ANSWERS/HINTS 339
File Name : C:\Computer Station\Maths-IX\Chapter\Answers (16–12–2005) PM65
EXERCISE 10.5
1. 45° 2. 150°, 30° 3.10°
- 80° 5. 110° 6.✂ BCD = 80° and ✂ ECD = 50°
- Draw perpendiculars AM and BN on CD (AB || CD and AB < CD). Show
☎ AMD ✝^ ☎ BNC. This gives ✂ C = ✂ D and, therefore, ✂ A + ✂ C = 180°.
EXERCISE 10.6 (Optional)- Let O be the centre of the circle. Then perpendicular bisector of both the chords will
be same and passes through O. Let r be the radius, then r^2 =112
2
✁
✄✞ ✆✟ + x^2=
2
(^5) (6 ) 2
2
✄ ✁✆ ✠ ✡x
✞ ✟
, where x is length of the perpendicular from O on the chord of
length 11 cm. This gives x = 1. So, r =^55
2
cm. 3. 3 cm.
- Let ✂ AOC = x and ✂ DOE = y. Let ✂ AOD = z. Then ✂ EOC = z and x + y + 2z = 360°.
✂ ODB = ✂ OAD + ✂ DOA = 90° –
1
2
z + z = 90° +1
2
z. Also ✂ OEB = 90° +1
2
z8. ✂ ABE = ✂ AD E,✂ ADF = ✂ ACF =
1
2
✂ C.
Therefore, ✂ EDF =✂ ABE + ✂ ADF =1
2
(✂ B + ✂ C) =
1
2
(180° – ✂ A) = 90° –
1
2
(^) ✂ A.
- Use Q. 1, Ex. 10.2 and Theorem 10.8.
10.Let angle-bisector of ✂ A intersect circumcircle of ☎ ABC at D. Join DC and DB. Then
✂ BCD = ✂ BAD =1
2
✂ A and✂ DBC = ✂ DAC =1
2
✂ A. Therefore, ✂ BCD =✂ DBC or, DB = DC. So, D lies on the perpendicular bisector of BC.EXERCISE 12.1
1.^322 , 900 3 cm
4
a 2. Rs 1650000 3. 20 2 m^2- 21 11 cm^2 5. 9000 cm^2 6. 915cm^2