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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
PQR and XYZ are triangles. The perimeter of each triangle is 12 cm. PQR is an equilateral XY=4 cm and YZ=ZX. Both the triangles are __________A. congruentB. similar but not congruentC. similarD. Both (a) and (c ) |
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Answer» Correct Answer - D Side of PQR `=(12)/(3)=4cm ` Given XY=4 cm and XY+YZ+ZX =12 cm `implies YZ+ZX=8 cm` YZ=ZX=4 cm `:.` PQR and XYZ are congurent . Both the triangles are similar. Option (d) follows . Hence the correct option is (d). |
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| 2. |
Distance between two circles with R and r is d units. If `d^2=R^2+r^2` , then the two circles __________(intersect at one point/ do not intersect /intersect at two distinct points). |
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Answer» Correct Answer - intersect at two distinct points N/A |
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| 3. |
If two circles touch each otner externally, then the number of transverse common tangents is __________. |
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Answer» Correct Answer - 1 N/A |
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| 4. |
If two circles intersect at two distinct points , then the number of common tangents is __________. |
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Answer» Correct Answer - 2 N/A |
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| 5. |
There are no congruent figures which are similar. (True/False). |
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Answer» Correct Answer - False N/A |
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| 6. |
Identify the pairs of shapes which are similar and congruent. |
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Answer» Similar shapes: (i) W and L Congruent shapes: (i) Z and I |
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| 7. |
Check whether given sides are the sides of right – angled triangles, using Pythagoras theorem,8, 15, 1712, 13, 1530, 40, 509, 40, 4124, 45, 51 |
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Answer» 1. 8, 15, 17 Take a = 8, b = 15 and c = 17 Now a2 + b2 = 82 + 152 = 64 + 225 = 289 172 = 289 = c2 ∴ a2 + b2 = c2 By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle. Answer: yes. 2. 12, 13, 15 Take a = 12 b = 13 and c = 15 Now a2 + b2 = 122 + 132 = 144 + 169 = 313 152 = 225 ≠ 313 By the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle. Answer: No. 3. 30, 40, 50 Take a = 30 b = 40 and c = 50 Now a2 + b2 = 302 + 402 = 900 + 1600 = 2500 c2 = 502 = 2500 ∴ a2 + b2 = c2 By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle. Answer: yes. 4. 9, 40, 41 Take a = 9 b = 40 and c = 41 Now a2 + b2 = 92 + 402 = 81 + 1600= 1681 c2 = 412 = 1681 ∴ a2 + b2 = c2 By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle. Answer: Yes. 5. 24, 45, 51 Take a = 24 b = 45 and c = 51 Now a2 + b2 = 242 + 452 = 576 + 2025 = 2601 c2 = 512 = 2601 a2 + b2 = c2 By the converse of Pythagoras theorem, the triangle with given measure is a right angled triangle. Answer: Yes. |
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| 8. |
In the figure, DA = DC and BA = BC. Are the triangles DBA and DBC congruent? Why? |
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Answer» Here AD = CD AB = CB DB = DB (common) ∆DBA ≅ ∆DBC [∵ By SSS Congruency] Also RHS rule also bind here to say their congruency. |
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| 9. |
State Pythagoras theorem. |
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Answer» In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. |
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| 10. |
The area of the trapezium is …….. |
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Answer» 1/2 x h x (a + b) sq. units |
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| 11. |
Say True or False:(i) In any triangle the Centroid and the Incentre are located inside the triangle.(ii) The centroid, orthocentre, and incentre of a triangle are collinear.(iii) The incentre is equidistant from all the vertices of a triangle. |
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Answer» (i) True (ii) True (iii) False |
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| 12. |
The distance between the parallel sides of a trapezium is called as ……… |
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Answer» The distance between the parallel sides of a trapezium is called as its height. |
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| 13. |
In an isosceles trapezium, the non-parallel sides are ……….. in length. |
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Answer» In an isosceles trapezium, the non-parallel sides are equal in length. |
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| 14. |
If the area and sum of the parallel sides are 60 cm2 and 12 cm, its height is ……….. |
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Answer» 10 cm Area of the trapezium = 1/2 x h (a + b) 60 = 1/2 x h x (12) h = (60×2)/12 = 10 cm |
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| 15. |
Can a rhombus, a square or a rectangle be called as a parallelogram? Justify your answer. |
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Answer» Yes, a rhombus, a square or a rectangle can be called as parallelogram as the opposite sides are equal and parallel and diagonals bisect each other in this figures. |
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| 16. |
From the figure.(i) name a pair of complementary angles(ii) name a pair of supplementary angles |
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Answer» (i) ∠FAE and ∠DAE are complementary (ii) ∠FAD and ∠DAC are supplementary |
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| 17. |
If ∆ GUT is isosceles and right angled, then ∠TUG is …………(a) 30°(b) 40°(c) 45°(d) 55° |
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Answer» (c) 45° ∠U ∠T = 45° (∆ GUT is an isosceles given) ∴ ∠TUG = 45° |
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| 18. |
Find the supplementary angle of(i) 70°(ii) 35°(iii) 165°(iv) 90°(v) 0°(vi) 180°(vii) 95° |
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Answer» How far we should go in the same direction to reach the straight angle (180°) is called supplementary angle. (i) Supplementary angle of 70° = 180° – 70° = 110° (ii) Supplementary angle of 35° is 180° – 35° = 145° (iii) Supplementary angle of 165° is 180° – 165° = 15° (iv) Supplementary angle of 90° is 180° – 90° = 90° (v) Supplementary angle of 0° is 180° – 0° = 180° (vi) Supplementary angle of 180° is 180° – 180° = 0° (vii) Supplementary angle of 95° is 180° – 95° = 85° |
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| 19. |
In a `DeltaABC, angleB=65^(@) and angleC=80^(@)`, then the longest side is _________. |
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Answer» Correct Answer - AB `angleB=65^(@), angleC=80^(@)` `rArr angleA=35^(@)` `rArr "The side opposite to " 80^(@)` is the longest side that is AB. |
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| 20. |
If one angle of an isosceles triangle is 70°, then find the possibilities for the other two angles. |
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Answer» 70°, 40° (or) 55°, 55° |
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| 21. |
If one angle of an isosceles triangle is 124°, then find the other angles. |
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Answer» In an isosceles triangle, any two sides are equal. Also, two angles are equal. Sum of three angles of a triangle = 180° Given one angle = 124° Sum of other two angles = 180° – 124° = 56° Other angles are = 56/2 = 28° 28° and 28°. |
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| 22. |
The angles of a right angled triangle are(a) acute, acute, obtuse(b) acute, right, right(c) right, obtuse, acute(d) acute, acute, right |
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Answer» (d) acute, acute, right |
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| 23. |
In a right-angled isosceles triangle, the measures of the angles are _______. |
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Answer» Correct Answer - `45^(@), 45^(@), 90^(@)` In a right-angled isosceles triangle, the angles are `45^(@), 45^(@) and 90^(@)` |
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| 24. |
What are the angles of an isosceles right-angled triangle? |
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Answer» Since it is a right-angled triangle One of the angles is 90° Other two angles are equal because it is an isosceles triangle. Other two angles must be 45° and 45° Angles are 90°, 45°, 45°. |
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| 25. |
The lengths of two sides of an isosceles triangle are 5 cm and 12 cm. The length of the third side is _______A. 12cmB. 5 cmC. 17cmD. 10cm |
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Answer» Correct Answer - A The length of two sides of an isosceles triangle are 5 cm and 12 cm . The third side is either 5 cm or 12 cm long. But only 12 cm satisfies the inequalities of triangle. Hence, the correct option is (a). |
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| 26. |
The following steps are involved in finding the third side of an isosceles triangle whose two sides are 6 cm and 12 cm. Arrange them in seqential order. (A) But the difference between two sides is less than the third side. (B) Since the given triangle is isosceles, the possible measureof the third side is either 6 cm or 12 cm . (C ) `implies` The measure of the third side is 12 cm. (D) `:.` 6 cm cannot be the measure of the third side.A. BDACB. BCADC. BADCD. BACD |
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Answer» Correct Answer - C (B),(A),(D),and (C ) is the required sequential order. Hence, the correct option is (c ). |
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| 27. |
In the given `Delta ABC, bar(AD), bar(BE)` and `bar(CF)` are the medians. G is centroid. What is the ratio of the area of `Delta BGD` and `Delta GCE`? |
| Answer» The three medians divide the triangle into six triangles of equal area. Hence, the ratio of the area of `Delta BGD` to that of `Delta GCE` is `1:1` | |
| 28. |
The angles of a quadrilateral are ¡n the ratio 1 : 2 : 3 : 4. Find all the angles. Let each ratio be x. |
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Answer» Then the angles are x°, 2x°, 3x°, 4x° x° + 2x° + 3x° + 4x° = 360° 10x° = 360° x° = \(\frac{360}{10}\) = 36° |
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| 29. |
The arc DTE is called ______of the circle. |
| Answer» Correct Answer - semicircle | |
| 30. |
In the figure above, AB || CD. EF and FG are the bisectors of `angleBEG and angleDGE`, respectively. `angleFEG=angleFGE+10^(@)`. Find `angleFGE`.A. `20^(@)`B. `25^(@)`C. `40^(@)`D. `35^(@)` |
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Answer» Correct Answer - C `AB||CD` `angleBEG+angleEGD=180^(@)` EF and FG bisect ` angleBEG and angleDGE,` respectively. `:. 2 angleFEG+2 angleFGE=180^(@)` `angleFEG+angleFGE=90^(@)" "`(1) `angleFEG=angleFGE+10^(@)" "`(2) On solving Eq. (1) and (2), we get `angleFGE=40^(@)`. Hence, the correct option is ( c). |
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| 31. |
Show that each diagonal of a parallelogram divide it into two congruent triangles. The following are the steps involved in showing the above result. Arrange them in sequential order. A) In `triangleABC` and `triangleCDA`, AB=DC and BC=AD `(therefore` opposite angles of parallelogram) AC=AC (common side). B) Let ABCD be a parallelogram. Join AC. C) By SSS congruence property, `triangleABC ~=triangleCDA`. D) Similarly, BD divides the triangle into two congruent triangles.A. BACDB. BDACC. BADCD. BDCA |
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Answer» Correct Answer - A BACD is the required sequential order. |
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| 32. |
One angle of a parallelogram is `30^(@)` more than twice its adjacent angles. Find the measure of its adjacent angle.A. `50^(@)`B. `60^(@)`C. `70^(@)`D. `80^(@)` |
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Answer» Correct Answer - A Let the adjacent angle be `x^(@)`. `:.` Given angle is `(2x^(@)+30^(@))` We know that the sum of the adjacent angles of a prallelogram is `180^(@)`. `implies (x^(@)+2x^(@)+30^(@))=180^(@)` `(3x^(@))=180^(@)-30^(@)` `x^(@)=(150^(@))/(3)` `:. x^(@)=50^(@)` Hence, the correct option is (a) . |
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| 33. |
The measure of one of the parallelogram is `70^(@)`. Find the measures of the angles of the parallelogram . The following steps are involved in solving the above problem. Arrange them in sequential order. (A) `70^(@)+x=180^(@)impliesx=110^(@)` (B) Let the angle adjacent to `70^(@)` be x. (C ) The sum of the measures of adjacent angle of a parallelogram is `180^(@)`. (D) The measure of the angles of the parallelogram are `70^(@),110^(@),70^(@) and 110^(@)`A. CBDAB. BCADC. BCDAD. CDAB |
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Answer» Correct Answer - B (B),(C ),(A) and (D) is the required sequential order. Hence , the correct option is (b) |
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| 34. |
Fill in the blank,The number of lines of symmetry in a regular hexagon is __________. |
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Answer» The number of lines of symmetry in a regular hexagon is 6. |
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| 35. |
If one angle of a cyclic quadrilateral is 75°, then the opposite angle is (1) 100° (2) 105° (3) 85° (4) 90° |
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Answer» (2) 105° 180° – 75° = 105° |
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| 36. |
Which of the letters of the English alphabet have only one line of symmetry ? |
| Answer» Correct Answer - A,B,C,D,E,K,M,I,U,V,W,Y | |
| 37. |
The number of line segments in(a) 1(b) 2(c) 3(d) 4 |
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Answer» (c) 3 AB, AC, BC |
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| 38. |
Measure the following line segments. |
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Answer» XY = 2.4 cm; AB = 3.4 cm; EF = 4 cm; PQ = 3 cm |
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| 39. |
Is a triangle possible with the angles 90°, 90° and 0°? Why? |
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Answer» No, a triangle cannot have more than one right angle. |
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| 40. |
Name all the line segments. |
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Answer» AB, AE, EB, CD, CE and ED |
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| 41. |
Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular) |
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Answer» (i) Parallel lines (ii) Parallel lines (iii) Parallel lines and Perpendicular lines (iv) Intersecting lines |
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| 42. |
From the following figure, identify, (i) Pairs of intersecting lines (ii) Parallel lines (iii) Concurrent lines |
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Answer» (i) F and G, F and H, F and I, G and I, H and I, J and H, J and G, J and F, K and G, K and H (ii) `J" || "I, J" || "K, I" || "K` (iii). F, G, H and I |
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| 43. |
Will the lengths of line segment AB and line segment BC make the length of line segment AC in Fig.? |
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Answer» The correct answer is Yes |
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| 44. |
In AD⊥BC, prove that AB2 + CD2 = BD2 + AC2. |
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Answer» From ΔADC, we have AC2 = AD2 + CD2 … (1) (Pythagoras theorem) From ΔADB, we have AB2 = AD2 + BD2 … (2) (Pythagoras theorem) Subtracting (1) from (2) we have, AB2 – AC2 = BD2 – CD2 AB2 + CD2 = BD2 + AC2 |
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| 45. |
In figure O is any point inside a rectangle ABCD. Prove that OB2 + OD2 = OA2 + OC2. |
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Answer» Through O, draw PQ||BC so that P lies on AB and Q lies on DC. Now, PQ||BC PQ ⊥ AB and PQ ⊥ DC (∵ ∠B = 90° and ∠C = 90°) So, ∠BPQ = 90° and ∠CQP = 90° Therefore BPQC and APQD are both rectangles. Now from ΔOPB, OB2 = BP2 + OP2 … (1) Similarly from ΔOQD, OD2 = OQ2 + DQ2 … (2) From ΔOQC, we have OC2 = OQ2 + CQ2 … (3) ΔOAP, we have OA2 = AP2 + OP2 … (4) Adding (1) and (2) OB2 + OD2 = BP2 + OP2 + OQ2 + DQ2 (As BP = CQ and DQ = AP) = CQ2 + OP2 + OQ2 + AP2 = CQ2 + OQ2 + OP2 + AP2 = OC2 + OA2 [From (3) and (4)] |
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| 46. |
Two sides of a triangle are 5 cm and 12 cm long. The measure of third sides is an integer in cm. if the tirangles is an obtuse tirangle, then how many such triangles are possible ?A. 9B. 8C. 7D. 6 |
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Answer» Correct Answer - D Two sides of a triangle 5 cm and 12 cm. Let `a=5 cm and b=12 cm` Let the third side be x cm `therefore 12-5 lt x lt 12 +5` `rArr7 lt xlt 17` `therefore` Possible integer values for x are 8,9,10,11,12, 13,14,15,and 16. Case 1: If b is the following side then `b^2gt a^2+x^2 ` `rArr 12^2gt 5^2+x^2` `rArr 144=25gt x^2` `rArr x^2lt 119` `therefore x` can be , 8,9 or 10. Case 2: IF x is the longest side, then `x^2gt a^2+b^2` `rArr x^2gt 5^2+12^2rArrx^2rArr169` `therefore x` can be `14,15 or 16` `therefore` Number of possible triangles. `=6 (because "The measurements third side is an integer is cm")`. |
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| 47. |
In the above figure, if `ST||QR` and `PS:PQ=2: 5 and TR=15 cm`, then PT=_____________. |
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Answer» Correct Answer - 10 cm N/A |
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| 48. |
In the diagram above, A,B,P and Q are points of contacts of direct common tangents of the two circles. If `angle ACB` is `120^@` , then find the angle between the two tangents and angle made by PQ at the centre of same circle. |
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Answer» Correct Answer - `60^@, 120^@` N/A |
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| 49. |
In the given figure, C is a point on the line segment BD. Find the measurements of `angleACB` and `angleABC.` |
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Answer» `(angleACB,angleACD)` is linear pair. `rArr angleACB+125^(@)=180^(@)` `angleACB=180^(@)-125^(@)=55^(@)` `angleABC+angleCAB=angleACD "(Exterior angle)"` `angleABC+95^(@)=125^(@)` `angleABC=125^(@)-95^(@)=30^(@)` |
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| 50. |
In the above figure, ACB is a straight line and `angleACD:angleDCB=2:1` . Find `angleDCB`. |
| Answer» Correct Answer - `60^(@)` | |