Rs Aggarwal 2017 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 2017 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 2017 Solutions. All Rs Aggarwal 2017 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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