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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):
24 | |
25 | |
15 | |
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:
$ \displaystyle 2\sqrt{3}$ | |
$ \displaystyle \frac{12\sqrt{3}}{7}$ | |
$ \displaystyle \frac{15\sqrt{3}}{8}$ | |
$ \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?
10: 3 | |
2: 1 | |
7: 3 | |
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 |
32o | |
84o | |
64o | |
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)
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:
100o | |
150o | |
80o | |
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)
By alternate segment theorem ∠ABC = ∠TCA = 50o
Thus ∠ BOA =2{180-50-(180-100)}=100o
Correct option is (a)
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