Students often refer to Kerala State Syllabus SCERT Class 8 Maths Solutions and Class 8 Maths Chapter 7 Bisectors Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 8 Maths Chapter 7 Solutions Bisectors
Class 8 Kerala Syllabus Maths Solutions Chapter 7 Bisectors Questions and Answers
Bisectors Class 8 Questions and Answers Kerala Syllabus
Bisectors of Lines (Page No. 107)
Draw the figures below using only a ruler and a compass.
Question 1.
Square of sides 4\(\frac {1}{4}\) centimetres.
Answer:
Draw a line 8.5 cm long.
Draw its perpendicular bisector.
Each portion is of length 4\(\frac {1}{4}\) cm.
Marking half the length of the line on the perpendicular bisector.
Complete the square.
Question 2.
Rectangle of sides 5\(\frac {1}{4}\) centimeters and 3\(\frac {1}{4}\) centimeters.
Answer:
Draw a line AB of length, \(5 \frac{1}{4}+3 \frac{1}{4}=8 \frac{1}{2}\) cm
Draw the perpendicular bisector of the line.
Half of the length of the line is 4.25 cm.
Mark the midpoint as M.
Mark a point P on AB which is 1cm away from the point M.
AP = 5\(\frac {1}{4}\) cm and BP = 3\(\frac {1}{4}\) cm.
Draw a perpendicular from P to AB and mark a point C on it such that PB = PC.
Draw a square APCD.
Question 3.
Equilateral triangle of side 2.75 centimetres.
Answer:
Draw a line of length 2.75 × 4 = 11 cm.
Draw the perpendicular bisector of the line.
Divides it into two equal parts.
Agai,n draw the perpendicular bisector for the one part, \(\frac {1}{4}\) is a part of 11 cm.
The length of one part is 2.75 cm.
Draw an equilateral triangle with this length as its sides.
Question 4.
Triangle of area 9 square centimetres and one side 4.5 centimetres.
Answer:
The area of the square is 2 × 9 = 18 sq.cm.
One side is 4.5 cm, and the other side is 4 cm.
One of its diagonals makes two equal triangles.
The area of a triangle is 9 sq cm, and one side is 4.5 cm.
A line AP of length 8.5 cm.
Mark the point B such that AB = 4.5 cm
Draw a perpendicular to the line through B and mark a point C on it such that BP = BC.
Draw the square ABCD.
Draw the diagonal of AC and mark the triangle as ABC.
Rhombuses (Page No. 111, 112)
Question 1.
Draw a rhombus with each pair of lengths given below for the diagonals:
(i) 6 centimetres, 4 centimetres
(ii) 6.5 centimetres, 4 centimetres
(iii) 6 centimetres, 4.5 centimetres
Answer:
(i) Draw a line 6 cm long.
Draw its perpendicular bisector.
The diagonals of a rhombus are perpendicular bisectors of each other.
(ii) Draw a line 6.5 cm long.
Draw its perpendicular bisector.
Draw the rhombus.
(iii) Draw a line 6 cm long.
Draw its perpendicular bisector.
Draw the rhombus.
Question 2.
The picture shows the quadrilateral formed by joining the midpoints of a rectangle:
(i) Are the diagonals of this quadrilateral parallel to the sides of the rectangle? Why?
(ii) Is this quadrilateral a rhombus? Why?
Answer:
(i) Square ABCD. AD and BC are equal and parallel.
These are perpendicular to AB.
So the line PR is parallel to AB and CD.
Similarly, the line SQ is parallel to AD and BC.
The diagonals of PQRS are parallel to the sides of ABCD.
(ii) Since P, Q, R, and S are the midpoints of the sides, the diagonals of PQRS are perpendicular bisectors of each other.
PQRS is a rhombus.
Question 3.
Draw a rhombus with diagonals 6.5 centimetres and 4.5 centimetres.
Answer:
Draw a square with side lengths of 6.5 cm and 4.5 cm.
Draw the perpendicular bisectors of length and breadth.
These lines pass through the midpoint of the sides.
Join the midpoints of the sides.
Question 4.
Prove that each diagonal of a rhombus bisects the angles at the vertices it joins.
Answer:
Consider the rhombus ABCD.
AC and BD are the diagonals.
Diagonals are the perpendicular bisectors of each other.
The four sides are equal.
In triangle ACD, AD = CD.
So the angles opposite to these sides are equal.
That is x = y.
Diagonals bisect the angle at the vertices they join.
Question 5.
Draw a square with diagonals 7 centimetres.
Answer:
Draw a line 7 cm long.
Draw the perpendicular bisector of the line.
Mark the midpoint as O.
Draw a circle considering the line as the diameter.
Complete the square.
Question 6.
Draw this picture.
Answer:
Bisector of an Angle (Page No. 115)
Question 1.
Draw the angles below:
(i) \(37 \frac{1}{2}^{\circ}\)
(ii) \(62 \frac{1}{2}^{\circ}\)
Answer:
Draw an angle of 75°.
Draw the arc in such a way that it cuts the sides at a fixed radius with the corner at the centre.
An isosceles triangle is obtained by joining the points that intersect the sides by the arc.
The perpendicular from the vertex is the angle bisector.
For this, consider the centre as the points that intersect the sides and draw arcs of the same radius that intersect each other.
Complete the bisector by drawing a line that joins this point with the vertex.
Question 2.
Use only a ruler and a compass to draw the following:
(i) The angles below.
(a) 45°
(b) 135°
(c) 75°
(d) 15°
(ii) The triangle with one side of length 6 centimetres and angle \(67 \frac{1}{2}^{\circ}\) and \(22 \frac{1}{2}^{\circ}\) at its ends.
Answer:
(i) (a) 45°
Draw a line and mark a point on it.
Draw the perpendicular to the line through the point.
Draw the angle bisector for the angle formed at 90°.
This forms a 45° angle.
(b) 135°
Draw a line and mark a point on it.
Draw the perpendicular to the line through the point.
Draw the angle bisector for the angle formed at 90°.
Draw an angle 45°
90 + 45 = 135.
(c) 75°
First, draw a 90° angle.
Inside this angle, draw a 60° angle.
Draw the angle bisector for the 30° angle formed along with these.
15 + 60 = 75
(d) 15°
Draw a 60° angle and mark the angle bisector of this.
Draw a 30° angle and mark the angle bisector of this.
We get a 15° angle.
(ii) Draw a 6 cm long line.
Draw an angle of 135° on one end and 45° on the other end.
Use only the compass and scale for the construction (above construction).
Draw a triangle with the angle bisectors of this.
Class 8 Maths Chapter 7 Kerala Syllabus Bisectors Questions and Answers
Class 8 Maths Bisectors Questions and Answers
Question 1.
Which of the following is a quadrilateral in which all the angles are equal?
(a) Rhombus
(b) Rectangle
(c) Parallelogram
(d) Square
Answer:
(d) Square
All angles are 90 °.
Question 2.
Two statements are given below:
(i) The diagonals of all parallelograms bisect each other perpendicularly.
(ii) The diagonal of a rhombus bisects each other perpendicularly.
(a) (i) and (ii) are true
(b) Only (i) is true
(c) Only (ii) is true
(d) (i) and (ii) are not true
Answer:
(c) Only (ii) is true.
Question 3.
If a 145° angle is divided equally, then what is the measure of one part?
(a) 75°
(b) 70°
(c) \(40 \frac{1^{\circ}}{2}\)
(d) \(72 \frac{1^{\circ}}{2}\)
Answer:
(d) \(72 \frac{1^{\circ}}{2}\)
Question 4.
What are the measures needed to draw a fixed rhombus?
(a) Only the length of its sides
(b) One diagonal and one side
(c) Two diagonals
(d) Perimeter
Answer:
(c) Two diagonals
Question 5.
The line l1 is perpendicular to l2. The line l2 is perpendicular to l3.
(i) l1 and l3 are perpendicular lines.
(ii) l1 and l3 are parallel lines.
(a) (i) and (ii) are true
(b) (i) and (ii) are false
(c) (i) is true (ii) is false
(d) (i) is false (ii) is true
Answer:
(d) (i) is false (ii) is true
Question 6.
Draw a 7.5-centimeter-long line and construct a right triangle whose perpendicular sides are 3.75 centimeters long.
Answer:
Draw a 7.5 cm long line.
Draw the perpendicular bisector of this line.
Complete the right triangle by drawing the half length of the line on the perpendicular line.
Question 7.
Draw an isosceles triangle whose length of one side is 9.5 centimetres, and the height of this side is 4.75 centimetres.
Answer:
Draw a line of length 9.5 cm.
Draw the perpendicular bisector of this line.
Mark the half-length of the line on the top of its perpendicular bisector.
From this point, join both ends of the line and form a triangle.
Question 8.
Draw a square with a perimeter is 23 centimeters.
Answer:
Draw a line of length 11.5 centimeters.
Draw the perpendicular bisector of the line.
Mark the half-length of the line on the perpendicular bisector.
Complete the square.
One side = \(\frac {11.5}{2}\) cm
Perimeter = 4 × \(\frac {11.5}{2}\) = 23 cm
Question 9.
Draw a line 7.5 cm long and divide it into two equal parts. Draw an equilateral triangle with a perimeter of 11.25 cm.
Answer:
Draw a line of length 7.5 cm.
Draw a perpendicular bisector of the line.
Length of one side = \(\frac {7.5}{2}\) cm
Draw an equilateral triangle by considering the half length as the side.
Question 10.
Draw a square with the length of its diagonals as 6.5 centimetres.
Answer:
Draw a line of length 6.5 cm.
Draw the perpendicular bisector of this line.
Take the midpoint of the line as the center and draw a circle with half the length of the line as its radius.
Draw a square by joining the ends of the line with the points on the perpendicular bisector where the circle touches it.
Question 11.
Draw a rhombus with the length of its diagonals as 6 centimetres and 8.5 centimetres.
Answer:
Draw the perpendicular bisector for one of its diagonals.
Draw the rhombus by marking the length of the other diagonal on the perpendicular bisector.
Question 12.
Draw \(22 \frac{1}{2}^{\circ}\) cone using scale and compass.
Answer:
Draw a line and mark a point on it.
Draw a perpendicular line through the point.
A 90° angle is formed. Draw the angle bisector.
A 45° angle is formed. Draw the angle bisector.
\(22 \frac{1}{2}^{\circ}\) angle is formed.
Question 13.
Draw a right triangle with two angles of 45° and a length of its hypotenuse of 4.75 centimetres.
Answer:
Draw a line of length 9.5 cm.
Draw its perpendicular bisector.
The part of the line on one side of the perpendicular bisector will be 4.75 cm.
The angle between the line and the perpendicular bisector is 90°.
Draw the angle bisector.
Draw a perpendicular from one end of the line to the angle bisector.
It forms a right triangle.
Question 14.
Draw a right triangle with one angle 30° and the length of its hypotenuse as 4.25 centimetres.
Answer:
Draw a line of length 8.5 cm.
Draw its perpendicular bisector.
The half length of the line is 4.25 cm.
Draw an angle of 60° to the vertex that intersects the bisector.
Draw a perpendicular from one end of the line to the side of its angle.
It forms a 30° – 60° – 90° right triangle.
The hypotension will be 4.25 cm.
Question 15.
The length of the side of an isosceles triangle is 4.25 centimetres. Draw the triangle with the height of its side as 7 centimetres.
Answer:
Draw a line of length 9.5 cm.
Draw the perpendicular bisector of the line.
One side of the bisector is 4.25 cm.
Draw the perpendicular bisector for the 4.25 cm line.
Mark the height on the perpendicular bisector and complete the triangle.
Class 8 Maths Chapter 7 Notes Kerala Syllabus Bisectors
→ A perpendicular line that passes through the midpoint of a line is called a perpendicular bisector. Any point on the perpendicular bisector of a line is at the same distance to its endpoints.
→ The perpendicular bisector of the line is the line joining the points that are at equal distance from the endpoints of a line.
→ Rhombuses are the quadrilaterals with opposite sides parallel. A parallelogram with four sides of the same length is called rhombus.
→ For a rhombus, the diagonals are perpendicular bisectors of each other. The converse is also true. The quadrilateral with its diagonals are perpendicular bisectors of each other is a rhombus.
→ If the diagonals of a quadrilateral are of the same length and are perpendicular bisectors of each other, then it forms a square.
→ The diagonals of a rectangle are equal and bisect each other. But they will not be a perpendicular bisector.
→ Any point on the bisector of an angle is at the same perpendicular distance from the sides of the angle.
→ Similarly, any point at the same perpendicular distance from two sides of an angle lies on the bisector of that angle.
As the word implies, bisectors divide equally. All lines that divide a line into equal parts are called bisectors of the line. We can draw many bisectors to a line, but a line which is perpendicular to a line is called the perpendicular bisector. In this chapter, we are discussing the geometrical concepts related to the construction of a perpendicular bisector. This concept is used in many situations. The basic idea behind the construction of the geometrical shape rhombus is that the diagonals are perpendicular bisectors of each other. In this chapter, we also discuss the bisector of the angle. The geometrical concepts of bisectors are essential for the further study of triangles and circles.
Bisectors of Lines
The point on a line at the same distance from both ends of a line is the midpoint of the line.
For example, a line of length 6 cm is equally spaced from both ends of the line, and the point on the line is 3 cm from both ends.
But what about the point that is 4 centimetres away from both sides?
To find that point, take 4 cm on a compass and mark it from both ends.
The point where they meet is 4 cm from the endpoints.
In this way, a point can be drawn at a distance of 5 cm from both ends of the line.
In this way, many points at the same distance can be drawn from both sides.
The line they are attached to will go through the midpoint of the line drawn first.
What can we understand from this?
To draw the perpendicular bisector of the line, mark the points at the same distance from both endpoints.
Let’s see how we can express this concept geometrically.
P is a point on the perpendicular bisector of the line AB.
M is the midpoint of AB.
AMP and BMP are right-angled triangles.
Using Pythagoras’ theorem,
AM2 + MP2 = AP2, BM2 + MP2 = BP2
AM = BM, and MP is a common line, so AP2 = BP2, AP = BP.
Worksheet – 1
Question 1.
Draw a 7.5-centimetre-long line and construct the perpendicular bisector.
Answer:
Draw a 7.5 cm line AB.
Considering the endpoints A and B as the centres, mark two arcs that cut each other.
Take a little distance more and draw another two arcs that intersect each other.
Draw a line through the point where the arcs intersect each other.
This is the perpendicular bisector of the first line.
Rhombuses
The shapes with four sides are called quadrilaterals.
The quadrilaterals with opposite sides that are parallel will be one of the following.
In this, we learn about rhombuses. And understand the characteristics of others.
The opposite sides of all these forms are parallel. But a different shape is formed when certain other characteristics are combined.
A diagonal is a line that joins the opposite corners.
The diagonals of a square and a rhombus are equal.
The diagonals of a quadrilateral and a rhombus are not equal, but they bisect each other.
Diagonals of a rhombus bisect each other perpendicularly.
Let us look at the characteristics of a square and a rhombus.
The sides are equal. Diagonals are perpendicular bisectors.
A rhombus can be drawn using the idea that the diagonals are perpendicular bisectors.
Worksheet – 2
Draw a rhombus with one angle of 40° and one side 5 centimetres.
Answer:
Draw an angle of 40°.
Draw a circle with a radius of 5 centimetres and consider the vertex as the centre.
Mark the points that cut the sides of the angle by the circle.
Draw a line parallel to the side of the angle through these points.
The sides of the angle and the lines together form a parallelogram.
Worksheet – 3
Draw a rhombus with the length of its diagonal as 6 centimetres.
Answer:
Draw a 6 cm long line AC.
Draw the perpendicular bisector of this line.
Draw a circle by considering the midpoint of the line as its centre.
Mark the points where the circle divides the perpendicular bisectors as C and D.
Draw a rectangle ABCD.
ABCD is a rhombus.
When did it become a square?
Bisector of an Angle
It’s a view from the perpendicular bisector to the angle bisector.
We know about the isosceles triangles. Two sides are equal.
If we draw a perpendicular bisector to the third side, it will pass through the opposite corner.
This line will divide the opposite angle into two equal parts. So this line is also called the angle bisector.
Worksheet – 4
Draw an angle of 40° and construct its angle bisector.
Answer:
Draw a 40° angle.
Draw a circle by considering the vertex A as the centre.
The circle cuts the sides of the angle at the points B and C.
Draw the triangle ABC. The triangle ABC is isosceles.
Draw the perpendicular bisector of BC. This line is the bisector of angle A.
