Free Practical Geometry 02 Practice Test - 6th grade
Question 1
Draw a circle with center O and radius 6cm. Draw two non-intersecting circles with radii 3cm each and centers as P and Q respectively passing through O. Now, join P and Q. Draw the diameter of the bigger circle RS such that it is perpendicular to PQ. Join PQRS.
PQRS is a:
Rhombus
Rectangle
Square
Trapezium
SOLUTION
Solution : A
Draw the circle with center O using compass and ruler. Now, since smaller circles are non intersecting and passing through the point O, there is only one possibility.
After joining PQ, drawing diameter RS perpendicular to PQ and joining PQRS, you will get:
PQRS is a rhombus.
Question 2
At what time shown in the options, the hour hand and the minute hand of your clock will be perpendicular to each other?
09:00
06:15
03:30
11:00
SOLUTION
Solution : A
At 09:00, hour hand and minute hand are at right angles to each other. At all the other instances, the minute hand has moved, so the hour hand would have moved too. Hence, the angle formed will not be 90∘ in those cases.
Question 3
Which of the following is the correct representation of line segment AB?
AB
¯¯¯¯¯¯¯¯AB
←→AB
→AB
SOLUTION
Solution : B
Line segments are represented with bar on top of their name, which contains two letters which are the two ends.
Question 4
Which one of the following angles can not be constructed using compass and ruler?
SOLUTION
Solution : C
We can draw 60∘ and bisect it to get 30∘.
Similarly, we can bisect 30∘ and get 15∘.
Also, we can draw 90∘ and bisect it to get 45∘.
But, 50∘ cannot be drawn using compass and ruler to get its bisected angle of 25∘.
Question 5
If you have two (45∘-45∘-90∘) set squares. Which of the following shapes is impossible to produce using them?
Square
Parallelogram
Pentagon
Trapezium
SOLUTION
Solution : D
We know that all squares are rhombuses and parallelograms. We can also make a pentagon as shown in the figure. But we can never get a trapezium using 2 set squares with angles (45∘-45∘-90∘).
Question 6
Draw a line AB. At A, draw an arc of length 3cm using compass such that it intersects AB at O. With the same spread of compass, put the compass pointer at O and make an arc that intersects the previous arc at P. Join A and P. Using the protractor, measure ∠PAB. What is the value of ∠PAB.
15∘
30∘
60∘
75∘
SOLUTION
Solution : C
If you follow all the steps, you would get this:
If you measure ∠PAB, the value which you would get is 60∘.
Question 7
If ∠ABC = 90∘, ∠ABC is also known as a/an
SOLUTION
Solution :∠ABC is also known as a right angle if it is equal to 90∘.
Question 8
Draw a circle with radius 4cm and center P. Draw another circle of same radius and with center at Q such that it intersects the previous circle. Join PQ. Now join the points of intersection and name this line segment as AB. Mark the point of intersection of AB and PQ as O.
∠AOP =
SOLUTION
Solution :If you draw step by step as given in the question, you would get:
We can clearly see that ∠AOP = 90∘.
Question 9
Draw a line segment AB of length 6.4cm. Mark a point C on the line segment such that AC=3cm. It is impossible to copy line segment CB without using a ruler or a compass or a divider.
True
False
SOLUTION
Solution : B
It is possible to copy line segment CB without using a ruler or a compass or a divider by tracing it on another paper. It may not be that accurate, but it is possible.
Question 10
It is not possible to draw the perpendicular bisector of a line segment using a protractor ruler with no markings on it.
True
False
SOLUTION
Solution : B
Take a line segment say AB. Make angles 60∘ on both the ends such that the lines along those angles intersect at C. Taking AB as the base side, draw 30∘ angle and a line along it such that it intersects AB at D. Point D bisects AB and ∠ADB is 90∘. So, AD is the perpendicular bisector of AB.
