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PrepBank · বিষয়ভিত্তিক প্রশ্ন

Geometry: Mensuration, Trigonometry

মোট প্রশ্ন২,০৮৫এই পাতা১০০প্রতি পাতা১০০
প্রশ্ন দেখুনঅনুশীলন
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

Geometry: Mensuration, Trigonometry

PrepBank · পাতা ১৫ / ২১ · ১,৪০১–১,৫০০ / ২,০৮৫

১,৪০১.
tan360° - 2sin60° = ?
  1. ক
    √3
  2. খ
    1/√3
  3. গ
    2
  4. ঘ
    2√3
ব্যাখ্যা

Question: tan360° - 2sin60° = ?

Solution:
Given that,
tan360° - 2sin60°
= (√3)3 - 2(√3/2)
= 3√3 - √3
= 2√3

১,৪০২.
If θ be a positive acute angle satisfying cos2θ + cos4θ = 1, then the value of tan2θ + tan4θ is?
  1. ক
    3/2
  2. খ
    0
  3. গ
    1
  4. ঘ
    1/2
  5. ঙ
    None of these
ব্যাখ্যা

Question: If θ be a positive acute angle satisfying cos2θ + cos4θ = 1, then the value of tan2θ + tan4θ is?

Solution: 
Given that, 
cos2θ + cos4θ = 1
⇒ cos4θ = 1 - cos2θ
⇒ cos4θ = sin2θ ; [sin2θ = 1 - cos2θ] 
⇒ cos2θ.cos2θ = sin2θ
⇒ cos2θ = sin2θ/cos2θ
⇒ cos2θ = tan2θ

Now, 
cos2θ + cos4θ = 1
⇒ cos2θ + (cos2θ)2 = 1
⇒ tan2θ + (tan2θ)2 = 1
∴ tan2θ + tan4θ = 1

১,৪০৩.
If a pole 15m high casts a shadow 5√3m long on the ground, then the elevation of the sun is-
  1. ক
    ক) 15°
  2. খ
    খ) 30°
  3. গ
    গ) 45°
  4. ঘ
    ঘ) 60°
ব্যাখ্যা
Question: If a pole 15m high casts a shadow 5√3m long on the ground, then the elevation of the sun is-

Solution:

খুঁটির উচ্চতা AB = 15m
খুঁটির ছায়ার দৈর্ঘ্য BC =5√3m

ΔABC 
tanθ = লম্ব/ভূমি 
tanθ = AB/BC
tanθ =15/5√3
tanθ =√3
tanθ = tan60°
θ = 60°
১,৪০৪.
To represent a family budget on a circle graph, how many degrees of the circle should be used to represent an item that is 20% of the total budget?
  1. ক
    ক) 76°
  2. খ
    খ) 72°
  3. গ
    গ) 60°
  4. ঘ
    ঘ) 20°
ব্যাখ্যা
Question: To represent a family budget on a circle graph, how many degrees of the circle should be used to represent an item that is 20% of the total budget?

Solution: 
সম্পূর্ণ বৃত্ত = ৩৬০°

তাহলে যে উপাদান ২০% জায়গা দখল করবে তার উৎপন্ন কোণ = ৩৬০° এর ২০%
= ৩৬০° এর ১/৫
= ৭২°
১,৪০৫.
If the area of a small pizza is 49π in2, what size pizza box would best fit the small pizza?
  1. ক
    ক) 10 in
  2. খ
    খ) 12 in
  3. গ
    গ) 14 in
  4. ঘ
    ঘ) 9 in
ব্যাখ্যা
area of a pizza (circle) = πr2 = 49π
Or, r2 = 49π/π = 49
Or, r = 7
So, the size of the pizza box would be =  2r = 2 × 7 = 14 in
১,৪০৬.
Each side of a rhombus is 17 cm, and one diagonal measures 30 cm. What is the length of the other diagonal?
  1. ক
    12 cm
  2. খ
    18 cm
  3. গ
    22 cm
  4. ঘ
    16 cm
ব্যাখ্যা
প্রশ্ন: Each side of a rhombus is 17 cm, and one diagonal measures 30 cm. What is the length of the other diagonal?
(একটি রম্বসের প্রতিটি বাহুর দৈর্ঘ্য ১৭ সে. মি. এবং একটি কর্ণের দৈর্ঘ্য ৩০ সে. মি. হলে রম্বসটির অপর কর্ণের দৈর্ঘ্য কত?)

সমাধান:
রম্বসের প্রতিটি বাহুর দৈর্ঘ্য সমান এবং কর্ণদ্বয় পরস্পরকে সমকোণে সমদ্বখণ্ডিত করে ।

তাহলে, AB = AD = BC = CD = ১৭ সে. মি. এবং কর্ণ AC = ৩০ সে. মি. হলে।

OA = ৩০/২ = ১৫ সে. মি.
AOB সমকোণী ত্রিভুজ হতে -
⇒ AB2 = OA2 + OB2
⇒ ১৭২ = ১৫২ + OB2
⇒ OB2 = ১৭২ - ১৫২
⇒ OB2 = ২৮৯ - ২২৫
⇒ OB2 = ৬৪
⇒ OB = √৬৪
∴ OB = ৮  

অপর কর্ণ, BD = OD + OB = OB + OB = (৮ + ৮) = ১৬ সে. মি.
১,৪০৭.
How many poles can be erected along fence of 200 feet at equal distance of 20 feet?
  1. ক
    ক) 10
  2. খ
    খ) 11
  3. গ
    গ) 12
  4. ঘ
    ঘ) 14
ব্যাখ্যা
Question: How many poles can be erected along fence of 200 feet at equal distance of 20 feet?

Solution: 
প্রথম ২০ ফুট এর জন্য খুঁটি লাগবে ২ টি 
বাকি ১৮০ ফুট এর জন্য খুঁটি লাগবে = ১৮০/২০ = ৯ টি

∴ মোট খুঁটি লাগবে = ৯ + ২ = ১১ টি
১,৪০৮.
A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm. Then what is the length of RS?
  1. ক
    3
  2. খ
    6
  3. গ
    9
  4. ঘ
    12
ব্যাখ্যা
Question: A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm. Then what is the length of RS?

Solution: 

If a circle touches all four sides of quadrilateral PQRS then,
PQ+ RS = SP+ RQ

So,
11 + RS = 8+ 12
⇒ RS = 20 - 11
⇒ RS = 9
১,৪০৯.
The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volume is-
  1. ক
    4 : 9
  2. খ
    9 : 4
  3. গ
    20 : 27
  4. ঘ
    20 : 25
ব্যাখ্যা
Question: The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. The ratio of their volume is-

Solution:
Let the radius of both cylinders be 2x and 3x.
Let the height of both cylinders be 5y and 3y.

Ratio of the volume of two cylinders = {π × (2x)2 × (5y)}/{π × (3x)2 × (3y)}
= (4x2 × 5)/(9x2 × 3)
= 20/27

∴ Ratio = 20 : 27
১,৪১০.
66 cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:
  1. ক
    64 m
  2. খ
    82 m
  3. গ
    74 m
  4. ঘ
    84 m
ব্যাখ্যা
Question: 66 cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in metres will be:

Solution:
Let, the length of the wire be h
Radius = 1/2 mm = 1/20 cm

ATQ,
πr2h = 66
⇒ (22/7) × (1/20)2 × h = 66
⇒ h = (66 × 7 × 20 × 20)/22
⇒ h = 8400 cm
⇒ h = 8400/100 m
∴ h = 84 m
১,৪১১.
The length of a rectangular plot is 10 meters less than three times its breadth. If the cost of fencing the plot at Tk 50 per meter is Tk 15000, what is the length of the plot in meters?
  1. ক
    100 meters
  2. খ
    110 meters
  3. গ
    90 meters
  4. ঘ
    120 meters
ব্যাখ্যা

Question: The length of a rectangular plot is 10 meters less than three times its breadth. If the cost of fencing the plot at Tk 50 per meter is Tk 15000, what is the length of the plot in meters?

Solution:
Let the breadth of the plot be x meters.
Then, the length of the plot is 3x - 10 meters.

Perimeter of the rectangle = 2 × (Length + breadth)
= 2 × (x + 3x - 10 )
= 2 × (4x - 10 )
= 8x - 20

Given,
Cost of fencing per meter = Tk 50
Total cost = Tk 15000

So, Perimeter × 50 = 15000
⇒ (8x - 20) × 50 = 15000
⇒ 8x - 20 = 15000/50
⇒ 8x = 300 + 20
⇒ 8x = 320 
∴ x = 40

∴ Breadth = 40 meters.
and length = 3x - 10 = 3 × 40 - 10 = 110 meters.

১,৪১২.
A triangle has a perimeter 13. The two shorter side, have integer lengths equal to x and x + 1. Which of the following could be the length of the other side.
  1. ক
    2
  2. খ
    6
  3. গ
    8
  4. ঘ
    None
ব্যাখ্যা
Question: A triangle has a perimeter 13. The two shorter side, have integer lengths equal to x and x + 1. Which of the following could be the length of the other side.

Solution:
The shorter sides have integer lengths equal to x and x + 1
Let the longest side be 'a'
∴ a + x + (x + 1) = 13
⇒ a + 2x = 12 .......(1)

We know that the sum of the lengths of the shorter sides has to be more than the length of the longer one
Looking at the options, we can't have 8 or 10 as values for 'a'

Similarly, we can't have 2 as values for 'a' as it wouldn't be the longest side then.

So, the correct length of other side is 6
১,৪১৩.
The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electric pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
  1. ক
    5 metre
  2. খ
    8 metre
  3. গ
    10 metre
  4. ঘ
    12 metre
ব্যাখ্যা
Question: The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electric pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?

Solution:

Let,
AB be the tower and CD be the electric pole
Then ∠ACB = 60°, ∠EDB = 30° and AB = 15 m
Let,
CD = h.
Then BE = (AB – AE)
= (AB - CD) =
(15 - h)

We have
AB/AC = tan 60° = √3
⇒ AC = AB/√3 = 15/√3

And,
BE/DE = tan30° = 1/√3
⇒ DE = (BE × √3) = √3 (15 - h)  [DE = AC = 15/√3]
⇒ 15/√3 = √3 (15 - h)
⇒ 3(15 - h) = 15
⇒ 3h = 45 - 15
⇒ 3h = 30
∴ h = 10 m
১,৪১৪.
In a right angled triangle, two sides are of the same length. Which of the options is one of the angles of that triangle?
  1. ক
    ক) 30°
  2. খ
    খ) 40°
  3. গ
    গ) 45°
  4. ঘ
    ঘ) 50°
ব্যাখ্যা
প্রশ্ন: In a right angled triangle, two sides are of the same length. Which of the options is one of the angles of that triangle?

সমাধান: 
If two sides of the right angled triangle are equal, it is a right angled isosceles triangle. So, the angles of the triangle are 45°, 45° and 90°
১,৪১৫.
tanA√(1 - sin2A) = ?
  1. ক
    1
  2. খ
    sinA
  3. গ
    cosA
  4. ঘ
    0
ব্যাখ্যা
Question: tanA√(1 - sin2A) = ? 

Solution: 
tanA√(1 - sin2A)
= tanA√(cos2A)
= (sinA/cosA) × cosA
=  sinA
১,৪১৬.
If 7sin2θ + 3cos2θ = 4, then tan30° = ?
  1. ক
    1/√3
  2. খ
    1/3
  3. গ
    1/√2
  4. ঘ
    2/3
ব্যাখ্যা
Question: If 7sin2θ + 3cos2θ = 4, then tan30° = ?

Solution:
7sin2θ + 3cos2θ = 4
⇒ 7sin2θ + 3(1 - sin2θ) = 4
⇒ 7sin2θ + 3 - 3sin2θ = 4
⇒ 4sin2θ = 1
⇒ sin2θ = 1/4
⇒ sinθ = 1/2
⇒ sinθ = sin30°
⇒ θ = 30°
⇒ tanθ = tan30°
∴ tanθ = 1/√3
১,৪১৭.
The lengths of the two sides of a right-angled triangle, adjacent to right angle are 8 cm and 15 cm respectively. Find the area of the triangle.
  1. ক
    60 cm2
  2. খ
    48 cm2
  3. গ
    36 cm2
  4. ঘ
    30 cm2
ব্যাখ্যা
Question: The lengths of the two sides of a right-angled triangle, adjacent to right angle are 8 cm and 15 cm respectively. Find the area of the triangle.

Solution:

Let,
the sides adjacent to right angle are a = 15 cm and b = 8 cm

We know,
The area = (1/2) × ab
= (1/2) × 15 × 8
= 60 cm2
১,৪১৮.
A water tank is 30 m long, 20 m wide and 12 m deep. It is made of iron sheet which is 3 m wide. The tank is open at the top. If the cost of iron sheet is TK. 10 per meter, what is the total cost of iron sheet required to build the tank?
  1. ক
    Tk. 6000
  2. খ
    Tk. 8000
  3. গ
    Tk. 9000
  4. ঘ
    Tk. 10000
  5. ঙ
    None of these
ব্যাখ্যা
Question: A water tank is 30 m long, 20 m wide and 12 m deep. It is made of iron sheet which is 3 m wide. The tank is open at the top. If the cost of iron sheet is TK. 10 per meter, what is the total cost of iron sheet required to build the tank?

Solution:
Length of water tank = 30 m
Width of water tank = 20 m
Depth of water tank = 12 m

Area of water tank = 2(lb + bh +hl) - lb
= 2 (30 × 20 + 20 × 12 + 12 × 30) - 30 × 20
= 2 (600 + 240 + 360) - 600
= 2 (1200) - 600
= 2400 - 600
= 1800 m2

Area of iron sheet (L × B) = area of tank = 1800 m2
⇒ L × B = 1800
⇒ L × 3 = 1800
∴ L = 1800/3 = 600 m

Given that, the cost of iron sheet is Tk. 10 per meter.
So, the total cost of iron sheet to build the tank = 10 × 600 = Tk. 6000
১,৪১৯.
A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m cube) is:
  1. ক
    ক) 4120 m cube
  2. খ
    খ) 4140 m cube
  3. গ
    গ) 5140 m cube
  4. ঘ
    ঘ) 5120 m cube
ব্যাখ্যা

l = (48 - 16)m = 32 m, [because 8+8 = 16]
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m cube
= 5120 m cube.

১,৪২০.
If θ be an acute angle and 7sin2θ + 3cos2θ = 4, then the value of tanθ is? 
  1. ক
    0
  2. খ
    1/√3
  3. গ
    1/√2
  4. ঘ
    1
ব্যাখ্যা

Question: If θ be an acute angle and 7sin2θ + 3cos2θ = 4, then the value of tanθ is?

Solution:
7sin2θ + 3cos2θ = 4
⇒ 7sin2θ + 3(1 - sin2θ) = 4
⇒ 7sin2θ + 3 - 3sin2θ = 4
⇒ 4sin2θ = 1
⇒ sin2θ = 1/4
⇒ sinθ = 1/2
⇒ sinθ = sin30°
∴ θ = 30°

∴ tanθ = tan30° = 1/√3

১,৪২১.
The sum of the circumference of a circle and the perimeter of a rectangle is 130 cm. The area of the rectangle is 104 cm2 and the length of the rectangle is 13 cm. What is the area of the circle?
  1. ক
    ক) 516 cm2
  2. খ
    খ) 616 cm2
  3. গ
    গ) 816 cm2
  4. ঘ
    ঘ) 216 cm2
ব্যাখ্যা

Given, Length of rectangle, L = 13 cm
breadth, B = 104/13 = 8 cm
So, its perimeter = 2(13+8) = 42
ATQ, 2πr = 130 - 42
Or, 2πr = 88
Or, r = 14
∴ Area of the circle = πr2 = 22/7 × 142 = 616 cm2

১,৪২২.
A square and a circle have the same perimeter. The side of the length of square is 44 cm, what is the area of the circle?
  1. ক
    1456 sq. cm.
  2. খ
    375 sq. cm.
  3. গ
    2464 sq. cm.
  4. ঘ
    1864 sq. cm.
ব্যাখ্যা

Question: A square and a circle have the same perimeter. The side of the length of square is 44 cm, what is the area of the circle?

Solution:
Perimeter of the square = 4 × side length
= 4 × 44 cm
= 176 cm

As per the question, the square and circle have the same perimeter.
∴ Circumference of the circle = 176 cm
We know that, Circumference of the circle = 2πr
∴ 2πr = 176
⇒ r = 176 / (2π)
⇒ r = 88 / π
⇒ r = 88 / (22/7)
⇒ r = 88 × 7 / 22
⇒ r = 4 × 7
⇒ r = 28 cm

Area of the circle = πr2
= (22/7) × 282
= (22/7) × (28 × 28)
= 22 × 4 × 28
= 2464 sq. cm

∴ The area of the circle is 2464 sq. cm.

১,৪২৩.
A right circular cylinder just encloses a sphere. If p is the surface area of the sphere and q is the curved surface area of the cylinder, then which one of the following is correct?
  1. ক
    P = q
  2. খ
    2p = 3
  3. গ
    2p = q
  4. ঘ
    p = 2q
ব্যাখ্যা
Question: A right circular cylinder just encloses a sphere. If p is the surface area of the sphere and q is the curved surface area of the cylinder, then which one of the following is correct?

Solution:
When a right circular cylinder just encloses a sphere as shown below, the radius of the sphere and cylinder are equal and the height of cylinder is equal to the diameter of the sphereLet, radius of sphere = radius of cylinder = r
Here, height of cylinder = 2r
Now, Curved surface area of sphere = 4π × (radius)2
= p = 4πr2 Curved surface area of cylinder = 2 × radius × height
⇒  q = 2πr × 2r = 4πr2
∴ p = q
১,৪২৪.
When the diameter of a circle is tripled, the area of the circle will be increased by how many time?
  1. ক
    ক) 3
  2. খ
    খ) 6
  3. গ
    গ) 9
  4. ঘ
    ঘ) 12
ব্যাখ্যা
Question: When the diameter of a circle is tripled,  the area of the circle will be increased by how many time?

Solution: 
সমাধান : 
ধরি,
বৃত্তের ব্যাসার্ধ r 
বৃত্তের ব্যাস = 2r
∴বৃত্তের ক্ষেত্রফল = πr2

ব্যাস তিনগুণ বৃদ্ধি পেলে হবে 6r   
∴ব্যাসার্ধ =6r/2 = 3r   
∴ঐ বৃত্তের ক্ষেত্রফল হবে π(3r)2 = 9πr2  
 
বৃত্তের ক্ষেত্রফল ৯ গুণ  পাবে।
১,৪২৫.
The right circular cone of height 24 cm has a volume of 1232 cm3, then the area of its curved surface is?
  1. ক
    ক) 513 cm3 
  2. খ
    খ) 530 cm3 
  3. গ
    গ) 550 cm3 
  4. ঘ
    ঘ) 570 cm3 
ব্যাখ্যা
Question: The right circular cone of height 24 cm has a volume of 1232 cm3, then the area of its curved surface is? 

Solution: 
Volume of the cone = (1/3) × r2 × h = 1232
⇒ (1/3) × (22/7) × r2 × 24 = 1232 
⇒ r2 = (1232 × 7 × 3)/(22 × 24)
⇒ r2 = 49
∴ r = 7 

So, slant height l = √(7)2 + (24)2 = √625 = 25 
So, curved surface area = πrl = (22/7) × 7 × 25 = 550 cm3 
১,৪২৬.
The midpoint of the line joining (10, 2) and (4, 8) is -
  1. ক
    (8, 4)
  2. খ
    (6, 6)
  3. গ
    (7, 5)
  4. ঘ
    (5, 7) 
ব্যাখ্যা

Question: The midpoint of the line joining (10, 2) and (4, 8) is -

Solution:
The formula for the midpoint of (x1, y1) and (x2, y2) is
= ((x1 + x2)/2 , (y1 + y2)/2)

∴ The midpoint of the line joining (10, 2) and (4, 8) is
= (14/2 , 10/2)
= (7, 5)

১,৪২৭.
If sin A = 1/2 , then the value of cotA is -
  1. ক
    ক) 1/√3
  2. খ
    খ) √3
  3. গ
    গ) 1
  4. ঘ
    ঘ) √3/2
ব্যাখ্যা
Question: If sin A = 1/2 , then the value of cotA is -

Solution:
দেওয়া আছে,
sinA = 1/2

আমরা জানি,
cos2A = 1 - sin2A
= 1 - (1/2)2
= 1 - (1/4)
= (4 - 1)/4
= 3/4
∴ cosA = √(3/4)
= √3/2

এখন,
cotA = cosA/sinA
= (√3/2)/(1/2)
= √3
১,৪২৮.
A 9 cm × 8 cm × 15 cm block of ice is being melted to make identical ice cubes. If surface area any two sides of the cubes has to be 72cm2 how many ice cubes can be made?
  1. ক
    3
  2. খ
    5
  3. গ
    7
  4. ঘ
    9
  5. ঙ
    11
ব্যাখ্যা
Question: A 9 cm × 8 cm × 15 cm block of ice is being melted to make identical ice cubes. If surface area any two sides of the cubes has to be 72cm2 how many ice cubes can be made?

Solution:
বড় ব্লকের আয়তন = 9 × 8 × 15 = 1080 cm3

একটি ঘন আইস কিউবের সব পাশ সমান ⇒ প্রতিটি পাশে ক্ষেত্রফল হবে a2

তাহলে দুই পাশের মোট পৃষ্ঠের ক্ষেত্রফল,
2a2 = 72
⇒ a2 = 72/2 = 36
⇒ a2 = 62
∴ a = 6 cm

এবং প্রতিটি আইস কিউবের আয়তন = a3 = 63 = 216 cm3

∴ মোট আইস কিউবের সংখ্যা = 1080/216 = 5টি
১,৪২৯.
In an acute angled triangle ABC, if cos 2(A + B - C) = 0 and cot (B + C - A) = √3,then the value of angle ∠B is-
  1. ক
    ক) 105°/2
  2. খ
    খ) 75°/2
  3. গ
    গ) 45°/2
  4. ঘ
    ঘ) 55°/2
ব্যাখ্যা
দেয়া আছে,
cos 2(A + B - C) = 0
cos 2(A + B - C) = cos 90°
2(A + B - C) = 90°
A + B - C = 45°................. (1)

cot (B + C - A) = √3
B + C - A = cot 30°
B + C - A = 30°....................(2)

(1) + (2) ⇒
A + B - C + B + C - A = 45° + 30°
2B = 75°
B =  75°/2
১,৪৩০.
The ratio of length and breadth of a rectangular park is 4 : 2. If a cat running along the boundary of the park at the speed of 18 km/hr completes one round in 10 minutes, find the area of the park in square meters.
  1. ক
    50000 sq. m.
  2. খ
    45000 sq. m.
  3. গ
    68000 sq. m.
  4. ঘ
    55000 sq. m.
  5. ঙ
    None of these
ব্যাখ্যা
Question: The ratio of length and breadth of a rectangular park is 4 : 2. If a cat running along the boundary of the park at the speed of 18 km/hr completes one round in 10 minutes, find the area of the park in square meters.

Solution:
One round of the park is equal to the perimeter of the park.
So, by completing one round, the cat covers a distance equal to the perimeter of the park.
Now,
Distance or perimeter = speed × time
= 18 × (10/60)
= 3 km
= 3000 meters

Let Length = 4x and breadth = 2x
So, Perimeter:
2(4x + 2x) = 3000
⇒ 8x + 4x = 3000
⇒ 12x = 3000
∴ x = 3000/12 = 250 meters

So, Length = 4 × 250 = 1000 meters
And, Breadth = 2 × 250 = 500 meters

Area = Length × Breadth
= 1000 × 500
= 500000 sq. m.
১,৪৩১.
A rectangular water tank is 8 m high, 6 m long and 2.5m wide. How many liters of water can it hold?
  1. ক
    1,30,000 litre
  2. খ
    1,10,000 litre
  3. গ
    1,25,000 litre
  4. ঘ
    1,20,000 litre
ব্যাখ্যা
Question: A rectangular water tank is 8 m high, 6 m long and 2.5m wide. How many liters of water can it hold?

Solution:
Volume = length × width × height 
= (6 × 2.5 × 8) m3
= 120 m3 

1 m3 = 1000 litre
120 m3 = 120 × 1000 litre
= 1,20,000 litre
১,৪৩২.
A rectangular field will be fenced on three sides, leaving one side of 20 feet uncovered. If the area of the field is 600 square feet, how many feet of fencing is required?
  1. ক
    65 feet
  2. খ
    72 feet
  3. গ
    80 feet
  4. ঘ
    88 feet
  5. ঙ
    90 feet
ব্যাখ্যা

Question: A rectangular field will be fenced on three sides, leaving one side of 20 feet uncovered. If the area of the field is 600 square feet, how many feet of fencing is required?

Solution:
আয়তাকার মাঠের ক্ষেত্রফল = 600 বর্গ ফুট
যে পাশে বেড়া দেওয়া হবে না তার দৈর্ঘ্য = 20 ফুট
অতএব, আয়তাকার মাঠের অন্য পাশের দৈর্ঘ্য = ক্ষেত্রফল/একপাশের দৈর্ঘ্য
= 600 / 20 = 30 ফুট

চতুর্দিকে বেড়া দেওয়ার প্রয়োজন নেই, কারণ একপাশ উন্মুক্ত থাকবে।
যে তিনটি পাশে বেড়া দিতে হবে, তাদের দৈর্ঘ্য হবে (30 + 20 + 30) ফুট।
সুতরাং, প্রয়োজনীয় বেড়ার মোট দৈর্ঘ্য = 30 + 20 + 30 = 80 ফুট।

১,৪৩৩.
How many cubes of 3 cm edge can be cut out of a cube of 18 cm edge?
  1. ক
    ক) 125
  2. খ
    খ) 179
  3. গ
    গ) 216
  4. ঘ
    ঘ) 232
ব্যাখ্যা
Question: How many cubes of 3 cm edge can be cut out of a cube of 18 cm edge? 

Solution: 
Number of cubes 
= (18 × 18 × 18)/(3 × 3 × 3) = 216
১,৪৩৪.
What is the minimum value of 2sin2θ + 3cos2θ?
  1. ক
    0
  2. খ
    2
  3. গ
    1
  4. ঘ
    none of these
ব্যাখ্যা
Question: What is the minimum value of 2sin2θ + 3cos2θ?

Solution: 
Let,
x = 2sin2θ + 3cos2θ
or, x = 2sin2θ + 2cos2θ + cos2θ
or, x = 2(sin2θ + cos2θ) + cos2θ
or, x = 2 + cos2θ     [sin2θ + cos2θ = 1]

the minimum value of x depends on the minimum value of cos2θ.
Since the minimum value of cos2θ is 0, the minimum value of x is 2.
১,৪৩৫.
In a trapezoid, the lengths of the two parallel bases are 12 and 20 units, respectively. If the height of the trapezoid is 5, find the area of the trapezoid. 
  1. ক
    70 square units
  2. খ
    50 square units
  3. গ
    80 square units
  4. ঘ
    None
ব্যাখ্যা

Question: In a trapezoid, the lengths of the two parallel bases are 12 and 20 units, respectively. If the height of the trapezoid is 5, find the area of the trapezoid.

Solution:
Given that,
Trapezoid with bases a = 12 and b = 20
Height, h = 5

We know,
Area of trapezoid = (1/2) × (sum of bases) × height
= (1/2) × (a + b) × h
= (1/2) × (12 + 20) × 5
= (1/2) × 32 × 5
= 80

So, the area of the trapezoid is 80 square units.

১,৪৩৬.
The difference between the length and breadth of a rectangle is 30m . If its perimeter is 500m, then its area is:
  1. ক
    ক) 15400 m²
  2. খ
    খ) 14000 m²
  3. গ
    গ) 15000 m²
  4. ঘ
    ঘ) 16000 m²
ব্যাখ্যা
We have: l - b = 30 and
2(l + b) = 500
or l + b = 250

Solving the two equations,
we get:
l = 140 and b = 110
∴ Area
= (l x b)
= (140 x 110) m²
= 15400 m²
১,৪৩৭.
The ratio of the angles of a triangle is 2 : 3 : 4. What is the smallest angle in degrees?
  1. ক
    20
  2. খ
    40
  3. গ
    60
  4. ঘ
    50
ব্যাখ্যা
Question: The ratio of the angles of a triangle is 2 : 3 : 4. What is the smallest angle in degrees?
 
Solution: 
ত্রিভুজের কোণগুলোর অনুপাত =  2 : 3 : 4
ধরি 
কোণগুলো = 2x , 3x  4x
 
প্রশ্নমতে,
2x + 3x + 4x = 180°
বা, 9x  = 180°
বা, x = 180°/9
x = 20°
 
∴ ক্ষুদ্রতম কোণ = 2 × 20° = 40°
১,৪৩৮.
যদি sinx = 1 - cosy, x = 30° এবং y সুক্ষ্মকোণ হয়, তবে y এর মান কত?
  1. ক
    ক) 30°
  2. খ
    খ) 45°
  3. গ
    গ) 60°
  4. ঘ
    ঘ) 90°
ব্যাখ্যা
প্রশ্ন: যদি sinx = 1 - cosy, x = 30° এবং y সুক্ষ্মকোণ হয়, তবে y এর মান কত? 

সমাধান:  
sinx = 1 - cosy
⇒ cosy = 1 - sinx
⇒ cosy = 1 - sin30°
⇒ cosy = 1 - (1/2)
⇒ cosy = 1/2
⇒ cosy = cos60°
∴ y = 60°
১,৪৩৯.
If the volume of a sphere is 2304π cm3, what is the surface area of the sphere?
  1. ক
    742π cm2
  2. খ
    286π cm2
  3. গ
    576π cm2
  4. ঘ
    1052π cm2
ব্যাখ্যা

Question: If the volume of a sphere is 2304π cm3, what is the surface area of the sphere?

Solution:
দেওয়া আছে,
গোলকের আয়তন, V = 2304π cm3

আমরা জানি,
গোলকের আয়তন, V = (4/3)πr3
⇒ (4/3)πr3 = 2304π
⇒ r3 = 2304 × (3/4)
⇒ r3 = 576 × 3
⇒ r3 = 1728
⇒ r = 12 সেমি

এখন,
গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল, A = 4πr2
⇒ A = 4π(12)2
⇒ A = 4π × 144
⇒ A = 576π cm2

অতএব, গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল হলো 576π cm2।

১,৪৪০.
A boy of height 1.5 m is walking away from the base of a lamp post at a speed of 0.8 m/sec. Find the height of the lamp post from the ground, if the shadow of the boy is 2.0 m after walking for 4 sec.
  1. ক
    2.3 m
  2. খ
    2.7 m
  3. গ
    3.5 m
  4. ঘ
    3.9 m
ব্যাখ্যা

Question: A boy of height 1.5 m is walking away from the base of a lamp post at a speed of 0.8 m/sec. Find the height of the lamp post from the ground, if the shadow of the boy is 2.0 m after walking for 4 sec.

Solution:

Given that,
Height of the boy = 1.5 m
Speed of the boy = 0.8 m/s
Distance travelled by boy in 4 sec = 0.8 × 4 = 3.2 m
Total distance of shadow of boy and distance from base of lamp post = 2.0 + 3.2 = 5.2 m

Let the height of lamp post be 'h' m
According to question,
⇒ 1.5/2.0 = h/5.2
⇒ h = (5.2 × 1.5)/2.0
⇒ h = 3.9 m

So, The height of the lamp post is 3.9 meters.

১,৪৪১.
If the length of a rectangle is increased by 10% and its breadth is decreased by 10%, the change in its area will be-
  1. ক
    10% increase
  2. খ
    10% decrease
  3. গ
    1% increase
  4. ঘ
    1% decrease
  5. ঙ
    No change
ব্যাখ্যা
Question: If the length of a rectangle is increased by 10% and its breadth is decreased by 10%, the change in its area will be-

Solution:
Let
the length and breadth be 100 unit and 100 unit respectively
∴ The area before change = (100 × 100) = 10000 square unit

The length after change = 100 + 10% of 100 = 100 + 10 = 110 unit
The breadth after change = 100 - 10% of 100 = 100 - 10 = 90 unit

The area after change = 110 × 90 = 9900 square unit

∴ Percentage change = [(10000 - 9900)/10000] × 100%
= (1/100) × 100%
= 1% 

∴ The area of the new rectangle is decreased by 1%.
১,৪৪২.
A room of size 5m × 3m and height 3m requires walls and ceiling painting. What is the area to be painted?
  1. ক
    63 sq. m
  2. খ
    70 sq. m
  3. গ
    64 sq. m
  4. ঘ
    90 sq. m
ব্যাখ্যা
Question: A room of size 5m × 3m and height 3m requires walls and ceiling painting. What is the area to be painted?

Solution: 
Area of Wall = ( 5 + 3 + 5 + 3 ) m. wall length × 3 m height
= 48 sq.m.
Area of Ceiling = 15 sq.m.

hence total painting area of walls and ceiling = 48 sq m + 15 sq m = 63 sq m
63 square meter area of walls and ceiling to be painted. 
১,৪৪৩.
The area of a rectangle is 60cm2 and one of its sides is 6 cm long. What will be its perimeter?
  1. ক
    ক) 20 cm
  2. খ
    খ) 24 cm
  3. গ
    গ) 32 cm
  4. ঘ
    ঘ) 36 cm
ব্যাখ্যা
দেয়া আছে,
আয়তক্ষেত্রের ক্ষেত্রফল 60 বর্গ  সে.মি.
আয়তক্ষেত্রের এক পাশের দৈর্ঘ্য 6 সে.মি.
আয়তক্ষেত্রের অপর পাশের দৈর্ঘ্য 60/6 সে.মি.
                                                     =10 সে.মি.

আয়তক্ষেত্রের পরিসীমা = 2 (6 +10) সে.মি.
                                     = 32 সে.মি.
১,৪৪৪.
If tanθ =√3 and sinθ = √3/2. Find the value of cosθ.
  1. ক
    √3/2
  2. খ
    1
  3. গ
    1/2
  4. ঘ
    1/√3
  5. ঙ
    2/√3
ব্যাখ্যা

Question: If tanθ =√3 and sinθ = √3/2. Find the value of cosθ.

Solution:
Given,
tanθ =√3 and sinθ = √3/2

We know,
tanθ = sinθ/cosθ
⇒√3 = (√3/2)/cosθ
⇒ cosθ = (√3/2)/√3
∴ cosθ = 1/2   

১,৪৪৫.
The area of the floor of an office is 324 square meters. How many unbroken tiles of dimension (6 × 18) cm2 will be required to cover the floor completely?
  1. ক
    1500
  2. খ
    15000
  3. গ
    3000
  4. ঘ
    30000
ব্যাখ্যা
Question: The area of the floor of an office is 324 square meters. How many unbroken tiles of dimension (6 × 18) cm2 will be required to cover the floor completely?

Solution:
Given that,
Area of the floor of an office = 324 sq. meter
= (324 × 10000) cm2
= 3240000 cm2

and,
Area of ​​unbroken tiles = (6 × 18) cm2
= 108 cm2

So, 
Total tiles required to cover the floor = 3240000/108
= 30000
১,৪৪৬.
The perimeter of rectangle is 260 meters. The breadth is 3/7 part of the leangth. What is the leangth?
  1. ক
    68m
  2. খ
    72m
  3. গ
    80m
  4. ঘ
    91m
ব্যাখ্যা
Question: The perimeter of rectangle is 260 meters. The breadth is 3/7 part of the leangth. What is the leangth?

সমাধান:
ধরি,
দৈর্ঘ্য = ৭ক 
এবং প্রস্থ = ৩ক

আমরা জানি,
আয়তক্ষেত্রের পরিসীমা = ২(দৈর্ঘ্য + প্রস্থ)

প্রশ্নমতে,
২(৭ক + ৩ক) = ২৬০
⇒ (৭ক + ৩ক) = ১৩০
⇒ ১০ক = ১৩০
∴ ক = ১৩

সুতরাং, দৈর্ঘ্য = ৭ × ১৩ = ৯১ মিটার
১,৪৪৭.
If the diagonal of a rectangle is 17 cm long its perimeter is 46 cm, find the area of the rectangle.
  1. ক
    ক) 96 cm2
  2. খ
    খ) 120 cm2
  3. গ
    গ) 144 cm2
  4. ঘ
    ঘ) 156 cm2
ব্যাখ্যা

Let length = x and breadth = y. Then,
2(x + y)= 46 or
x + y = 23 and
x2 + y2 = (17)2
= 289.

Now, (x + y)2 = (23)2
⇒ (x2 + y2) + 2xy = 529
⇒ 289 + 2xy = 529
⇒ xy = 120.

∴ Area = xy = 120 cm2

১,৪৪৮.
If a 24 meters ladder is placed against a 12 meters wall such that it just reaches the top of the wall, then what is the angle of elevation? 
  1. ক
    ক) 30°
  2. খ
    খ) 45°
  3. গ
    গ) 60°
  4. ঘ
    ঘ) 90°
ব্যাখ্যা
Question: If a 24 meters ladder is placed against a 12 meters wall such that it just reaches the top of the wall, then what is the angle of elevation? 

Solution: 


AC = 24 meters
AB = 12 meters
∠ACB = θ

Now,
sinθ = AB/AC
⇒ sinθ = 12/24
⇒ sinθ = 1/2
⇒ sinθ = sin 30°
∴ θ = 30°
১,৪৪৯.
If cos A + cos2A = 1, then the value of the expression (cos2A + cosA) is –
  1. ক
    ক) 1
  2. খ
    খ) 1/2
  3. গ
    গ) 2
  4. ঘ
    ঘ) 3
ব্যাখ্যা
Given, cos A + cos2A = 1
Or, cos2A + cosA = 1
১,৪৫০.
A polygon that has 10 sides is called _____ .
  1. ক
    ক) Heptagon
  2. খ
    খ) Decagon
  3. গ
    গ) Hexagon
  4. ঘ
    ঘ) Octagon
ব্যাখ্যা
যে বহুভুজের বাহুর সংখ্যা 10 টি তাকে Decagon বলে।
যে বহুভুজের বাহুর সংখ্যা 6 টি তাকে Hexagon বলে।
যে বহুভুজের বাহুর সংখ্যা 7 টি তাকে Heptagon বলে। 
যে বহুভুজের বাহুর সংখ্যা ৪ টি তাকে Octagon বলে।
১,৪৫১.
An equilateral triangle has a perimeter of 24 meters. What is its area?
  1. ক
    15√3 m2
  2. খ
    16√3 m2
  3. গ
    17√2 m2
  4. ঘ
    9 m2
ব্যাখ্যা
Question: An equilateral triangle has a perimeter of 24 meters. What is its area?
(সমবাহু ত্রিভুজের পরিসীমা ২৪ মিটার হলে এর ক্ষেত্রফল কত হবে?)

Solution:
দেওয়া আছে,
সমবাহু ত্রিভুজের পরিসীমা ২৪ মিটার
সমবাহু ত্রিভুজের এক বাহুর দৈর্ঘ্য = ২৪/৩ মিটার = ৮ মিটার

সমবাহু ত্রিভুজের ক্ষেত্রফল =(√৩/৪) × ৮২ বর্গমিটার
= (√৩/৪) × ৬৪ বর্গমিটার
= ১৬√৩ বর্গমিটার
১,৪৫২.
∠P and ∠Q are complementary to each other. If ∠P = 20° + 4x and ∠Q = 6x, find the value of ∠Q.
  1. ক
    42°
  2. খ
    48°
  3. গ
    70°
  4. ঘ
    52°
ব্যাখ্যা

Question: ∠P and ∠Q are complementary to each other. If ∠P = 20° + 4x and ∠Q = 6x, find the value of ∠Q.

Solution:
Here,
∠P = 20° + 4x and ∠Q = 6x

For complementary angles,
∠P + ∠Q = 90°
⇒ (20° + 4x) + 6x = 90°
⇒ 20° + 4x + 6x = 90°
⇒ 20° + 10x = 90°
⇒ 10x = 90° - 20°
⇒ 10x = 70°
∴ x = 7°

So, ∠Q = 6 × 7° = 42°

১,৪৫৩.
A right angled triangle, whose perpendicular sides measure 1.8 cm and 2.4 cm, is inscribed in a circle. What is the circumference of the circle (in cm)?
  1. ক
    π
  2. খ
    3π
  3. গ
    2π
  4. ঘ
    4π
  5. ঙ
    5π
ব্যাখ্যা

None of the above
Radius of the circle = r
So hypotenuse = 2r
So, 2r = √(2.42 + 1.82)
Or, 2r = √9
Or, 2r = 3
the circumference of the circle = 2πr = 3π

১,৪৫৪.
A cylindrical pencil sharpened at one edge is the combination of-
  1. ক
    ক) a cone and a cylinder
  2. খ
    খ) frustum of a cone and a cylinder
  3. গ
    গ) a hemisphere and a cylinder
  4. ঘ
    ঘ) two cylinders
  5. ঙ
    ঙ) None of these
ব্যাখ্যা
A cylindrical pencil sharpened at one edge is the combination of a cone and a cylinder.

১,৪৫৫.
A 50-meter cable is attached from the top of a vertical pole down to the ground. If the cable makes an angle of 30 degrees with the ground, find the height of the pole.
  1. ক
    20 meters
  2. খ
    25 meters
  3. গ
    30 meters
  4. ঘ
    35 meters
  5. ঙ
    None of these
ব্যাখ্যা

Question: A 50-meter cable is attached from the top of a vertical pole down to the ground. If the cable makes an angle of 30 degrees with the ground, find the height of the pole.

Solution: 

ধরি,
উচ্চতা(Height), AB = h

দেয়া আছে,
AC = 50m                    
∠ACB = 30°

∴ sin30°= AB/AC
⇒ 1/2 = h/50
⇒ h = 50 × 1/2
∴ h = 25 m

১,৪৫৬.
Find the area of the triangle formed by the point located at (5, 2), (- 9, - 3) and (- 3, - 5).
  1. ক
    ক) 27 square unit
  2. খ
    খ) 29 square unit
  3. গ
    গ) 31 square unit
  4. ঘ
    ঘ) 30 square unit
ব্যাখ্যা
Question: Find the area of the triangle formed by the point located at (5, 2), (- 9, - 3) and (- 3, - 5).

Solution: 
ত্রিভুজের তিনটি প্রান্তবিন্দু দেয়া থাকলে,
ত্রিভুজটির ক্ষেত্রফল = 1/2 {x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)}
= (1/2) {5 × 2 + -9 × -7 + -3 × 5}
= 1/2(10 + 63 - 15)
= 29 বর্গএকক
১,৪৫৭.
The value of sin265° + sin225° + cos235° + cos255° is?
  1. ক
    ক) 0
  2. খ
    খ) 1
  3. গ
    গ) 2
  4. ঘ
    ঘ) - 2
ব্যাখ্যা
Question: The value of sin265° + sin225° + cos235° + cos255° is?

Solution:
sin265° + sin225° + cos235° + cos255°
= sin265° + sin2(90° - 65°) + cos235° + cos2(90° - 35°)
= (sin265° + cos265°) + (cos235° + sin235°)
= 1 + 1
= 2

১,৪৫৮.
A ladder is leaning against a wall. It makes a 60° angle with the wall. If the distance between foot of ladder and wall is 7.5 meters, find the length of the ladder.
  1. ক
    22.5 m
  2. খ
    27 m
  3. গ
    14.5 m
  4. ঘ
    15 m
ব্যাখ্যা

Question: A ladder is leaning against a wall. It makes a 60° angle with the wall. If the distance between foot of ladder and wall is 7.5 meters, find the length of the ladder.

Solution:

Let BC be the wall and AC be the ladder.
∠BAC = 60° and AB = 7.5 meter
In ΔABC,
cos60° = AB/AC
⇒ 1/2 = 7.5/AC
⇒ AC = 7.5 × 2
∴ AC = 15

১,৪৫৯.
50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m³, then the rise in the water level in the tank will be:
  1. ক
    ক) 20 cm
  2. খ
    খ) 25 cm
  3. গ
    গ) 35 cm
  4. ঘ
    ঘ) 50 cm
ব্যাখ্যা

Total volume of water displaced = (4 x 50) m3 = 200 m3.
Rise in water level =200/(40 x 20)m 0.25 m = 25 cm.

১,৪৬০.
In a right triangle, the length of one of the legs is 3 and the length of the hypotenuse is 5. What is the length of the other leg?
  1. ক
    4 unit
  2. খ
    5 unit
  3. গ
    6 unit
  4. ঘ
    7 unit
ব্যাখ্যা
Question: In a right triangle, the length of one of the legs is 3 and the length of the hypotenuse is 5. What is the length of the other leg?

Solution: 
সমকোণী ত্রিভুজের অতিভুজ = 5
সমকোণ সংলগ্ন এক বাহু = 3
সমকোণ সংলগ্ন অপর বাহু = a 

প্রশ্নমতে 
a2 + 32 = 52
⇒ a2 + 9 = 25
⇒ a2 = 25 - 9
⇒ a2 = 16
⇒ a2 = 42
⇒ a = 4
১,৪৬১.
The surface area of a sphere is the same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:
  1. ক
    4 cm
  2. খ
    5 cm
  3. গ
    6 cm
  4. ঘ
    8 cm
ব্যাখ্যা
Question: The surface area of a sphere is the same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

Solution: 
The surface area of sphere = 4πr12
The curved surface area of cylinder =2πr2h
diameter = 12 cm
radius r2 = 6 cm

⇒ 4πr12 = 2πr2h
⇒ r12 = (6 × 12)/2
⇒ r12 = 36
⇒ r1 = 6

radius of the sphere 6 cm
১,৪৬২.
The volume of a sphere with radius r is (4/3)πr3 and the surface area is 4πr2. If a spherical balloon has a volume of 2304π cubic centimeters, what is the surface area of the balloon in square centimeters?
  1. ক
    225π
  2. খ
    361π
  3. গ
    576π
  4. ঘ
    None
ব্যাখ্যা
Question: The volume of a sphere with radius r is (4/3)πr3 and the surface area is 4πr2. If a spherical balloon has a volume of 2304π cubic centimeters, what is the surface area of the balloon in square centimeters?

Solution:
Here,
radius = (4/3)πr3
surface area is = 4πr2

Now,
Volume = (4/3)πr3 = 2304π
⇒ r3 = (2304 × 3)/4
⇒ r3 = 6912/4
⇒ r3 = 1728
⇒ r = 12

So, the surface area would be = 4πr2
= 4 × π × 122
= 4 × π × 144
= 576π
১,৪৬৩.
The length of two parallel sides of a trapezium are 43 cm and 58 cm respectively, and the distance between the parallel sides is 18 cm. Find the area of the trapezium.
  1. ক
    ক) 949 cm2
  2. খ
    খ) 939 cm2
  3. গ
    গ) 929 cm2
  4. ঘ
    ঘ) 909 cm2
ব্যাখ্যা
Area of the Trapezium = (1/2) × (Sum of the parallel sides) × (Distance between parallel sides)
                                    = (1/2) × (43 + 58) × 18
                                    = (1/2) × 101 × 18
∴ Area of the Trapezium = 909 cm2
১,৪৬৪.
If tanθ = 5/12, then find the value of sinθ.
  1. ক
    5/12
  2. খ
    12/5
  3. গ
    0
  4. ঘ
    13/5
  5. ঙ
    5/13
ব্যাখ্যা

Question: If tanθ = 5/12, then find the value of sinθ.

Solution:
Given, tanθ = Perpendicular/Base = 5/12
Here, base = 12 and perpendicular = 5
Let hypotenuse = x

Now, according to Pythagorean theorem,
x2 = 52 + 122
⇒ x2 = 25 + 144
⇒ x = √169
⇒ x = 13

∴ sinθ = Perpendicular/Hypotenuse = 5/13

১,৪৬৫.
The area of a square is 1 hectare. Calculate the perimeter of the garden in metre?
  1. ক
    100 meter
  2. খ
    200 meter
  3. গ
    300 meter
  4. ঘ
    400 meter
ব্যাখ্যা
Question: The area of a square is 1 hectare. Calculate the perimeter of the garden in metre?

Solution:
দেওয়া আছে,
বর্গক্ষেত্রের ক্ষেত্রফল = ১ হেক্টর
= ১০০০০ বর্গমিটার
বর্গক্ষেত্রের একবাহু, a = √১০০০০ মিটার
= ১০০ মিটার

∴ বর্গক্ষেত্রের পরিসীমা = ৪a = (৪ × ১০০) মিটার
= ৪০০ মিটার
১,৪৬৬.
The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree is:
  1. ক
    ক) 30°
  2. খ
    খ) 45°
  3. গ
    গ) 60°
  4. ঘ
    ঘ) 90°
ব্যাখ্যা
Question: The angle of elevation of the sun, when the length of the shadow of a tree √3 times the height of the tree is:

Solution:

Let,
AB = height of tree
BC= Shadow of tree
angle of elevation = C
∴  BC = √3 AB

We know,
tanC = AB/BC
⇒ tanC = AB/√3AB
⇒ tanC = 1/√3
⇒ tanC = tan30°
∴ C = 30°
১,৪৬৭.
The volume of a right circular cylinder is 100π cubic units and its height is 5 units. What is the circumference of its base?
  1. ক
    10π
  2. খ
    2√5π
  3. গ
    20π
  4. ঘ
    4√5π
ব্যাখ্যা
Question: The volume of a right circular cylinder is 100π cubic units and its height is 5 units. What is the circumference of its base?

Solution:
একটি সিলিন্ডারের উচ্চতা h একক ও ব্যাসার্ধ r একক হলে,
উক্ত সিলিন্ডারের আয়তন = πr2h ঘন একক

প্রশ্নমতে,
⇒ πr2 × h = 100π
⇒ πr2 × 5 = 100π
⇒ r2 = 20
∴ r = √20 = 2√5

সুতরাং বৃত্তের পরিধি = 2πr = 2π × 2√5 = 4√5π
১,৪৬৮.
A rectangular prism has dimensions 8 cm, 4 cm, and 2 cm. Calculate the volume of the prism.
  1. ক
    45 cm3
  2. খ
    50 cm3
  3. গ
    64 cm3
  4. ঘ
    70 cm3
ব্যাখ্যা
Question: A rectangular prism has dimensions 8 cm, 4 cm, and 2 cm. Calculate the volume of the prism.

Solution: 
The volume of a rectangular prism can be found using the formula:
Volume = length × width × height
= 8 × 4 × 2 cm3
= 64 cm3
১,৪৬৯.
The angle measure of base angles of an isosceles triangle are represented by x and the vertex angle is 3x+15. Find the measure of base angle.
  1. ক
    ক)  33°
  2. খ
    খ)  35°
  3. গ
    গ)  34°
  4. ঘ
    ঘ)  36°
ব্যাখ্যা
ATQ, x + x + 3x + 15 = 180°
Or, 5x = 165°
Or, x = 33°
১,৪৭০.
The slope of the line 4x - 2y = 8 is not the same as the slope of which one of the following lines? ​
  1. ক
    4x - 2y = 12
  2. খ
    2x - y = - 5
  3. গ
    y = 2x - 1
  4. ঘ
    x + 2y = 6
ব্যাখ্যা

Question: The slope of the line 4x - 2y = 8 is not the same as the slope of which one of the following lines?
​
​Solution:
​প্রথমে, প্রদত্ত রেখাটির ঢাল নির্ণয় করতে হবে। 
​রেখাটির সমীকরণকে y = mx + c তে রূপান্তর করতে হবে। এখানে 'm' হলো ঢাল (Slope)।

প্রদত্ত রেখার সমীকরণ: 4x - 2y = 8
⇒ - 2y = - 4x + 8
⇒ y = (- 4/- 2)x + (8/- 2)
⇒ y = 2x - 4
∴ এই রেখাটির ঢাল (m) হলো 2.

এবার, প্রদত্ত অপশনগুলোর প্রত্যেকটির ঢাল নির্ণয় করি:
ক) 4x - 2y = 12
⇒ -2y = -4x + 12
⇒ y = 2x - 6
∴ ঢাল 2

খ) 2x - y = -5
⇒ -y = -2x - 5
⇒ y = 2x + 5
∴ ঢাল 2

গ) y = 2x - 1
∴ ঢাল 2

ঘ) x + 2y = 6
⇒ 2y = -x + 6
⇒ y = (-1/2)x + 3
∴ ঢাল - 1/2

সুতরাং, দেখা যাচ্ছে যে শুধুমাত্র অপশন (ঘ) এর রেখার ঢাল মূল রেখার ঢাল থেকে ভিন্ন।

১,৪৭১.
If y = sin(sinx) then what is the value of dy/dx?
  1. ক
    cos(sinx)(sinx)(tanx)
  2. খ
    cosx.cos(cosx)
  3. গ
    (cos2x)2
  4. ঘ
    cosx.cos(sinx)
  5. ঙ
    cosx.cos(tanx)
ব্যাখ্যা

Question: If y = sin(sinx) then what is the value of dy/dx?

Solution:

১,৪৭২.
Given the area of the rhombus is 120-meter square then find the length of one of the diagonals if the other diagonal is of length 12 m.
  1. ক
    25 cm
  2. খ
    30 cm
  3. গ
    20 m
  4. ঘ
    15 cm
ব্যাখ্যা
Question: Given the area of the rhombus is 120-meter square then find the length of one of the diagonals if the other diagonal is of length 12 m.

Solution: 
Since we know that,
Area of Rhombus = (1/2) × Diagonal1 × Diagonal2 

Putting all the known values, we get
120 = (1/2) × Diagonal1 × Diagonal2
⇒ 240 = 12 × Diagonal2
⇒  Diagonal2 = 240/12
∴ Diagonal2 = 20 m
১,৪৭৩.
If a right-angled isosceles triangle has base 6 cm, then height is:
  1. ক
    4 cm
  2. খ
    8 cm
  3. গ
    6 cm
  4. ঘ
    2 cm
ব্যাখ্যা

Question: If a right-angled isosceles triangle has base 6 cm, then height is:
(Officer Cash 2022 অনুযায়ী)

Solution:
(Right-angled isosceles triangle) সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি = 6 cm.
সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি ও উচ্চতা সমান।
ভূমি = উচ্চতা = 6 cm.
∴ উচ্চতা = 6 cm

১,৪৭৪.
What would be the measure of the diagonal of a square whose area is equal to 578cm2?
  1. ক
    ক) 34cm
  2. খ
    খ) 38cm
  3. গ
    গ) 32cm
  4. ঘ
    ঘ) 42cm
ব্যাখ্যা
ধরি 
বর্গের এক বাহুর দৈর্ঘ্য a সে.মি.
প্রশ্নমতে,
a2 = 578
a = √578
a = 24.04

বর্গের কর্ণের দৈর্ঘ্য = a√2 = 24.04√2 = 33.99cm ≈ 34cm
১,৪৭৫.
If θ be an acute angle and 5sin2θ + 3cos2θ = 4, then the value of tanθ is?
  1. ক
    √2
  2. খ
    1/√2
  3. গ
    1
  4. ঘ
    0
ব্যাখ্যা

প্রশ্ন: If θ be an acute angle and 5sin2θ + 3cos2θ = 4, then the value of tanθ is?

সমাধান:
5sin2θ + 3cos2θ = 4
⇒ 5sin2θ + 3(1 - sin2θ) = 4
⇒ 5sin2θ + 3 - 3sin2θ = 4
⇒ 2sin2θ = 1
⇒ sin2θ = 1/2
⇒ sinθ = √(1/2)
⇒ sinθ = 1/√2
⇒ sinθ = sin45°
⇒ θ = 45°

∴ tanθ = tan45° = 1

১,৪৭৬.
A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paisa per sq. metre.
  1. ক
    ক) Tk. 570
  2. খ
    খ) Tk. 620
  3. গ
    গ) Tk. 680
  4. ঘ
    ঘ) Tk. 750
ব্যাখ্যা

Area of the plot = (110 × 65)m2
= 7150m2
Area of the plot excluding the path = [(110 - 5) × (65 - 5)]m2
= 6300 m2

∴ Area of the path = (7150 - 6300) m2
= 850 m2

Cost of graveling the path = Tk. {850 × (80/100)}
= Tk. 680

১,৪৭৭.
ln a square ABCD, diagonals AC and BD intersect at O. The angle bisector of ∠CAB meets BD and BC at F and G, respectively. OF : CG is equal to-
  1. ক
    1 : 2
  2. খ
    1 : 3
  3. গ
    2 : 3
  4. ঘ
    3 : 1
  5. ঙ
    2 : 1
ব্যাখ্যা
Question: ln a square ABCD, diagonals AC and BD intersect at O. The angle bisector of ∠CAB meets BD and BC at F and G, respectively. OF : CG is equal to-

Solution:

ABCD is a square
AC = √2AB
AO = OC = AC/2 = √2AB/2 = AB/√2
ΔAOF ∼ ΔABG

[By AA property]
AO/AB = OF/BG
(AB/√2)/AB = OF/BG
1/√2 = OF/BG
BG = √2OF . . . . . . (i)
AG is angle bisector of ΔABC
AB/AC = BG/GC =1/√2
[angle bisector theorem]
BG = (1/√2)GC . . . . . . (ii)

Compare (i) and (ii)
√2OF = (1/√2)GC
OF : CG = 1 : 2
১,৪৭৮.
By what percent the volume of a cube increases if the length of each edge was increased by 50%?
  1. ক
    ক) 50%
  2. খ
    খ) 125%
  3. গ
    গ) 237.5%
  4. ঘ
    ঘ) 273.5%
ব্যাখ্যা

Let original edge = a,
Then, original volume = a3

New edge = (150/100)a
= 3a/2
New volume = (3a/2)3
= 27a3/8

Increase in volume = (27a3)/8 - (a3)
= 19a3/8

∴ Increase% = {(19a3/8) × (1/a3) × 100}%
= 237.5%

১,৪৭৯.
If the height of a cone is doubled and its base diameter is trebled, then the ratio of the volume of the resultant cone to that of the original cone is? 
  1. ক
    ক) 9 : 1
  2. খ
    খ) 13 : 2
  3. গ
    গ) 4 : 11
  4. ঘ
    ঘ) 18 : 1
ব্যাখ্যা
Question: If the height of a cone is doubled and its base diameter is trebled, then the ratio of the volume of the resultant cone to that of the original cone is? 

Solution:  
Let the original radius and height of the cone be r and h respectively.
Then, new radius = 3r and new height = 2h 
∴ New volume/Original volume = ((1/3) × π × (3r)2 × 2h) : ((1/3) × π × r2 × h) = 18 : 1 
১,৪৮০.
A parallelogram has a base of 30m and height is 10m long. Then its area is-
  1. ক
    200 m2
  2. খ
    250 m2
  3. গ
    300 m2
  4. ঘ
    320 m2
ব্যাখ্যা
Question: A parallelogram has a base of 30m and  height is 10m. Then its area is-

Solution: 
area = base × height
= 30 × 10
= 300 m2
১,৪৮১.
From the top of a lighthouse 60 m high above sea level, the angle of depression of a boat is 45°. How far is the boat from the foot of the lighthouse?
  1. ক
    40 m
  2. খ
    60 m
  3. গ
    30 m
  4. ঘ
    55 m
ব্যাখ্যা

Question: From the top of a lighthouse 60 m high above sea level, the angle of depression of a boat is 45°. How far is the boat from the foot of the lighthouse?

Solution:

Let the height of the lighthouse above sea be AC and it is given 60 m.
Angle of depression = 45°

Boat is at point B so the distance between the base of lighthouse A and Boat is AB.

 So, tan 45° = AC / AB
⇒ 1 = 60 / AB
⇒ AB = 60 m

∴ The boat is 60 m away from the foot of the lighthouse.

১,৪৮২.
How far apart are the centers of two circles with diameters of 16 cm and radii of 6 cm, when they touch each other externally? 
  1. ক
    14 cm
  2. খ
    16 cm
  3. গ
    22 cm
  4. ঘ
    10 cm
ব্যাখ্যা

Question: How far apart are the centers of two circles with diameters of 16 cm and radii of 6 cm, when they touch each other externally? 

Solution:
আমরা জানি,
দুইটি বৃত্ত পরস্পরকে বহিঃস্পর্শ করলে কেন্দ্রদ্বয়ের মধ্যবর্তী দূরত্ব বৃত্ত দুইটির ব্যাসার্ধের যোগফলের সমান।

এখানে,
১ম বৃত্তের ব্যাসার্ধ = 16/2 = 8 সে.মি.
২য় বৃত্তের ব্যাসার্ধ = ৬ সে.মি.

∴ কেন্দ্রদ্বয়ের মধ্যবর্তী দূরত্ব = (8 + 6) সে.মি.
= 14 সে.মি.

১,৪৮৩.
If the radius of a circle is reduced by 40%, its circumference is reduced by-
  1. ক
    60%
  2. খ
    40%
  3. গ
    35%
  4. ঘ
    30%
ব্যাখ্যা
Question:  If the radius of a circle is reduced by 40%, its circumference is reduced by-

Solution: 
If radius of a circle is r, circumference 2πr where 2π is constant 
So, if radius is changed, the circumference will change by the same amount.

The radius of a circle is reduced by 40%,then its circumference is reduced by 40%
১,৪৮৪.
If θ lies in the first quadrant and cos2θ - sin2θ = 1/2. then the value of tan22θ + sin23θ is-
  1. ক
    3
  2. খ
    4
  3. গ
    2
  4. ঘ
    None of the above
ব্যাখ্যা

Question: If θ lies in the first quadrant and cos2θ - sin2θ = 1/2. then the value of tan22θ + sin23θ is-

Solution:
cos2θ - sin2θ = 1/2
⇒ cos2θ = 1/2
⇒ cos2θ = cos60°
⇒ 2θ = 60°
⇒ θ = 30°

Now, tan22θ + sin23θ
= tan260° + sin290°
= 3 + 1
= 4

১,৪৮৫.
A ladder is placed against a wall such that its foot is at a distance of 2.5 from the wall and its top reaches a window 6m above the ground. Find the length of the ladder.
  1. ক
    ক) 6.5m
  2. খ
    খ) 8.5m
  3. গ
    গ) 4.5m
  4. ঘ
    ঘ) 2.5m
ব্যাখ্যা


ধরি 
মইয়ের দৈর্ঘ্য x মিটার 
এখানে 
AC = x = মই 
AB = দেয়াল 

পিথাগোরাসের সূত্র অনুসারে 
x2 = 62 + 2.52
x2 = 36 + 6.25 
x2 = 42.25 
x = √42.25 
x = 6.5
১,৪৮৬.
Find the volume of the cylinder having a radius of 5 units and a height of 8 units?
  1. ক
    314.57 Cubic units
  2. খ
    628.57 Cubic units
  3. গ
    125.71 Cubic units
  4. ঘ
    None of these
ব্যাখ্যা
Question: Find the volume of the cylinder having a radius of 5 units and a height of 8 units?

Solution:
We have, 
Radius,r = 5 units
Height,h = 8 units

Volume of the cylinder, V = πr2h cubic units.
V = (22/7) × 52 × 8
V = 22/7 × 25 × 8
V = 628.57 Cubic units.

Hence, the volume of the cylinder is 628.57 cubic units.
১,৪৮৭.
If the radius of a circle is doubled, its area is increased by-
  1. ক
    ক) 200%
  2. খ
    খ) 100%
  3. গ
    গ) 300%
  4. ঘ
    ঘ) 50%
ব্যাখ্যা
Let
Radius of the circle = r
then Area of the circle = πr2
if we doubled the radius then radius = 2r
and Area of the circle = π(2r)2 = 4πr2 

Increased Area = 4πr2 - πr2=3πr2 
Increase % = (3πr2/πr2) ×100}% 
                  = 300%
১,৪৮৮.
If the angles of a triangle are in the ratio 1 : 2 : 6, what is the measure of the middle angle (in degrees)?
  1. ক
    75°
  2. খ
    20°
  3. গ
    80°
  4. ঘ
    60°
  5. ঙ
    None of these
ব্যাখ্যা
Question: If the angles of a triangle are in the ratio 1 : 2 : 6, what is the measure of the middle angle (in degrees)?

Solution:
Given that,
The angles of a triangle are in the ratio 1 : 2 : 6
Let,
x, 2x, 6x

We know that,
Sum of angles in a triangle = 180°

Now
x + 2x + 6x = 180°
⇒ 9x = 180°
⇒ x = 180°/9 = 20°
∴ x = 20°

∴ middle angle = 2x = 2 × 20 = 40°
১,৪৮৯.
There is an equilateral triangle with a square inscribed inside it. One of the sides of the square lies on a side of the equilateral triangle. What is the ratio of the area of the square to that of the equilateral triangle?
  1. ক
    6 : (5 + 7√3)
  2. খ
    24 : (24 + 7√3)
  3. গ
    12 : (12 + 7√3)
  4. ঘ
    18 : (12 + 7√3)
ব্যাখ্যা
Question: There is an equilateral triangle with a square inscribed inside it. One of the sides of the square lies on a side of the equilateral triangle. What is the ratio of the area of the square to that of the equilateral triangle?

Solution: 


let, side of equilateral triangle is a 
area =  (√3/4)a2 

let side of square r
BD = CE = r/tan60 = r/√3
BC = a =  r +  r/√3 +  r/√3 = (2r + √3r)/√3
= r (2 + √3)/√3

area of triangle = (√3/4) {r (2 + √3)/√3}2
= r2(√3/4) (7 + 4√3)/3
= r2 (7√3 + 12)/12

area of square = r2 

ratio = r2 : r2 (7√3 + 12)/12
= 12 : (12 + 7√3)
১,৪৯০.
If sec θ = 5/4, then what is the value of sinθ?
  1. ক
    3/5
  2. খ
    8/3
  3. গ
    3/4
  4. ঘ
    4/5
ব্যাখ্যা

Question: If sec θ = 5/4, then what is the value of sinθ?

Solution:
এখানে,
secθ = 5/4 = অতিভুজ/ভূমি
∴ অতিভুজ = 5, ভূমি = 4

পিথাগোরাসের উপপাদ্য অনুসারে, লম্ব নির্ণয় করি,
লম্ব = √(অতিভুজ2 - ভূমি2)
= √(52 - 42)
= √(25 - 16)
= √9
= 3

এখন,
sinθ = লম্ব/অতিভুজ 
= 3/5

সুতরাং, sinθ = 3/5।

১,৪৯১.
If the side of a square is increased by 20%, by what percent will the area be increased?
  1. ক
    21%
  2. খ
    42%
  3. গ
    44%
  4. ঘ
    20%
ব্যাখ্যা

Question: If the side of a square is increased by 20%, by what percent will the area be increased?

Solution:
Let the original side length = 10 units.

∴ Area = 10 × 10 = 100 square units

Again, 
After a 20% increase, the new side length = 10 + (20% of 10)
= 10 + 2 = 12 units

∴ New area = 12 × 12 = 144 square units

∴ Increase in area = (144 - 100) square units
= 44 square units

∴ Percentage increase in area = (44/100) × 100%
= 44%

So the area will increase by 44%

১,৪৯২.
A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. what is the area of the circle?
  1. ক
    1250 sq. cm
  2. খ
    616 sq. cm
  3. গ
    154 sq. cm
  4. ঘ
    88 sq. cm
ব্যাখ্যা
Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. what is the area of the circle?

Solution:
Perimeter of the rectangle =2(18 + 26)
= 88 cm

∴ Circumference of circle = 88 cm
⇒ 2πr = 88
⇒ r = 88/2π
⇒ r = (88 × 7)/(2 × 22)
∴ r = 14 cm

∴ Area of circle = πr2
= (22/7) × (14)2
= 616 sq. cm
১,৪৯৩.
Find 
  1. ক
    1
  2. খ
    1/4
  3. গ
    1/2
  4. ঘ
    2
ব্যাখ্যা

Question: find 

Solution: 

১,৪৯৪.
If C is the midpoint of the points A(2, - 1) and B(8, 5), find the length of AC.
  1. ক
    8√2
  2. খ
    3√2
  3. গ
    3√5
  4. ঘ
    5
ব্যাখ্যা

Question: If C is the midpoint of the points A(2, - 1) and B(8, 5), find the length of AC.

Solution:
দেওয়া আছে,
 A(2, - 1) এবং B(8, 5), 
এবং C হলো AB-এর মধ্যবিন্দু।

দূরত্বের সূত্র ব্যবহার করে AB-এর দৈর্ঘ্য নির্ণয় করি,
AB = √(x2 - x1)2 + (y2 - y1)2)
= √(8 - 2)2 + {5 - (-1)}2
= √(62 + 62)
= √(36 + 36)
= √72
= √(36 × 2)
= 6√2

C হলো AB এর মধ্যবিন্দু, তাই AC = AB/2
 = 6√2/2
 = 3√2

১,৪৯৫.
If the perimeter of a certain rectangle is 76 m and its area is 360 m2, then what is the length of its shortest side?
  1. ক
    ক) 13
  2. খ
    খ) 15
  3. গ
    গ) 18
  4. ঘ
    ঘ) 10
ব্যাখ্যা

We know, Perimeter = 2(l+b) = 76
⇒ (l+b) = 38
⇒ l = 38 - b .....(1)
Given, the area, lb = 360 .....(2)
From (2)
(38 – b)b = 360
⇒ 38b – b2 = 360
⇒ b2 – 38b + 360 = 0
⇒ (b – 20)(b – 18) = 0
So, b = 20 or b = 18
When b = 20, l = 18 and when b = 18, l = 20.
∴ the shorter side is 18.

১,৪৯৬.
A cylindrical water tank has a radius of 35 inches and a height of 120 inches. Calculate the total surface area.
  1. ক
    35120 sq. inches.
  2. খ
    34100 sq. inches.
  3. গ
    32321 sq. inches.
  4. ঘ
    34155 sq. inches.
ব্যাখ্যা
Question: A cylindrical water tank has a radius of 35 inches and a height of 120 inches. Calculate the total surface area.

Solution:
Water tank is cylindrical in nature.
Total Surface Area of a cylinder is given by, 2πr(h + r)

∴ Total Surface Area = 2 × (22/7) × 35(120 + 35)
= 2 × 22 × 5 × 155
= 34100

∴ Total Surface Area = 34100 sq. inches.
১,৪৯৭.
The difference between the length and the perimeter of a rectangle is 100 cms. What is the breadth of the rectangle?
  1. ক
    80 cms
  2. খ
    60 cms
  3. গ
    100 cms
  4. ঘ
    Data Inadequate
ব্যাখ্যা
Question: The difference between the length and the perimeter of a rectangle is 100 cms. What is the breadth of the rectangle?

Solution:
Let the length of the rectangle be 'x' and breadth of the rectangle be 'y'
According to the question:
2(x + y) - x = 100
⇒ 2x + 2y - x = 100
⇒ x + 2y = 100

From this we cannot find 'y' (breadth), so the given data is inadequate.
১,৪৯৮.
The area of the right angle triangle is 80 square cm one of its leg is 16 cm long. Find the length of the other leg.
  1. ক
    23 cm
  2. খ
    36 cm
  3. গ
    10 cm
  4. ঘ
    28 cm
ব্যাখ্যা
Area of the triangle = 1/2 × base × height
⇒ 80 = 1/2 × 16 × other leg
So,
other leg = (80 × 2)/16
= 10 cm
১,৪৯৯.
A cube of side 4 cm is cut into cubes of side 1 cm. Calculate the total surface area of all the small cubes.
  1. ক
    ক) 24 cm2
  2. খ
    খ) 384 cm2
  3. গ
    গ) 96 cm2
  4. ঘ
    ঘ) 196 cm2
ব্যাখ্যা
Question: A cube of side 4 cm is cut into cubes of side 1 cm. Calculate the total surface area of all the small cubes.

Solution: 
Side of original cube = 4 cm.

Number of cubes of side 1 cm
= (Volume of the bigger cube)/(Volume of the smaller cube)
= (4 × 4 × 4)/(1 × 1 × 1)
= 64

Now, the surface area of one cube = 6(side)2
= 6 × 12
= 6 cm2

∴ The surface area of 64 cubes = 6 × 64 cm2
= 384 cm2
১,৫০০.
The diameter of a circle is 14 cm. What is the circumference of the circle?
  1. ক
    44 m
  2. খ
    0.44 m
  3. গ
    22 cm
  4. ঘ
    2.2 cm
ব্যাখ্যা
Question: The diameter of a circle is 14 cm. What is the circumference of the circle?

Solution: 
Radius of the circle r = 14/2 = 7
The circumference of the circle = 2πr
= 2 × (22/7) × 7
= 44 cm
= 44/100 m
= 0.44 m