Concise Mathsclass 6 Answers Archives

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry

November 10, 2023May 18, 2022 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry (Including Constructions on Symmetry)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry (Including Constructions on Symmetry)

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 6 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 6 with Free PDF download option. Selina Publishers Concise Mathematics for Class 6 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Concept of Symmetry (Linear Symmetry) : Consider a plane mirror mm’. If your face F is at a distance ‘d’ before the mirror ; the image F, of your face is also formed behind the mirror at distance “d” from it.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 1
Now, if the Figure 1, given alongside, is folded about the mirror mm’, the object F and its image F’ coincide. Since on folding the figure about mm’ its two parts exactly coincide, we say that the figure is symmetrical about the mirror line mm’. For the reason, the mirror line mm’ is called line of symmetry of the whole figure including face F, its image F’ and the mirror mm’.
Keep in Mind :
in order to check, whether a given figure is symmetrical about a line in it or not; fold the figure about that line. If part of the figure, which lies on one side of the line exactly coincides with the part of the figure on the other side of the line, then the figure is symmetrical about that line.
For Example :
(i) A line segment is symmetrical about its perpendicular bisector
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 2
(ii) An angle (with equal arms) is symmetrical about its bisector

(iii) An isosceles triangle has only one line of symmetry. The line of symmetry is the bisectors of the angle contained by equal sides.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 4
(iv) An equilateral triangle has three lines of symmetry. The bisectors of the angles are the lines of symmetry.

(v) A kite shaped figure has only one line of symmetry.

(vi) A figure of the shape of an arrow-head has only one line of symmetry.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 7
(vii) The letter ‘A’ has only one line of symmetry.

(viii) The letter ‘H’ has two lines of symmetry.

(ix) A circle has an infinite number of lines of symmetry.
Every line, which passes through the centre of the circle, is a line of symmetry.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 10

Revision Exercise Symmetry Exercise 30 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State, whether true or false :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 11
Solution:
(If both the semi-circles of B are equal)
(i) True
(ii) True
(iii) False
(iv) False
(v) False
(vi) True
(vii) True
(viii) False.

Question 2.
Construct a triangle ABC, in which AB = AC = 5 cm and BC 6 cm. Draw all its lines of Symmetry.
Solution:
Steps of Construction :
(i) Draw a line BC = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 12
(ii) With B as centre and radius 5 cm draw
(iii) With C as centre and radius 5 cm draw on arc which cut the previous arc at a.
(iv) Join AB and AC.
(v) ∆ABC is the required triangle line of symmetry is shown in the figure.

Question 3.
Examine each of the following figures carefully, draw line(s) of symmetry in which ever figure possible :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 13
Solution:

Question 4.
Construct a triangle XYZ, in which XY = YZ = ZX = 4.5 cm. Draw all its lines of symmetry.
Solution:
Steps of Construction :
(i) Draw line XY = 4.5 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 16
(ii) With X as a centre and radius 4.5 cm draw an arc at Z.
(iii) With Y as a centre and radius 4.5 cm draw an arc which cuts the previous arc at Z.
(iv) Join XZ and YZ.
∆XYZ is the required triangle and lines of symmetries are shown in the figure.

Question 5.
Construct a triangle ABC, in which AB = BC = 4 cm and ∠ABC = 60°. Draw all its lines of symmetry.
Solution:
Steps of Construction :
(i) Draw a line AB = 4 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 17
(ii) At B draw an angle of 60° with the help of compass.
(iii) With B as centre and radius upon cut BC = 4 cm.
(iv) Join AC.
∆ABC is the required triangle and line of symmetry is shown in centre.

Question 6.
Draw the line(s) of symmetry for each figure drawn below :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 18
Solution:

Question 7.
In each of the following case, construct a point that is symmetric to the given point P with respect to the given line AB.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 30 Revision Exercise Symmetry image - 20
Solution:

(i) Draw PQ perpendicular to AB from PO produced cut OQ = OP.
Point Q is symmetric to the given point P with respect to the given line AB.
(ii) Draw PO perpendicular to AB from P and produce PO to Q such that OQ = PO.
Point Q is symmetric to the given point P with respect to the given line AB.
(iii) Draw PQ perpendicular to AB from PO produced cut OQ = OP.
Point Q is symmetric to the given point P with respect to the given line AB.

Question 8.
Mark two points A and B 5.5 cm apart. Draw a line PQ so that A and B are symmetric with respect to the line PQ. Give a special name to line PQ.
Solution:
Steps of Construction :

Take points A and B 5.5 cm apart draw perpendicular bisect of the line segment AB. The perpendicular bisector PQ is the required line of symmetry with respect to A and B.
PQ is the perpendicular bisector of AB.

Question 9.
For each letter of the English alphabet, draw the maximum possible number of lines of symmetry.
Solution:

Question 10.
Draw all the possible lines of symmetry for each figure given below :

Solution:
The line or lines of symmetry of tile given figure are drawn as given below :
(i) A parallelogram has no line of symmetry.
(ii) A rectangle has two lines of symmetry.
(iii) A square has form lines of symmetry.

(iv) A semicircle has one line of symmetry.
(v) A quadrant has one line of symmetry.

Question 11.
For each shaded portion given below, draw all the possible lines of symmetry :
Solution:
The shaded portion of each figure given has the line or lines of symmetry as given below

(i) A minor segment has one line of symmetry.
(ii) A major segment has one line of symmetry.
(iii) quadrant has one line of symmetry.
(iv) The given figure has one line of symmetry.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle

November 10, 2023May 18, 2022 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. A circle is a round enclosed figure, whose mid-point is called its centre.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -1
2. The line segment joining the centre to any point on the circle is called a radius. A centre has infinite radii and all radii of a circle are equal.
3. A line segment which contains the centre of the circle and whose ends points lie on the circle is called diameter of the circle. Diameters of a circle are also equal.
4. Parts of a circle: A circle has three parts (i) Interior (ii) Exterior and (iii) Circle itself.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -2
5. Concentric circles: Two or more circles having the same centre but different radii are called concentric circles.
6. Chord of a circle: A line which divides the circle into two parts is called chord of the circle. Diameter is the longest chord of the circle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -3
7. Segment of a circle: When a chord divides the circle into two unequal parts, the bigger part is called the major segment and smaller part is called the minor segment.
8. Arc of a circle: A part of circumference of a circle is called an arc of the circle. Arc greater than half circle is called the major arc and less than half circle is called the minor arc.
9. Sector of a circle: A diameter divides the circle into two equal parts and each part is called a semicircle. Sector greater than a semi-circle is called the major sector and less than semi-circle is called the minor sector of the circle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -4

The Circle Exercise 29A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Use the figure given below to fill in the blanks :
(i) R is the …… of the circle.
(ii) Diameter of a circle is …… .
(iii) Tangent to a circle is … .
(iv) EF is a …… of the circle.
(v) …… is a chord of the circle.
(vi) Diameter = 2 x …..
(vii) ……. is a radius of the circle.
(viii) If the length of RS is 5 cm, the length of PQ = ……
(ix) If PQ is 8 cm long, the length of RS =…..
(x) AB is a ….. of the circle
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -5
Solution:
(i) center
(ii) PQ
(iii)AB
(iv) secant
(v) CD
(vi) radius
(vii) RS
(viii) 10 cm
(ix) 4 cm
(x) tangent.

Question 2.
Draw a circle of radius 4.2 cm. Mark its centre as O. Take a point A on the circumference of the circle. Join AO and extend it till it meets point B on the circumference of the circle,
(i) Measure the length of AB.
(ii) Assign a special name to AB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -6
(i) By measurement AB = 8.4 cm.
(ii) ∴ AB is the diameter of the circle.

Question 3.
Draw circle with diameter :
(i) 6 cm
(ii) 8.4 cm.
In each case, measure the length of the radius of the circle drawn.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -7

Question 4.
Draw a circle of radius 6 cm. In the circle, draw a chord AB = 6 cm.
(i) If O is the centre of the circle, join OA and OB.
(ii) Assign a special name to ∆AOB
(iii) Write the measure of angle AOB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -8

Question 5.
Draw a circle of radius 4.8 cm and mark its centre as P.
(i) Draw radii PA and PB such that ∠APB = 45°.
(ii) Shade the major sector of the circle
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -9
PA is the radius of circle. i.c., PA = 4.8 cm.
(i) ∠APB = 45° in which P is the centre of the circle and PA and PB are radii of circle.
(ii) Major sector of circle is shaded in the figure.

Question 6.
Draw a circle of radius 3.6 cm. In the circle, draw a chord AB = 5 cm. Now shade the minor segment of the circle.
Solution:
Shaded portion of circle is the minor segment of the circle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -10

Question 7.
Mark two points A and B ,4cm a part, Draw a circle passing through B and with A as a center
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -11
Solution:
In the figure, A is the centre of the circle and AB = 4 cm [radius of circle]

Question 8.
Draw a line AB = 8.4 cm. Now draw a circle with AB as diameter. Mark a point C on the circumference of the circle. Measure angle ACB.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -12
By measurement ∠ACB =90

The Circle Exercise 29B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Construct a triangle ABC with AB = 4.2 cm, BC = 6 cm and AC = 5cm. Construct the circumcircle of the triangle drawn.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -13
(i) Draw ∆ABC in which AB = 4.2 cm. BC = 6 cm. and AC = 5 cm.
(ii) Draw the perpendicular bisectors of any two sides of the triangle. Let these intersect at O.
(iii) Taking O as centre and OA or OB or OC as radius draw a circle.
This circle will pass through vertices A, B and C.

Question 2.
Construct a triangle PQR with QR = 5.5 cm, ∠Q = 60° and angle R = 45°.
Construct the circumcircle cif the triangle PQR.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -14
(i) Draw a ∆PQR in which QR = 5.5 cm, ∠Q = 60° and ∠R = 45°.
(ii) Draw the arc bisector of PQ and PR which intersect at O.
(iii) Taking O as centre and radius OP or OQ or OR draw a circle.
This circle will pass through vertices P, Q and R.

Question 3.
Construct a triangle ABC with AB = 5 cm, ∠B = 60° and BC = 6. 4 cm.
Draw the incircle of the triangle ABC. Sol. Steps of Construction :
Solution:
Steps of Construction:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -15
(i) Draw a line AB = 5 cm.
(ii)B as a centre draw an angle with the help of compass ∠B = 60°. Cut the line with an arc BC = 6 4 cm.
(iii) Join AC.
(iv) Now, from A and B cut the bisector of ∠A and ∠B, which intersect each other at point D.
(v) With D as a centre draw an incircle which touches all the three sides of AABC.

Question 4.
Construct a triangle XYZ in which XY = YZ= 4.5 cm and ZX = 5.4 cm. Draw the circumcircle of the triangle and measure its circumradius.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -16
(i) Draw a triangle XYZ in which XY = YZ = 4.5 cm and ZX = 5.4 cm.
(ii) Draw the bisectors of XZ and YZ which meet at O.
(iii) With O as centre and radius OX or OY or OZ draw a circle.
This circle will pass through X, Y and Z.

Question 5.
Construct a triangle PQR in which, PQ = QR = RP = 5.7 cm. Draw the incircle of the triangle and measure its radius.
Solution:
Steps of Construction :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -17
(i) Draw an equilateral ∆ RPQ in which PQ = QR = RP = 5.7 cm each.
(ii) From P and Q cut the bisector of ∠P and ∠Q, which intersect each other at point O.
(iii) With P as a centre draw an incircle which touches all the three sides of ∆RPQ.

The Circle Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
The centre of a circle is at point O and its radius is 8 cm. State the position of a point P (point P may lie inside the circle, on the circumference of the circle, or outside the circle), when :
(a) OP = 10.6 cm
(b) OP = 6.8 cm
(c) OP = 8 cm
Solution:
(a) Draw circle each of radius 8 cm. With centre O
In figure (i) draw OP = 10.6 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -18
We see that point P lies outside the circle as OP > radius of the circle
(b) In figure (ii) OP = 6.8 cm. We see that P lies inside the circle as OP < radius of the circle.
(c) In figure, OP = 8 cm. We see that P lies on the circle as OP = radius of the circle.

Question 2.
The diameter of a circle is 12.6 cm. State, the length of its radius.
Solution:
Diameter of the circle = 12.6 cm
∴Radius = \(\frac { 1 }{ 2 }\) diameter = \(\frac { 1 }{ 2 }\) x 12.6 cm
= 6.3 cm

Question 3.
Can the length of a chord of a circle be greater than its diameter ? Explain.
Solution:
No, the length of chord cannot be greater than the diameter of the cirlce as the diameter of a circle is the greatest chord of that circle.

Question 4.
Draw a circle of diameter 7 cm. Draw two radii of this circle such that the angle between these radii is 90°. Shade the minor sector obtained. Write a special name for this sector.
Solution:
Draw a circle with diameter = 7 cm
OA and OB are the radii of the circle such that ∠AOB = 90°
Now shade the minor sector AOB This is the quadrant of the circle
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 29 The Circle IMAGE -19

Question 5.
State, which of following statements are true and which are false :
(i) If the end points A and B of the line segment lie on the circumference of a circle, AB is a diameter.
(ii) The longest chord of a circle is its diameter.
(iii) Every diameter bisects a circle and each part of the circle so obtained is a semi-circle.
(iv) The diameters of a circle always pass through the same point in the circle.
Solution:
(i) False, as AB may be diameter or may not be, it can be chord.
(ii) True, diameter of a circle is the longest chord.
(iii) True.
(iv) True, all the diameter of a circle pass through the same point i.e., centre, of the circle.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

November 10, 2023May 18, 2022 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

1. Polygon : A closed plane geometrical figure, bounded by atleast three line segments, is called a
polygon.
The adjoining figure is a polygon as it is :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 1
(i) Closed
(ii) bounded by five line segments AB, BC, CD, DE and AE.
Also, it is clear from the given polygon that:
(i) the line segments AB, BC, CD, DE and AE intersect at their end points.
(ii) two line segments, with a common vertex, are not collinear i.e. the angle at any vertex is not 180°.
A polygon is named according to the number of sides (line-segments) in it:

Note : No. of sides :	3	4	5	6
Name of polygon :	Triangle	Quadrilateral	Pentagon	Hexagon

2. Sum of Interior Angles of a Polygon
1. Triangle : Students already know that the sum of interior angles of a triangle is always 180°.
i.e. for ∆ ABC, ∠B AC + ∠ABC + ∠ACB = 180°
⇒ ZA + ZB + ZC = 180°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 2

2. Quadrilateral : Consider a quadrilateral ABCD as shown alongside.
If diagonal AC of the quadrilaterals drawn, the quadrilateral will be divided into two triangles ABC and ADC.
Since, the sum of interior angles of a triangle is 180°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 3
∴ In ∆ ABC, ∠ABC + ∠BAC +∠ACB = 180°
And, in ∆ ADC ∠DAC + ∠ADC + ∠ACD = 180°
Adding we get:
∠ABC + ∠BAC +∠ACB + ∠DAC + ∠ADC + ∠ACD = 180° +180°
⇒(∠BAC + ∠DAC) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC = 360°
⇒∠BAD + ∠ABC + ∠BCD + ∠ADC = 360°
⇒ ∠A + ∠B + ∠C + ∠D = 360°
Alternative method : On drawing the diagonal AC, the given quadrilateral is divided into two triangles. And, we know the sum of the interior angles of a triangle is 180°.
∴ Sum of interior angles of the quadrilateral ABCD
= Sum of interior angles of ∆ ABC + sum of interior angles of ∆ ADC = 180°+ 180° = 360°

3. Pentagon : Consider a pentagon ABCDE as shown alongside.
On joining CA and CE, the given pentagon is divided into three triangles ABC, CDE and ACE.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 4
Since, the sum of the interior angles of a triangle is 180°
Sum of the interior angles of the pentagon ABCDE = Sum of interior angles of (∆ ABC + ∆ CDE + ∆ACE)
= 180° + 180° + 180° = 540°

4. Hexagon :
It is clear from the given figure that the sum of the interior angles of the hexagon ABCDEF.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 5
= Sum of inteior angles of
(∆ABC + ∆ ACF + ∆ FCE + ∆ ECD)
= 180° + 180° + 180° + 180° = 720°

3. Using Formula : The sum of interior angles of a polygon can also be obtained by using the following formula:
Note : Sum of interior angles of a polygon = (n – 2) x 180°
where, n = number of sides of the polygon.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 6

Polygons Exercise 28A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
State, which of the following are polygons :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 7
Solution:
Only figure (ii) and (iii) are polygons.

Question 2.
Find the sum of interior angles of a polygon with :
(i) 9 sides
(ii) 13 sides
(iii) 16 sides
Solution:
(i) 9 sides
No. of sides n = 9
∴Sum of interior angles of polygon = (2n – 4) x 90°
= (2 x 9 – 4) x 90°
= 14 x 90°= 1260°
(ii) 13 sides
No. of sides n = 13
∴ Sum of interior angles of polygon = (2n – 4) x 90° = (2 x 13 – 4) x 90° = 1980°
(iii) 16 sides
No. of sides n = 16
∴ Sum of interior angles of polygon = (2n – 4) x 90°
= (2 x 16 – 4) x 90°
= (32 – 4) x 90° = 28 x 90°
= 2520°

Question 3.
Find the number of sides of a polygon, if the sum of its interior angles is :
(i) 1440°
(ii) 1620°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 8

Question 4.
Is it possible to have a polygon, whose sum of interior angles is 1030°.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 10

Question 5.
(i) If all the angles of a hexagon arc equal, find the measure of each angle.
(ii) If all the angles of an octagon are equal, find the measure of each angle,
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 12

Question 6.
One angle of a quadrilateral is 90° and all other angles are equal ; find each equal angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 13

Question 7.
If angles of quadrilateral are in the ratio 4 : 5 : 3 : 6 ; find each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 14

Question 8.
If one angle of a pentagon is 120° and each of the remaining four angles is x°, find the magnitude of x.
Solution:
One angle of a pentagon = 120°
Let remaining four angles be x, x, x and x
Their sum = 4x + 120°
But sum of all the interior angles of a pentagon = (2n – 4) x 90°
= (2 x 5 – 4) x 90° = 540°
= 3 x 180° = 540°
∴ 4x+120o° = 540°
4x = 540° – 120°
4x = 420
x = \(\frac { 420 }{ 4 }\) ⇒ x = 105°
∴Equal angles are 105° (Each)

Question 9.
The angles of a pentagon are in the ratio 5 : 4 : 5 : 7 : 6 ; find each angle of the pentagon.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 15

Question 10.
Two angles of a hexagon are 90° and 110°. If the remaining four angles arc equal, find each equal angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 16

Polygons Exercise 28B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
In case of regular polygon, with
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 17
Solution:

Question 2.
Find the number of sides in a regular polygon, if its each interior angle is :
(i) 160°
(ii) 150°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 21

Question 3.
Find number of sides in a regular polygon, if its each exterior angle is :
(i) 30°
(ii) 36°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 22

Question 4.
Is it possible to have a regular polygon whose each interior angle is :
(i) 135°
(ii) 155°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 24

Question 5.
Is it possible to have a regular polygon whose each exterior angle is :
(i) 100°
(ii) 36°
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 26

Question 6.
The ratio between the interior angle and the exterior angle of a regular polygon is 2 : 1. Find :
(i) each exterior angle of this polygon.
(ii) number of sides in the polygon.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 28 Polygons IMAGE - 28

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral

November 10, 2023May 18, 2022 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS
4.Quadrilateral: A quadrilateral is a plane figure enclosed by four sides. It has four sides, four interior angles and four vertices.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 1
In quadrilateral ABCD, shown alongside:
(i) four sides are : AB, BC, CD and DA.
(ii) four angles are : ∠ABC,∠BCD, ∠CDA and ∠DAB ; which are numbered∠1, ∠2, ∠3 and ∠4 respectively.
(iii) four vertices are : A, B, C and D.

5. Diagonals of a Quadrilateral : The line segments joining the opposite vertices of a quadrilateral are called its diagonals.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 2
The given figure shows a quadrilateral PQRS with diagonals PR and QS.
6. Types of Quadrilaterals :
1. Trapezium: A trapezium is a quadrilateral in which one pair of opposite sides are parallel.

The figure, given alongside, shows a trapezium as its sides AB and DC are parallel i.e. AB || DC.
When the non-parallel sides of the trapezium are equal in length, it is called an isosceles trapezium.
The given figure shows a trapezium ABCD whose non-parallel sides AD and BC are equal in length i.e. AD = BC; therefore it is an isosceles trapezium.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 4
Also, in an isosceles trapezium :

(i) base angles are equal:
i.e. ∠A = ∠B and ∠D =∠C
(ii) diagonals are equal
i.e. AC = BD.

2.Parallelogram : A parallelogram is a quadrilateral, in which both the pairs of opposite sides are parallel.
The quadrilateral ABCD, drawn alongside, is a parallelogram; since, AB is parallel to DC and AD is parallel to BC i.e.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 6
AB || DC and AD || BC.
Also, in a parallelogram ABCD:
(i) opposite sides are equal:
i.e. AB = DC and AD = BC.
(ii) opposite angles are equal:
i.e. ∠ABC = ∠ADC and ∠BCD = ∠BAD
(iii) diagonals bisect each other :
i.e. OA = OC = \(\frac { 1 }{ 2 }\) AC and OB = OD = \(\frac { 1 }{ 2 }\) BD.

7. Some special types of Parallelograms
(a) Rhombus : A rhombus is a parallelogram in which all its sides are equal.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 9
∴In a rhombus ABCD :
(i) opposite sides are parallel:
i.e. AB||DC and AD||BC.
(ii) all the sides are equal:
i.e. AB = BC = CD = DA.
(iii) opposite angles are equal:
i.e. ∠A = ∠C and ∠B = ∠D.
(iv) diagonals bisect each other at right angle :
i.e. OA= OC = \(\frac { 1 }{ 2 }\) AC ; OB = OD = \(\frac { 1 }{ 2 }\) BD.
and ∠AOB= ∠BOC = ∠COD = ∠AOD = 90°
(v) diagonals bisect the angles at the vertices :
i.e. ∠1 =∠2 ;∠3 = ∠4 ; ∠5 = ∠6 and ∠7 =∠8.

(b) Rectangle : A rectangle is a parallelogram whose any angle is 90°.
A rectangle is also defined as a quadrilateral whose each angle is 90°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 10
Note : If any angle of a parallelogram is 90° ; automatically its each angle is 90° ; the reason being that the opposite angles of a parallelogram are equal.
Also, in a rectangle:
(i) opposite sides are parallel.
(ii) opposite sides are equal.
(iii) each angle is 90°.
(iv) diagonals are equal.
(v) diagonals bisect each other.

(c) Square : A square is a parallelogram, whose all side are equal and each angle is 90°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 11
A square can also be defined as :
(i) a rhombus whose any angle is 90°.
(ii) a rectangle whose all sides are equal.
(iii) a quadrilateral whose all sides are equal and each angle is 90°.
∴ If ABCD is a square :
(i) all its sides are equal, i.e. AB = BC = CD = DA
(ii) each angle of it is 90°.
i.e. ∠A = ∠B = ∠C = ∠D = 90°.
Also,
(iii) diagonals are equal.
i.e. AC = BD.
(iv) diagonals bisect each other at 90°.
i.e. OA = OC =\(\frac { 1 }{ 2 }\) AC;OB = OD = \(\frac { 1 }{ 2 }\) BD
and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°.
Since, diagonals AC and BD are equal; therefore ; OA = OC = OB = OD.
(v) diagonals bisect the angles at the vertices
i.e. ∠1 = ∠2 = 45° [∵ ∠1 + ∠2 = 90°]
Similarly; ∠3 = ∠4 = 45° ;
∠5 – ∠6 = 45° and ∠7 = ∠8 = 45°.

Quadrilateral Exercise 27A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Two angles of a quadrilateral are 89° and 113°. If the other two angles are equal; find the equal angles.
Solution:
Let the other angle = x°
According to given,
89° + 113° + x° + x° = 360°
2x° = 360° – 202°
2x° = 158°
x° = \(\frac { 158 }{ 2 }\) =79°
∴other two angles = 79° each

Question 2.
Two angles of a quadrilateral are 68° and 76°. If the other two angles are in the ratio 5 : 7; find the measure of each of them.
Solution:
Two angles are 68° and 76°
Let other two angles be 5x and 7x
∴ 68°+76°+5x + 7x = 360°
12x + 144° = 360°
12x = 360° – 144°
12x= 216°
x = 18°
angles are 5x and 7x
i.e. 5×18° and 7×18° i.e. 90° and 126°

Question 3.
Angles of a quadrilateral are (4x)°, 5(x+2)°, (7x-20)° and 6(x+3)°. Find
(i) the value of x.
(ii) each angle of the quadrilateral.
Solution:
Angles of quadrilateral are,
(4x)°, 5(x+2)°, (7x-20)° and 6(x+3)°.
4x+5(x+2)+(7x-20)+6(x+3) = 360°
4x+5x+10+7x-20+6x+18 = 360° 22x+8 = 360°
22x = 360°-8°
22x = 352°
x = 16°
Hence angles are,
(4x)° = (4×16)° = 64°,
5(x+2)° = 5(16+2)° = 90°,
(7x-20)° = (7×16-20)° = 92°
6(x+3)° = 6(16+3) = 114°

Question 4.
Use the information given in the following figure to find :
(i) x
(ii) ∠B and ∠C
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 12
Solution:
∵ ∠A = 90° (Given)
∠B = (2x+4°)
∠C = (3x-5°)
∠D = (8x – 15°)
∠A + ∠B + ∠C + ∠D = 360°
90° + (2x + 4°) + (3x – 5°) + (8x – 15°) = 360°
90° + 2x + 4° + 3x – 5° + 8x – 15° = 360°
⇒ 74° + 13x = 360°
⇒13x = 360° – 74°
⇒ 13x = 286°
⇒ x = 22°
∵ ∠B = 2x + 4 = 2. x 22° + 4 – 48°
∠C = 3x – 5 = 3×22° -5 = 61°
Hence (i) 22° (ii) ∠B = 48°, ∠C = 61°

Question 5.
In quadrilateral ABCD, side AB is parallel to side DC. If ∠A : ∠D = 1 : 2 and ∠C : ∠B = 4:5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 13
∵∠A : ∠D = 1:2
Let ∠A = x and ∠B = 2x
∵∠C : ∠B = 4 : 5 Let ∠C = 4y and ∠B = 5y
∵AB || DC
∠A + ∠D = 180° x + 2x = 180°
3x = 180° x = 60°
∴A = 60°
∠D = 2x = 2 x 60 = 120° Again ∠B + ∠C = 180°
5y + 4y= 180°
9y = 180°
y = 20°
∴∠B = 5y- = 5 x 20 = 100°
∠C = 4y = 4 x 20 = 80°
Hence ∠A = 60° ; ∠B = 100° ; ∠C = 80° and ∠D = 120°

Question 6.
From the following figure find ;
(i) x,
(ii) ∠ABC,
(iii) ∠ACD.
Solution:
(i) In Quadrilateral ABCD,
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 14
x + 4x + 3x + 4x + 48° = 360°
12x = 360° – 48°
12x =312
(ii) ∠ABC = 4x
4 x 26 = 104°
(iii) ∠ACD = 180°-4x-48°
= 180°-4×26°-48°
= 180°-104°-48°
= 180°-152° = 28°

Question 7.
Given : In quadrilateral ABCD ; ∠C = 64°, ∠D = ∠C – 8° ;
∠A = 5(a+2)° and ∠B = 2(2a+7)°.
Calculate ∠A.
Solution:
∵∠C = 64° (Given)
∴∠D = ∠C- 8°
= 64°- 8°
= 56°
∠A = 5 (a + 2)°
∠B = 2(2a + 7)°
Now ∠A + ∠B + ∠C + ∠D = 360°
5(a + 2)° + 2(2a + 7)° + 64°+ 56° = 360°
5a + 10 + 4a + 14° + 64° + 56° = 360°
9a + 144° = 360°
9a = 360°-144°
9a = 216°
a = 24°
∴∠A = 5(a + 2)
= 5(24 + 2)
= 130°

Question 8.
In the given figure :
∠b = 2a + 15
and ∠c = 3a+5; find the values of b and c.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 15
Solution:
∠b = 2a + 15
& ∠c = 3a + 5
∵Sum of angles of quadrilateral = 360°
70° + a + 2a + 15 + 3a + 5 – 360°
6a+90° = 360°
6a = 270°
a = 45°
∴ b = 2a+15= 2×45+15 = 105°
c = 3a+5 = 3×45+5 = 140°
105° and 140°

Question 9.
Three angles of a quadrilateral are equal. If the fourth angle is 69°; find the measure of equal angles.
Solution:
Let each equal angle be
x° x + x + x + 69° = 360°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 16
3x = 360°-69 3x =291 x = 97°
Each equal angle = 97°

Question 10.
In quadrilateral PQRS, ∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other. Is PS also parallel to QR ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 17
∵∠P : ∠Q : ∠R : ∠S = 3 : 4 : 6 : 7
Let ∠P = 3x
∠Q = 4x
∠R = 6x & ∠S = 7x
∴∠P+∠Q+∠R+∠S = 360°
3x + 4x + 6x + lx = 360°
20x = 360°
x = 18°
∴ ∠P = 3x = 3×18 = 54°
∠Q = 4x = 4×18 = 72°
∠R = 6x = 6×18 = 108°
∠S = 7x = 7×18 = 126°
∠Q + ∠R = 72°+108° = 180° or ∠P +∠S = 54°+126° = 180°
Hence PQ || RS
As ∠P + ∠Q = 72° +54° = 126°
Which is * 180°.
∴PS and QR are not parallel.

Question 11.
Use the information given in the following figure to find the value of x.
Solution:
Take A, B, C, D as the vertices of quadrilateral and BA is produced to E (say).
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 18
Since ∠EAD = 70°
∴∠DAB = 180° – 70° = 110° [∵ EAB is a straight line and AD stands on it]
∴∠EAD + ∠DAB = 180°
∴110° + 80° + 56° + 3x = 360°
[∵ sum of interior angles of a quadrilateral = 360°]
∴3x = 360° – 110° – 80° – 56° + 6°
3x = 360° – 240° = 120°
∴x = 40°

Question 12.
The following figure shows a quadrilateral in which sides AB and DC are parallel.
If ∠A : ∠D = 4 : 5, ∠B = (3x – 15)° and ∠C = (4x + 20)°, find each angle of the quadrilateral ABCD.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 19
Solution:
Let ∠A = 4x
∠D = 5x
Since ∠A + ∠D = 180° [∵ AB || DC]
∴4x + 5x = 180°
⇒ 9x = 180° ⇒x = 20°
∴∠A = 4 (20) = 80°, ∠D = 5 (20) = 100° Again ∠B + ∠C = 180° [ ∵ AB || DC]
∴ 3x – 15° + 4x + 20° = 180°
7x = 180°-5°
⇒ 7x = 175° ⇒ x = 25°
∴∠B = 75°-15° = 60° and ∠C = 4(25) + 20 = 100° + 20° = 120°

Quadrilateral Exercise 27B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
In a trapezium ABCD, side AB is parallel to side DC. If ∠A = 78° and ∠C = 120°, find angles B and D.
Solution:
∵ AB || DC and BC is transversal
∴∠B and ∠C, ∠A and ∠D are Cointerior angles with their sum = 180°
i.e. ∠B + ∠C = 180°
⇒ ∠B + 120° = 180°
⇒ ∠B = 180° – 120°
⇒ ∠B = 60°
Also ∠A + ∠D = 180°
⇒ 78° + ∠D = 180°
⇒ ∠D = 180° – 78°
∠D = 102°

Question 2.
In a trapezium ABCD, side AB is parallel to side DC. If ∠A = x° and ∠D = (3x – 20)°; find the value of x.
Solution:
∵AB || DC and BC is transversal
∴∠A and ∠B are Co-interior angles with their sum = 180°
i.e. ∠A + ∠D = 180°
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 20
⇒ x° + (3x – 20)° = 180°
⇒ x° + 3x° – 20° = 180°
⇒ 4x° = 180° + 20°
x° = \(\frac { 200 }{ 4 }\) = 50°
∴Value of x = 50°

Question 3.
The angles A, B, C and D of a trapezium ABCD are in the ratio 3 : 4 : 5 : 6.
Le. ∠A : ∠B : ∠C : ∠D = 3:4: 5 : 6. Find all the angles of the trapezium. Also, name the two sides of this trapezium which are parallel to each other. Give reason for your answer
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 21

Question 4.
In an isosceles trapezium one pair of opposite sides are ….. to each Other and the other pair of opposite sides are ….. to each other.
Solution:
In an isosceles trapezium one pair of opposite sides are parallel to each other and the other pair of opposite sides are equal to each other.

Question 5.
Two diagonals of an isosceles trapezium are x cm and (3x – 8) cm. Find the value of x.
Solution:
∵The diagonals of an isosceles trapezium are of equal length
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 23
∴3x – 8 = x
⇒ 3x – x = 8 cm
⇒ 2x = 8 cm
⇒ x = 4 cm
∴ The value of x is 4 cm

Question 6.
Angle A of an isosceles trapezium is 115° ; find the angles B, C and D.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 24

Question 7.
Two opposite angles of a parallelogram are 100° each. Find each of the other two opposite angles.
Solution:
Given : Two opposite angles of a parallelogram are 100° each
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 25
∵ Adjacent angles of a parallelogram are supplementary,
∴∠A + ∠B = 180°
⇒ 100° + ∠B = 180°
⇒ ∠B = 180° – 100°
⇒ ∠B = 80°
Also, opposite angles of a parallelogram are equal
∴∠D = ∠B = 80°
∴∠B = ∠D = 80°

Question 8.
Two adjacent angles of a parallelogram are 70° and 110° respectively. Find the other two angles of it.
Solution:
Given two adjacent angles of a parallelogram are 70° and 110° respectively.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 26
Since, we know that opposite angles of a parallelogram are equal
∴∠C = ∠A = 70° and ∠D = ∠B = 110°

Question 9.
The angles A, B, C and D of a quadrilateral are in the ratio 2:3: 2 : 3. Show this quadrilateral is a parallelogram.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 27

Question 10.
In a parallelogram ABCD, its diagonals AC and BD intersect each other at point O.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 29
If AC = 12 cm and BD = 9 cm ; find; lengths of OA and OD.
Solution:

Question 11.
In parallelogram ABCD, its diagonals intersect at point O. If OA = 6 cm and OB = 7.5 cm, find the length of AC and BD.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 32
Solution:

Question 12.
In parallelogram ABCD, ∠A = 90°
(i) What is the measure of angle B.
(ii) Write the special name of the paralleogram.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 34

Question 13.
One diagnol of a rectangle is 18 cm. What is the length of its other diagnol?
Solution:
∵ In a rectangle, diagnols are equal
⇒ AC = BD
Given, one diagnol of a rectangle = 18cm
∴ Other diagnol of a rectangle will be = 18cm
i.e. AC = BD = 18cm.

Question 14.
Each angle of a quadrilateral is x + 5°. Find :
(i) the value of x
(ii) each angle of the quadrilateral.
Give the special name of the quadrilateral taken.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 36
Solution:

Question 15.
If three angles of a quadrilateral are 90° each, show that the given quadrilateral is a rectangle.
Solution:
The given quadrilateral ABCD will be a rectangle, if its each angle is 90°
Since, the sum of interior angles of a quadrilateral is 360°.
∴∠A +∠B + ∠C + ∠D = 360°
⇒ 90° + 90° + 90° + ∠D = 360°
⇒ 270° + ∠D = 360°
⇒ ∠D = 360° – 270°
⇒ ∠D = 90°
Since, each angle of the quadrilateral is 90°.
∴The given quadrilateral is a rectangle.

Question 16.
The diagnols of a rhombus are 6 .cm and 8 cm. State the angle at which these diagnols intersect.
Solution:
The diagnols of a Rhombus always intersect at 90°.

Question 17.
Write, giving reason, the name of the figure drawn alongside. Under what condition will this figure be a square.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 39
Solution:
Since, all the sides of the given figure are equal.
i.e. AB = BC = CD = DA = 6 cm
∴ The given figure is a rhombus.
This figure shall be considered as a square, if any angle is 90°.

Question 18.
Write two conditions that will make the adjoining figure a square.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 27 Quadrilateral image - 40
Solution:
The conditions that will make the ad-joining figure a square are :
(i) All the sides must be equal.
(ii) Any angle is 90°.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles

November 10, 2023May 18, 2022 by Prasanna

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles (Including Types, Properties and Constructions)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles (Including Types, Properties and Constructions)

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IMPORTANT POINTS

1. Collinear Points: Three or more points which lie on the same straight line, are called collinear points.
2. Non-Collinear Points: Three or more points which do not lie on the same line, are called non- col linear points.
3. Triangle: By joining the three non-collinear points, a triangle is formed or A triangle is a figure which is enclosed by three lines segments. In the figure, ABC is a triangle.
4. Parts of triangle: A triangle has six parts, three sides and three angles which are on the vertices of the triangle.
5. Sum of angles of a triangle: The sum of the three angles of a triangle is 180°.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 1
6. Exterior angle of a triangle: If one side of a triangle is produced then the exterior angle is formed. Exterior angle of a triangle is equal to sum of its interior opposite angles. In other words, we can say that exterior angle of a triangle is greater than each of its interior opposite angles. In the figure.
∠ACE is exterior angle and ∠ACE = ∠A + ∠B and also ∠ACE > ∠A and ∠ACF > ∠B.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 2

7. Classification of triangles :
(A) According to their sides.
(i) Equilateral Triangle: If three sides of a triangle are equal, it is called an equilateral triangle.
(ii) Isosceles Triangle: If any two sides of a triangle are equal, then it is called an isosceles triangle.
(iii) Scalene Triangle: If no two sides of the triangle are equal. Then it is called a scalene triangle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 3
(B) According to Angles :
(i) Acute-angled Triangle: A triangle whose each angle is acute, is called an acute angled triangle.
(ii) Right-angled Triangle: A triangle whose one angle is a right angle i.e. 90°, is called a right angled triangle.
(iii) Obtused-angled Triangle: A triangle whose one angle is an obtused angle, is called an obtused-angled triangle.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 4

8. Some special properties of a triangle:
(i) If one angle of a triangle is equal to the sum of other two cycles, the angle is a right angled.
(ii) If the acute angles of a right angled triangle are equal, then the triangle is a right angled isosceles triangle and its each acute angle will be of 45°.
(iii) Sum of two sides of a triangle is greater than the third side.
(iv) There can be only one right angle in a triangle.
(v) There can be only one obtuse angle in a triangle.
(vi) Side opposite to greater angle is greater.
(vii) Angle opposite to shorter side is shorter.

Triangles Exercise 26A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
In each of the following, find the marked unknown angles :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 5

Solution:
(i) Since, sum of all angles of triangle = 180°
Hence, 70° + 72° + z = 180°
⇒ 142°+ z = 180° ”
⇒ z= 180°-142°
z = 38°
(ii) Since, sum of all angles of a triangle = 180°
1st Triangle 50° + 80° + b = 180°
⇒ 130°+ &= 180°
⇒b= 180° – 130°
b = 50°
IInd Triangle 40° + 45° + a = 180°
⇒ 85° + a = 180°
⇒ a = 180° -85
a = 95°
(iii) 60° + 45° + 20° + x = 180°
⇒ 125° + x = 180°
⇒ x = 180° – 125° => x = 55°

Question 2.
Can a triangle together have the following angles ?
(i) 55°, 55° and 80°
(ii) 33°, 74° and 73°
(iii) 85°, 95° and 22°.
Solution:
(i) Sum of all angles of a triangle = 180° Here, 55° + 55° + 80° = 180°
190° ≠ 180°
No.
(ii) 33°+ 74°+ 73°= 180°
180°= 180°
Yes.
(iii) 85° + 95° + 22° = 180°
202° ≠ 180°
No.

Question 3.
Find x, if the angles of a triangle are:
(i) x°, x°, x°
(ii) x°, 2x°, 2x°
(iii) 2x°, 4x°, 6x°
Solution:
(i) Since, sum of all the angles of a triangle =180
x° + x° + x° = 180
⇒ 3x° = 180
⇒ x° = \(\frac { 180 }{ 3 }\)
x = 60
(ii) x° + 2x° + 2x° = 180
5x° = 180
x° = \(\frac { 180 }{ 5 }\)
x° = 36
(iii) 2x° + 4x° + 6x° =180
12x° = 180
x° = \(\frac { 180 }{ 12 }\)
x° = 15

Question 4.
One angle of a right-angled triangle is 70°. Find the other acute angle.
Solution:
We know that, sum of angles of a triangle = 180°.
Let, the acute angle be ‘x’
∴ x + 90° + 70° = 180°
⇒ x+ 160° = 180°
⇒ x= 180°-160°
⇒ x = 20°
∴The acute angle is 20°.

Question 5.
In ∆ABC, ∠A = ∠B = 62° ; find ∠C.
Solution:
∠A + ∠B + ∠C= 180°
⇒ 62° + 62° + ∠C = 180°
⇒ 124° + ∠C = 180°
⇒ ∠C = 180° – 124°
⇒∠C = 56°

Question 6.
In ∆ABC, C = 56°C = 56° ∠B = ∠C and ∠A = 100° ; find ∠B.
Solution:
∠A + ∠B + ∠C = 180°
⇒ 100° + ∠B + ∠B = 180°
⇒ 2∠B = 180° 100°
∠B = \(\frac { 80 }{ 2 }\)°
∠B = 40°
∠C = ∠B = 40°

Question 7.
Find, giving reasons, the unknown marked angles, in each triangle drawn below:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 7

Solution:
We know that,
Exterior angle of a triangle is always equal to the sum of its two interior opposite angles (property)
(i) ∴ 110° = x + 30° (by property)
⇒x=110°-30° x = 80°
(ii) x+115° = 180°
(linear property of angles)
⇒x = 180°- 115° ⇒x = 65°
∴115° = x + y
⇒ 115° = 65° + _y ⇒ y= 115° – 65° =50°
y = 50°
(iii) 110° = 2x + 3x
5x – 110°
x = \(\frac { 110 }{ 5 }\)°
x = 22°
∴2x = 2 x 22 = 44°
3x = 3 x 22 = 66°

Question 8.
Classify the following triangles according to angle :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 9

Solution:
(i) Since, it has an obtuse angle of 120° Hence, it is obtuse angled triangle.
(ii) Since, all the angle of triangle is less than 90°.
Hence, it is an acute angled triangle.
(iii) Since ∠MNL = 90°, and sum of two acute angle ∠M + ∠N = 30° + 60° = 90°.
Hence, it is a right angled triangle.

Question 9.
Classify the following triangles according to sides :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 11
Solution:
(i) Since, two sides-are equal.
Hence, Isosceles triangle.
(ii) Since, all the three sides are unequal.
Hence, Scalene, triangle.
(iii) Since, all the three sides are unequal Hence, Scalene triangle.
(iv) All the three sides are equal.
Hence, equilateral triangle.

Triangles Exercise 26B – Selina Concise Mathematics Class 6 ICSE Solutions

Construct traingle ABC, when :
Question 1.
AB = 6 cm, BC = 8 cm and AC = 4 cm.
Solution:
Steps of Construction:
(1) Draw a line AB = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 12
(2) compasses and taking B as centre, draw an arc of 8 cm radius.
(3) With A as centre, draw an arc of 4 cm radius, which cuts the previous arc at C.
(4) Join AC and BC.
Triangle ABC, obtained, is the required triangle.

Question 2.
AB = 3.5 cm, AC = 4.8 cm and BC = 5.2 cm.
Solution:
Steps of Construction :
(1) Draw a line AB = 3.5 cm.
(2) Using compasses and taking B as centre, draw an arc of 5.2 cm radius.
(3) With A as centre, draw an arc of 4.8
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 13
(4) Join AC and BC

Question 3.
AB = BC = 5 cm and AC = 3 cm. Mea¬sure angles A and C. Is ∠A = ∠C?
Solution:
Steps of Construction :
(1) Draw a line AB = 5 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 14
(2) Using compasses and taking B as cen¬tre, draw an arc of 5 cm radius.
(3) With A as centre, draw an arc of 3 cm radius, which cuts the previous arc at C.
(4) Join AC and BC.

Question 4.
AB = BC = CA = 4.5 cm. Measure all the angles of the triangle. Are they equal ?
Solution:
Steps of Construction :
(1) Draw a a line AB =4.5
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 15
(2) Using compasses and taking BC as centre, draw an arc of 4.5 cm radius.
(3) With AC as centre, draw an arc of 4-5 cm radius, which cuts the previous arc at C.
(4) Join AC and BC.
(5) Measurement, ∠A = ∠B = ∠C = 60°.

Question 5.
AB = 3 cm, BC = 7 cm and ∠B = 90°.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 3 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 16
(2) With the help of compasses, construct ∠ABC = 90°.
(3) With B as centre, draw an arc of 7 cm length which cuts BP at C.
(4) Join A and C.
(5) Triangle ABC, so obtained, is the required triangle.

Question 6.
AC = 4.5 cm, BC = 6 cm and ∠C = 60°.
Solution:
Steps of Construction :
(1) Draw a line AC = 4.5 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 17
(2) With the help of compasses, construct ∠ACB = 60°.
(3) With C as centre, draw an arc of 6 cm radius, which cuts CB at C.
(4) Join B and A.

Question 7.
AC = 6 cm, ∠A = 60“ and ∠C = 45°. Measure AB and BC.
Solution:
Steps of Construction :
(1) Draw a line segment AC = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 18.
(2) At A construct an angle ∠A = 60°.
(3) At C construct an angle ∠C = 45°.
(4) AD and CE intersect each other at B.
(5) ∴ ∆ABC is the required triangle.
(6) On measuring AB = 4-4 cm, BC = 5.4 cm.

Question 8.
AB = 5.4 cm, ∠A = 30° and ∠B = 90°. Measure ∠C and side BC.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 5.4 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 19
(2) At A construct an angle ∠A = 30°.
(3) At B construct an angle ∠B = 90°.
(4) AD and BE intersect each other at C.
(5) ∴ ∆ABC is the required triangle.
(6) On measuring ∠C = 60° side BC = 31 cm.

Question 9.
AB = 7 cm, ∠B = 120° and ∠A = 30°. Measure AC and BC.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 7 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 20
(2) At A construct an angle ∠A = 30°.
(3) At C construct an angle ∠C = 45°.
(4) AE and BD intersect each other at C.
(5) ∴ ∆ABC is the required triangle.
(6) On measuring length of AC = 12cm and BC = 7 cm respectively.

Question 10.
BC = 3 cm, AC = 4 cm and AB = 5 cm. Measure angle ACB. Give a special name to this triangle.
Solution:
Steps of Construction :
(1) Draw a line segment AB = 5 cm
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 21
(2) From A, with the help of compass cut arc AC = 4cm
(3) From point B cut an arc BC = 3 cm.
(4) Join AC and BC.
(5) AABC is required right-angled triangle.
Measuring ∠ACB = 90°

Triangles Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
If each of the two equal angles of an isosceles triangle is 68°, find the third angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 22

Question 2.
One of the angles of a triangle is 110°, the two other angles are equal. Find their value.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 23

Question 3.
The angles of a triangle are in the ratio 3:5: 7. Find each angle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 25

Question 4.
The angles of a triangle are (2x – 30°),(3x – 40°) and (\(\frac { 5 }{ 2 }\)x + 10°) Find the value of x .
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 26

Question 5.
In each of the following figures, triangle ABC is equilateral and triangle PBC is isosceles. If PBA = 20°; find in each case:
(a) angle PBC.
(b) angle BPC.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 28
Solution:

Question 6.
Construct a triangle ABC given AB = 6 cm, BC = 5 cm and CA = 5.6 cm. From vertex A draw a perpendicular on to side BC. Measure the length of this perpendicular.
Solution:
Steps of Construction :
(1) Draw a line AB = 6 cm.
(2) Using compass, taking A and B as centre draw arcs of 5 cm and 5.6 cm respectively, which cut each other at C.
(3) Join AC and BC.
(4) Now, from vertex A draw a bisector AD towards BC.
On measuring length AD = 5 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 31

Question 7.
Construct a triangle PQR, given PQ = 6 cm, ∠P = 60° and ∠Q = 30°. Measure angle R and the length of PR.
Solution:
Steps of Construction :
(1) Draw a line PQ = 6 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 32
(2) Using compass taking P as centre draw an angle ∠P = 60°.
(3) Using compass taking Q as a centre draw an angle ∠Q = 30°.
(4) PS and QT intersect each other R.
(5) ∆RPQ is the required triangle.
On measuring; ∠R = 90°, length of PR = 3 cm.

Question 8.
Construct a triangle ABC given BC = 5 cm, AC = 6 cm and ∠C = 75°. Draw the bisector of the interior angle at A. Let this bisector meet BC at P ; measure BP.
Solution:
Steps of Construction :
(1) Draw BC = 5 cm.
(2) With the help of compass from centre C. Draw an angle ∠C = 75°.
(3) From CD, cut an arc AC = 6 cm.

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 33
(4) JoinAB.
(5) From A draw an bisector AP.
(6) On measuring BP = 2.6 cm.

Question 9.
Using ruler and a pair compass only, construct a triangle XYZ given YZ = 7 cm, ∠XYZ = 60° and ∠XZY = 45°. Draw the bisectors of angles X and Y.
Solution:
Steps of Construction :
(1) Draw a line YZ = 7 cm.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 34
(2) Y as a centre draw an ∠XYZ = 60°.
(3) Z as a centre draw an ∠XZY = 45°.
(4) From X and Y as centre draw bisector of ∠X and ∠Y, which meet at point O.

Question 10.
Using ruler and a pair compass only, construct a triangle PQR, given PQ = 5.5 cm, QR = 7.5 cm and RP = 6 cm. Draw the bisectors of the interior angles at P, Q and R. Do these bisectors meet at the same point ?
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 35

Steps of Construction :
(1) Draw a line PQ = 5.5 cm.
(2) From Q as a centre draw an arc QR = 7.5 cm.
(3) From P as a centre draw an arc PR = 6 m, which intersects previous arc at R.
(4) Join PR and QR.
(5) Now, draw interior bisectors of ∠P, ∠QR ∠R which meets each other at point S.

Question 11.
One angle of a triangle is 80° and the other two are in the ratio 3 : 2. Find the unknown angles of the triangle.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 36
Let angle ∠A of a triangle ABC = 80°
But sum of three angles of a triangle = 180°
Sum of remaining two angles = 180° – 80° = 100°
Ratio in their two angles = 3:2
Let second angle = 3x
and third angle = 2x
3x + 2x = 100°
5x = 100 x = \(\frac { 100 }{ 5 }\) = 20°
second angle ∠B = 3x = 3 x 20° = 60°
and third angle ∠C = 2x = 2 x 20° = 40°
Hence other two angles are 60° and 40°.

Question 12.
Find the value of x if ∠A = 32°, ∠B = 55° and obtuse angle AED = 115°.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 26 Triangles image - 37
In the figure, ∠A = 32, ∠B = 55°
∠AED =115°
In ∆ABC
Exterior ∠ACD = ∠A + ∠B = 32° + 55° = 87°
Similarly in ∆CDE
Ext. ∠AED = ∠D + ∠ACD
⇒ 115° = x + 87° ⇒ x = 115°- 87° = 28°
Hence x = 28°