RD Sharma solution class 7 chapter 17 Construction Exercise 17.4

Exercise 17.4

Pages-17.6

Question 1:

Construct ∆ ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.

Answer 1:

Steps of construction:

1. Draw a line segment BC of length 4 cm.
2. Draw $\angle CBX \mathrm{such} \mathrm{that} \angle CBX\hspace{0.17em}=50°$.
3. Draw $\angle BCY$ with Y on the same side of BC as X such that $\angle$BCY = 70$°$.
4. Let CY and BX intersect at A.
5. ABC is the required triangle.

               

Question 2:

Draw ∆ ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.

Answer 2:

$$\begin{gathered} \angle ABC+\angle BCA+\angle CAB=180° \\ \angle BCA=180°-\angle ABC-\angle CAB \\ \angle BCA=180°-100°=80° \end{gathered}$$

Steps of construction:

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

    
      

Question 3:

Draw ∆ PQR in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.

Answer 3:

Steps of construction:

1. Draw a line segment QR = 4.5 cm.
2. Draw $\angle RQX = 80° and \angle QRY =55°$.
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.

Question 4:

Construct ∆ ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°.

Answer 4:

Steps of construction:

1. Draw a line segment AB = 6.4 cm.
2. Draw $\angle BAX = 45°$.
3. Draw $\angle$ABY with Y on the same side of AB as X such that $\angle$ABY = 60$°$.

Let AX and BY intersect at C; ABC is the required triangle.

Question 5:

Draw ∆ ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.

Answer 5:

We can see that ∠A+∠B+∠C = 180°. Therefore ∠C = 180$°$ − 60$°$ − 90$°$ = 30°.

Steps of construction:

1. Draw a line segment AC = 6 cm.
2. Draw $\angle ACX=30°$.
3. Draw $\angle$CAY with Y on the same side of AC as X such that $\angle$CAY = 90$°$.
4. Join CX and AY. Let these intersect at B.
5. ABC is the required triangle where angle $\angle$ABC = 60$°$.