Question

(a) Construct $\Delta PQR$, given $\overline{PQ}=3.5cm,\overline{PR}=4.5cm$, and $\overline{QR}=5.5cm$

(b) Construct ΔPQR, when PQ = 4 cm, QR = 6 cm and ∠PQR = 60${}^{\circ }$. [4 MARKS]

Answer 1

Each part: 2 Marks

(a) Steps of construction:

Draw $\overline{PQ}=3.5cm$

With P as the centre, draw an arc with the radius equal to 4.5 cm.

With Q as the centre, draw another arc with radius 5.5 cm to cut the previous arc at R.

Join R to P and Q.

PQR is the required triangle.



Thus, we see that if three sides of a triangle are given and the sum of any two sides is always more than the third side, then a triangle can be constructed. This is known as SSS criterion for construction of triangles.


(b) Steps of construction:

Step 1: Draw a line segment QR = 6 cm.

Step 2: Construct an angle of 60${}^{\circ }$ at point Q.

Step 3: Draw an arc on the ray QX with Q as the centre and the radius equal to 4 cm.

Step 4: Name the point where the arc cuts the ray QX, as P.

Step 5: Join points P and R.

PQR is the required triangle.