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- Question
- In a ? ABC, the bisectors of ? B and ? C intersect each other at a point O. Then ? BOC is equal to :
Options- A. 90 ° - 1 ?A 2
- B. 120° + 1 ?A 2
- C. 90 ° + 1 ?A 2
- D. 120° - 1 ?A 2
- Correct Answer
- 90 ° + 1 ?A 2
Explanation
- 1. In a ? ABC, the sides AB and AC are produced to P and Q respectively. The bisectors of ? OBC and ? QCB intersect at a point O. Then ? BOC is equal to:
Options- A. 90 ° + 1 ?A 2
- B. 90 ° - 1 ?A 2
- C. 120° + 1 ?A 2
- D. 120° - 1 ?A 2 Discuss
- 2. In the figure, BD and CD are angle bisectors of ? ABC and ? ACE, respectively. Then ? BDC is equal to :
Options- A. ?BAC
- B. 2?BAC
- C. 1 ?BAC 2
- D. 1 ?BAC 3 Discuss
- 3. ABCD is a parallelogram and X, Y are the mid-points of sides AB and CD respectively. Then quadrilateral AXCY is a :
Options- A. parallelogram
- B. rhombus
- C. square
- D. rectangle Discuss
- 4. Of all the chords of a circle passing through a given point in it, the smallest is that which :
Options- A. Is trisected at the point
- B. Is bisected at the point
- C. Passes through the centre
- D. None of these Discuss
- 5. In a circular lawn, there is a 16 m long path in the form of a chord. If the path is 6 m away from the center of the lawn, then find the radius of the circular lawn.
Options- A. 16 m
- B. 6 m
- C. 10 m
- D. 8 m Discuss
- 6. In the fig. XY || AC and XY divides triangular region ABC into two part equal in area.
Then AX is equal to : AB
Options- A. 1 ?2
- B. ?2 + 2 ?2
- C. 1 2
- D. ?2 - 2 ?2 Discuss
- 7. Two poles of ht. a and b meters are p meters apart ( b > a). The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is :
Options- A. a+b ab
- B. p a+b
- C. ab a+b
- D. a+b p Discuss
- 8. If log 2 = 0.3010, then the number of digits in 2 64 is?
Options- A. 18
- B. 19
- C. 20
- D. 21 Discuss
- 9. The value of $ \frac{\log_{a}{x}}{\log_{ab}{x}} - \log_{a}{b}$ is?
Options- A. 0
- B. 1
- C. a
- D. ab Discuss
- 10. The value of log 23 x log 32 x log 34 x log 43 is?
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
Plane Geometry problems
Search Results
Correct Answer: 90 ° - 1 ?A 2
Explanation:
Correct Answer: 1 ?BAC 2
Explanation:
Correct Answer: parallelogram
Explanation:
Correct Answer: None of these
Explanation:
Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle.
Let CD is another chord passes through point M.
We have to prove that AB < CD.
Now join OM and draw OL perpendicular to CD.
In right angle triangle OLM,
OM is the hypotenuse.
So OM > OL
? chord CD is nearer to O in comparison to AB.
? CD > AB
? AB < CD
So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point.
Hence required answer will be option D .
Correct Answer: 10 m
Explanation:
From given figure , we can see that
Let AB = 16 m , OM = 6 m , AM = BM = 8 m and OA = r
In triangle AMO ,
By pythagoras theorem ,
OA2 = AM2 + OM2
OA2 = 82 + 62 = 64 + 36
OA2 = 100 ? OA = ? 100 = 10
? OA = r = 10 m
The radius of the circular lawn = 10 m
Correct Answer: ?2 - 2 ?2
Explanation:
Correct Answer: ab a+b
Explanation:
Correct Answer: 20
Explanation:
Required answer = [64 log10 2] + 1
= [ 64 x 0.3010 ] + 1
= 19.264 + 1
= 19 + 1
= 20
Correct Answer: 1
Explanation:
? loga x = ( logabx) / (logaba)
? The given expression = [[(logabx) / (logaba)] / [( logabx )] - ( logab)
= (1/logaba) - logab = logaab - logab = loga(ab/b)
= logaa = 1
Correct Answer: 1
Explanation:
Given Exp.= log23 x log 32 x log34 x log43
= (log3 / log2) x ( log2 / log3) x (log4 / log3) x (log3 / log4) = 1
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