Answer:
- 13. (a) 120°, 120°, (b) 8.94 cm
- 14. (a) 17.9 cm, (b) 22.4 cm
Step-by-step explanation:
- Refer to the attached sketch (not to scale)
Question 13
(a) The center of the circle is also the circumcenter of the triangle PQR.
Since ΔPQR is equilateral, all sides are equal, therefore opposite angles are also congruent:
Sum of the three angles is 360° as cover the full circle.
Each angle measure is:
- ∠POQ = ∠QOR = ∠POR = 360°/3 = 120°
(b) Each side is 16 cm and the radius is 12 cm.
The distance of the midpoint M of PQ from the center O is perpendicular bisector of ΔPOQ.
The segment MO is:
- MO² = PO² - PM² = PO² - (PQ/2)²
- MO² = 12² - (16/2)² = 80
- MO = √80 = 8.94 cm (rounded)
Question 14
ΔXYZ is isosceles with:
- XY = XZ = 20 cm
- YZ = 18 cm
(a) The altitude of ΔXYZ is a perpendicular bisector of YZ. h = XM is the altitude
Find h:
- h² = XZ² - ZM² = XZ² - (ZY/2)²
- h² = 20² - (18/2)² = 319
- h = √319
- h = 17.9 cm (rounded)
(b) Note the ∠ZXM is half of the ∠ZOM as both the angles intercept half of the arc ZY and ∠ZOM is a central angle.
Find ∠ZXM:
- sin ∠ZXM = ZM/ZX = 9 / 20
- m∠ZXM = arcsin (9/20) = 26.7° (rounded)
Find ∠ZOM:
- m∠ZOM = 2*m∠ZXM = 2*26.7 = 53.4°
Find the measure of the radius:
- sin ∠ZOM = ZM/r
- sin 53.4° = 9/r
- r = 9 / sin 53.4°
- r = 11.2 cm (rounded)
Find the measure of the diameter:
- d = 2r = 2*11.2 cm = 22.4 cm
