Answer:
2. C) 21.2 yd 3. B) angle IQL 4. B) 55 degree
Step-by-step explanation:
2. The area of an equilateral triangle is given by,
A = [tex]\frac {\sqrt{3} \times a^{2}}{4}[/tex]------(1) where a = length of each side.
Here, a = 7 yd
so , putting a = 7 yd in (1) , we get,
A = [tex]\frac {\sqrt{3} \times 49}{4}[/tex] sq. yd.
[tex]\simeq 21.2[/tex] sq. yd.
3. B) angle IQL
4. Let, the centre of the circle be at O,
then, ∠DOB = 105° and ∠BOC = 145° (given)
so, ∠OBD = [tex]\frac{180^{\circ} - \angle DOB}{2}[/tex]
= [tex]\frac{180^{\circ} - 105^{\circ}}{2}[/tex]
= 37.5° [since the sum of three angles of a triangle is 180° and OB and OD are two radii of the same circle]
and, similarly, ∠OBC = [tex]\frac{180^{\circ} - \angle COB}{2}[/tex]
= [tex]\frac{180^{\circ} - 145^{\circ}}{2}[/tex]
= 17.5° [since the sum of three angles of a triangle is 180° and OB and OC are two radii of the same circle]
so, ∠CBD = ∠OBD + ∠OBC
= 37.5° + 17.5°
= 55°
