RD Sharma Solutions Class 7 Constructions Exercise 17.4

RD Sharma Solutions Class 7 Chapter 17 Exercise 17.4

RD Sharma Class 7 Solutions Chapter 17 Ex 17.4 PDF Free Download

Exercise 17.4

Q1. Construct \(\triangle ABC\) in which BC= 4 cm, \(\angle B=50^{\circ}\) and \(\angle C=70^{\circ}\).

1

Steps of construction:

  1. Draw a line segment BC of length 4 cm.
  2. Draw \(\angle CBX\) such that \(\angle CBX=50^{\circ}\).
  3. Draw \(\angle BCY\) with Y on the same side of BC as X such that \(\angle BCY=70^{\circ}\).
  4. Let CY and BX intersect at A.
  5. ABC is the required triangle.

Q2. Draw \(\triangle ABC\) in which BC= 8 cm, \(\angle B=50^{\circ}\) and \(\angle A=50^{\circ}\).

\(\angle ABC+\angle BCA+\angle CAB=180^{\circ}\)

\(\angle BCA=180^{\circ}-\angle CAB-\angle ABC\)

\(\angle BCA=180^{\circ}-100^{\circ}=80^{\circ}\)

2

Steps of construction:

  1. Draw a line segment BC of length 8 cm.
  2. Draw \(\angle CBX\) such that \(\angle CBX=50^{\circ}\).
  3. Draw \(\angle BCY\) with Y on the same side of BC as X such that \(\angle BCY=80^{\circ}\).
  4. Let CY and BX intersect at A.

Q3. Draw \(\triangle ABC\) in which \(\angle Q=80^{\circ}\), \(\angle R=55^{\circ}\) and QR = 4.5 cm. Draw the perpendicular bisector of side QR.

3

Steps of construction:

  1. Draw a line segment QR = 4.5 cm.
  2. Draw \(\angle RQX=80^{\circ}\) and \(\angle QRY=55^{\circ}\).
  3. Let QX and RY intersect at P so that PQR is the required triangle.
  4. With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
  5. With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
  6. Join MN

MN is the required perpendicular bisector of QR.

Q4. Construct \(\triangle ABC\) in which AB = 6.4 cm, \(\angle A=45^{\circ}\) and \(\angle B=60^{\circ}\)

4

Steps of construction:

Draw a line segment AB = 6.4 cm.

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

Draw \(\angle ABY\) with Y on the same side of AB as X such that \(\angle ABY=60^{\circ}\).

Let AX and BY intersect at C.

ABC is the required triangle.

Q5. Draw \(\triangle ABC\) in which AC= 6 cm, \(\angle A=90^{\circ}\) and \(\angle B=60^{\circ}\).

\(\angle A+\angle B+\angle C=180^{\circ}\)

Therefore \(\angle C=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}\)

5

Steps of construction:

  1. Draw a line segment AC = 6 cm.
  2. Draw \(\angle ACX=30^{\circ}\).
  3. Draw \(\angle CAY\) with Y on the same side of AC as X such that \(\angle CAY=90^{\circ}\).
  4. Join CX and AY. Let these intersect at B.

ABC is the required triangle where angle \(\angle ABC=60^{\circ}\)..


Practise This Question

If the discount is 10%, an item sold at  ₹ 9 , has a marked price of _______