# KSEEB Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4

Students can Download Chapter 14 Practical Geometry Ex 14.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Maths Chapter 14 Practical Geometry Ex 14.4

Question 1.
Draw any line segment $$\overline{\mathbf{A B}}$$. Mark any point M on it. Through M, draw a perpendicular to $$\overline{\mathbf{A B}}$$. (use ruler and compasses)
Solution:
(1) Draw the given line segment $$\overline{\mathbf{A B}}$$ and mark an point M on it.
(2) With M as centre and a convenient radius, construct an arc intersecting the line segment $$\overline{\mathbf{A B}}$$ at two points C and D.

(3) With C and D as centres and a radius greater than CM, construct two arcs. Let these be intersecting each other at E.
(4) Join EM. $$\overline{\mathrm{EM}}$$ is perpendicular to AB

Question 2.
Draw a line segment $$\overline{\mathrm{AB}}$$ Take any point R not on it. Through R, draw a perpendicular to $$\overline{\mathrm{AB}}$$. (use ruler and set-square)
Solution:
(1) Take the given line segment $$\overline{\mathrm{PQ}}$$ and mark any point R outside $$\overline{\mathrm{PQ}}$$.
(2) Place a set square on $$\overline{\mathrm{PQ}}$$ such that one arm of its right angle aligns along $$\overline{\mathrm{PQ}}$$.

(3) Place the ruler along the edge opposite to the right P angle of the set square.
(4) Hold the ruler fixed. Slide the set square along the ruler till the point R touches the other arm of the set square.
(5) Draw a line along this edge of the set square which will be passing through R. It is the required line, which is perpendicular to $$\overline{\mathrm{PQ}}$$.

Question 3.
Draw a line l and a point X on it. Through X, draw a line segment $$\overline{\mathrm{XY}}$$ perpendicular to l.
Now draw a perpendicular to $$\overline{\mathrm{XY}}$$ at Y. (use ruler and compasses)
Solution :
(1) Draw a line l and mark a point X on it.
(2) Taking X as centre and with a convenient radius, draw an arc intersecting line 1 at two points A and B.
(3) With A and B as centres and a radius more than AX, construct two arcs intersecting each other at Y.

(4) Join XY. $$\overline{\mathrm{XY}}$$ is perpendicular to l.
Similarly, a perpendicular to $$\overline{\mathrm{XY}}$$ at point Y can be drawn. The line $$\overline{\mathrm{ZY}}$$ is perpendicular t0 $$\overline{\mathrm{XY}}$$ at Y.

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