Page 211 - Start Up Mathematics_7

3. Construct ∠BCY = 45° by bisecting a right angle as shown (Fig. 2).
4. Let A be the point of intersection of BX and CY.
ABC is the required triangle.
A

60° 45°

B 5.6 cm C

Example 6: Construct a ∆ PQR in which PQ = 5 cm, ∠PQR = 105° and ∠QRP = 50°.
R
Solution: Construction: Draw a rough sketch of ∆ PQR and mark the 50°
given dimensions.

Steps of Construction: 105°
1. Find the measure of angle QPR using angle sum P 5 cm Q
property, i.e., ∠QPR = 180° – (105° + 50°) = 180° – 155° = 25°.

2. Draw a line segment PQ = 5 cm. P 5 cm Q

3. Using a protractor, draw ∠PQX = 105º (Fig. 1).

X X
R Y

105° 105°
25°
P 5 cm Q P 5 cm Q
Fig. 1 Fig. 2
4. Using a protractor, draw ∠QPY = 25° (Fig. 2).
5. Let R be the point of intersection of QX and PY. PQR is the required triangle.

Example 7: Examine whether you can construct ∆ DEF such that EF = 7.2 cm, ∠E = 110° and
∠F = 80°. Justify your answer.

Solution: We cannot construct such a triangle as the sum of two of its angles exceeds 180º.

Case IV: RHS criterion
Construction of a right triangle when its hypotenuse and one side are given A
Let’s construct a right triangle ABC in which BC = 3.9 cm, ∠B = 90º and 5 cm

hypotenuse AC = 5 cm.
Construction: Draw a rough sketch of ∆ ABC and mark the given dimensions. B 3.9 cm C

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