basic properties of circle

23.
(a)
Let r cm be the radius of the circle.

Draw a line OHK⊥both PQ and RS, and cut PQ at H and RS at K.
i.e. OH = 2.5 cm

In ΔOHQ :
OQ² = OH² + HQ² (Pythagorean theorem)
r² = 2.5² + (12/2)²
r² = 42.25
r = 6.5

Hence, the radius of the circle = 6.5 cm

(b)
In ΔOKS :
OS² = OK² + KS² (Pythagorean theorem)
(6.5 cm)² = OK² + (5/2 cm)²
OK² = 36 cm²
OK = 6 cm

Distance between PQ and RS
= (6 - 2.5) cm
= 3.5cm


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19.
In ΔPAQ : Since AP = PQ, then ∠PQB = ∠PAB
In ΔBPQ : Since BP = BQ, then ∠QPB = ∠PQB = ∠PAB
∠APB= 90° (∠ in semicircle)

In ΔAPQ :
∠PAB+ (∠APB+ ∠QPB)+ ∠PQB= 180° (∠ sum of Δ)
∠PAB+ (90° + ∠PAB) + ∠PAB = 180°
∠PAB= 30°


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20.
∠ABC+ ∠ADC= 180° (opp. ∠s of concyclic quad.)
50° + (40° + ∠BDC) = 180°
∠BDC= 90°

Hence, ∠BDC is the angle in semicircle, and thus BC is a diameter of the circle.


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25.
In ΔORQ :
OR = OQ (radii of the same circle)
Then ΔORQ is an isosceles Δ with ∠QRO = ∠RQO
Hence, ∠RQO = 50°

∠RQO+ ∠PSR= 180° (opp. ∠s of concyclic quad.)
50° + x = 180°
x = 130°

In ΔORQ :
∠POR= ∠QRO+ ∠RQO(ext. ∠ of Δ)
∠POR= 50° + 50°
∠POR= 100°

In ΔOPR :
OP = OB (radii of the same circle)
Then ΔOPR is an isosceles Δ with ∠PRO = ∠RPO
∠PRO+ ∠RPO+ ∠POR= 180° (∠ sum of Δ)
∠PRO+ ∠PRO+ 100° = 180° (∠ sum of Δ)
∠PRO= 40°

∠PSR+ ∠ORS= 180° (int. ∠s, PS // OR)
x + (y + ∠PRO) = 180°
130° + y + 40° = 180°
y = 10°