急!急! F4 Properties Circles q1

13.
PQ
= 20 - 5 cm
= 15 cm

MQ
= 15 x (1/2) cm
= 7.5 cm


ON = OM (given)
OS = OQ (radii of the same circle)
∠ONS = ∠OMQ = 90° (given)
ΔONS ≡ ΔOMQ (RHS)
NS = MQ (corr. sides of cong. Δs)

NS = MQ = 7.5 cm

RS
= 7.5 x 2 cm
= 15 cm


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21.
(a)
Denote O as the centre of the circle.

In ΔOAN :
OA² = AN² + ON² (Pythagoreantheorem)
(34/2)² = (16/2)² + ON²
289 = 64 + ON²
ON² = 225
ON = 15

The distance from the centre to AB = 15 cm

(b)
NP
= OP - ON
= (34/2) - 15 cm
= 2 cm


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23.
(a)
Let H be the mid-point of PQ.
Then, ∠OHQ = 90° and PQ = 12 x (1/2) cm = 6 cm

In ΔOHQ :
OQ² = OH² + HQ² (Pythagoreantheorem)
OQ² = 2.5² + 6² cm²
OQ² = 6.25 + 36 cm²
OQ² = 42.25 cm²
OQ = 6.5 cm

The radius of the circle = 6.5 cm

(b)
Let K be the mid-point of RS.
Then, ∠OKS = 90° and KS = 5 x (1/2) cm = 2.5 cm

In ΔOKS :
OS² = OK² + KS²
(6.5)² = OK² + (2.5)² cm²
42.25 = OK² + 6.25 cm²
OK² = 36 cm²
OK = 6 cm

Distance between PQ and RS
= OK - OH
= 6 - 2.5 cm
= 3.5 cm