the centre, draw an arc of 3.5 cm. Name that point as A.
Step 4: With C as the centre, draw an arc of 4.6 cm. Name that point as D.
Step 5: Join A and D.
Then, ABCD 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 QR5.6 cm
Step 2: Make Q=45 and R=90
Step 3: With Q as the centre, draw an arc of 6 cm. Name that point as P.
Step 4: With R as the centre, draw an arc of 2.7cm. Name that point as S.
Step 6: Join P and S.
Then, PQRS 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 AB5.6 cm
Step 2: Make A=50 and B=105
Step 3: With B as the centre, draw an arc of 4cm.
Step 3: Sum of all the angles of the quadrilateral is 360.
A+B+C+D=36050+105+C+80=360235+C=360C=360-235C=125
Step 5: With C as the centre, make C equal to 125.
Step 6: Join C and D.
Step 7: Measure D=80
Then, ABCD 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 PQ5cm
Step 2:
P+Q+R+S=360100+Q+100+75=360275+Q=360Q=360-275Q=85
Step 3: Make P=100 and Q=85
Step 3: With Q as the centre, draw an arc of 6.5 cm.
Step 4: Make R=100
Step 6: Join R and S.
Step 7: Measure S=75
Then, PQRS 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 AB=4cm
Step 2: Make B=90
Step 3: AC2=AB2+BC252=42+BC225-16=BC2BC=3cm
With B as the centre, draw an arc equal to 3 cm.
Step 4: Make C=90
Step 5: With A as the centre and radius equal to 5.5 cm, draw an arc and name that point as D.
Then, ABCD 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 5.2cm
Step 2: With B as the centre, draw an arc of 4.7 cm.
Step 3: With A as the centre, draw another arc of 7.6 cm, 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 5.2cm.
Step 6: With A as the centre, draw another arc of 4.7 cm, 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= 4.3cm
Step 2: With B as the centre, draw an arc of 6.8 cm.
Step 3: With A as the centre, draw another arc of 4cm, 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 4.3cm.
Step 6: With B as the centre, draw another arc of 4 cm, 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 PQR=60
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= 5cm
Step 2: Make an BCD=120
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 5cm, 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 AB4.4cm
Step 2: With A as the centre and radius 2.8cm, draw an arc.
Step 3: With B as the centre and radius 3.5cm, 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 AC3.8cm
Step 2: Bisect AC at O.
Step 3: Make COX=60 
Produce XO to Y.
Step 4:
OB = 12(4.6) cmOB=2.3 cmand OD =12(4.6) cmOD=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 = 11cm 
Step 2: Make A=90 B=90
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 6.4cm 
Step 2: Make A=90 B=90
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 AC5.8 cm
Step 2: Draw the perpendicular bisector XY of AC, meeting it at O.
Step 3:
From O:OB=12(5.8) cm = 2.9 cmOD=12(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 QR3.6cm 
Step 2: Make  Q=90R=90
Step 3:
 PR2=PQ2+QR262=PQ2+3.62PQ2=36-12.96PQ2=23.04PQ=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.
F
rom 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= 6cm
Step 2:Draw a perpendicular bisector(XY) of AC, which bisects AC at O.
Step 3:
OB = 12(8) cmOB=4cmand OD =12(8) cmOD=4cm
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 AB4cm
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 AB7.2 cm 
Step2: Draw ABY=60° BAX=120°
Sum of the adjacent angles is 180°.
BAX+ABY=180°=>BAX=180°-60°=120°
Step 3:  
Set off AD (7.2 cm) along AX and BC ( 7.2 cm) along BY.
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  ABX=75
Step 3: With B as the centre, draw an arc at 4cm. Name that point as C.
Step 4:  ABCD

 ABX+BCY=180°             BCY=180° -75°=105°
Make  BCY=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, ABX, 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:
 ABDCABX+BCY=180°BCY=180°-60°=120°

Draw an angle,  BCY,  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)°, (2x)°, (3x)° and (4x)°.
Sum of the angles of a quadrilateral is 360.
x+2x+3x+4x=36010x=360x=36010x=36

(2x)°=(2×36)=72(3x)°=(3×36)=108(4x)°=(4×36)=144

The angles of the quadrilateral are 36, 72, 108 and 144.

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 be (2x)° and (3x)°.
Sum of any two adjacent angles of a parallelogram is 180.

 2x+3x =1805x=180x=36

(2x)°=(2×36)°=72(3x)°=(3×36)°=108

Measures of the angles are 72 and 108.

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

 2(l+b)=1802(4x+5x)=1802(9x)=18018x=180x=10
 
Length=4x cm=4×10=40 cmBreadth=5x cm=5×10=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:

AOB and CODOAB=OCD     (alternate angle)ODC=OBA    (alternate angle)DOC=AOB     (vertically opposite angles)AOBCOBAO=COOB=OD

Therefore, the diagonals bisects at O.

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

Consider CODand COB:
CD=CB         (all sides of a rhombus are equal)CO=CO         (common side)OD=OB         (point O bisects BD)

COD COB
COD = COB          (corresponding parts of congruent triangles)

Further, COD+COB=180°     (linear pair)

COD=COB=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.
The diagonal and the side of a rhombus form right triangles.

In AOB:
AB2=AO 2+OB2       =82+62       =64+36       =100AB=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) 37o, 143o, 37o 143o

Opposite angles of a parallelogram are equal.
   3x-2=50-x3x+x=50+24x=52x=13               

Therefore, the first and the second angles are:
3x-2=2×13-2=3750-x=50-13=37
Sum of adjacent angles in a parallelogram is 180.
Adjacent angles = 180-37=143

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)°, (3x)°, (7x)° and (9x)°.

Sum of the angles of the quadrilateral is 360.

x+3x+7x+9x=36020x=360x=18

Angles: (3x)°=(3×18)=54(7x)°=(7×18)°=126(9x)°=(9×18)°=162

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.
Diagonal2= Length2+ Breadth2102=82+x2100-64=x2x2=36x=6 cm

Breadth of the rectangle = 6 cm

Page No 202:

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.


PQ=QR(2x+3)=(3x-5)=>2x-3x=-5-3=>x=8 cm

Page No 202:

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:

 A2andB2 or A2=(90-A2)Sum of the angles of a triangle is 180°.A2+90-A2+O=180O=180-90O=90

Hence, the two bisectors intersect at right angles.

Page No 202:

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 = n(n-3)2              (where n is the number of sides)=6(6-3)2=9

Page No 202:

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=180(n-2)n135=180(n-2)n135n=180n-360360=180n-135nn=8

It has 8 sides.

Page No 202:

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

(ii) Sum of all interior angles = (n-2)×180°

(iii) Number of diagonals = n(n-3)2

Page No 202:

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.

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

Page No 202:

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 =180°n-360°nInterior angle =(180°×8)-360°8                       =135°
(ii) Sum of the interior angles of a regular hexagon = (6-2)×180=720

(iii) Each exterior angle of a regular polygon is 60.
 36060=6
Therefore, the given polygon is a hexagon.

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

Let there be n sides of a polygon.

72n=360n=36072n=5

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

(v) A pentagon has 5 diagonals.

 If n is the number of sides, the number of diagonals = n(n-3)25(5-3)2=5



Page No 203:

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.

Page No 203:

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 XPQ=120, YQP=60
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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