Draw a line segment AB 6.1 cm. Draw a perpendicular bisector of it and mark the mid point of AB as M. With M as centre draw a circle whose diameter is AB. Pls tell me the ans quick those who tell right answer i will mark them as brainlist
Answers
Answer:
Draw a circle of radius 3.2 cm.
Answer:
Steps of construction:
(a) Open the compass for the required radius of 3.2 cm.
(b) Make a point with a sharp pencil where we want the centre of circle to be.
(c) Name it O.
(d) Place the pointer of compasses on O.
(e) Turn the compasses slowly to draw the circle.
Hence, it is the required circle.
Class_6_Practical_Geometry_Construction_Of_A_Circle_With_Radius_3.2cm
Question 2:
With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Answer:
Steps of construction:
(a) Marks a point O with a sharp pencil where we want the centre of the circle.
(b) Open the compasses 4 cm.
Class_6_Practical_Geometry_Two_Cocentric_Circles
(c) Place the pointer of the compasses on O.
(d) Turn the compasses slowly to draw the circle.
(e) Again open the compasses 2.5 cm and place the pointer of the compasses on D.
(f) Turn the compasses slowly to draw the second circle.
Hence, it is the required figure.
Question 3:
Draw a circle and any two of its diameters. If you join the ends of these diameters, what
is the figure obtained if the diameters are perpendicular to each other? How do you check
your answer?
Answer:
(i) By joining the ends of two diameters, we get a rectangle. By measuring, we find AB = CD = 3
cm, BC = AD = 2 cm, i.e., pairs of opposite sides are equal and also
angle A = angle B = angle C = angle D = 90 degree
i.e. each angle is of 90 degree. Hence, it is a rectangle.
Class_6_Practical_Geometry_Construction_Of_A_Circle1
(ii) If the diameters are perpendicular to each other, then by joining the ends of two
diameters, we get a square.
By measuring, we find that AB = BC = CD = DA = 2.5 cm, i.e., all four sides are equal
Also, angle A = angle B = angle C = angle D = 90 degree
i.e. each angle is of 90 degree. Hence, it is a square.
Class_6_Practical_Geometry_Construction_Of_A_Circle2
Question 4:
Draw any circle and mark points A, B and C such that:
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.
Answer:
(i) Mark a point O with sharp pencil where we want centre of the circle.
(ii) Place the pointer of the compasses at O. Then move the compasses slowly to draw a circle.
(a) Point A is on the circle.
(b) Point B is in interior of the circle.
(c) Point C is in the exterior of the circle
Class_6_Practical_Geometry_Construction_Of_A_Circle
Question 5:
Let A, B be the centers of two circles of equal radii; draw them so that each one of them passes through the centre of the other.
Let them intersect at C and D. Examine whether AB and CD are at right angles.
Answer:
Draw two circles of equal radii taking A and B as their centre such that one of them passes
through the centre of the other. They intersect at C and D. Join AB and CD.
Yes, AB and CD intersect at right angle as angle COB is 90 degree.
Class_6_Practical_Geometry_Two_Circles_With_SameRadii
(iv) Join PQ. Then PQ is the perpendicular bisector of AB.
(v) We observe that this perpendicular bisector of AB passes through the centre C of the circle.
Class_6_Practical_Geometry_Construction_Of_A_Circle_With_Diameter
Question 8:
Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Answer:
(i) Draw the circle with O and radius 4 cm.
(ii) Draw any two chords AB and CD in this circle.
(iii) Taking A and B as centres and radius more than half AB,
draw two arcs which intersect each other at E and F.
(iv) Join EF. Thus EF is the perpendicular bisector of chord CD.
(v) Similarly draw GH the perpendicular bisector of chord CD.
(vi) These two perpendicular bisectors meet at O, the centre of the circle.
Question 9:
Draw any angle with vertex O. Take a point A on one of its arms and B on another such that OA = OB. Draw the perpendicular bisectors of OA and OB.
Let them meet at P. Is PA = PB?
Answer:
Steps of construction:
(i) Draw any angle with vertex O.
(ii) Take a point A on one of its arms and B on another
such that OA = OB.
(iii) Draw perpendicular bisector of OA and OB.
(iv) Let them meet at P. Join PA and PB.
(v) With the help of divider, we check that PA = PB.