4) Using ruler and compass only, construct a triangle DEF in which DE = 5.2 cm,
∠D = 50° and ∠E = 75°.
(i) Construct perpendicular bisectors of any two sides and mark the point of
intersection of perpendicular bisectors as O.
(ii) Construct a circle with point O as centre and radius OD.
Answers
Answer:
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Explanation:
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.
Steps of construction:
(i) Draw a circle with centre C and radius 3.4 cm.
(ii) Draw any chord AB.
(iii) Taking A and B as centres and radius more than half of AB,
draw two arcs which cut each other at P and Q.
(iv) Join PQ. Then PQ is the perpendicular bisector of AB.
(v) This perpendicular bisector of AB passes through the centre C of the circle.
Question 7:
Repeat Question 6, if AB happens to be a diameter.
Answer:
Steps of construction:
(i) Draw a circle with centre C and radius 3.4 cm.
(ii) Draw its diameter AB.
(iii) Taking A and B as centres and radius more than half of it,
draw two arcs which intersect each other at P and Q.
(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.
Exercise 14.6
Question 1:
Draw angle POQ of measure 750 and find its line of symmetry.
Answer:
Class_6_Practical_Geometry_Angle_75_And_Find_Its_LineOfSymmetry
Steps of construction:
(a) Draw a line l and mark a point O on it.
(b) Place the pointer of the compasses at O and draw an arc
of any radius which intersects the line l at A.
(c) Taking same radius, with centre A, cut the previous arc at B.
(d)Join OB, then angle BOA = 600.
(e) Taking same radius, with centre B, cut the previous arc at C.
(f) Draw bisector of angle BOC. The angle is of 900. Mark it at D. Thus, angle DOA = 900
(g) Draw OP as bisector of angle DOB.
Thus, angle POA = 750
Question 2:
Draw an angle of measure 1470 and construct its bisector.
Answer:
Class_6_Practical_Geometry_Angle_147_And_Bisect_It
Steps of construction:
(a) Draw a ray OA.
(b) With the help of protractor, construct angle AOB = 1470.
(c) Taking centre O and any convenient radius, draw an arc
which intersects the arms OA and OB at P and Q respectively.
(d) Taking P as centre and radius more than half of PQ, draw an arc.
(e) Taking Q as centre and with the same radius, draw another arc which intersects the
previous at R.
(f) Join OR and produce it.
(g) Thus, OR is the required bisector of angle AOB.