{"id":77582,"date":"2022-02-12T16:44:37","date_gmt":"2022-02-12T11:14:37","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=77582"},"modified":"2025-05-29T10:03:44","modified_gmt":"2025-05-29T04:33:44","slug":"ncert-solutions-for-class-6-maths-practical-geometry-exercise-14-5","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-material\/cbse-notes\/class-6\/maths\/ncert-solutions-for-class-6-maths-practical-geometry-exercise-14-5\/","title":{"rendered":"NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.5"},"content":{"rendered":"

NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.5<\/h2>\n

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry<\/a> Ex 14.5<\/strong><\/p>\n

Exercise 14.5<\/strong><\/p>\n

Ex 14.5 Class 6 Maths Question 1.
\nDraw AB of length 7.3 cm and find its axis of symmetry.
\nSolution:
\nStep I: Draw \\(\\overline { AB }\\) = 7.3 cm
\n\"NCERT
\nStep II: Taking A and B as centre and radius more than half of \\(\\overline { AB }\\), draw two arcs which intersect each other at C and D.
\nStep III: Join C and D to intersect \\(\\overline { AB }\\) at E. Thus, CD is the perpendicular bisector or axis of symmetry of \\(\\overline { AB }\\).<\/p>\n

Ex 14.5 Class 6 Maths Question 2.
\nDraw a line segment of length 9.5 cm and construct its perpendicular bisector.
\nSolution:
\nStep I: Draw a line segment \\(\\overline { PQ }\\) =9.5 cm
\n\"NCERT
\nStep II: With centres P and Q and radius more than half of PQ, draw two arcs which meet each other at R and S.
\nStep III: Join R and S to meet \\(\\overline { PQ }\\) at T.
\nThus, RS is the perpendicular bisector of PQ.<\/p>\n

Ex 14.5 Class 6 Maths Question 3.
\nDraw the perpendicular bisector of \\(\\overline { XY }\\) whose length is 10.3 cm.
\n(a) Take any point P on the bisector drawn. Examine whether PX = PY.
\n(b) If M is the midpoint of \\(\\overline { XY }\\) . What can you say about the length of MX and MY?
\nSolution:
\nStep I: Draw a line segment \\(\\overline { XY }\\) = 10.3 cm.
\n\"NCERT
\nStep II : With centre X and Y and radius more than half of XY, draw two arcs which meet each other at U and V.
\nStep III: Join U and V which meets \\(\\overline { XY }\\) at M.
\nStep IV: Take a point P on \\(\\overline { UV }\\) .
\n(a) On measuring, PX = PY = 5.6 cm.
\n(b) On measuring, \\(\\overline { MX }\\) = \\(\\overline { MY }\\) = \\(\\frac { 1 }{ 2 }\\) XY = 5.15 cm.<\/p>\n

Ex 14.5 Class 6 Maths Question 4.
\nDraw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
\nSolution:
\nStep I: Draw a line segment \\(\\overline { AB }\\) = 12.8 cm
\n\"NCERT
\nStep II : With centre A and B and radius more than half of AB, draw two arcs which meet each other at D and E.
\nStep III : Join D and E which meets \\(\\overline { AB }\\) at C which is the midpoint of \\(\\overline { AB }\\).
\nStep IV : With centre A and C and radius more than half of AC, draw two arcs which meet each other at F and G.
\nStep V: Join F and G which meets \\(\\overline { AC }\\) at H which is the midpoint of \\(\\overline { AC }\\) .
\nStep VI : With centre C and B and radius more than half of CB, draw two arcs which meet each other at J and K.
\nStep VII : Join J and K which meets \\(\\overline { CB }\\) at L which is the midpoint of \\(\\overline { CB }\\) .
\nThus, on measuring, we find
\n\\(\\overline { AH }\\) = \\(\\overline { HC }\\) = \\(\\overline { CL }\\) = \\(\\overline { LB }\\) = 3.2 cm.<\/p>\n

Ex 14.5 Class 6 Maths Question 5.
\nWith \\(\\overline { PQ }\\) of length 6.1 cm as diameter, draw a circle.
\nSolution:
\nStep I: Draw \\(\\overline { PQ }\\) = 6.1 cm
\nStep II: Draw a perpendicular bisector of \\(\\overline { PQ }\\) which meets \\(\\overline { PQ }\\) at R i.e. R is the midpoint of \\(\\overline { PQ }\\).
\n\"NCERT
\nStep III : With centre R and radius equal to \\(\\overline { RP }\\) , draw a circle passing through P and Q.
\nThus, the circle with diameter \\(\\overline { PQ }\\) = 6.1 cm is the required circle.<\/p>\n

Ex 14.5 Class 6 Maths Question 6.
\nDraw a circle with centre C and radius 3.4 cm. Draw any chord \\(\\overline { AB }\\) . Construct the perpendicular bisector of \\(\\overline { AB }\\) and examine if it passes through C.
\nSolution:
\nStep I: Draw a circle with centre C and radius 3.4 cm.
\nStep II: Draw any chord \\(\\overline { AB }\\).
\nStep III : Draw the perpendicular bisector of \\(\\overline { AB }\\) which passes through the centre C.
\n\"NCERT<\/p>\n

Ex 14.5 Class 6 Maths Question 7.
\nRepeat Question number 6, if \\(\\overline { AB }\\) happens to be a diameter.
\nSolution:
\nStep I: Draw a circle with centre C and radius 3.4 cm.
\nStep II : Draw a diameter AB of the circle.
\n\"NCERT
\nStep III : Draw a perpendicular bisector of AB which passes through the centre C and on measuring, we find that C is the midpoint of \\(\\overline { AB }\\) .<\/p>\n

Ex 14.5 Class 6 Maths Question 8.
\nDraw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
\nSolution:
\nStep I: Draw a circle with centre 0 and radius 4 cm.
\n\"NCERT
\nStep II: Draw any two chords \\(\\overline { AB }\\) and \\(\\overline { CD }\\) of the circle.
\nStep III : Draw the perpendicular bisectors of \\(\\overline { AB }\\) and \\(\\overline { CD }\\) i.e. I and m.
\nStep IV : On producing the two perpendicular bisectors meet each other at the centre O of the circle.<\/p>\n

Ex 14.5 Class 6 Maths Question 9.
\nDraw 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 \\(\\overline { OA }\\) and \\(\\overline { OB }\\) . Let them meet at P. Is PA = PB?
\nSolution:
\nStep I: Draw an angle XOY with O as its vertex.
\nStep II : Take any point A on OY and B on OX, such that OA + OB.
\n\"NCERT
\nStep III : Draw the perpendicular bisectors of OA and OB which meet each other at a point P.
\nStep IV : Measure the lengths of \\(\\overline { PA }\\) and \\(\\overline { PB }\\). Yes, \\(\\overline { PA }\\) = \\(\\overline { PB }\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.5 NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"Class 6 Maths Practical Geometry Exercise 14.5","_yoast_wpseo_title":"","_yoast_wpseo_metadesc":"Class 6 Maths Practical Geometry Exercise 14.5. Free PDF download of NCERT Solutions solved by expert Maths teachers on Infinitylearn.com","custom_permalink":"study-material\/cbse-notes\/class-6\/maths\/ncert-solutions-for-class-6-maths-practical-geometry-exercise-14-5\/"},"categories":[92,19,21],"tags":[],"table_tags":[],"acf":[],"yoast_head":"\nNCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.5 - Infinity Learn by Sri Chaitanya<\/title>\n<meta name=\"description\" content=\"Class 6 Maths Practical Geometry Exercise 14.5. 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