SELINA Solution Class 9 Chapter 15 Construction of Polygons (Using ruler and compass only) Exercise 15

Question 1

Construct a quadrilateral ABCD, when:
AB = 3.2 cm, BC = 5.2 cm, CD = 6.2 cm, DA = 4.2 cm and BD = 5.2 cm.

Sol:

Steps:
1. Draw AB = 3.2 cm.
2. With A as a centre draw an arc at D and with B as a centre and radius 5.2 cm draw an arc at D.
3. Join AD and DB.
4. With D and B as a centre taking radius 6.2 cm and 5.2 cm draw an arc at C. Now join BC and DC.

ABCD is the required quadrilateral.

Question 2

Construct a quadrilateral ABCD, when:
AB = 7.2 cm, BC = 5.8 cm, CD = 6.3 cm, AD = 4.3 cm and angle A = 75°.

SoL:

Steps:
1. Draw AB = 7.2cm
2. Through A draw AP such that ∠A = 75°.
3. From AP cut AD = 4.3 cm
4. With D and B as center and radii 6.2 cm and 5.8 cm respectively, draw arcs cutting each other at C.
5. Join DC and BC.

ABCD is the required quadrilateral.

Question 3

Construct a quadrilateral ABCD, when:
∠ A = 90^o, AB = 4.6 cm, BD = 6.4 cm, AC = 6.0 cm and CD = 4.2 cm.

Sol:

Steps:
1. Draw AB = 4.6 cm.
2. Through A, draw AP such that Angle A = 90°.
3. With B as a center and radii 6.4 cm draw an arc at D on AP.
4. With D and A as a center and radii, 4.2 cm and 6 cm draw an arc cutting each other at C.
5. Now join BD, AC, and CB.

ABCD is the required quadrilateral.

Question 4

Construct a quadrilateral ABCD, when:
AB = 3.8 cm, AC = 4.8 cm, AD = 2.8 cm, angle A = 105° and angle B = 60°.

Sol:

Steps:
1. Draw AD = 2.8 cm
2. Draw AB = 3.8 cm and ∠A = 105°.
3. Draw BP such that ∠B = 60°.
4. With A as a center and radii 4.8 cm draw an arc cutting BP at C.
5. Join AC, AD.

ABCD is the required quadrilateral.

Question 5

Construct a quadrilateral ABCD, when:
BC = 7.5 cm AC = 5.8 cm, AD = 3.6 cm, CD = 4.2 cm and angle A = 120°.

Sol:

Steps:
1. Draw AD = 3.6 cm
2. draw AP such that ∠A = 120°.
3. With A and D as a center and radii 5.8 cm and 4.2 cm draw arcs cutting each other at C.
4. Now join AC and CD.
5. Now with C as center and radii 7.5 cm draw an arc at B on AP.
6. Now join CB.

ABCD is the required quadrilateral.

Question 6

Construct a quadrilateral ABCD, when:
AD = AB = 4 cm, BC = 2.8 cm, CD = 2.5 cm and angle BAD = 45°.

Sol:

Steps:
1. Draw AD = 4cm.
2. Draw AP such that ∠A = 45°.
3. With A as a center with radii, 4 cm draw an arc at B on AP.
4. Now taking B and D as a center and radii 2.8 cm and 2.8 cm draw arcs cutting each other at C.
5. Now join BC and CD.

ABCD is the required quadrilateral.

Question 7

Construct a quadrilateral ABCD, when:
AB = 6.3 cm, BC = CD = 4.2 cm and ∠ABC = ∠BCD = 90°.

Sol:

Steps :
1. draw AB = 6.3 cm.
2. Draw BP such that ∠ABP = 90°.
3. With B as a center and radii, 4.2 cm draw an arc AP at C.
4. With C as a center draw a line CD with radii 4.2 cm. draw a line such that ∠BCD = 90°.
5. Now join AD.

ABCD is the required quadrilateral.

Question 8

Construct a parallelogram ABCD, when:
AB = 4.4 cm, AD = 6.2 cm and AC = 4.8 cm.

Sol:

Steps:
1. Draw AD = 6.2 cm.
2. Draw triangle ACD.
3. Then draw triangle ABC.

ABCD is the required parallelogram.

Question 9

Construct a parallelogram ABCD, when:
Diagonal AC = 6.4 cm, diagonals BD = 8.2 cm and angle between the diagonals = 60°.

Sol:

Steps:
1. Draw AC = 6.4 cm.
2. Draw line BOD such that ∠DOC = 60° and
OB = OD = ${\displaystyle \frac{1}{2}\text{BD}=\frac{1}{2}}$ x 8.2 = 4.1 cm
3. Join AB, BC, CD, and DA.

ABCD is the required parallelogram.

Question 10

Construct a parallelogram ABCD, when:
AB = 5.8 cm, diagonal AC = 8.2 cm and diagonal BD = 6.2 cm.

Sol:

Steps:
1. Since diagonal of a parallelogram bisect each other, construct OAB such that;

OA = ${\displaystyle \frac{1}{2}\text{AC}=\frac{1}{2}}$ x 8.2 cm = 4.1 cm.

OB = ${\displaystyle \frac{1}{2}\text{BD}=\frac{1}{2}}$ x 6.2 cm = 3.1 cm.
and AB = 5.8 cm.

2. Produce AO up to C, such that OC = OA = 4.1 cm and BO up to D, such that DO = OB = 3.1 cm.
3. Join AD, DC, and CB.

ABCD is the required parallelogram.

Question 11

Construct a parallelogram ABCD, when:
AB = 6.0 cm, AD = 5.0 cm and ∠A = 45°.

Sol:

Steps:
1. Draw AB = 6 cm.
2. Draw AD with radii 5 cm with an angle of 45°.
3. With D and B as a center and radii 6 cm and 5 cm draw arcs cutting each other at C.
4. Now join DC and BC.

ABCD is the required parallelogram.

Question 12

Construct a parallelogram ABCD, when:
Base AB = 6.5 cm, BC = 4 cm and the altitude corresponding to AB = 3.1 cm.

Sol:

Steps :
1. Draw AB = 6.5 cm.
2. At B, draw BP ⊥ AB.
3. From BP cut BE = 3.1 cm.
4. Through E draw a perpendicular to BP to get QR parallel to AB.
5. With B as a center and radius = AC = 4cm, draw an arc that cuts QR at C.
6. With A as a center and radius = AD = 4 cm, draw an arc that cuts QR at D.

ABCD is the required parallelogram.

Question 13

Construct a parallelogram ABCD, when:
AB = 4.5 cm, ∠B = 120° and the distance between AB and DC = 3.0 cm.

Sol:

Steps:
1. Draw AB = 4.5 cm.
2. At B, draw BP ⊥ AB.
3. From BP cut BE = 3 cm.
4. Through E draw a perpendicular to BP to get QR parallel to AB.
5. With B as a center draw an arc that cuts QR at C.
6. With A as a center draw an arc that cuts QR at D.
7. Now join AD and BC.

ABCD is the required parallelogram.

Question 14

Construct a parallelogram ABCD, when:
Base BC = 5.6 cm, diagonal BD = 6.5 cm and altitude = 3.2 cm.

Sol:

Steps:
1. Draw BC = 5.6 cm.
2. At C, draw CX perpendicular to BC.
3. with C as a centre and taking radius 3.2 cm to draw an arc to cut CX at Y.
4. Through Y draw a straight line PQ parallel to BC.
5. With B as a centre and radius, 6.5 cm draw an arc to meet PQ at D.
6. With D as a centre and radius equal to 5.6 cm, draw an arc to meet PQ at A.
7. Join BA, BD and CD.

ABCD is the required parallelogram.

Question 15

Construct a rectangle ABCD, when:
Its sides are 6.0 cm and 7.2 cm.

Sol:

Since each angle of a rectangle is 90° and the opposite sides are equal. Therefore,
Steps:
1. Draw BC = 7.2 cm.
2. With B as a centre draw a line BX taking as a 90°.
3. Now taking radius 6 cm to draw an arc at A.
4. From the point, A draw a line AY parallel to BC.
5. With A as a center taking radius 7.2 cm to draw an arc at D.
6. Now join CD.

ABCD is the required rectangle.

Question 16

Construct a rectangle ABCD, when:
One side = 4 cm and one diagonal is 5 cm. Measure the length of the other side.

Sol:

Steps:
1. Draw BC = 4 cm.
2. With C as a center and radius 5 cm draw an arc at A.
3. Now join AB and AC.
4. With A as a center draw an arc at D.
5. Now join AD and CD.

ABCD is the required rectangle.

Question 17

Construct a rectangle ABCD, when:
One diagonal = 6.0 cm and the acute angle between the diagonals = 45^o.

Sol:

Steps:
1. Draw AC = 6cm.
2. Draw the right triangle ACB.
3. Draw the right triangle ADB.
4. Join DC.

ABCD is the required rectangle.

Question 18

Construct a rectangle ABCD, when:
Area = 24 cm² and base = 4.8 cm².

Sol:

Given that the base = 4.8 cm² and Area = 24 cm²
We know that the area of the rectangle = base x Height.
Therefore,
24 = 4.8 x height
Height = 5

With base = 4.8 cm² and height = 5 cm²,
the rectangle is shown below:

Steps:
1. Draw base AB = 4.8 cm².
2. With A and B as a center draw an arc taking radius 5 cm² at D and C.
3. Now join AD, BC, and DC.

ABCD is the required rectangle.

Question 19

Construct a rectangle ABCD, when:
Area = 36 cm² and height = 4.5 cm.

Sol:

Given that the height = 4.5 cm and Area = 36 cm²
We know that the area of rectangle = base x Height
Therefore,
36 = base x 4.5
Base = 8 cm
With height = 4.5 and base = 8 cm, the rectangle is shown below:
Steps:
1. Draw base AB = 8 cm.
2. With A and B as a center draw arc taking radius = 4.5 cm at D and C.
3. Now join AD, BC, and DC.

ABCD is the required rectangle.

Question 20

Construct a trapezium ABCD, when:
AB = 4.8 cm, BC = 6.8 cm, CD = 5.4 cm, angle B = 60^o and AD // BC.

Sol:

Steps:
1. Draw BC = 6.8 cm.
2. With B as a center and radii 4.8 cm draw an arc at A such that ∠B = 60°.
3. From point draw a line AP such that AP || BC.
4. With C as a center and radii 5.4 cm draw an arc at D on the line AP.
5. Now join AB, CD.

ABCD is the required trapezium.

Question 21

Construct a trapezium ABCD, when:
AB = CD = 3.2 cm, BC = 6.0 cm, AD = 4.4 cm and AD // BC.

Sol:

Steps:
1. Draw BC = 6 cm.
2. From BC cut BE = AD = 4.1 cm.
3. Draw triangle DEC such that DE = AB = 3.2 cm and CD = 3.2cm.
4. Taking B and D as a center and radii 3.2 cm and 4.1 cm respectively, draw arcs cutting each other at A.
5. Join AB and AD.

ABCD is the required trapezium.

Question 22

Construct a rhombus ABCD, when:
Its one side = 6 cm and ∠A = 60^o.

Sol:

Steps:
1. Draw a line AB = 6 cm.
2. At A, we construct ∠BAP = 60°. 
3. From AP, we cut at D taking AD = 6 cm.
4. Through B, we draw BQ || AD.
5. Through D, we draw DC || AB to cut BQ at C.

ABCD is the required rhombus.

Question 23

Construct a rhombus ABCD, when:
One side = 5.4 cm and one diagonal is 7.0 cm.

Sol:

Steps:
1. We construct the segment AC = 7 cm.
2. With A as a centre and radius 5.4 cm, we draw an arc extending on both sides of AC.
3. With C as a centre and same radius as in step 2, we draw an arc extending on both sides of AC to cut the first arc at B and D.
4. Join AB, BC, CD and DA.

ABCD is the required rhombus.

Question 24

Construct a rhombus ABCD, when:
Diagonal AC = 6.3 cm and diagonal BD = 5.8 cm.

Sol:

Steps:
1. Draw AC = 6.3 cm.
2. Draw perpendicular bisector to AC which cuts AC at O.
3. From this perpendicular cut OD and OB such that,

OD = OB = ${\displaystyle \frac{1}{2}\text{BD}=\frac{1}{2}}$ x 5.8 = 2.9 cm.

4. Join AB, BC, CD, and DA.

ABCD is the required rhombus.

Question 25

Construct a rhombus ABCD, when:
One side = 5.0 cm and height = 2.6 cm.

Sol:

Steps:
1. Draw AB = 5 cm.
2. At B, draw BP ⊥ AB.
3. From BP, cut BE = 2.6 cm = height.
4. Through E draw a perpendicular to CP to get QR parallel to AB.

With A and B as a centre and radii 5 cm draw arcs cutting QR at D andC.

ABCD is the required rhombus.

Question 26

Construct a rhombus ABCD, when: A = 60° and height = 3.0 cm.

Sol:

Steps:
1. Draw a line AP.
2. Now draw a line AF such that ∠A = 60°.
3. At S draw a perpendicular SE of length 3 cm such that it cut at AF at D.
4. Through D draw a line QR parallel to AP.
5. Now taking the radius same as AD draws an arc at B on AP.
6. Now through and B taking radius same as AD and AB draw arcs cutting each other at C.
7. Now join BC.

ABCD is the required rhombus.

Question 27

Construct a rhombus ABCD, when: Diagonal AC = 6.0 cm and height = 3.5 cm.

SoL:

Steps:

1. draw a line AP.

2. now draw AC = 6cm and CP = 3.5cm.

3. Now draw a line BC such that AB = BC.

4. Now at C draw a line CY parallel to AP.

5. At point C and A, taking radius same as AB draw arcs cutting each other at D.

6. Now join AD.

ABCD is the required rhombus.

Question 28

Construct a square ABCD, when: One side = 4.5 cm.

Sol:

Steps:

1. Draw a line segment AB = 4.5cm

2. Draw AP ⊥ AB.

3. From AP cut off AD = 4.5cm.

4. With B as a center and radius 4.5 cm draw an arc.

5. With D as center and radius 4.5 cm draw another arc cutting the former arc at C.

6. Join BC and CD.

ABCD is the required square.

Question 29

Construct a square ABCD, when: One diagonal = 5.4 cm.

Sol:

We know that the diagonals of a square are equal and bisect each other at right angles.

Steps:

1. draw AC = 5.4cm.

2. Draw the right bisector XY of AC, meeting AC at O.

3. From O, set off OB = ${\displaystyle \frac{1}{2}}$ (5.4) = 2.7cm along OY and along OX.

4. Join AB, BC, CD, and DA.

ABCD is the required square. 

Question 30

Construct a square ABCD, when: Perimeter = 24 cm.

Sol:

The perimeter of a square
P = 4a
Where a is the length of each side
We have Perimeter = 24 cm
Therefore,
24 = 4a 
4 =6
Therefore the sides of the squares are of length 6 cm. 

Steps:

1. Draw a line segment Ab = 6cm.

2. Draw AP ⊥ AB.

3. From AP cut off AD = 6cm.

4. With B as a center and radius, 6 cm draw an arc.

5. With D as center and radius, 6 cm draw another arc cutting the former arc at C.

6. Join BC and CD.

ABCD is the required square.

Question 31

Construct a rhombus, having given one side = 4.8 cm and one angle = 75^o.

Sol:

Steps:

1.draw a line AB = 4.8cm

2. At A Draw AX such that ∠BAX = 75°.

3. With A as a center and measurement equal to AB cut off an arc at D on AX.

4. Using the same radius taking D and B as centers cut off arcs, which will intersect at C.

5. Join CD and CB.

ABCD is the required rhombus.

Question 32.1

Construct a regular hexagon of side 2.5 cm

Sol:

The length of the side of a regular hexagon is equal to the radius of its circumcircle.

Steps of construction:

1. Draw a circle of radius 2.5 cm

2. Taking any point A on the circumference of the circle as the centre, draw arcs of same radii (i.e. 2.5 cm) which cut the circumference at B and F.

3. With B and F as centres, again draw two arcs of same radii which cut the circumference at C and E respectively.

4. With C or E as the centre, draw one more arc of the same radius which cuts the circumference at point D.

In this way, the circumference of the circle is divided into six equal parts.

5. Join AB, BC, CD, DE, EF and FA.

ABCDEF is the required regular hexagon.

Question 32.2

Construct a regular hexagon of side 3.2 cm

Sol:

The length of the side of a regular hexagon is equal to the radius of its circumcircle.

Steps of construction:

1. Draw a circle of radius 3.2 cm

2. Taking any point A on the circumference of the circle as a centre, draw arcs of same radii (i.e. 3.2 cm) which cut the circumference at B and F.

3. With B and F as centres, again draw two arcs of same radii which cut the circumference at C and E respectively.

4. With C or E as a centre, draw one more arc of the same radius which cuts the circumference at point D.
In this way, the circumference of the circle is divided into six equal parts.

5. Join AB, BC, CD, DE, EF and FA.

ABCDEF is the required regular hexagon. 

Question 33

Using ruler and compasses only, construct the quadrilateral ABCD, having given AB = 5 cm, BC = 2.5 cm, CD = 6 cm. angle BAD = 90^o and the diagonal AC = 5.5 cm.

Sol:

Steps:

1. draw AB = 5cm.

2. Now draw ∠XAB such that it is 90°.

3. Taking A and B as a center and radius 2.5 cm and 5.5 cm draw arcs cut off at C.

4. Now join BC and AC.

5. Taking C as a center and radius 6 cm draw arcs at D on AX.

ABCD is the required quadrilateral.

Question 34

Using ruler and compasses only, construct a trapezium ABCD, in which the parallel sides AB and DC are 3.3 cm apart; AB = 4.5 cm, angle A = 120^o BC = 3.6 cm and angle B is obtuse.

Sol:

Steps:
1. Draw AB = 4.5cm.

2. now draw ∠BAS = 120° and draw EA ⊥ AB such that AX = 3.3cm.

3. Through X draws a line QR which is parallel to AB which cuts AS at D.
4. Through B draw an arc taking radius 3.6 cm at C on PQ.

5. Join CB.

ABCD is the required trapezium. 

Question 35

Using ruler and compasses only, construct the quadrilateral ABCD, having given AB = 5 cm, BC = 2.5 cm CD = 6 cm, ∠BAD = 90^o and diagonal BD = 5.5 cm.

Sol:

Steps:

1.Draw AB = 5cm.

2. From A draw a line AY such that ∠A =90°.

3. Taking B as a center with radius 5.5 cm draw an arc at D on AY.

4. With D and B as center and radii, 6 cm and 2.5 cm draw arcs cutting each other at C.

5. Join DC and BC.

ABCD is the required quadrilateral.

Question 36

Using ruler and compasses only, construct a parallelogram ABCD using the following data: AB = 6 cm, AD = 3 cm and ∠DAB = 45^o. If the bisector of ∠DAB meets DC at P,
prove that ∠APB is a right angle.

Sol:

Steps:

1. draw AB = 6cm.

2. With A as a center draw a line AX such that ∠BAX = 45°.

3. With A as a center and radii, 3 cm draw an arc on AD.

4. now with D and B as a center and radii 6 cm and 3 cm draw arcs cutting each other at C.

5. Join DC and BC.

ABCD is the required parallelogram.
Here
∠PAB = ∠APD                ...[ Alternate angles ]
∠CPB = ∠PBA                 ...[ Alternate angles ] 

Now,
∠DPA + ∠APB + ∠CPB = 180° …… (i)

Also, considering APB,
∠PAB + ∠PBA + ∠APB = 180° …… (ii)

Therefore, from (i) and (ii)
∠APB = 90°
Hence proved.

Question 37

The perpendicular distance between the pair of opposite sides of a parallelogram are 3 cm and 4 cm, and one of its angles measures 60^o. Using a ruler and compasses only,
construct the parallelogram.

Sol:

Steps:
1. Draw a baseline AQ.

2. A take some random distance in compass and draw one are below and above the line. Now without changing the distance in compass draw one are below and above the line. These arcs intersect each other above and below the line. Draw the line passing through these intersecting points, you will get perpendicular to the line AQ.

3. Take distance of 4 cm in compass and mark an arc on the perpendicular above the line. Draw a line parallel to line AQ passing through this arc.

4. From point A measure an angle of 60 degrees and draw the line which intersects the above-drawn line at some point label it as D.

5. Using the procedure given in step 2 again draw a perpendicular to line AD.

6. Take distance of 3 cm in compass and mark an arc on the perpendicular above the line. Draw a line parallel to line AD passing through this arc which intersects the line AQ at some point label it as B and to other lines at point C.

ABCD is the required parallelogram.  

Question 38

Draw parallelogram ABCD with the following data:
AB = 6 cm, AD = 5 cm and ∠DAB = 45^o.
Let AC and DB meet in O and let E be the mid-point of BC. Join OE.
Prove that:
(i) OE // AB
(ii) OE = ${\displaystyle \frac{1}{2}}$ AB.

Sol:

To draw the parallelogram follows the steps:
1. First, draw a line AB of measure 6cm. Then draw an angle of measure 45° at point A such that ∠DAB = 45° and AD = 5cm.

2. Now draw a line CD parallel to the line AB of measure 6cm. Then join BC to construct the parallelogram as shown below: 

3. Now it is given that E is the mid-point of BC. We join OE. Now we are to prove that OE || AB and  OE = ${\displaystyle \frac{1}{2}}$ AB.

4. Since O is the mid-point of AC and E is the midpoint of BC, therefore the line is parallel to AB and  OE = ${\displaystyle \frac{1}{2}}$ AB

Question 39

Using ruler and compasses only, construct a rectangle each of whose diagonals measure 6 cm and the diagonals interest at an angle of 45^o.

Sol:

To draw the rectangle follows the steps:
1. First draw a line AC of measure 6cm.
2. Then draw the perpendicular bisector of AC  through O.
3. At O draw an angle of measure 45°. Then produce OD  of measure 3cm and OB of measure 3cm each. 
4. Now join AD, AB, BC,  and CD to form the rectangle. 

S.chand Class 8 Maths Solution Chapter 15 Visualising Solid Shapes Exercise 15

 Exercise 15

Question 1

<fig>

1. The adjoining net is made up of two equilateral triangles and three rectangles.

(i) Name the solid it represents.

(ii) How many faces, edges and vertices does this solid have ?

2. For each solid, count the number of faces, vertices and edges. Check if the Euler's rule is true for each one.

<fig>

3. Name the solids that have :

(i) 4 faces

(ii) 1 curved surface

(iii) 6 faces

(iv) 5 faces and 5 vertices

(v) 8 triangular faces

(vi) 6 rectangular faces and 2 hexagonal faces.

4. Name the different plane shapes needed to draw the net of:

(i) a cube

(ii) a triangular prism

(iii) a triangular pyramid

(iv) a cylinder

5.Given here is the net of an octahedron with numbers 0 to 7 on the faces.Enlarge these on a cardboard to make a model

(i) How many faces, edges and vertices does this solid have ?

(ii) Check whether Euler's rule is true for the octahedron or not.

6. Draw the 2-D representation of a

<fig>

7. On an isometric dotted paper draw the following shapes:

(i) cube

(ii) cuboid

(iii) hexagonal prism

(iv) pentagonal prism

8. Name the solid that would be formed by each net :

<fig>

Multiple Choice Questions (MCQs)

9. Which of the following solid shapes has the side

<fig>

10. Manya is building prism using straws and balls of clay. How many straws does she need to build a pentagonal prism ?

(a) 18

(b) 15

(c) 20

(d) 13

High Order Thinking Skills (HOTS)

11. How many triangles will the net of hexagonal pyramid contain ?

RS Aggarwal solution class 8 chapter 15 Quadrilaterals Exercise 15

Exercise 15

Page-185

Q1 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 1:

Fill in the blanks:
(i) A quadrilateral has ......... sides.
(ii) A quadrilateral has ......... angles.
(iii) A quadrilateral has ......... vertices, no three of which are .........
(iv) A quadrilateral has ......... diagonals.
(v) A diagonal of a quadrilateral is a line segment that joins two ......... vertices of the quadrilateral.
(vi) The sum of the angles of a quadrilateral is.........

Answer 1:

(i) 4
(ii) 4
(iii) 4, co-linear
(iv) 2
(v) opposite
(vi) 360°

Page-186

Q2 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 2:

In the adjoining figure, ABCD is a quadrilateral.
(i) How many pairs of adjacent sides are there? Name them.
(ii) How many pairs of opposite sides are there? Name them.
(iii) How many pairs of adjacent angles are there? Name them.
(iv) How many pairs of opposite angles are there? Name them.
(v) How many diagonals are there? Name them.

Answer 2:

(i) There are four pairs of adjacent sides, namely (AB,BC), (BC,CD), (CD,DA) and (DA,AB).
(ii) There are two pairs of opposite sides, namely (AB,DC) and (AD,BC).
(iii) There are four pairs of adjacent angles, namely $\left(\angle A,\angle B\right), \left(\angle B, \angle C\right), \left(\angle C,\angle D\right) and \left(\angle D,\angle A\right)$.
(iv) There are two pairs of opposite angles, namely $\left(\angle A,\angle C\right) and \left(\angle B,\angle D\right)$.
(v) There are two diagonals, namely AC and BD.

Q3 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 3:

Prove that the sum of the angles of a quadrilateral is 360°.

Answer 3:

      

Let ABCD be a quadrilateral.
Join A and C.

Now, we know that the sum of the angles of a triangle is 180°.

For $$\begin{gathered} △ABC: \\ \angle 2+\angle 4+\angle B={180}^o ... \left(1\right) \end{gathered}$$

For $$\begin{gathered} △ADC: \\ \angle 1+\angle 3+\angle D={180}^o ... \left(2\right) \end{gathered}$$

Adding (1) and (2):
$\left(\angle 1+\angle 2+\angle 3+\angle 4\right)+\angle B+\angle D={360}^o$

or $\angle A+\angle B+\angle C+\angle D={360}^o$

Hence, the sum of all the angles of a quadrilateral is 360°.

Q4 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 4:

The three angles of a quadrilateral are 76°, 54° and 108°. Find the measure of the fourth angle.

Answer 4:

Sum of all the four angles of a quadrilateral is 360°.



The fourth angle measures 122°.

Q5 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 5:

The angles of a quadrilateral are in the ratio 3 : 5 : 7 : 9. Find the measure of each of these angles.

Answer 5:

​

Q6 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 6:

A quadrilateral has three acute angles, each measuring 75°. Find the measure of the fourth angle.

Answer 6:

Sum of the four angles of a quadrilateral is 360°.

If the unknown angle is $x$°, then:



The fourth angle measures 135°.

Q7 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 7:

Three angles of a quadrilateral are equal and the measure of the fourth angle is 120°. Find the measure of each of the equal angles.

Answer 7:

Let the three angles measure $x°$ each.

Sum of all the angles of a quadrilateral is 360°.



Each of the equal angles measure 80°.

Q8 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 8:

Two angles of a quadrilateral measure 85° and 75° respectively. The other two angles are equal. Find the measure of each of these equal angles.

Answer 8:

Let the two unknown angles measure $x°$ each.
Sum of the angles of a quadrilateral is 360°.

 $$\begin{gathered} \therefore 85+75+x+x=360 \\ 160+2x=360 \\ 2x=360-160=200 \\ x=100 \end{gathered}$$

Each of the equal angle measures 100°.

Q9 | Ex-15 | Class 8 | RS AGGARWAL | chapter 15 | Quadrilaterals 

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Question 9:

In the adjacent figure, the bisectors of ∠A and ∠B meet in a point P. If ∠C = 100° and ∠D = 60°, find the measure of ∠APB.

Answer 9:

Sum of the angles of a quadrilateral is 360°.


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