RD Sharma Solutions Class 6 Geometrical Constructions Exercise 19.2

RD Sharma Class 6 Solutions Chapter 19 Ex 19.2 PDF Free Download

RD Sharma Solutions Class 6 Chapter 19 Exercise 19.2

Exercise 19.2

Q 1. How many lines can be drawn which are perpendicular to a given line and pass through a given point lying  

(i) outside it ?

Solution: The perpendicular line from a given point to a given line is the shortest distance between them. Only one shortest distance is possible.

Thus, only one perpendicular line is possible from the given point (outside the line) to a given line.

(ii) on it?

Solution: At any point on the line, we can draw only one perpendicular line.

Thus, on the given line on a point, we can draw only one perpendicular line.

Q 2. Draw a line PQ. Take a point R on it. Draw a line perpendicular to PQ and passing through R.

(i) Using ruler and set square

Solution:

Step I Draw a line PQ and take a point R on it.

Step II Place the ruler with one of its long edges lying on the given line PQ.

Step III Hold the ruler and place above it a set-square, such that its one arm of its right angle in contact with the line PQ .

Step IV Slide the set-square along the edge of the ruler until the vertex of its right angle in contact with the ruler.

Step V Without the position of the ruler, remove the set-square and draw a line XY through point R.

Thus, XY is the required line perpendicular to line PQ passing through R as shown in figure below;

RD sharma class 6 maths solution chapter 19 exercise 19.2 ans 1

(ii) Using ruler and compasses

Solution: 

Step I Draw a line PQ and take a point R on it.

Step II With R as centre and a convenient radius construct an arc touching the line PQ at two points A and B.

Step III With A and B as centres and a radius greater than RA construct two arcs cutting each other at S.

Step IV Join RS and extend it in both directions to get a line perpendicular to PQ.

Thus XY is the line required which is perpendicular to PQ and passes through R.

RD sharma class 6 maths solution chapter 19 exercise 19.2 Qs 1

 

Q 3. Draw a line l, take a point A not lying on l. Draw a line m such that \(m \perp l\) and passing through A. 

(i) Using ruler and set square 

Solution:

Step I Draw a line l and take a point A outside it.

Step II Place a set square PQR such that its one arm PQ of the right angle is along the line l.

Step III Hold the set square fixed and a ruler along the edge PR opposite to the right angle of the set square.

Step IV Holding the ruler fixed, slide the set-square along the ruler until the other arm QR reaches the point A.

Step V Without disturbing the position of the set-square, draw a line m along the edge QR.

The line m is the required line perpendicular to line l.

RD sharma class 6 maths solution chapter 19 exercise 19.2 ans 3

(ii) Using ruler and compass.

Step I With A as centre, draw an arc PQ, which intersects line l at points P and Q.

Step II Without disturbing the compass and taking P and Q as centres, we construct two arcs such that they intersect each other at point say B, on the other side.

Step III Join point A and B and extend it in both directions.

The required perpendicular line is AB as shown in the figure fig;

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Q 4. Draw a line AB and take two points C and E on opposite sides of AB. Through C, draw CD ⊥ AB and through E draw EF ⊥ AB. Using (i) ruler and set square (ii) ruler and compass.

Solution: (i) ruler and set square

Step I Draw a line AB and take two points C and E on the opposite sides of the line AB.

Step II On the side of E, place a set-square PQR, such that its one arm PQ of the right angle is along the line AB. Without disturbing the position of the set-square, place a ruler along its edge PR.

Step III Holding the ruler fixed, slide the set-square along the ruler until the other arm QR reaches the point C.

Step IV Without disturbing the position of the set-square, draw a line CD, where D is a point on AB.

CD is the required line and CD⊥ AB. We repeat the same process starting with taking set-square on the side of E, we draw a line EF  AB.

(ii) ruler and compass.

Step I Draw a line AB and take two points C and E on its opposite sides.

Step II With C as centre, draw an arc PQ, which intersects line AB at P and Q.

Step III Without disturbing the compass and taking P and Q as centres, construct two arcs, such that they intersect each other at H.

Step IV Join points H and C. HC crosses line AB at D. Thus we got CD AB.

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Step V Similarly, take E as centre and draw an arc RS.

Step VI Taking R and S as centres, draw two arcs which intersect each other at G.

Step VII Join points G and E. GE crosses AB at F.

We have EF ⊥ AB.

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Q5. Draw a line segment AB of length 10 cm. Mark a point P on AB such that AP = 4 cm. Mark a point P on AB such that AP = 4 cm. Draw a line through P perpendicular to AB.

Solution:

Step I We draw line L and take a point A on it.

Step II Using a ruler and a compass, we mark a point B, 10 cm from A, on the line L. AB is the required line segment of 10 cm.

Step III Again, we mark a point P, which is 4 cm from A, in the direction of B.

Step IV With P as centre, take a radius of 4 cm and construct an arc intersecting the line L at two points A and E.

Step V With A and E as centres, take a radius of 6 cm and construct two arcs intersecting each other at R.

Step VI We join PR and extend it. PR is the required line, which is perpendicular to AB.

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Q 6. Draw a line segment PQ of length of length 12 cm. Mark a point O outside this segment. Draw a line through O perpendicular to PQ.

Solution:

Step I Draw a line L and take a point P on it.

Step II Using a ruler and a compass, mark a point Q on the line L, where PQ = 12 cm.

Step III Mark a point Q outside PQ.

Step IV Now, with O as centre, draw an arc of appropriate radius such that the arc cuts the line at points A and B.

Step V Taking A and B as centres, construct two arcs such that they intersect each other at C. Join the points O and C.

Thus, OC is the required line which is perpendicular to PQ.

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Q 7. Using a protractor, draw ∠BAC of measure 700. On side AC, take a point P, such that AP = 2cm. From P draw a line perpendicular to AB.

Solution:

Step I Draw a line segment AC.

Step II Take a protractor and place it on the segment AC such that segment AC coincides with the line of diameter of protractor and middle of this line coincides with point A.

Step III Counting from the right side, mark the point as B at the point of 70° of the protractor and draw AB. (iii) Now, measuring 2 cm from A on AC, mark a point P.

Step IV With P as centre, draw an arc intersecting line 1 at points E and F.

Step V Using the same radius and E and F as centres, construct two arcs that intersect at point G on the other side. Join the point P and G.

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Q 8. Draw a line segment AB of length 8 cm. At each end of this line segment, draw a line perpendicular to AB. Are these two lines parallel?

Solution:

Step I Take a convenient radius with A as centre and draw an arc intersecting the line at points W and X.

Step II With W and X as centres and radius greater than AW, construct two arcs intersecting each other at M.

Step III Join AM and extend it in both directions to P and Q.

Step IV Take a convenient radius with B as centre and draw an arc intersecting the line at points Y and Z.

Step V With Y and Z as centres and a radius greater than YB, construct two arcs intersecting each other at N.

Step VI Join BN and extend it in both directions to S and R.

Let the lines perpendicular at A and B be PQ and RS, respectively.

Since, ∠QAB = 90° and ∠ ABR = 90°

Therefore, ∠ QAB = ∠ ABR 

When two parallel lines are intersected by a third line, the two alternate interior angles are equal.

Since, ∠QAB = ∠ ABR 

Therefore, PQ and RS are parallel.

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Q 9. Using a protractor, draw ∠BAC of measure 450. Take a point P in the interior of ∠ BAC. From P draw line segments PM and PN such that PM ⊥ AB and PN ⊥ AC. Measure ∠ MPN.

Solution:

Step I Draw a line segment A on the line L .

Step II Take a protractor and place it on the segment AC such that AC coincides with the line of the diameter of the protractor and the middle point of the line coincides with point A.

Step III Counting from the right side, mark a point as B at the point of 45° of protractor and draw a line segment AB.

Step IV Take a convenient radius with P as centre, construct an arc intersecting the line segments AB at T and Q and AC at R and S.

Step V Using the same radius and with T and Q as centres, construct two arcs intersecting at G on the other side.

Step VI Using the same radius and with R and S as centres, construct two arcs intersecting at H on the other side.

Step VII Join PG and PH which intersects AB and AC at M and N, respectively.

On measuring MPN using a protractor, we get it equal to 135°.

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Q 10. Draw an angle and label it as ∠ BAC. Draw a bisector ray AX and take point P on it. From P draw line segments PM and PN, such that PM ⊥ AB and PN ⊥ AC, where M and N are respectively, points on rays AB and AC. Measure PM and PN. Are the two lengths equal?

Solution:

Step I Draw BAC on the line segment AC.

Step II With a convenient radius and A as centre, draw an arc from AB and AC.

Step III The points where arc cuts AB and AC, take both points as centres and draw two small arcs intersecting at X. Now, draw AX.

Step IV Take a point P on the ray AX. Take a convenient radius with P as centre and construct an arc intersecting the line segments AB at T and Q and AC at R and S, respectively.

Step V Using the same radius and with T and Q as centres, construct two arcs intersecting at G on the other side.

Step VI Using the same radius and with R and S as centres, construct two arcs intersecting at H on the other side.

Step VII Join PG and PH, which intersects AB and AC at M and N, respectively.

On measuring PM and PN using a ruler, we find that both are equal.

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