# Free Understanding Elementary Shapes Subjective Test 01 Practice Test - 6th grade

### Question 1

A ball is thrown from point A and it reaches point B which is 7m away. There is another point C which lies somewhere on the line AB. Given that the ball follows a straight path, and AC = 9m, BC = 2m. Which is the point that lies in between other two? Draw the path of the ball. [1 MARK]

#### SOLUTION

Solution :

Given that

Length of line AB = 7m

Length of line AC = 9m

Length of line BC = 2m

We can clearly see that the length of line AC > BC.

Also, AB + BC = AC

∴ Point B lies between A and C.AB + BC = AC.

### Question 2

What fraction of a clockwise revolution does the hour hand turn through when it rotates from: [2 MARKS]

a) 3 to 6

b) 6 to 12

#### SOLUTION

Solution :Fractions: 1 Mark each

We may observe that in one complete clockwise revolution, the hour hand will rotate through 360∘

a) We have to find out the fraction of a clockwise revolution that an hour hand turns through when it rotates from 3 to 6.

Now, when the hour hand rotates from 3 to 6 it will rotate through 1 right angle.

∴ The fraction of revolution = 90∘360∘ = 14 revolution

b) Here the hour hand rotates from 6 to 12, which is to right angles.

∴ The fraction of revolution = 180∘360∘ = 12 revolution

### Question 3

Given a △ABC, and the sum of two angles of the triangle is 90∘. What type of triangle is it? [2 MARKS]

#### SOLUTION

Solution :Triangle: 2 Marks

Given that:

The sum of two angles of the triangle is 90∘.

We know that the sum of all the three angles in a triangle is 180∘.

∴ The other angle = 180-90 = 90∘

Now we know that if one of the angles in a triangle is 90∘, then the triangle is a right-angled triangle.

Hence, the given triangle is a right-angled triangle.

### Question 4

Given that the lengths of the sides of a triangle are 6m, 6m, 8m. What is the type of triangle? What are the various properties of such triangles? [3 MARKS]

#### SOLUTION

Solution :Type of triangle: 1 Mark

Properties: 2 Marks

The sides of the triangle are given as 6m, 6m, 8m.

Now two sides of the triangle are equal 6m, 6m.

We know that if a triangle has two sides equal then the triangle is isosceles.

Hence, the given triangle is an isosceles triangle.

Also if two sides of a triangle are equal then their respective angles are also equal.

### Question 5

The line l is perpendicular to line m.

a) Is CE= EG?

b) Does PE bisect CG?

c) Identify any two line segments for which PE is the perpendicular bisector.

[3 MARKS]

#### SOLUTION

Solution :Each Option: 1 Mark

Given that:

Line l is perpendicular to line m.

a) We can clearly see from the above diagram that CE = 5-3 = 2 units

Also, EG = 7-5 = 2 units

∴ CE = EG = 2 units

b)As we have seen above CE = EG

So, E is the mid-point of line CG.

∴ Line PE bisects CG.

c) We can see from the above figure that

BE = 5-2 = 5 units

HE = 8-5 = 3 units

∴ E is the mid-point of the line BH.

Also given that line m is perpendicular to the line l.

∴ PE is the perpendicular bisector of the line segment BH.

Also, we have seen above that PE is also the perpendicular bisector of line segment CG.

∴ The required line segments are BH and CG.

### Question 6

In a moving car, there are 6 and a half rotations of a wheel per second. How many right angles did the wheel rotate through in one second? [3 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Answer: 1 Mark

Given that:

The wheel rotates 6 and a half times in one second.

We know that in one revolution there are 4 right angles.

∴ In 6 revolutions, the number of right angles = 6×4 = 24

∴ In half revolution, the number of right angles = 12×4 = 2

The total number of right angles = 24 + 2 = 26.

The wheel went through 26 right angles.

### Question 7

A man runs in a particular path in such a way that it is a quadrilateral. The opposite sides and opposite angles of the path are equal. Also, the opposite sides are parallel. Given that the adjacent sides are not equal, what is the quadrilateral called? [4 MARKS]

#### SOLUTION

Solution :Steps: 3 Marks

Quadrilateral's name: 1 Mark

Given that:

A man runs in a particular path, in such a way that it is a quadrilateral.

Also, the opposite sides are equal and opposite angles are equal.

Also, the opposite sides are parallel.

So the given quadrilateral can either be a rectangle, parallelogram, or a square.

But the adjacent sides are not equal.

Therefore it will either be a parallelogram or a rectangle.

If all the angles of the quadrilateral are 90∘, then the quadrilateral is a rectangle, otherwise, it is a parallelogram.

### Question 8

Write the name of the:

a. vertices

b. edges and

c. faces of the prism shown in the figure.

[4 MARKS]

#### SOLUTION

Solution :Each option: 1 Mark

Steps: 1 Mark

A vertex is a point where two edges meet.

An edge is a line segment where two faces meet.

A closed plane formed by line segments is called the face.

a)As we can clearly see in the above figure the various vertices are A, B, C, D, E, F.

b) The various edges are AB, AC, AE, BC, BD, CF, DE, DF.

c) The various faces are DEF, ACFE, BDFC, BAED, BAC.

### Question 9

Match the geometric shapes written below to its corresponding diagram.

a. perpendicular bisector

b. bisector

c. only bisector

d. only perpendicular

[4 MARKS]

#### SOLUTION

Solution :Matching: 1 Mark

a. Perpendicular bisector divides the line segment into two equal parts making a right angle. So, figure (ii) shows the perpendicular bisector.

b. In figure (ii) and (iii), the line segments are divided into two equal parts. So, they show bisectors.

c. In figure (iii), the line segment is divided into two equal parts only. So, it shows the only bisector.

d. In figure (i), the line segment is intersected at 90∘. So, it shows only a perpendicular.

### Question 10

In the given fig.,

∠AOD is a/an _____angle.

∠COA is a/an _____angle.

∠AOE ia a/an _____angle.

∠EOC ia a/an _____angle.

[4 MARKS]

#### SOLUTION

Solution :Each blank: 1 Mark

∠AOD = 40+20+30 = 90∘ = A right angle

∠COA = 20+30 = 50∘ = An acute angle

∠AOE = 40+40+20+30 = 130∘ = An obtuse angle

∠EOC = 40+40 = 80∘ = An acute angle