Exercise 14.4
Question 1: Draw any line segment . Mark any point M on it. Through M, draw a perpendicular to . (Use ruler and compasses)
Answer 1:
Steps of construction:
(i) Draw a line segment and mark a point M on it.
(ii) Taking M as centre and a convenient radius, construct an arc intersecting the line segment at points C and D respectively.
(iii) By taking centres as C and D and radius greater than CM, construct two arcs such that they intersect each other at point E.
(iv) Join EM. Now is perpendicular to
Question 2: Draw any line segment . Take any point R not on it. Through R, draw a perpendicular to . (Use ruler and setsquare)
Answer 2:
(i) Draw a given line segment and mark a point R outside the line segment
(ii) Place a set square on such that one of its right angles arm aligns along
(iii) Now, place the ruler along the edge opposite to right angle of set square.
(iv) Hold the ruler fixed. Slide the set square along the ruler such that the point R touches the other arm of set square.
(v) Draw a line along this edge of set square which passes through point R. Now, it is the required line perpendicular to
Question 3: Draw a line l and a point X on it. Through X, draw a line segment perpendicular to l. Now draw a perpendicular to to Y. (use ruler and compasses)
Answer 3:
Steps of construction:
(i) Draw a line l and mark a point X on it.
(ii) By taking X as centre and with a convenient radius, draw an arc intersecting the line l at points A and B respectively.
(iii) With A and B as centres and a radius more than AX, construct two arcs such that they intersect each other at point Y.
(iv) Join XY. Here is perpendicular to l
Similarly, by taking C and D as centres and radius more than CY, construct two arcs intersecting at point Z. Join ZY. The line is perpendicular to at Y
Exercise 14.5
Question 1: Draw of length 7.3 cm and find its axis of symmetry.
Answer 1: Axis of symmetry of line segment will be the perpendicular bisector of . So, draw the perpendicular bisector of .
Steps of construction:
(i) Draw a line segment = 7.3 cm
(ii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
(iii) Join CD. Then CD is the axis of symmetry of the line segment AB.
Question 2:
Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Answer 2:
Steps of construction:
(i) Draw a line segment = 9.5 cm
(ii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
(iii) Join CD. Then CD is the perpendicular bisector of .
Question 3: Draw the perpendicular bisector of whose length is 10.3 cm.
(a) Take any point P on the bisector drawn.
Examine whether PX = PY.
(b) If M is the midpoint of , what can you say about the lengths MX and XY?
Answer 3:
Steps of construction:
(i) Draw a line segment = 10.3 cm
(ii) Taking X and Y as centres and radius more than half of AB, draw two arcs which intersect each other at C and D.
(iii) Join CD. Then CD is the required perpendicular bisector of .
Now:
(a) Take any point P on the bisector drawn. With the help of divider we can check that
(b) If M is the midpoint of
Question 4:
Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Answer 4:
Steps of construction:
(i) Draw a line segment AB = 12.8 cm
(ii) Draw the perpendicular bisector of which cuts it at C. Thus, C is the midpoint of .
(iii) Draw the perpendicular bisector of which cuts it at D. Thus D is the midpoint of.
(iv) Again, draw the perpendicular bisector of which cuts it at E. Thus, E is the midpoint of .
(v) Now, point C, D and E divide the line segment in the four equal parts.
(vi) By actual measurement, we find that
Question 5:
With of length 6.1 cm as diameter, draw a circle.
Answer 5:
Steps of construction:
(i) Draw a line segment = 6.1 cm.
(ii) Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the midpoint of .
(iii) Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment .
Question 6:
Draw a circle with centre C and radius 3.4 cm. Draw any chord . Construct the perpendicular bisector and examine if it passes through C.
Answer 6:
Steps of construction:
(i) Mark any point C on the sheet
(ii) Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, turn compasses slowly to draw the circle. This is the required circle of 3.4 cm radius.
(iii) Mark any chord in the circle
(iv) Now, taking A and B as centres, draw arcs on both sides of . Let these intersect each other at points D and E.
(v) Join DE. Now DE is the perpendicular bisector of AB.
If is extended, it will pass through point C.
Question 7: Repeat Question 6, if happens to be a diameter.
Answer 7:
Steps of construction:
(i) Mark any point C on the sheet.
(ii) Adjust the compasses up to 3.4 cm and by putting the pointer of compasses at point C, Turn the compasses slowly to draw the circle. This is the required circle of 3.4 cm
(iii) Now mark any diameter in the circle.
(iv) Now taking A and B as centres, draw arcs on both sides of with radius more than . Let these intersect each other at points D and E.
(v) Join DE, which is perpendicular bisector of AB.
Now, we may observe that is passing through the centre C of the circle.
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 8:
Steps of construction:
(i) Mark any point C on the sheet. Now adjust the compasses up to 4 cm and by placing the pointer of compasses at point C, turn the compasses slowly to draw the circle. This is the required circle of 4 cm radius
(ii) Take any two chords in the circle
(iii) By taking A and B as centres and radius more than half of draw arcs on both sides of AB. The arcs are intersecting each other at point E and F. Join EF which is perpendicular bisector of AB.
(iv) Again take C and D as centres and radius more than half of draw arcs on both sides of CD such that they are intersecting each other at points G, H. Join GH which is perpendicular bisector of CD
We may observe that when EF and GH are extended they meet at the point O, which is the centre of 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 and . Let them meet at P. Is PA = PB?
Answer 9:
Steps of construction:
(i) Draw any angle with vertex as O.
(ii) By taking O as centre and with convenient radius, draw arcs on both rays of this angle. Let these points are A and B
(iii) Now take O and A as centres and with radius more than half of OA, draw arcs on both sides of OA. Let these intersects at points C and D respectively. Join CD
(iv) Similarly we may find which is perpendicular bisector of . These perpendicular bisectors intersects each other at point P. Now measure PA and PB. They are equal in length.
Exercise 14.6
Question 1:
Draw ∠ POQ of measure 75° and find its line of symmetry.
Answer 1:
Steps of construction:
(i) Draw a line l and mark a point O on it.
(ii) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line l at A.
(iii) Taking same radius, with centre A, cut the previous arc at B.
(iv) Join OB, then ∠ BOA = 60 .
(v) Taking same radius, with centre B, cut the previous arc at C.
(vi) Draw bisector of ∠BOC. The angle is of 90 . Mark it at D. Thus, ∠DOA = 90^{o}
(vii) Draw as bisector of ∠DOB. Thus, ∠POA = 75^{o.}
Question 2: Draw an angle of measure 147^{o} and construct its bisector.
Answer 2:
Steps of construction:
Following steps are followed to construct an angle of measure 147^{0} and its bisector
(i) Draw a line l and mark point O on it. Place the centre of protractor at point O and the zero edge along line l
(ii) Mark a point A at an angle of measure 147^{0}. Join OA. Now OA is the required ray making 147^{0} with line l
(iii) By taking point O as centre, draw an arc of convenient radius. Let this intersect both rays of angle 147^{0} at points A and B.
(iv) By taking A and B as centres draw arcs of radius more than 1 / 2 AB in the interior angle of 147^{0}. Let these intersect each other at point C. Join OC.
OC is the required bisector of 147^{0} angle
Question 3:
Draw a right angle and construct its bisector.
Answer 3:
Steps of construction:
(i) Draw a line PQ and take a point O on it.
(ii) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.
(iii) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.
(iv) Join OC. Thus, ∠COQ is the required right angle.
(v) Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.
(vi) Join OD. Thus, is the required bisector of ∠COQ.
Question 4:
Draw an angle of measure 153^{o} and divide it into four equal parts.
Answer 4:
Steps of construction:
Following steps are followed to construct an angle of measure 153^{0} and its bisector
(i) Draw a line l and mark a point O on it. Place the centre of protractor at point O and the zero edge along line l
(ii) Mark a point A at the measure of angle 153^{0}. Join OA. Now OA is the required ray making 153^{0} with line l
(iii) Draw an arc of convenient radius by taking point O as centre. Let this intersects both rays of angle 153^{0} at points A and B.
(iv) Take A and B as centres and draw arcs of radius more than 1 / 2 AB in the interior of angle of 153^{0}. Let these intersect each other at C. Join OC
(v) Let OC intersect major arc at point D. Draw arcs of radius more than 1 / 2 AD with A and D as centres and also D and B as centres. Let these are intersecting each other at points E and F respectively. Now join OE and OF
OF, OC, OE are the rays dividing 153^{0} angle into four equal parts.
Question 5:
Construct with ruler and compasses, angles of following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°
Answer 5:
Steps of construction:
(a) 60°
(i) Draw a ray .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Join OQ.
Thus, ∠BOA is required angle of 60° .
(b) 30^{o}
(i) Draw a ray .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Join OQ. Thus, ∠BOA is required angle of 60^{o}.
(v) Put the pointer on P and mark an arc.
(vi) Put the pointer on Q and with same radius, cut the previous arc at C. Thus, ∠COA is required angle of 30^{o}.
(c) 90°
(i) Draw a ray .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects at X.
(iii) Taking X as centre and same radius, cut previous arc at Y.
(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
(vi) Join OS and produce it to form a ray OB.
Thus, ∠BOA is required angle of 90^{0}.
(d) 120°
(i) Draw a ray .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Taking Q as centre and same radius cut the arc at S.
(v) Join OS.
Thus, ∠AOD is required angle of 120°.
(e) 45^{o}
(i) Draw a ray .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects at X.
(iii) Taking X as centre and same radius, cut previous arc at Y.
(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
(vi) Join OS and produce it to form a ray OB. Thus, ∠BOA is required angle of 90°.
(vii) Draw the bisector of ∠BOA.
Thus, ∠MOA is required angle of 45°.
(f) 135°
(i) Draw a line PQ and take a point O on it.
(ii) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.
(iii) Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.
(iv) Join OR. Thus, ∠QOR = ∠POQ = 90°.
(v) Draw the bisector of ∠POR.
Thus, ∠QOD is required angle of 135°.
Question 6:
Draw an angle of measure 45^{o} and bisect it.
Answer 6:
Steps of construction:
Following steps are followed to construct an angle of measure 45^{0} and its bisector.
(i) Using the protractor ∠POQ of 45^{0} measure may be formed on a line l
(ii) Draw an arc of convenient radius with centre as O. Let this intersects both rays of angle 45^{0} at points A and B
(iii) Take A and B as centres, draw arcs of radius more than 1 / 2 AB in the interior of angle of 45^{0}. Let these intersect each other at C. Join OC
OC is the required bisector of 45^{0} angle
Question 7:
Draw an angle of measure 135^{o} and bisect it.
Answer 7:
Steps of construction:
Following steps are followed to construct an angle of measure 135^{0} and its bisector.
(i) By using a protractor ∠POQ of 135^{0} measure may be formed on a line l
(ii) Draw an arc of convenient radius by taking O as centre. Let this intersect both rays of angle 135^{0} at points A and B respectively.
(iii) Take A and B as centres, draw arcs of radius more than 1 / 2 AB in the interior of angle of 135^{0}.
Let these intersect each other at C. Join OC.
OC is the required bisector of 135^{0} angle
Question 8:
Draw an angle of 70^{o}. Make a copy of it using only a straight edge and compasses.
Answer 8:
Steps of construction:
(i) Draw an angle 70^{o} with protractor, i.e., ∠POQ = 70^{o}
(ii) Draw a ray
(iii) Place the compasses at O and draw an arc to cut the rays of ∠POQ at L and M.
(iv) Use the same compasses, setting to draw an arc with A as centre, cutting AB at X.
(v) Set your compasses setting to the length LM with the same radius.
(vi) Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.
(vii) Join AY. Thus, ∠YAX = 70^{o}
Question 9:
Draw an angle of 40^{o}. Copy its supplementary angle.
Answer 9:
Steps of construction:
(i) Draw an angle of 40° with the help of protractor, naming ∠AOB.
(ii) Draw a line PQ.
(iii) Take any point M on PQ.
(iv) Place the compasses at O and draw an arc to cut the rays of ∠AOB at L and N.
(v) Use the same compasses setting to draw an arc O as centre, cutting MQ at X.
(vi) Set your compasses to length LN with the same radius.
(vii) Place the compasses at X and draw the arc to cut the arc drawn earlier Y.
(viii) Join MY.
Thus, ∠QMY = 40° and ∠PMY is supplementary of it.
120 videos301 docs39 tests

1. What are the basic tools required for practical geometry? 
2. How can we construct an angle bisector using a compass and ruler? 
3. Can we construct a triangle if we know the lengths of its three sides? 
4. How can we construct a perpendicular bisector of a line segment? 
5. Can we construct a square if we know the length of its diagonal? 
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