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NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry

Exercise 14.4
Question 1: Draw any line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. Mark any point M on it. Through M, draw a perpendicular to NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. (Use ruler and compasses)
Answer 1:
Steps of construction:
(i) Draw a line segmentNCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry and mark a point M on it.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
(ii) Taking M as centre and a convenient radius, construct an arc intersecting the line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry at points C and D respectively.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(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.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
(iv) Join EM. Now NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometryis perpendicular toNCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
Question 2: Draw any line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. Take any point R not on it. Through R, draw a perpendicular to NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. (Use ruler and set-square)
Answer 2:
(i) Draw a given line segmentNCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry and mark a point R outside the line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
(ii) Place a set square on NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry such that one of its right angles arm aligns along NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iii) Now, place the ruler along the edge opposite to right angle of set square.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iv) Hold the ruler fixed. Slide the set square along the ruler such that the point R touches the other arm of set square.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(v) Draw a line along this edge of set square which passes through point R. Now, it is the required line perpendicular toNCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry


Question 3:  Draw a line l and a point X on it. Through X, draw a line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry perpendicular to l. Now draw a perpendicular to NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry to Y. (use ruler and compasses) 
Answer 3: 
Steps of construction:
(i) Draw a line l and mark a point X on it.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(ii) By taking X as centre and with a convenient radius, draw an arc intersecting the line l at points A and B respectively.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(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.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iv) Join XY. Here NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry is perpendicular to l
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical GeometrySimilarly, by taking C and D as centres and radius more than CY, construct two arcs intersecting at point Z. Join ZY. The line NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry is perpendicular to NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry at Y
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
Exercise 14.5
Question 1: Draw NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry of length 7.3 cm and find its axis of symmetry.
Answer 1: Axis of symmetry of line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry will be the perpendicular bisector of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. So, draw the perpendicular bisector of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
Steps of construction:
(i) Draw a line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry = 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 NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry = 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 NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry .

NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry


Question 3: Draw the perpendicular bisector of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry whose length is 10.3 cm.
(a) Take any point P on the bisector drawn.
Examine whether PX = PY.
(b) If M is the mid-point of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry, what can you say about the lengths MX and XY?
Answer 3:
Steps of construction:

NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical GeometryNCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(i) Draw a line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry = 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 NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry .
Now:
(a) Take any point P on the bisector drawn. With the help of divider we can check that NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
(b) If M is the mid-point of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry


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:
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(i) Draw a line segment AB = 12.8 cm
(ii) Draw the perpendicular bisector of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry which cuts it at C. Thus, C is the midpoint of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry.
(iii) Draw the perpendicular bisector of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry which cuts it at D. Thus D is the midpoint of.
(iv) Again, draw the perpendicular bisector of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry which cuts it at E. Thus, E is the mid-point of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry.
(v) Now, point C, D and E divide the line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry in the four equal parts.
(vi) By actual measurement, we find that NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry


Question 5:
With NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry of length 6.1 cm as diameter, draw a circle.
Answer 5:
Steps of construction:
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
(i) Draw a line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry = 6.1 cm.
(ii) Draw the perpendicular bisector of PQ which cuts, it at O. Thus O is the mid-point of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry .
(iii) Taking O as centre and OP or OQ as radius draw a circle where diameter is the line segment NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry .


Question 6:
Draw a circle with centre C and radius 3.4 cm. Draw any chord NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. Construct the perpendicular bisector NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry

(iii) Mark any chord NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry in the circle
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iv) Now, taking A and B as centres, draw arcs on both sides of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. Let these intersect each other at points D and E.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(v) Join DE. Now DE is the perpendicular bisector of AB.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
If NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry is extended, it will pass through point C.

Question 7: Repeat Question 6, if NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iii) Now mark any diameter NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry in the circle.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iv) Now taking A and B as centres, draw arcs on both sides of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry with radius more than NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry.  Let these intersect each other at points D and E.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(v) Join DE, which is perpendicular bisector of AB.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
Now, we may observe that NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(ii) Take any two chords NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry in the circle
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iii) By taking A and B as centres and radius more than half of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iv) Again take C and D as centres and radius more than half of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry

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 NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry and NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. Let them meet at P. Is PA = PB?
Answer 9:
Steps of construction:
(i) Draw any angle with vertex as O.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(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
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(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
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(iv) Similarly we may find NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry which is perpendicular bisector of NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry. These perpendicular bisectors NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry intersects each other at point P. Now measure PA and PB. They are equal in length.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry


Exercise 14.6
Question 1:
Draw ∠ POQ of measure 75° and find its line of symmetry.
Answer 1:
Steps of construction:
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(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 = 90o 
(vii) Draw NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry as bisector of ∠DOB. Thus, ∠POA = 75o.


Question 2: Draw an angle of measure 147o  and construct its bisector.
Answer 2:
Steps of construction:
Following steps are followed to construct an angle of measure 1470 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 1470. Join OA. Now OA is the required ray making 1470 with line l
(iii) By taking point O as centre, draw an arc of convenient radius. Let this intersect both rays of angle 1470 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 1470. Let these intersect each other at point C. Join OC.
OC is the required bisector of 1470 angle
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry


Question 3:
Draw a right angle and construct its bisector.
Answer 3:
Steps of construction:
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(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, NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry is the required bisector of ∠COQ.


Question 4:
Draw an angle of measure 153o and divide it into four equal parts.
Answer 4:
Steps of construction:
Following steps are followed to construct an angle of measure 1530 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 1530. Join OA. Now OA is the required ray making 1530 with line l
(iii) Draw an arc of convenient radius by taking point O as centre. Let this intersects both rays of angle 1530 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 1530. 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 1530 angle into four equal parts.
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry

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°
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(i) Draw a ray NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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) 30o

NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(i) Draw a ray NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry at P.
(iii) Taking P as centre and same radius, cut previous arc at Q.
(iv) Join OQ. Thus, ∠BOA is required angle of 60o.
(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 30o.

(c) 90°
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(i) Draw a ray NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry.
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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 900.

(d) 120°
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(i) Draw a ray NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry.
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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) 45o 
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(i) Draw a ray NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry .
(ii) Taking O as centre and convenient radius, mark an arc, which intersects NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry 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°

NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry

(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 NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry the bisector of ∠POR.
Thus, ∠QOD is required angle of 135°.

Question 6:
Draw an angle of measure 45o and bisect it.
Answer 6:
Steps of construction:
Following steps are followed to construct an angle of measure 450 and its bisector.
(i) Using the protractor ∠POQ of 450 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 450 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 450. Let these intersect each other at C. Join OC
OC is the required bisector of 450 angle
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
Question 7:
Draw an angle of measure 135o and bisect it.
Answer 7:
Steps of construction:
Following steps are followed to construct an angle of measure 1350 and its bisector.
(i) By using a protractor ∠POQ of 1350 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 1350 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 1350.
Let these intersect each other at C. Join OC.
OC is the required bisector of 1350 angle
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
Question 8:
Draw an angle of 70o. Make a copy of it using only a straight edge and compasses.
Answer 8:
Steps of construction:
(i) Draw an angle 70o with protractor, i.e., ∠POQ = 70o
(ii) Draw a ray NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry
(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 = 70o

Question 9:
Draw an angle of 40o. Copy its supplementary angle.
Answer 9:
Steps of construction:
NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical GeometryNCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry(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.

The document NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry is a part of the Class 6 Course Mathematics (Maths) Class 6.
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FAQs on NCERT Solutions for Class 8 Maths chapter 14 - (Ex: 14.4 to 14.6): Practical Geometry

1. What are the basic tools required for practical geometry?
Ans. The basic tools required for practical geometry are a compass, a ruler, a protractor, and a set square. These tools are essential for constructing various geometrical shapes and measuring angles and lengths accurately.
2. How can we construct an angle bisector using a compass and ruler?
Ans. To construct an angle bisector using a compass and ruler, follow these steps: 1. Draw an angle using a compass and ruler. 2. Place the compass on the vertex of the angle and draw two arcs that intersect both sides of the angle. 3. Using a ruler, draw a straight line that connects the vertex of the angle to the point where the two arcs intersect. 4. The line drawn in step 3 is the angle bisector.
3. Can we construct a triangle if we know the lengths of its three sides?
Ans. Yes, we can construct a triangle if we know the lengths of its three sides. This is known as the SSS (side-side-side) criterion. To construct a triangle using the SSS criterion, draw a line segment of the given length for each side, and then connect the endpoints of all three line segments to form the triangle.
4. How can we construct a perpendicular bisector of a line segment?
Ans. To construct a perpendicular bisector of a line segment, follow these steps: 1. Draw a line segment using a ruler. 2. Place the compass on one endpoint of the line segment and draw an arc that intersects the line segment. 3. Without changing the compass width, place the compass on the other endpoint of the line segment and draw another arc that intersects the line segment. 4. Using a ruler, draw a straight line that connects the two points where the arcs intersect the line segment. 5. The line drawn in step 4 is the perpendicular bisector of the line segment.
5. Can we construct a square if we know the length of its diagonal?
Ans. Yes, we can construct a square if we know the length of its diagonal. This is known as the diagonal method. To construct a square using the diagonal method, draw the diagonal of the square using a ruler. Then, place the compass on one endpoint of the diagonal and draw an arc that intersects the diagonal. Without changing the compass width, place the compass on the other endpoint of the diagonal and draw another arc that intersects the diagonal. Connect the two points where the arcs intersect the diagonal to form a side of the square. Repeat this process for the other three sides of the square.
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