 # RD Sharma Solutions Class 7 Constructions Exercise 17.4

## RD Sharma Solutions Class 7 Chapter 17 Exercise 17.4

#### Exercise 17.4

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

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

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

$\angle BCA=180^{\circ}-100^{\circ}=80^{\circ}$ 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. 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 $\angle B=60^{\circ}$ 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 $\angle C=180^{\circ}-60^{\circ}-90^{\circ}=30^{\circ}$ 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 $\angle ABC=60^{\circ}$..