Question

(a) Find the sum of all the exterior angles of a triangle.
(b) Is it possible to draw a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm respectively? [4 MARKS]

Answer 1

(a) Steps: 1 Mark
Correct answer: 1 Mark

(b) Reason: 1 Mark
Correct answer: 1 Mark


(a)

In the figure, 1, 2 and 3 are the interior angles whereas 4, 5 and 6 are exterior angles. Using linear pair axiom, we can say that:
$\angle 1+\angle 4={180}^{\circ }$

$\angle 3+\angle 6={180}^{\circ }$

$\angle 2+\angle 3={180}^{\circ }$

Adding all of them, we get:

$\angle \left(\right(1+4)+(3+6)+(2+5\left)\right)={540}^{\circ }$

Rearranging the terms, we get:

$\angle \left(\right(1+2+3)+(4+5+6\left)\right)={540}^{\circ }$

$\angle (1+2+3)={180}^{\circ }$ (Angle Sum Property of a triangle)

So, $\angle \left(\right(4+5+6\left)\right)={540}^{\circ }-{180}^{\circ }={360}^{\circ }$

So, the sum of the exterior angles of a triangle is ${360}^{\circ }$


(b) Suppose such a triangle is possible. Then the sum of the lengths of any two sides would be greater than the length of the third side. Let's check this.
Is 4.5 + 5.8 > 10.2? Yes
Is 5.8 + 10.2 > 4.5? Yes
Is 10.2 + 4.5 > 5.8? Yes

Therefore, the triangle is possible.