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Geometry and Mensuration: Level 3 Test 6

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Question 1
The length of the common chord of two circles of radii 15 cm and 20 cm whose centers are 25 cm apart, is (in cm):
A
24
B
25
C
15
D
20
Question 1 Explanation: 
$ \begin{array}{l}Let\,\,the\,\,total\,\,length\,\,be\,\,2x\,\,units\\\frac{1}{2}x\times 25=\frac{1}{2}\times 15\times 20\\x=12units\\thus\,\,2x=24\,units\\Thus\,\,the\,\,correct\,\,option\,\,is\,\,a\end{array}$
Question 2
In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB =4, AC= 3 and ∠A= 60o, then, the length of AD is:
A
$ \displaystyle 2\sqrt{3}$
B
$ \displaystyle \frac{12\sqrt{3}}{7}$
C
$ \displaystyle \frac{15\sqrt{3}}{8}$
D
$ \displaystyle \frac{6\sqrt{3}}{7}$
Question 2 Explanation: 
$ \begin{array}{l}let\,\,the\,\,length\,\,of\,\,AD\,\,be\,\,x\,\,units\\\frac{1}{2}\times 4x\times Sin30=\frac{4}{7}\times 4\times 3\times Sin60\\x=\frac{12}{7}\sqrt{3}\\Correct\,\,option\,\,is\,\,b\end{array}$
Question 3
In the figure (not drawn to scale) given below, P is a point on AB such that AP: PB= 4: 3. PQ is parallel to AC and QD is parallel to CP. In ∠ARC, ∠ARC= 90o, and in ΔPSQ, ∠PSQ= 90o. The length of QS is 6 cms. What is ratio AP: PD?
119
A
10: 3
B
2: 1
C
7: 3
D
8: 3
Question 3 Explanation: 
$ \displaystyle \begin{array}{l}AP:\text{ }PB\text{ }=\text{ }4:3\\Thus\text{ }AP=\text{ }4x\text{ }and\text{ }PB=3x\\Again,\text{ }PD:DB\text{ }=\text{ }CQ:AB\text{ }=\text{ }4:3\\PD=\frac{4}{7}\times BP=\frac{4}{7}\times \frac{3}{4}AP=\frac{3}{7}AP\end{array}$
Question 4
In the figure (not drawn to scale) given below, if AD = CD = BC, and BCE = 96o, how much is DBC?120
A
32o
B
84o
C
64o
D
Cannot be determined
Question 4 Explanation: 
Let angle CAD= DCA = x
and thus CDB = 2x and CBD = 2x.
Thus, 180-2x+x+96= 180
=> 2x= 64
Correct option is (c)
Question 5
In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O, The chord BA is extended to a points T such that CT becomes a tangent at the circle  at point  C. If ∠ATC= 30o and ∠ACT= 50o, then the angle BOA is:
A
100o
B
150o
C
80o
D
Not possible to determine
Question 5 Explanation: 
Since angle ∠CAT = 180-50-30 = 100
By alternate segment theorem ∠ABC = ∠TCA = 50o
Thus ∠ BOA =2{180-50-(180-100)}=100o
Correct option is (a)
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