RD Sharma Solutions Class 7 Constructions Exercise 17.2

RD Sharma Class 7 Solutions Chapter 17 Ex 17.2 PDF Free Download

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2

Exercise 17.2

1. Draw \(\triangle ABC\) in which AB = 5.5 cm. BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.

Solution:

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2-1

Construction:

  1. First, draw a line segment AB which is 5.5 cm long.
  2. Cut an arc of radius 6 cm from B using a compass.
  3. Now, cut another arc of radius 7 cm from C, such that, it intersects the previously drawn arc at C.
  4. Draw lines from A to C and B to C, so that it joins AC and BC to obtain the desired triangle.
  5. Draw two arcs of radius more than half of BC, keeping B as the centre, on both sides of BC.
  6. Similarly, draw two arcs of radius more than half of BC, keeping C as the centre, on both sides of BC such that it intersects the arcs drawn in the previous step at X and Y.
  7. By joining XY, we obtain the perpendicular bisector of BC.

 

2. Draw \(\triangle PQR\) in which PQ = 3 cm, QR. 4 cm and RP= 5 cm. Also, draw the bisector of \(\angle Q\)

Solution:

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2-2

Construction:

  1. First, draw a line segment PQ which is 3 cm long.
  2. Cut an arc of radius 4 cm from Q using a compass.
  3. Now, cut another arc of radius 5 cm from P, such that, it intersects the previously drawn arc at R.
  4. Draw lines from P to R and Q to R, so that it joins PR and OR to obtain the required triangle.
  5. Cut arcs of equal radius from Q,  such that it intersects PQ at M and QR at N, respectively.
  6. Cut arcs of equal radius from both M and N, such that it intersects at point S.
  7. Join QS and extend it further to produce the angle bisector of angle PQR.
  8. Now, PQS and angle SQR = \(45^{\circ}\) each.

 

3. Draw an equilateral triangle one of whose sides is of length 7 cm.

Solution:

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2-3

Construction:

  1. First, draw a line segment AB which is 7 cm long.
  2. Draw an arc with centre A and radius 7 cm..
  3. Draw another arc with centre B and radius 7 cm such that it intersects the previously drawn arc at C.
  4. To get the required triangle, join the points A to C and B to C, i.e., AC and BC .

 

4. Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.

Solution:

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2-4

Construction:

  1. First, draw a line segment PR which is 7 cm long.
  2. Draw an arc with centre P and radius 5 cm..
  3. Draw another arc with centre R and radius 4 cm such that it intersects the previously drawn arc at Q.
  4. To get the required triangle, join the points P to Q and R to Q, i.e., PQ and QR .
  5. Draw two arcs of radius more than half of PR, keeping P as the centre, on both sides of PR.
  6. Similarly, draw two arcs of same radius, keeping R as the centre, on both sides of PR such that it intersects the arcs drawn in the previous step at M and N.
  7. Therefore, the required perpendicular bisector of the largest side is MN.

 

5. Draw a triangle ABC with AB = 6 cm, BC = 7 cm and CA = 8 cm. Using ruler and compass alone, draw (i) the bisector AD of \(\angle A\) and (ii) perpendicular AL from A on BC. Measure LAD.

Solution:

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2-5

Construction:

  1. Draw a line segment BC of length 7 cm.
  2. With centre B, draw an arc of radius 6 cm.
  3. With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.
  4. Join AC and BC to get the required triangle.
  5. From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.
  6. From E and F, cut arcs of equal radius intersecting at point H.
  7. Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.
  8. From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).
  9. From P and Q, cut arcs of equal radius intersecting at M.
  10. Join AM cutting BC at L.
  11. AL is the perpendicular to the line BC.
  12. Therefore, Angle LAD = \(15^{\circ}\).

 

6. Draw \(\triangle DEF\) such that DE= DF= 4 cm and EF = 6 cm. Measure \(\angle E\)  and \(\angle F\).

Solution:

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2-6

Construction:

  1. First, draw a line segment EF which is 6 cm long.
  2. Draw an arc with centre E and radius 4 cm..
  3. Draw another arc with centre F and radius 4 cm such that it intersects the previously drawn arc at D.
  4. To get the required triangle, join the points E to D and F to D, i.e., ED and DF .
  5. Now, we get, \(\angle E\)= \(\angle F\)= \(40^{\circ}\).

 

7. Draw any triangle ABC. Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.

Solution:

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.2-7

Construction:

  1. Draw a triangle ABC with each side = 6 cm.
  2. Draw an arc from A on either side of line AB.
  3. With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.
  4. Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.
  5. With B as centre, draw an arc cutting BC and BA at M and N, respectively.
  6. With centre D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.
  7. With centre Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.
  8. Join XD and extend it to intersect AC at E.
  9. The required parallel line=DE.

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