RD Sharma Solutions Class 7 Constructions Exercise 17.3

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.3

RD Sharma Class 7 Solutions Chapter 17 Ex 17.3 PDF Free Download

Exercise 17.3

Q1. Draw \(\triangle ABC\)  in which AB = 3 cm, BC= 5 cm and \(\angle Q=70^{\circ}\).

1

Steps of construction:

  1. Draw a line segment AB of length 3 cm.
  2. Draw \(\angle XBA =70^{\circ}\).
  3. Cut an arc on BX at a distance of 5 cm at C.
  4. Join AC to get the required triangle.

 

 

Q2. Draw \(\triangle ABC\)  in which \(\angle A=70^{\circ}\)., AB = 4 cm and AC= 6 cm. Measure BC.

2

Steps of construction:

  1. Draw a line segment AC of length 6 cm.
  2. Draw \(\angle XAC =70^{\circ}\).
  3. Cut an arc on AX at a distance of 4 cm at B.
  4. Join BC to get the desired triangle.
  5. We see that BC = 6 cm.

 

 

Q3. Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is \(45^{\circ}\).

3

Steps of construction:

Draw a line segment PQ of length 3 cm.

Draw \(\angle QPX =45^{\circ}\).

Cut an arc on PX at a distance of 3 cm at R.

Join QR to get the required triangle.

 

Q4. Draw \(\triangle ABC\)  in which \(\angle A=120^{\circ}\), AB = AC =3 cm. Measure \(\angle B\) and \(\angle C\).  

4

Steps of construction:

  1. Draw a line segment AC of length 3 cm.
  2. Draw \(\angle XAC =120^{\circ}\).
  3. Cut an arc on AX at a distance of 3 cm at B.
  4. Join BC to get the required triangle.

By measuring, we get \(\angle B\)= \(\angle C\)= \(30^{\circ}\).

 

Q5. Draw \(\triangle ABC\)  in which \(\angle C=90^{\circ}\) and AC = BC = 4 cm.

5

Steps of construction:

  1. Draw a line segment BC of length 4 cm.
  2. At C, draw \(\angle BCY=90^{\circ}\).
  3. Cut an arc on CY at a distance of 4 cm at A.
  4. Join AB. ABC is the required triangle.

 

 

Q6. Draw a triangle ABC in which BC= 4 cm, AB = 3 cm and \(\angle B=45^{\circ}\). Also, draw a perpendicular from A on BC.

6

Steps of construction:

  1. Draw a line segment AB of length 3 cm.
  2. Draw an angle of \(45^{\circ}\) and cut an arc at this angle at a radius of 4 cm at C.
  3. Join AC to get the required triangle.
  4. With A as centre, draw intersecting arcs at M and N.
  5. With centre M and radius more than half of MN, cut an arc on the opposite side of A.
  6. With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
  7. Join AE, it meets BC at D, then AE is the required perpendicular.

 

 

Q7. Draw a triangle ABC with AB = 3 cm, BC = 4 cm and \(\angle B=60^{\circ}\). Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure \(\angle COA\).

7

Steps of construction:

Draw a line segment BC = 4 cm.

Draw \(\angle CBX=60^{\circ}\).

Draw an arc on BX at a radius of 3 cm cutting BX at A.

Join AC to get the required triangle.

 

Angle bisector for angle A:

  1. With A as centre, cut arcs of the same radius cutting AB and AC at P and Q, respectively.
  2. From P and Q cut arcs of same radius intersecting at R.
  3. Join AR to get the angle bisector of angle A.

 

Angle bisector for angle C:

  1. With A as centre, cut arcs of the same radius cutting CB and CA at M and N, respectively.
  2. From M and N, cut arcs of the same radius intersecting at T
  3. Join CT to get the angle bisector of angle C.

Mark the point of intersection of CT and AR as 0.

Angle \(\angle COA=120^{\circ}\).