## 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:**

Construction:

- First, draw a line segment AB which is 5.5 cm long.
- Cut an arc of radius 6 cm from B using a compass.
- Now, cut another arc of radius 7 cm from C, such that, it intersects the previously drawn arc at C.
- Draw lines from A to C and B to C, so that it joins AC and BC to obtain the desired triangle.
- Draw two arcs of radius more than half of BC, keeping B as the centre, on both sides of BC.
- 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.
- 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:**

Construction:

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

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

**Solution:**

Construction:

- First, draw a line segment AB which is 7 cm long.
- Draw an arc with centre A and radius 7 cm..
- Draw another arc with centre B and radius 7 cm such that it intersects the previously drawn arc at C.
- 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:**

Construction:

- First, draw a line segment PR which is 7 cm long.
- Draw an arc with centre P and radius 5 cm..
- Draw another arc with centre R and radius 4 cm such that it intersects the previously drawn arc at Q.
- To get the required triangle, join the points P to Q and R to Q, i.e., PQ and QR .
- Draw two arcs of radius more than half of PR, keeping P as the centre, on both sides of PR.
- 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.
- 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:**

Construction:

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

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

**Solution:**

Construction:

- First, draw a line segment EF which is 6 cm long.
- Draw an arc with centre E and radius 4 cm..
- Draw another arc with centre F and radius 4 cm such that it intersects the previously drawn arc at D.
- To get the required triangle, join the points E to D and F to D, i.e., ED and DF .
- 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:**

Construction:

- Draw a triangle ABC with each side = 6 cm.
- Draw an arc from A on either side of line AB.
- 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.
- Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.
- With B as centre, draw an arc cutting BC and BA at M and N, respectively.
- 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.
- With centre Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.
- Join XD and extend it to intersect AC at E.
- The required parallel line=DE.