## GEOMETRY SET II MCQs

Total Questions : 60 | Page 1 of 6 pages
Question 1. Suresh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Suresh walks according to the following plan: He moves along the altitude-to-the-hypotenuse until he reaches the hypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitude to the hypotenuse of the larger one. He repeats this process forever. What is the total distance that Suresh walks?
1.    4825
2.    125
3.    12
4.    15
:
C

Let M be the end point of the altitude on the hypotenuse. Since, we are dealing with right angle
triangles, ΔMAC ~ΔABC, so AM = 125. Let N be the endpoint he reaches on side AC. Δ
MAC ~Δ NAM, So, MNAM =45. This means that each altitude that he walks gets shorter
by a factor of 45. The total distance is thus an infinite G.P. =[a1r] = 12
Question 2. There is a ship which is moving away from an iceberg 150 m high, takes 2 mins to change the angle of elevation of the top of the iceberg from 60 to 45. What is the speed of the ship?
1.    2.46 kmph
2.    1.3 kmph
3.    1.9 kmph
4.    1.7 kmph
Answer: Option C. -> 1.9 kmph
:
C

In a 45-45-90 triangle, sides are in the ratio 1:1:2
Thus OB= 150
In a 30-60-90 triangle, sides are in the ration 1:3:2
Thus OA = 503
Distance travelled in 2 mins = AB = 150- 503= 63.4 m
speed = (63.42)× (601000) = 1.9 kmph
Question 3. All sides of a regular pentagon are extended to form a star with vertices P,Q,R,S,T. What is the sum of the angles made at the vertices?___

:

Interior angle = [(2n-4)x90]/n = 108
2x= 360-(108x2) = 144
angle P = 180-144= 36
Total = 36x5= 180
Useful to remember: Sum of angles in a star= 180
Shortcut: Clearly the star point trisects the angle in a pentagon. Hence each angle = 108/3 = 36.
Sum of angles = 36*5 = 180
Alternatively: Sum of angles of a n point star = (n – 4) x 180, where n = number of sides of a polygon.
Here, n = 5.Sum of angles of a n point star = (5 – 4) x 180=180.
Question 4. Three balls of equal radius are placed such that they are touching each other. A fourth smaller ball is kept such that it touches the other three.Find the ratio of the radii of smaller to larger ball.
1.    (2−√3)√3
2.    3- 2√3
3.    3.5
4.    4√5
:
A

Option (a)
Let the centers of the large balls be x, y, z and radius R.
O is the centre of the smaller ball and radius r...
x, y, z form an equilateral triangle with side equal to 2R.
O is the centroid of this triangle.
Therefore ox=oy=oz=R+r= 23(height of the triangle xyz)
Height=(32)(2R) =3R
Therefore R+r = 23(3R) rR = (23)3
Shortcut:- Using the approximation technique used in class, the radius of the bigger circle: smaller circle is close to 0.2.
Question 5. If O is the centre of a circle and the radius of the circle is 4 units, find the value of length of AC.
1.    4( √3 + 1)
2.    9 √2
3.    10.5
4.    none of these
Answer: Option D. -> none of these
:
D
Option (d)
AC has to less than 8 as the diameter will be of 8 units. So the correct option is “D” as all are greater than 8
Question 6. What is the maximum number of times x circles of the same size can intersect each other?
1.    2x
2.    (x - 1)
3.    x(x-1)
4.    x(x+1)/2
:
C
Deducing a Pattern:
Taking two circles. The number of points of intersection is 2 at most.
If u consider 3 circles, the number of points of intersection = 6 ( the third circles will have 4 points of intersection with the other 2 circles and those two circles will have 2 points together).
Shortcut:- Assumption & Reverse Gear:
Taking two circles. The number of points of intersection is 2 at most. From this itself, option a, b and d can be eliminated giving answer c.
Question 7. A metal cuboid with sides in the ratio of 2:4:6 is melted to form small cubes of side 2 cm. If the sum of all the edges of the cuboid is 144 cm, then find the ratio of the total surface area of the cuboid to the total surface area of the cubes.
1.    2:1
2.    13:4
3.    11:54
4.    4:1
:
C
Sides are in the ratio 2:4:6
Volume of cuboid= 2x .4x . 6x = 48 x3
Volume of cube= 2.2.2=8
Number of cubes= 6x3
Length of all edges of the cuboid= 144 =4(2x+4x+6x) x=3cm
Ratio = 2[72+108+216]6(27) x 6 x 4
= 1154
Question 8. Find the distance between the incentre and the circumcentre of a triangle with circumradius 6 and inradius 2?
1.    3√2
2.    2√3
3.    √10
4.    None of these
:
B
Use Eulers triangle theorem which states that the distance,d between the incentre and circumcentre of a trinagle is given by d2 = R(R-2r) where R = circum radius, r= inradius
d2= R(R-2r) = 6(6 - 2 x 2) = 12
d=23
Question 9. Two travelers start walking from the same point at an angle of 1500 with each other at the rate of 4 kmph and 3 kmph. Find the distance between them after 2 hours
1.    2√25+8√3
2.    28-48√3
3.    2√25+12√3
4.    none of these
:
C
option c
let OA and OB be the paths traveled by the two travelers in 2 hours. Let BCÃ¢Â?Â´ AO at C. then BOC= 180-150= 30
In right angled triangle OCB, BC= 62=3 KM and OC= 33 km
In right angled triangle ACB, AB2= AC2 + BC2 = (8+3√3)2+32= 100+483
AB= 225+123
Shortcut
a2+b22abcosα; α is the angle between the paths a and b between the 2 people. 100+48is the answer
Question 10. With no wastage, small cubes of side 3 inches are formed from a cuboid with the dimensions 2:3:9 are formed. Find the ratio of the body diagonal of the cuboid to that of all of the cubes.
1.    1:3
2.    Ratio = √946a23√3
3.    Ratio = √65a23
4.    none of these
Answer: Option B. -> Ratio = √946a23√3
:
B
Number of cubes = 2a×3a×9a3×3×3 = 2a3
Diagonal of a cube = 4a2+9a2+81a2
Sum of the diagonals of cubes = 2a3 × 33
Ratio =4a2+9a2+81a22a3×33
The text passage contains multiple geometry questions with options and answers provided for some of them.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
## GEOMETRY SET II MCQs

Total Questions : 60 | Page 1 of 6 pages
Question 1. Suresh is standing on vertex A of triangle ABC, with AB = 3, BC = 5, and CA = 4. Suresh walks according to the following plan: He moves along the altitude-to-the-hypotenuse until he reaches the hypotenuse. He has now cut the original triangle into two triangles; he now walks along the altitude to the hypotenuse of the larger one. He repeats this process forever. What is the total distance that Suresh walks?
1.    4825
2.    125
3.    12
4.    15
:
C

Let M be the end point of the altitude on the hypotenuse. Since, we are dealing with right angle
triangles, ΔMAC ~ΔABC, so AM = 125. Let N be the endpoint he reaches on side AC. Δ
MAC ~Δ NAM, So, MNAM =45. This means that each altitude that he walks gets shorter
by a factor of 45. The total distance is thus an infinite G.P. =[a1r] = 12
Question 2. There is a ship which is moving away from an iceberg 150 m high, takes 2 mins to change the angle of elevation of the top of the iceberg from 60 to 45. What is the speed of the ship?
1.    2.46 kmph
2.    1.3 kmph
3.    1.9 kmph
4.    1.7 kmph
Answer: Option C. -> 1.9 kmph
:
C

In a 45-45-90 triangle, sides are in the ratio 1:1:2
Thus OB= 150
In a 30-60-90 triangle, sides are in the ration 1:3:2
Thus OA = 503
Distance travelled in 2 mins = AB = 150- 503= 63.4 m
speed = (63.42)× (601000) = 1.9 kmph
Question 3. All sides of a regular pentagon are extended to form a star with vertices P,Q,R,S,T. What is the sum of the angles made at the vertices?___

:

Interior angle = [(2n-4)x90]/n = 108
2x= 360-(108x2) = 144
angle P = 180-144= 36
Total = 36x5= 180
Useful to remember: Sum of angles in a star= 180
Shortcut: Clearly the star point trisects the angle in a pentagon. Hence each angle = 108/3 = 36.
Sum of angles = 36*5 = 180
Alternatively: Sum of angles of a n point star = (n – 4) x 180, where n = number of sides of a polygon.
Here, n = 5.Sum of angles of a n point star = (5 – 4) x 180=180.
Question 4. Three balls of equal radius are placed such that they are touching each other. A fourth smaller ball is kept such that it touches the other three.Find the ratio of the radii of smaller to larger ball.
1.    (2−√3)√3
2.    3- 2√3
3.    3.5
4.    4√5
:
A

Option (a)
Let the centers of the large balls be x, y, z and radius R.
O is the centre of the smaller ball and radius r...
x, y, z form an equilateral triangle with side equal to 2R.
O is the centroid of this triangle.
Therefore ox=oy=oz=R+r= 23(height of the triangle xyz)
Height=(32)(2R) =3R
Therefore R+r = 23(3R) rR = (23)3
Shortcut:- Using the approximation technique used in class, the radius of the bigger circle: smaller circle is close to 0.2.
Question 5. If O is the centre of a circle and the radius of the circle is 4 units, find the value of length of AC.
1.    4( √3 + 1)
2.    9 √2
3.    10.5
4.    none of these
Answer: Option D. -> none of these
:
D
Option (d)
AC has to less than 8 as the diameter will be of 8 units. So the correct option is “D” as all are greater than 8
Question 6. What is the maximum number of times x circles of the same size can intersect each other?
1.    2x
2.    (x - 1)
3.    x(x-1)
4.    x(x+1)/2
:
C
Deducing a Pattern:
Taking two circles. The number of points of intersection is 2 at most.
If u consider 3 circles, the number of points of intersection = 6 ( the third circles will have 4 points of intersection with the other 2 circles and those two circles will have 2 points together).
Shortcut:- Assumption & Reverse Gear:
Taking two circles. The number of points of intersection is 2 at most. From this itself, option a, b and d can be eliminated giving answer c.
Question 7. A metal cuboid with sides in the ratio of 2:4:6 is melted to form small cubes of side 2 cm. If the sum of all the edges of the cuboid is 144 cm, then find the ratio of the total surface area of the cuboid to the total surface area of the cubes.
1.    2:1
2.    13:4
3.    11:54
4.    4:1
:
C
Sides are in the ratio 2:4:6
Volume of cuboid= 2x .4x . 6x = 48 x3
Volume of cube= 2.2.2=8
Number of cubes= 6x3
Length of all edges of the cuboid= 144 =4(2x+4x+6x) x=3cm
Ratio = 2[72+108+216]6(27) x 6 x 4
= 1154
Question 8. Find the distance between the incentre and the circumcentre of a triangle with circumradius 6 and inradius 2?
1.    3√2
2.    2√3
3.    √10
4.    None of these
:
B
Use Eulers triangle theorem which states that the distance,d between the incentre and circumcentre of a trinagle is given by d2 = R(R-2r) where R = circum radius, r= inradius
d2= R(R-2r) = 6(6 - 2 x 2) = 12
d=23
Question 9. Two travelers start walking from the same point at an angle of 1500 with each other at the rate of 4 kmph and 3 kmph. Find the distance between them after 2 hours
1.    2√25+8√3
2.    28-48√3
3.    2√25+12√3
4.    none of these
:
C
option c
let OA and OB be the paths traveled by the two travelers in 2 hours. Let BCÃ¢Â?Â´ AO at C. then BOC= 180-150= 30
In right angled triangle OCB, BC= 62=3 KM and OC= 33 km
In right angled triangle ACB, AB2= AC2 + BC2 = (8+3√3)2+32= 100+483
AB= 225+123
Shortcut
a2+b22abcosα; α is the angle between the paths a and b between the 2 people. 100+48is the answer
Question 10. With no wastage, small cubes of side 3 inches are formed from a cuboid with the dimensions 2:3:9 are formed. Find the ratio of the body diagonal of the cuboid to that of all of the cubes.
1.    1:3
2.    Ratio = √946a23√3
3.    Ratio = √65a23
4.    none of these
Answer: Option B. -> Ratio = √946a23√3
:
B
Number of cubes = 2a×3a×9a3×3×3 = 2a3
Diagonal of a cube = 4a2+9a2+81a2
Sum of the diagonals of cubes = 2a3 × 33
Ratio =4a2+9a2+81a22a3×33
user
The text passage contains multiple geometry questions with options and answers provided for some of them.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS