Question

Constructa $\Delta PQR,$ in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another triangle whose sides are $\frac{4}{5}$ to,of the corresponding sides of $\Delta PQR$.

Answer 1

Steps of construction:

Step 1: Draw a line segment BC = 6 cm

Step 2: With B as the Center and radius equal to 5 cm draw an arc.

Step 3: With C as the center and radius equal to 7 cm draw another arc cutting the previous arc at A.

Step 4: Join AB and AC. Thus, ΔΔ ABC is obtained.

Step 5: Below BC draw another line BX.

Step 6: Mark 7 points $B_1{B}_2{B}_3{B}_4{B}_5{B}_6{B}_7$such that:

$B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5=B_5{B}_6=B_6{B}_7$

Step 7: Join B7C

Step 8: From B5, draw $B_{5}D||B_{7}C$

Step 9: Draw a line DE through D parallel to CA

Hence Δ BDE is the required triangle.