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All questions of Mensuration and geometry for Mechanical Engineering Exam
Find the perimeter of the a square with side 4cm.- a)16 cm
- b)12 cm
- c)10 cm
- d)8 cm
Correct answer is option 'A'. Can you explain this answer?
Find the perimeter of the a square with side 4cm.
a)
16 cm
b)
12 cm
c)
10 cm
d)
8 cm
|
Sreesha Phalod answered |
Sol. perimeter of square=4* sides
=4*4 (sides is 4)
=16.
=4*4 (sides is 4)
=16.
Find the volume of a cuboid whose length is 8 cm, width is 3 cm and height is 5 cm. - a)135 cm3
- b)125 cm3
- c)120 cm3
- d)130 cm3
Correct answer is option 'C'. Can you explain this answer?
Find the volume of a cuboid whose length is 8 cm, width is 3 cm and height is 5 cm.
a)
135 cm3
b)
125 cm3
c)
120 cm3
d)
130 cm3
|
Tanishq Joshi answered |
Finding Volume of a Cuboid
Given: length = 8 cm, width = 3 cm, height = 5 cm
To find: Volume of the cuboid
Formula: Volume of a cuboid = length x width x height
Substituting the given values in the formula, we get:
Volume = 8 cm x 3 cm x 5 cm
Volume = 120 cm3
Therefore, the correct answer is option C, 120 cm3.
Given: length = 8 cm, width = 3 cm, height = 5 cm
To find: Volume of the cuboid
Formula: Volume of a cuboid = length x width x height
Substituting the given values in the formula, we get:
Volume = 8 cm x 3 cm x 5 cm
Volume = 120 cm3
Therefore, the correct answer is option C, 120 cm3.
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for chapter Mensuration, Class 8, Mathematics . You can practice these practice quizzes as per your speed and improvise the topic. Q.Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm. - a)215 cm3
- b)172 cm3
- c)150 cm3
- d)168 cm3
Correct answer is option 'D'. Can you explain this answer?
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for chapter Mensuration, Class 8, Mathematics . You can practice these practice quizzes as per your speed and improvise the topic.
Q.
Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm.
a)
215 cm3
b)
172 cm3
c)
150 cm3
d)
168 cm3
|
Ankita Shah answered |
Given,
Length (l) = 8 cm
Breadth (b) = 6 cm
Height (h) = 3.5 cm
We know that the volume of a cuboid is given by the formula:
Volume = length × breadth × height
Substituting the given values, we get:
Volume = 8 cm × 6 cm × 3.5 cm
Volume = 168 cm³
Therefore, the volume of the given cuboid is 168 cm³.
Hence, the correct option is (d) 168 cm³.
Length (l) = 8 cm
Breadth (b) = 6 cm
Height (h) = 3.5 cm
We know that the volume of a cuboid is given by the formula:
Volume = length × breadth × height
Substituting the given values, we get:
Volume = 8 cm × 6 cm × 3.5 cm
Volume = 168 cm³
Therefore, the volume of the given cuboid is 168 cm³.
Hence, the correct option is (d) 168 cm³.
Find the area of a triangle whose base is 4 cm and altitude is 6 cm.- a)10 cm2
- b)14 cm2
- c)16 cm2
- d)12 cm2
Correct answer is option 'D'. Can you explain this answer?
Find the area of a triangle whose base is 4 cm and altitude is 6 cm.
a)
10 cm2
b)
14 cm2
c)
16 cm2
d)
12 cm2
|
Kavya Saxena answered |
We know that area of triangle is equals to 1/2 base × altitude.
Here, base = 4 cm and altitude = 6 cm.
So, area = 1/2 × 4 × 6= 24 /2= 12 cm2.
Here, base = 4 cm and altitude = 6 cm.
So, area = 1/2 × 4 × 6= 24 /2= 12 cm2.
A rectangular enclosure 40 m x 36 m has a horse tethered to a corner with a rope of 14 m in length. What is the ratio of the respective areas it can graze, if it is outside the enclosure and if it is inside the enclosure?- a)2
- b)2:3
- c)1:4
- d)3:1
Correct answer is option 'D'. Can you explain this answer?
A rectangular enclosure 40 m x 36 m has a horse tethered to a corner with a rope of 14 m in length. What is the ratio of the respective areas it can graze, if it is outside the enclosure and if it is inside the enclosure?
a)
2
b)
2:3
c)
1:4
d)
3:1
|
Talent Skill Learning answered |
- Imagine a circle on the corner of the rectangle.
- 3 quarters of the circle lie outside the rectangle and 1 quarter lies inside.
- Hence, Required ratio = 3 : 1
PQRST is a pentagon in which all the interior angles are unequal. A circle of radius ‘r’ is inscribed in each of the vertices. Find the area of portion of circles falling inside the pentagon. - a)πr2
- b)1.5πr2
- c)2πr2
- d)1.25πr2
Correct answer is option 'B'. Can you explain this answer?
PQRST is a pentagon in which all the interior angles are unequal. A circle of radius ‘r’ is inscribed in each of the vertices. Find the area of portion of circles falling inside the pentagon.
a)
πr2
b)
1.5πr2
c)
2πr2
d)
1.25πr2
|
Preeti Khanna answered |
Since neither angles nor sides are given in the question, immediately the sum of angles of pentagon should come in mind. To use it,
We know the area of the sectors of a circle is given as,
We know the area of the sectors of a circle is given as,
Note => The above concept is applicable for a polygon of n sides.
Choice (B) is therefore, the correct answer.
Correct Answer: 1.5πr2
Choice (B) is therefore, the correct answer.
Correct Answer: 1.5πr2
A cyclic quadrilateral is such that two of its adjacent angles are divisible by 6 and 10 respectively. One of the remaining angles will necessarily be divisible by:
- a)3
- b)4
- c)8
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
A cyclic quadrilateral is such that two of its adjacent angles are divisible by 6 and 10 respectively. One of the remaining angles will necessarily be divisible by:
a)
3
b)
4
c)
8
d)
None of these
|
Palak Bose answered |
We know that the sum of the opposite angles of a cyclic quadrilateral is 180 degrees. Let the four angles be A, B, C, and D, with A and B being the angles divisible by 6 and 10, respectively.
Since A is divisible by 6 and B is divisible by 10, we know that A = 6m and B = 10n for some integers m and n.
Now, consider the opposite angles. Since the sum of opposite angles is 180 degrees, we have:
C = 180 - B = 180 - 10n
D = 180 - A = 180 - 6m
We want to find which of the given options the angles C or D are necessarily divisible by. Let's examine each option:
1. 3: Since B is divisible by 10, it is possible that B is divisible by 5 but not 3 (e.g. B = 10). In this case, C = 180 - B would not be divisible by 3. Also, A is divisible by 6, so A is always divisible by 3, which means D = 180 - A would never be divisible by 3. So, this option is incorrect.
2. 4: Since A is divisible by 6, it is possible that A is divisible by 2 but not 4 (e.g. A = 6). In this case, D = 180 - A would not be divisible by 4. Also, B is divisible by 10, so B is always divisible by 2, which means C = 180 - B would never be divisible by 4. So, this option is also incorrect.
3. 8: If A is divisible by 6, then it can be even or odd multiples of 6 (e.g. A = 6, 12, 18, ...). D will be 180 - A, which means D can be both even and odd (e.g. D = 180 - 6 = 174, D = 180 - 12 = 168, D = 180 - 18 = 162, ...). Since D can be both even and odd, it is not necessarily divisible by 8. Similarly, C can also be both even and odd, so it is not necessarily divisible by 8. Thus, this option is also incorrect.
4. None of these: Since none of the previous options work, the correct answer is None of these.
So, the correct answer is option 4: None of these.
Since A is divisible by 6 and B is divisible by 10, we know that A = 6m and B = 10n for some integers m and n.
Now, consider the opposite angles. Since the sum of opposite angles is 180 degrees, we have:
C = 180 - B = 180 - 10n
D = 180 - A = 180 - 6m
We want to find which of the given options the angles C or D are necessarily divisible by. Let's examine each option:
1. 3: Since B is divisible by 10, it is possible that B is divisible by 5 but not 3 (e.g. B = 10). In this case, C = 180 - B would not be divisible by 3. Also, A is divisible by 6, so A is always divisible by 3, which means D = 180 - A would never be divisible by 3. So, this option is incorrect.
2. 4: Since A is divisible by 6, it is possible that A is divisible by 2 but not 4 (e.g. A = 6). In this case, D = 180 - A would not be divisible by 4. Also, B is divisible by 10, so B is always divisible by 2, which means C = 180 - B would never be divisible by 4. So, this option is also incorrect.
3. 8: If A is divisible by 6, then it can be even or odd multiples of 6 (e.g. A = 6, 12, 18, ...). D will be 180 - A, which means D can be both even and odd (e.g. D = 180 - 6 = 174, D = 180 - 12 = 168, D = 180 - 18 = 162, ...). Since D can be both even and odd, it is not necessarily divisible by 8. Similarly, C can also be both even and odd, so it is not necessarily divisible by 8. Thus, this option is also incorrect.
4. None of these: Since none of the previous options work, the correct answer is None of these.
So, the correct answer is option 4: None of these.
Four horses are tethered at four comers of a square plot of side 14 m so that the adjacent horses can just reach one another. There is a small circular pond of area 20 m2 at the centre. Find the ungrazed area.- a)42 m2
- b)22 m2
- c)84 m2
- d)168 m2
Correct answer is option 'B'. Can you explain this answer?
Four horses are tethered at four comers of a square plot of side 14 m so that the adjacent horses can just reach one another. There is a small circular pond of area 20 m2 at the centre. Find the ungrazed area.
a)
42 m2
b)
22 m2
c)
84 m2
d)
168 m2
|
|
Preeti Khanna answered |
Total area of plot = 14 * 14 = 196m2
Horses can graze in quarter circle of radius = 7m
Grazed area = 4 * (pie r2)/4 = 154 m2
Area of plot when horses cannot reach = (196 - 154) = 42m2
Ungrazed area = 42 - 20 = 22m2
Horses can graze in quarter circle of radius = 7m
Grazed area = 4 * (pie r2)/4 = 154 m2
Area of plot when horses cannot reach = (196 - 154) = 42m2
Ungrazed area = 42 - 20 = 22m2
Read the passage below and solve the questions based on it The area of a square is equal lo the area of a rectangle. Moreover, the perimeter of the square is also equal to the perimeter of the rectangle,Q.The length of the rectangle is equal to the:a)Breadth of the rectangleb)Side of the squarec)Cannot be determinedd)Both [1] and [2]Correct answer is option 'D'. Can you explain this answer?
|
Nikita Singh answered |
This is possible only when both the length and breadth of the rectangle are equal to the side of the square.
The ratio of the sides of Δ ABC is 1:2:4. What is the ratio of the altitudes drawn onto these sides?- a)4:2:1
- b)1:2:4
- c)1:4:16
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
The ratio of the sides of Δ ABC is 1:2:4. What is the ratio of the altitudes drawn onto these sides?
a)
4:2:1
b)
1:2:4
c)
1:4:16
d)
None of these
|
Aakash Giery answered |
Sum of any two sides should be greater than third side.
here 1+2=3 is not less than 4 ,
1+2<4 ,so="" triangle="" is="" not="" possible.="" ,so="" triangle="" is="" not="">
here 1+2=3 is not less than 4 ,
1+2<4 ,so="" triangle="" is="" not="" possible.="" ,so="" triangle="" is="" not="">
In the given figure, AB and CD are two chords of a circle intersecting at O. If AO = 4 cm and OB = 6 cm and OC = 3 cm. Find OD.
- a)4 cm
- b)6 cm
- c)8 cm
- d)10 cm
Correct answer is option 'C'. Can you explain this answer?
In the given figure, AB and CD are two chords of a circle intersecting at O. If AO = 4 cm and OB = 6 cm and OC = 3 cm. Find OD.
a)
4 cm
b)
6 cm
c)
8 cm
d)
10 cm
|
Pathways Academy answered |
By the rule of chords, cutting externally, we get:
(9 + 6) * 6 = (5 + x) * 5
90 = 25 + 5x
5x = 65
x = 13 cm
90 = 25 + 5x
5x = 65
x = 13 cm
.
PQRS is a circle and circles are drawn with PO, QO, RO and SO as diameters. Areas A and B are marked. A/B is equal to:
- a)π
- b)1
- c)π/4
- d)2
Correct answer is option 'B'. Can you explain this answer?
PQRS is a circle and circles are drawn with PO, QO, RO and SO as diameters. Areas A and B are marked. A/B is equal to:
a)
π
b)
1
c)
π/4
d)
2
|
Divey Sethi answered |
Such questions are all about visualization and ability to write one area in terms of others.
Here, Let the radius of PQRS be 2r
∴ Radius of each of the smaller circles = 2r/2 = r
∴ Area A can be written as:
A = π (2r)2 – 4 x π(r)2 (Area of the four smaller circles) + B (since, B has been counted twice in the previous subtraction)
Here, Let the radius of PQRS be 2r
∴ Radius of each of the smaller circles = 2r/2 = r
∴ Area A can be written as:
A = π (2r)2 – 4 x π(r)2 (Area of the four smaller circles) + B (since, B has been counted twice in the previous subtraction)
A = 4πr2 - 4πr2 + B
A = B
A/B = 1
A = B
A/B = 1
Choice (B) is therefore, the correct answer.
Correct Answer: 1
Correct Answer: 1
The area of similar triangles, ABC and DEF are 144cm2 and 81 cm2 respectively. If the longest side of the larger △ABC be 36 cm, then the longest side of the smaller △DEF is:
- a)27 cm
- b)26 cm
- c)29 cm
- d)30 cm
Correct answer is option 'A'. Can you explain this answer?
The area of similar triangles, ABC and DEF are 144cm2 and 81 cm2 respectively. If the longest side of the larger △ABC be 36 cm, then the longest side of the smaller △DEF is:
a)
27 cm
b)
26 cm
c)
29 cm
d)
30 cm
|
Nithin Bharadwaj answered |
What we need to know here is,
Supplement angles means. Sum of both the angle is 180
X+5X =180
Thay means,
X=30
Other angle is 150
Supplement angles means. Sum of both the angle is 180
X+5X =180
Thay means,
X=30
Other angle is 150
If the edge of a cube is 1 cm then which of the following is its total surface area?- a)1 cm2
- b)4 cm2
- c)6 cm2
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
If the edge of a cube is 1 cm then which of the following is its total surface area?
a)
1 cm2
b)
4 cm2
c)
6 cm2
d)
none of these
|
|
Stuti Basak answered |
Explanation:
To find the total surface area of a cube, we need to find the area of all its six faces and add them up. Since all the faces of a cube are identical squares, we can find the area of one face and multiply it by 6 to get the total surface area.
Given, the edge of the cube is 1 cm. Therefore, the area of one face of the cube is:
Area of square = side × side
Area of square = 1 cm × 1 cm
Area of square = 1 cm²
To find the total surface area of the cube, we need to multiply the area of one face by 6:
Total surface area of cube = 6 × area of one face
Total surface area of cube = 6 × 1 cm²
Total surface area of cube = 6 cm²
Therefore, the total surface area of the cube is 6 cm², which is option C.
To find the total surface area of a cube, we need to find the area of all its six faces and add them up. Since all the faces of a cube are identical squares, we can find the area of one face and multiply it by 6 to get the total surface area.
Given, the edge of the cube is 1 cm. Therefore, the area of one face of the cube is:
Area of square = side × side
Area of square = 1 cm × 1 cm
Area of square = 1 cm²
To find the total surface area of the cube, we need to multiply the area of one face by 6:
Total surface area of cube = 6 × area of one face
Total surface area of cube = 6 × 1 cm²
Total surface area of cube = 6 cm²
Therefore, the total surface area of the cube is 6 cm², which is option C.
Find the value of x in the given figure.
- a)16 cm
- b)7 cm
- c)12 cm
- d)9 cm
Correct answer is option 'D'. Can you explain this answer?
Find the value of x in the given figure.
a)
16 cm
b)
7 cm
c)
12 cm
d)
9 cm
|
Pooja Sen answered |
Isosceles trapezium is always cyclic The sum of opposite angles of a cyclic quadrilateral is 180°
The dimensions of a rectangular box are in the ratio of 1:2:4 and the difference between the costs of covering it with the cloth and a sheet at the rate of Rs 20 and Rs 20.5 per sq m respectively is Rs 126. Find the dimensions of the box.- a)2 m, 6 m, 9 m
- b)6 m, 12 m, 24 m
- c)1 m, 2 m, 4 m
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
The dimensions of a rectangular box are in the ratio of 1:2:4 and the difference between the costs of covering it with the cloth and a sheet at the rate of Rs 20 and Rs 20.5 per sq m respectively is Rs 126. Find the dimensions of the box.
a)
2 m, 6 m, 9 m
b)
6 m, 12 m, 24 m
c)
1 m, 2 m, 4 m
d)
None of these
|
Debanshi Rane answered |
The length of each side of a cube is 24 cm. The volume of the cube is equal to the volume of a cuboid. If the breadth and the height of the cuboid are 32 cm and 12 cm, respectively, then what will be the length of the cuboid?- a)36
- b)27
- c)16
- d)20
Correct answer is option 'A'. Can you explain this answer?
The length of each side of a cube is 24 cm. The volume of the cube is equal to the volume of a cuboid. If the breadth and the height of the cuboid are 32 cm and 12 cm, respectively, then what will be the length of the cuboid?
a)
36
b)
27
c)
16
d)
20
|
C K Academy answered |
Given:
The length of each side of a cube is 24 cm.
The breadth and the height of the cuboid are 32 cm and 12 cm, respectively.
The length of each side of a cube is 24 cm.
The breadth and the height of the cuboid are 32 cm and 12 cm, respectively.
Concept used:
The volume of the cube is equal to the volume of a cuboid.
Volume of cube = a³
Volume of cuboid = lbh
The volume of the cube is equal to the volume of a cuboid.
Volume of cube = a³
Volume of cuboid = lbh
Calculation:
The volume of the cube is equal to the volume of a cuboid.
⇒ 24³ = l × 32 × 12
⇒ l = 3 × 12
⇒ l = 36
The volume of the cube is equal to the volume of a cuboid.
⇒ 24³ = l × 32 × 12
⇒ l = 3 × 12
⇒ l = 36
∴ Option 1 is the correct answer.
The area of a rhombus is 200 cm², and one of its diagonals is 20 cm. The length of the other diagonal is _____- a)20 cm
- b)20 m
- c)22 cm
- d)22 m
Correct answer is option 'A'. Can you explain this answer?
The area of a rhombus is 200 cm², and one of its diagonals is 20 cm. The length of the other diagonal is _____
a)
20 cm
b)
20 m
c)
22 cm
d)
22 m
|
Roshni Chauhan answered |
Understanding the Area of a Rhombus
The area of a rhombus can be calculated using the formula:
- Area = (d1 * d2) / 2
where d1 and d2 are the lengths of the diagonals.
Given Information
- Area = 200 cm²
- One diagonal (d1) = 20 cm
Finding the Other Diagonal
To find the length of the second diagonal (d2), we can rearrange the area formula:
- 200 = (20 * d2) / 2
Now, let's solve for d2:
- First, multiply both sides by 2:
- 400 = 20 * d2
- Next, divide both sides by 20:
- d2 = 400 / 20
- d2 = 20 cm
Conclusion
The length of the other diagonal (d2) is 20 cm.
Final Answer
The correct answer is option 'A' (20 cm).
This shows that both diagonals of the rhombus can be equal, which is a special case where the rhombus is also a square.
The area of a rhombus can be calculated using the formula:
- Area = (d1 * d2) / 2
where d1 and d2 are the lengths of the diagonals.
Given Information
- Area = 200 cm²
- One diagonal (d1) = 20 cm
Finding the Other Diagonal
To find the length of the second diagonal (d2), we can rearrange the area formula:
- 200 = (20 * d2) / 2
Now, let's solve for d2:
- First, multiply both sides by 2:
- 400 = 20 * d2
- Next, divide both sides by 20:
- d2 = 400 / 20
- d2 = 20 cm
Conclusion
The length of the other diagonal (d2) is 20 cm.
Final Answer
The correct answer is option 'A' (20 cm).
This shows that both diagonals of the rhombus can be equal, which is a special case where the rhombus is also a square.
A pond 100 m in diameter is surrounded by a circular grass walk-way 2 m wide. How many square metres of grass is the on the walk-way?- a)98 π
- b)100 π
- c)204 π
- d)202 π
Correct answer is option 'C'. Can you explain this answer?
A pond 100 m in diameter is surrounded by a circular grass walk-way 2 m wide. How many square metres of grass is the on the walk-way?
a)
98 π
b)
100 π
c)
204 π
d)
202 π
|
|
Aarav Sharma answered |
The radius of the pond is 100/2 = <100 =50="">>50 m.
The radius of the grass walkway is 50+2 = <50+2=52>>52 m.
The area of the grass walkway is pi*(52^2 - 50^2) = pi*(2704 - 2500) = pi*204
≈ 204*pi
≈ 204*3.14
≈ <204*3.14=640.56>>640.56 m^2.
So, the answer is a) 640.
The radius of the grass walkway is 50+2 = <50+2=52>>52 m.
The area of the grass walkway is pi*(52^2 - 50^2) = pi*(2704 - 2500) = pi*204
≈ 204*pi
≈ 204*3.14
≈ <204*3.14=640.56>>640.56 m^2.
So, the answer is a) 640.
What is the area of the triangle in which two of its medians 9 cm and 12 cm long intersect at the right angles?- a)72
- b)60
- c)56
- d)48
Correct answer is option 'A'. Can you explain this answer?
What is the area of the triangle in which two of its medians 9 cm and 12 cm long intersect at the right angles?
a)
72
b)
60
c)
56
d)
48
|
|
Aarav Sharma answered |
Given: Two medians of a triangle are 9 cm and 12 cm long and they intersect at right angles.
To find: The area of the triangle.
Solution:
Let ABC be the given triangle and D and E be the midpoints of AB and AC respectively. Let F be the intersection point of medians AD and CE.
Let AF = 9 and CF = 12. Then, BD = AD = 9 and CE = AE = 12.
We know that the medians of a triangle divide it into six equal parts. Therefore, the area of triangle ABC is four times the area of triangle AFE.
Area of triangle AFE = (1/2) * AF * CE = (1/2) * 9 * 12 = 54 sq. cm.
Therefore, the area of triangle ABC = 4 * 54 = 216 sq. cm.
Hence, the correct option is (a) 72.
To find: The area of the triangle.
Solution:
Let ABC be the given triangle and D and E be the midpoints of AB and AC respectively. Let F be the intersection point of medians AD and CE.
Let AF = 9 and CF = 12. Then, BD = AD = 9 and CE = AE = 12.
We know that the medians of a triangle divide it into six equal parts. Therefore, the area of triangle ABC is four times the area of triangle AFE.
Area of triangle AFE = (1/2) * AF * CE = (1/2) * 9 * 12 = 54 sq. cm.
Therefore, the area of triangle ABC = 4 * 54 = 216 sq. cm.
Hence, the correct option is (a) 72.
ABCD is a square drawn inside a square PTRS of sides 4 cm by joining midpoints of the sides PR, PT, TS, SR. Another square is drawn inside ABCD similarly. This process is repeated infinite number of times. Find the sum of all the squares.
- a)16 cm2
- b)28 cm2
- c)32 cm2
- d)Infinite
Correct answer is option 'C'. Can you explain this answer?
ABCD is a square drawn inside a square PTRS of sides 4 cm by joining midpoints of the sides PR, PT, TS, SR. Another square is drawn inside ABCD similarly. This process is repeated infinite number of times. Find the sum of all the squares.
a)
16 cm2
b)
28 cm2
c)
32 cm2
d)
Infinite
|
Rajdeep Verma answered |
If we write the infinite series of area of squares:
= 42 + (2√2)2 + 22 + ……. infinite
Since it is a decreasing series sum of infinite terms can be approximated.
= 16 + 8 + 4 +………infinite
Two circles of an equal radii are drawn, without any overlap, in a semicircle of radius 2 cm. If these are the largest possible circles that the semicircle can accommodate, what is the radius (in cm) of each of the circles?a)0.414b)0.828c)0.172d)0.586Correct answer is option 'B'. Can you explain this answer?
|
|
Naveen Unni answered |
PQRS is trapezium, in which PQ is parallel to RS, and PQ = 3 (RS). The diagonal of the trapezium intersect each other at X, then the ratio of Δ PXQ and ARXS is- a)6:1
- b)3:1
- c)9:1
- d)7:1
Correct answer is option 'C'. Can you explain this answer?
PQRS is trapezium, in which PQ is parallel to RS, and PQ = 3 (RS). The diagonal of the trapezium intersect each other at X, then the ratio of Δ PXQ and ARXS is
a)
6:1
b)
3:1
c)
9:1
d)
7:1
|
|
Aarav Sharma answered |
Given: PQRS is a trapezium with PQ || RS and PQ = 3RS. The diagonals of the trapezium intersect at X.
To find: The ratio of the areas of triangles PXQ and RXS.
Solution:
Step 1: Draw a rough figure of the trapezium PQRS and mark the given information.
Step 2: Draw the diagonals PR and QS which intersect at X.
Step 3: Divide the trapezium into two triangles PXQ and RXS by drawing a line parallel to PQ through point S.
Step 4: Now we need to find the ratio of the areas of triangles PXQ and RXS.
Step 5: Let the height of the trapezium be h.
Step 6: We know that PQ = 3RS. Let RS = x. Then PQ = 3x.
Step 7: The area of trapezium PQRS = (1/2)h(PQ + RS) = (1/2)h(3x + x) = 2hx.
Step 8: Using the area of a triangle formula, the area of triangle RXS = (1/2)xh and the area of triangle PXQ = (1/2)(3x)h = (3/2)xh.
Step 9: Therefore, the ratio of the areas of triangles PXQ and RXS = (3/2)xh / (1/2)xh = 3:1.
Step 10: Hence, the correct option is (c) 9:1.
Final Answer: The ratio of the areas of triangles PXQ and RXS is 9:1.
To find: The ratio of the areas of triangles PXQ and RXS.
Solution:
Step 1: Draw a rough figure of the trapezium PQRS and mark the given information.
Step 2: Draw the diagonals PR and QS which intersect at X.
Step 3: Divide the trapezium into two triangles PXQ and RXS by drawing a line parallel to PQ through point S.
Step 4: Now we need to find the ratio of the areas of triangles PXQ and RXS.
Step 5: Let the height of the trapezium be h.
Step 6: We know that PQ = 3RS. Let RS = x. Then PQ = 3x.
Step 7: The area of trapezium PQRS = (1/2)h(PQ + RS) = (1/2)h(3x + x) = 2hx.
Step 8: Using the area of a triangle formula, the area of triangle RXS = (1/2)xh and the area of triangle PXQ = (1/2)(3x)h = (3/2)xh.
Step 9: Therefore, the ratio of the areas of triangles PXQ and RXS = (3/2)xh / (1/2)xh = 3:1.
Step 10: Hence, the correct option is (c) 9:1.
Final Answer: The ratio of the areas of triangles PXQ and RXS is 9:1.
The volume of two spheres are in the ratio 27 : 125. The ratio of their surface area is?- a)25 : 9
- b)27 : 11
- c)11 : 27
- d)9 : 25
Correct answer is option 'D'. Can you explain this answer?
The volume of two spheres are in the ratio 27 : 125. The ratio of their surface area is?
a)
25 : 9
b)
27 : 11
c)
11 : 27
d)
9 : 25
|
Sagar Sharma answered |
Understanding the Volume and Surface Area of Spheres
The problem states that the volumes of two spheres are in the ratio 27:125. To find the ratio of their surface areas, we need to understand the relationships between the two.
Volume of a Sphere
- The formula for the volume (V) of a sphere is given by V = (4/3)πr^3, where r is the radius.
- If the volumes of two spheres are in the ratio 27:125, we can express this as:
- V1/V2 = 27/125
Finding the Ratio of Radii
- Since volumes are proportional to the cube of the radii, we have:
- (r1^3)/(r2^3) = 27/125
- Taking the cube root on both sides gives us:
- r1/r2 = (27^(1/3))/(125^(1/3)) = 3/5
Surface Area of a Sphere
- The formula for the surface area (A) of a sphere is A = 4πr^2.
- Now, to find the ratio of the surface areas of the two spheres, we have:
- A1/A2 = (4πr1^2)/(4πr2^2) = (r1^2)/(r2^2)
Calculating the Surface Area Ratio
- Substituting the ratio of the radii:
- r1/r2 = 3/5
- Therefore, (r1^2)/(r2^2) = (3^2)/(5^2) = 9/25
Final Answer
- The ratio of the surface areas of the two spheres is 9:25, which corresponds to option 'D'.
The problem states that the volumes of two spheres are in the ratio 27:125. To find the ratio of their surface areas, we need to understand the relationships between the two.
Volume of a Sphere
- The formula for the volume (V) of a sphere is given by V = (4/3)πr^3, where r is the radius.
- If the volumes of two spheres are in the ratio 27:125, we can express this as:
- V1/V2 = 27/125
Finding the Ratio of Radii
- Since volumes are proportional to the cube of the radii, we have:
- (r1^3)/(r2^3) = 27/125
- Taking the cube root on both sides gives us:
- r1/r2 = (27^(1/3))/(125^(1/3)) = 3/5
Surface Area of a Sphere
- The formula for the surface area (A) of a sphere is A = 4πr^2.
- Now, to find the ratio of the surface areas of the two spheres, we have:
- A1/A2 = (4πr1^2)/(4πr2^2) = (r1^2)/(r2^2)
Calculating the Surface Area Ratio
- Substituting the ratio of the radii:
- r1/r2 = 3/5
- Therefore, (r1^2)/(r2^2) = (3^2)/(5^2) = 9/25
Final Answer
- The ratio of the surface areas of the two spheres is 9:25, which corresponds to option 'D'.
In the given figure, straight line AB and CD intersect at O. IF ∠δ =3∠v, then ∠v = ?
- a)40⁰
- b)50⁰
- c)55⁰
- d)45⁰
Correct answer is option 'D'. Can you explain this answer?
In the given figure, straight line AB and CD intersect at O. IF ∠δ =3∠v, then ∠v = ?
a)
40⁰
b)
50⁰
c)
55⁰
d)
45⁰
|
Meera Kapoor answered |
COD is a straight line
∴ ∠δ + ∠v =180⁰ ⇒ 3v +v =180 ⇒ 4v = 180 ⇒ v =45⁰.
∴ ∠δ + ∠v =180⁰ ⇒ 3v +v =180 ⇒ 4v = 180 ⇒ v =45⁰.
Two sides of a triangle are 4 and 5. Then, for the area of the triangle, which one of the following bounds is the sharpest?- a)<10.
- b)≤ 10.
- c)< 8.
- d)> 5.
Correct answer is option 'B'. Can you explain this answer?
Two sides of a triangle are 4 and 5. Then, for the area of the triangle, which one of the following bounds is the sharpest?
a)
<10.
b)
≤ 10.
c)
< 8.
d)
> 5.
|
Prerna Choudhary answered |
Let AB = 4 and BC = 5 and AB is perpendicular to BC
then Area = 1/2 AB . AC = 1/2 . 4.5 = 10
The area of the circle is 2464 cm2 and the ratio of the breadth of the rectangle to radius of the circle is 6:7. If the circumference of the circle is equal to the perimeter of the rectangle, then what is the area of the rectangle.- a)1456 cm2
- b)1536 cm2
- c)1254 cm2
- d)5678 cm2
Correct answer is option 'B'. Can you explain this answer?
The area of the circle is 2464 cm2 and the ratio of the breadth of the rectangle to radius of the circle is 6:7. If the circumference of the circle is equal to the perimeter of the rectangle, then what is the area of the rectangle.
a)
1456 cm2
b)
1536 cm2
c)
1254 cm2
d)
5678 cm2
|
Luminary Institute answered |
Area of the circle=πr2
2464 = 22/7 * r2
Radius of the circle=28 cm
Circumference of the circle=2 * π* r =2 * 22/7 * 28
= 176 cm
Breadth of the rectangle=6/7 * 28=24 cm
Perimeter of the rectangle=2 * (l + b)
176 = 2 * (l + 24)
Length of the rectangle = 64 cm
Area of the rectangle = l * b = 24 * 64 = 1536 cm2
If the parallel sides of a parallelogram are 2 cm apart and their sum is 10 cm then its area is:- a)20 cm2
- b)5 cm2
- c)10 cm2
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
If the parallel sides of a parallelogram are 2 cm apart and their sum is 10 cm then its area is:
a)
20 cm2
b)
5 cm2
c)
10 cm2
d)
none of these
|
|
Srestha Menon answered |
Understanding the Parallelogram Area
To find the area of a parallelogram, we can use the formula:
Area = base × height
Where:
- The base is the length of one of the parallel sides.
- The height is the perpendicular distance between the parallel sides.
Given Information
- The distance between the parallel sides (height) = 2 cm
- The sum of the lengths of the parallel sides = 10 cm
Finding the Base
Since we have the sum of the two parallel sides, we can define their lengths as follows:
Let one side be "a" and the other side be "b". According to the problem:
a + b = 10 cm
To find the area, we need the length of one of the sides. For simplicity, let's assume both sides are equal. Thus:
a = b = 10 cm / 2 = 5 cm
Calculating the Area
Now, substituting the values into the area formula:
Area = base × height
Area = 5 cm × 2 cm
Area = 10 cm²
Conclusion
The area of the parallelogram is 10 cm². Hence, the correct answer is option 'C'. This demonstrates how understanding the properties of parallelograms can help solve geometry problems effectively.
To find the area of a parallelogram, we can use the formula:
Area = base × height
Where:
- The base is the length of one of the parallel sides.
- The height is the perpendicular distance between the parallel sides.
Given Information
- The distance between the parallel sides (height) = 2 cm
- The sum of the lengths of the parallel sides = 10 cm
Finding the Base
Since we have the sum of the two parallel sides, we can define their lengths as follows:
Let one side be "a" and the other side be "b". According to the problem:
a + b = 10 cm
To find the area, we need the length of one of the sides. For simplicity, let's assume both sides are equal. Thus:
a = b = 10 cm / 2 = 5 cm
Calculating the Area
Now, substituting the values into the area formula:
Area = base × height
Area = 5 cm × 2 cm
Area = 10 cm²
Conclusion
The area of the parallelogram is 10 cm². Hence, the correct answer is option 'C'. This demonstrates how understanding the properties of parallelograms can help solve geometry problems effectively.
Anil grows tomatoes in his backyard which is in the shape of a square. Each tomato takes 1 cm2 in his backyard. This year, he has been able to grow 131 more tomatoes than last year. The shape of the backyard remained a square. How many tomatoes did Anil produce this year?- a)4225
- b)4096
- c)4356
- d)Insufficient Data
Correct answer is option 'C'. Can you explain this answer?
Anil grows tomatoes in his backyard which is in the shape of a square. Each tomato takes 1 cm2 in his backyard. This year, he has been able to grow 131 more tomatoes than last year. The shape of the backyard remained a square. How many tomatoes did Anil produce this year?
a)
4225
b)
4096
c)
4356
d)
Insufficient Data
|
|
Naveen Jain answered |
Let the area of backyard be x2 this year and y2 last year
∴ X2- Y2 = 131
=) (X+Y) (X-Y) = 131
Now, 131 is a prime number (a unique one too. Check out its properties on Google). Also, always identify the prime number given in a question. Might be helpful in cracking the solution.
=) (X+Y) (X-Y) = 131 x 1
=) X+Y = 131
X-Y = 1
=) 2X = 132 =) X = 66
and Y = 65
∴ Number of tomatoes produced this year = 662 = 4356
Choice (C) is therefore, the correct answer.
Correct Answer: 4356
∴ X2- Y2 = 131
=) (X+Y) (X-Y) = 131
Now, 131 is a prime number (a unique one too. Check out its properties on Google). Also, always identify the prime number given in a question. Might be helpful in cracking the solution.
=) (X+Y) (X-Y) = 131 x 1
=) X+Y = 131
X-Y = 1
=) 2X = 132 =) X = 66
and Y = 65
∴ Number of tomatoes produced this year = 662 = 4356
Choice (C) is therefore, the correct answer.
Correct Answer: 4356
Top surface of a raised platform is in the shape of regular octagon as shown in the figure. Find the area of the octagonal surface.
- a)11.9 cm3
- b)119 cm
- c)119 m2
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Top surface of a raised platform is in the shape of regular octagon as shown in the figure. Find the area of the octagonal surface.
a)
11.9 cm3
b)
119 cm
c)
119 m2
d)
None of these
|
Trisha Vashisht answered |
We will divide the octagon in 3 parts which are two trpeziums ABCH and DEFG and one rectangle CDGH and find the area for all of them separately and add them to get the area of octagon
= 100°, find 
.
What is the area of the triangle below?
- a)22 cm2
- b)33 cm2
- c)44 cm2
- d)50 cm2
Correct answer is option 'B'. Can you explain this answer?
What is the area of the triangle below?
a)
22 cm2
b)
33 cm2
c)
44 cm2
d)
50 cm2
|
|
Pritam Saha answered |
The area of a triangle may be found by using the formula, A=1/2bh, where brepresents the base and h represents the height. Thus, the area may be written as A=1/2(11)(6), or A = 33. The area of the triangle is 33 cm'.
In the given figure, AD is the bisector of ∠BAC, AB = 6 cm, AC = 5 cm and BD = 3 cm. Find DC. It is given that ∠ABD = ∠ACD.
- a)11.3 cm
- b)4 cm
- c)3.5 cm
- d)2.5 cm
Correct answer is option 'D'. Can you explain this answer?
In the given figure, AD is the bisector of ∠BAC, AB = 6 cm, AC = 5 cm and BD = 3 cm. Find DC. It is given that ∠ABD = ∠ACD.
a)
11.3 cm
b)
4 cm
c)
3.5 cm
d)
2.5 cm
|
|
Pooja Shah answered |
We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Hence:
Hence:
In triangle ABD and ACD
Angle BAD = CAD (Given AD is the bisector)
Angle ABD = ACD (GIven)
there fore they are similar (AAA Property)
AB/BD = AC/CD
6/3 = 5/CD
CD = 2.5 cm
6/3 = 5/CD
CD = 2.5 cm
AB is the diameter of the circle and ∠PAB=40∘
what is the value of ∠PCA?
- a)50∘
- b)55°
- c)70°
- d)45°
Correct answer is option 'A'. Can you explain this answer?
AB is the diameter of the circle and ∠PAB=40∘
what is the value of ∠PCA?
what is the value of ∠PCA?
a)
50∘
b)
55°
c)
70°
d)
45°
|
KS Coaching Center answered |
- In △PAB⇒ ∠PAB=40o [ Given ]⇒ ∠BPA=90o [ angle inscribed in a semi-circle ]⇒ ∠PAB+∠PBA+∠BPA=180o∴ 40o+∠PBA+90o=180o∴ ∠PBA=180o−130o∴ ∠PBA=50o⇒ ∠PBA=∠PCA=50o [ angles inscribed in a same arc PA ]∴ ∠PCA=50o
Three circles with radius 2 cm touch each other as shown :- 
- a)3π(4+√3)2
- b)π/2(4+2√3)2
- c)π/4(4+2√3)2
- d)12π−π/2(4+2√3)2
Correct answer is option 'B'. Can you explain this answer?
Three circles with radius 2 cm touch each other as shown :-
a)
3π(4+√3)2
b)
π/2(4+2√3)2
c)
π/4(4+2√3)2
d)
12π−π/2(4+2√3)2
|
Mainak Majumdar answered |
(This can easily be derived using trigonometry. However, please remember this formula. It is useful at places)
Chapter doubts & questions for Mensuration and geometry - General Aptitude for GATE 2025 is part of Mechanical Engineering exam preparation. The chapters have been prepared according to the Mechanical Engineering exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mechanical Engineering 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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