Geometry: Mensuration, Trigonometry

PrepBank · পাতা ১২ / ২১ · ১,১০১–১,২০০ / ২,০৮৫

১,১০১.

The top and bottom of a flag on a building subtend angles of 60° and 30° respectively at a point B which is 48 meter away from the building. Find the height of the flag?

ক
32 m
খ
32√3 m
গ
18.49 m
ঘ
16 m

ব্যাখ্যা

Question: The top and bottom of a flag on a building subtend angles of 60° and 30° respectively at a point B which is 48 meter away from the building. Find the height of the flag?

Solution:

Let height of building be AC = X and height of flag be CD = h.

In ΔDAB
tan60° = (X + h)/48
⇒ √3 = (X + h)/48
⇒ X + h = 48√3
∴ h = 48√3 - X ..................(1)

In ΔCAB
tan30° = X/48
⇒ 1/√3 = X/48
∴ X = 48/√3

From (1) we get,
h = 48√3 - 48/√3
= (48 × 3 - 48)/√3
= (144 - 48)/√3
= 96/√3
= (32 × 3)/√3
= 32√3

১,১০২.

Square ABCD with a perimeter of 48 units. Find length of BD.

ক
12 units
খ
12√2 units
গ
9√2 units
ঘ
24 units

ব্যাখ্যা

Question: Square ABCD with a perimeter of 48 units. Find length of BD.

Solution:
Each side of the square must be 48/4 = 12 units
∴ 12² + 12² = (BD)²
⇒ BD² = 2 × 144
∴ BD = 12√2

১,১০৩.

Find the area of a right angled triangle whose hypotenuse is 10 cm and base 8 m.

ক
ক) 24 sq.cm
খ
খ) 34 sq.cm
গ
গ) 36 sq.cm
ঘ
ঘ) 48 sq.cm

ব্যাখ্যা

Length of the triangle = √(10² - 8²) = 6
∴ Area of the triangle = 1/2 × 8 × 6 = 24 cm²

Alternative approach,
ত্রিভুজটির লম্ব, AB হলে
AB² = AC² - BC²
= (10)² - (8)²
= 100 - 64
= 36
∴ AB = 6
∴ ত্রিভুজটির ক্ষেত্রফল = 1/2 × 8 × 6 = 24 cm²

১,১০৪.

The distance between two parallel tangents of a circle is 18 cm, then the radius of the circle is

ক
8 cm
খ
10 cm
গ
9 cm
ঘ
18 cm
ঙ
12 cm

ব্যাখ্যা

Distance between two parallel tangents = 18 cm
That means, diameter = 18 cm
Therefore, radius of the circle = 18/2 = 9 cm

১,১০৫.

Lubna has purchased a square sheet of plywood of 289 square feet area. To cover a wall she must cut of two feet from side. What is the area, in square feet, of the wall?

ক
ক) 225
খ
খ) 230
গ
গ) 235
ঘ
ঘ) 240

ব্যাখ্যা

প্রশ্ন: Lubna has purchased a square sheet of plywood of 289 square feet area. To cover a wall she must cut of two feet from side. What is the area, in square feet, of the wall?

সমাধান:
বর্গাকার প্লাইউডের ক্ষেত্রফল ২৮৯ বর্গফুট

এক বাহুর দৈর্ঘ্য = √২৮৯ ফুট
= ১৭ ফুট

দেয়াল ঢাকার জন্য পাশ থেকে ২ ফুট কেটে নেয়া হয়েছে।
অতএব, বর্গাকৃতি দেয়ালের এক বাহুর দৈর্ঘ্য = ১৭ - ২ ফুট
= ১৫ ফুট
∴ দেয়ালটির ক্ষেত্রফল = ১৫^২ বর্গফুট
= ২২৫ বর্গফুট

১,১০৬.

tanA√(1 - sin²A) = ?

ক
ক) 1/sinA
খ
খ) sinA
গ
গ) sin²A
ঘ
ঘ) 1/cosA

ব্যাখ্যা

Question: tanA√(1 - sin²A) = ?

Solution:
tanA√(1 - sin²A)
= tanA√(cos²A)
= (sinA/cosA) × cosA
= sinA

১,১০৭.

In a rectangle, the diagonal length is 15 and the width is 9. What is the perimeter of the rectangle?

ক
22
খ
32
গ
42
ঘ
12

ব্যাখ্যা

Question: In a rectangle, the diagonal length is 15 and the width is 9. What is the perimeter of the rectangle?

Solution:
Given:
Diagonal of rectangle, d = 15
Width, w = 9
Length = l

We know, by Pythagoras theorem,
l² + w² = d²
⇒ l² + 9² = 15²
⇒ l² + 81 = 225
⇒ l² = 225 - 81
⇒ l² = 144
∴ l = 12
∴ Length of the rectangle, l = 12

∴ Perimeter of a rectangle = 2(l + w)
= 2(12 + 9)
= 2 × 21
= 42

So the perimeter of the rectangle is 42.

১,১০৮.

In a cylinder, the radius is doubled and height is halved, the curved surface area will be-

ক
ক) Halved
খ
খ) Doubled
গ
গ) Same
ঘ
ঘ) Four times
ঙ
ঙ) None of these

ব্যাখ্যা

We know that the curved surface area of a cylinder is 2πrh
Given that, r = 2R, h= H/2
Hence, the CSA of new cylinder = 2π(2R)(H/2) = 2πRH
Therefore, the answer is “Same”.

১,১০৯.

The area of a triangle is equal to the area of a square whose each side is 60 metres.The height of the triangle is 100 metres. The base of the triangle will be-

ক
ক) 85 m
খ
খ) 65 m
গ
গ) 72 m
ঘ
ঘ) 80 m

ব্যাখ্যা

(1/2) × Base × Height = 60 × 60
⇒ (1/2) × Base × 100 = 3600
⇒ Base × 50 = 3600
⇒ Base = 3600/50
Base = 72

১,১১০.

Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Dan’s house and directly east of Karen’s house. If the cafe is 2 miles closer to Dan’s house than to Karen’s house, how many miles is the cafe from Karen’s house?

ক
9
খ
6
গ
7
ঘ
8

ব্যাখ্যা

Question: Dan and Karen, who live 10 miles apart meet at a cafe that is directly north of Dan’s house and directly east of Karen’s house. If the cafe is 2 miles closer to Dan’s house than to Karen’s house, how many miles is the cafe from Karen’s house?

Solution:

From the figure it is evident that

10² = x² + (x - 2)²

⇒ 2x² - 4x + 4 = 100
⇒ 2x² - 4x - 96 = 0

⇒ x² - 2x - 48=0
⇒ x² - 8x + 6x - 48 = 0
⇒ x(x - 8) + 6(x - 8) = 0
⇒ (x - 8)(x + 6) = 0
∴ x = 8 [neglecting the (- ve) value of x]

Hence x = 8

১,১১১.

The radius of a wheel is 14 cm. How many revolutions will it make in travelling 88 kilometers?

ক
150000
খ
180000
গ
200000
ঘ
100000

ব্যাখ্যা

Question: The radius of a wheel is 14 cm. How many revolutions will it make in travelling 88 kilometers?

Solution:
আমরা জানি,
চাকার পরিধি = 2πr = 2 × (22/7) × 14 = 88 সে. মি.

∴ মোট দূরত্ব = 88 কি. মি.
= 88 × 1000 × 100
= 8800000 সে. মি.

∴ ঘূর্ণন সংখ্যা = 8800000/88 = 100000 টি

১,১১২.

If the breadth of a rectangle with area A is increased by 20% and B is the difference between the area of the two rectangles then which of the following is true?

ক
A = 5B
খ
5B = A
গ
A = B
ঘ
A = 2B

ব্যাখ্যা

Let,
The length of the rectangle be l and breadth be b.
Then the original area of the rectangle = A = lb....(i)
Now,
the length of the new rectangle is l
And, the breadth of the new rectangle = 120/100 × b
= 6b/5
Then,
The area of the new one = (6/5)lb
Therefore,
The difference between two rectangles = B
= (6/5)lb - lb
= (1/5)lb...(ii)
From (i) and (ii),
B = (1/5)A [A = lb from equation no 1]
⇒ 5B = A.
Answer: 5B = A.

১,১১৩.

A wall measures 10 meters in length, 5 meters in height, and 30 cm in thickness. Each brick used for construction measures 20 cm × 10 cm × 5 cm. How many bricks are needed to build the wall?

ক
10,000
খ
25,000
গ
15,000
ঘ
5,000

ব্যাখ্যা

Question: A wall measures 10 meters in length, 5 meters in height, and 30 cm in thickness. Each brick used for construction measures 20 cm × 10 cm × 5 cm. How many bricks are needed to build the wall?

Solution:
Length = 10 m = 1000 cm
Height = 5 m = 500 cm
Thickness = 30 cm

∴ Volume of wall = Length × Height × Thickness
= 1000 × 500 × 30
= 15,000,000 cm³Volume of one brick = 20 × 10 × 5 = 1000 cm³∴ Number of bricks = Volume of wall ÷ Volume of one brick
= 15,000,000 ÷ 1000
= 15,000

১,১১৪.

A cylindrical water tank has a radius of 40 inches and a height of 150 inches. Calculate the surface area.

ক
37771.4 sq. inches
খ
57771.4 sq. inches
গ
27771.4 sq. inches
ঘ
47771.4 sq. inches

ব্যাখ্যা

Question: A cylindrical water tank has a radius of 40 inches and a height of 150 inches. Calculate the 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 × 40(150 + 40)
= 2 × 22/7 × 40 × 190
= 44/7 × 7600
= 334400/7

∴ Area = 47771.4 sq. inches.

১,১১৫.

An observer 2m tall is 10√3 m away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:

ক
ক) 10 m
খ
খ) 7 m
গ
গ) 15 m
ঘ
ঘ) 25 m
ঙ
ঙ) 12 m

ব্যাখ্যা

SR = PQ = 2 m
PS = QR = 10√3
tan 30° = TS/PS
1/√3 = TS/10√3
TS = 10√3/√3
= 10
TR = TS + SR
= 10 + 2
= 12 m.

১,১১৬.

A room measures 7.5 m in length and 3.2 m in width. If the cost of paving is Tk. 750 per square metre, what is the total cost?

ক
Tk. 17500
খ
Tk. 18000
গ
Tk. 17800
ঘ
Tk. 18200

ব্যাখ্যা

Question: A room measures 7.5 m in length and 3.2 m in width. If the cost of paving is Tk. 750 per square metre, what is the total cost?

Solution:
Given that,
Length = 7.5 m
Width = 3.2 m
And rate = Tk. 750 per square metre

Now,
Area of the room is = Length × Width
= 7.5 × 3.2
= 24 m²

Now, Total cost = Area × Rate
= 24 × 750
= Tk. 18000

১,১১৭.

Which trigonometric ratio is undefined in value?

ক
sin 90°
খ
cos 90°
গ
sec 0°
ঘ
cosec 0°

ব্যাখ্যা

Question: Which trigonometric ratio is undefined in value?

Solution:
sin90° = 1
cos90° = 0
sec0° = 1
cosec0° = ∞(Undefined)

১,১১৮.

Square ABCD with a perimeter of 64 units. Find length of AC.

ক
16 units
খ
8√2 units
গ
16√2 units
ঘ
8 units

ব্যাখ্যা

Question: Square ABCD with a perimeter of 64 units. Find length of AC.

Solution:
Each side of the square must be 64/4 = 16 units
∴ 16² + 16² = (AC)²
⇒ AC² = 2 × 256
⇒ AC = √(2 × 256)
∴ AC = 16√2

১,১১৯.

If the length of a rectangle is increased by 10% and its breadth is decreased by 10%, the change in its area will be-

ক
ক) 1% increase
খ
খ) 1% decrease
গ
গ) 10% increase
ঘ
ঘ) No change

ব্যাখ্যা

Let length and breadth be a and b respectively
⇒ The area before change = ab
The length after change = a + 10% of a
= (a + 10a/100)
= (10a + a)/10
= 11a/10
The breadth after change = b - 10% of b
= (b - 10b/100)
= b - b/10
= (10b - b)/10
=9b/10
⇒ The area after change = (11a/10)(9b/10)
⇒ The area after change = 99ab/100

⇒ Percentage change = [{ab - (99ab/100) }/ab] × 100%
= (1/100) × 100%
= 1%
∴ The area of the new rectangle is decreased by 1%.

১,১২০.

What is the value of sinA if cosA = √5/3?

ক
2/3
খ
3/2
গ
2/√3
ঘ
2/√5

ব্যাখ্যা

Question: What is the value of sinA if cosA = √5/3?

Solution:
cosA = ভূমি/অতিভুজ = √5/3

লম্ব = √{(অতিভুজ)২ - (ভূমি)২}
= √{(3)2 - (√5)2}
= 2

sinA = 2/3

১,১২১.

The angels of a pentagon are in ratio 9 : 10 : 12 : 14 : 15. What is the sum of measure of the smallest and largest angels?

ক
ক) 54
খ
খ) 81
গ
গ) 135
ঘ
ঘ) 216

ব্যাখ্যা

আমরা জানি, পঞ্চভূজের সমষ্টি = {(5 - 2) × 180}° = 540°
পঞ্চভুজের কোণের অনুপাতের সমষ্টি = 9 + 10 + 12 + 14 + 15 = 60
∴ পঞ্চভুজের বৃহত্তম এবং ক্ষুদ্রতম কোণের সমষ্টি = [{(9+15)/60} × 540]° = 216°

১,১২২.

If each side of the square is increased by 50%, what will be the ratio between the new area and the original area of the square?

ক
5 : 4
খ
9 : 4
গ
4 : 5
ঘ
4 : 9

ব্যাখ্যা

Question: If each side of the square is increased by 50%, what will be the ratio between the new area and the original area of the square?

Solution:
Let,
The side of original square is x
∴ The area of original square is x²

The side of new square is x + 50% of x = x + x/2 = 3x/2
∴ The area of new square is (9x²)/4

∴ The ratio between the new area and the original area of the square = (9x²)/4 : x²
= 9/4 : 1
= 9 : 4

১,১২৩.

Find the value of sin⁵θ + cosec⁵θ if sinθ + cosecθ = 2.

ক
32
খ
0
গ
1
ঘ
2
ঙ
None of the above

ব্যাখ্যা

Question: Find the value of sin⁵θ + cosec⁵θ if sinθ + cosecθ = 2.

Solution:
sinθ + cosecθ = 2
or, sinθ + 1/sinθ = 2
or, sin²θ + 1 = 2sinθ
or, sin²θ - 2sinθ + 1 = 0
or, (sinθ - 1)² = 0
or, sinθ - 1 = 0
∴ sinθ = 1

cosecθ = 1/sinθ = 1/1 = 1

∴ sin⁵θ + cosec⁵θ = (1)⁵ + (1)⁵
= 2

১,১২৪.

If isosceles ΔDEF has sides of length 11.5 and 13.7, which of the following could be the perimeter of the triangle ?

ক
ক) 2.1
খ
খ) 12.0
গ
গ) 25.2
ঘ
ঘ) 36.7

ব্যাখ্যা

Question: If isosceles ΔDEF has sides of length 11.5 and 13.7, which of the following could be the perimeter of the triangle ?

Solution:

ΔDEF এ
DE = DF = 11. 5
ΔDEF এর পরিসীমা = 11.5 + 11.5 + 13.7 = 36.7

আবার,
DE = DF = 13.7 হলে
ΔDEF এর পরিসীমা = 11.5 + 13.7 + 13.7 = 38.9 যা অপশনে নেই।
অতএব সঠিক উত্তর হবে 36.7

১,১২৫.

The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. What is the radius of the pilllar?

ক
ক) 5 m
খ
খ) 7 m
গ
গ) 9 m
ঘ
ঘ) 11 m

ব্যাখ্যা

Question: The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. What is the radius of the pilllar?

Solution:
let, radius r m and height h m

The curved surface area of a cylindrical pillar is 264 m2
2πrh = 264 m²

its volume is 924 m³
πr²h = 924

πr²h /2πrh = 924/264
⇒ r/2 = 924/264
∴ r = 7 m

১,১২৬.

In a right-angled triangle, if the two angles other than the right angle differ by 4 degrees, what is the value of the smaller angle?

ক
43°
খ
33°
গ
60°
ঘ
55°

ব্যাখ্যা

Question: In a right-angled triangle, if the two angles other than the right angle differ by 4 degrees, what is the value of the smaller angle?
(একটি সমকোণী ত্রিভুজের সমকোণ ব্যতীত অপর দুইটি কোণের পার্থক্য ৪ ডিগ্রি হলে ক্ষুদ্রতম কোণটি কত ডিগ্রি হবে?)

Solution:
ধরি,
ক্ষুদ্রতম কোণ = ক
বৃহত্তম কোণ = ক + ৪°

প্রশ্নমতে,
ক + ক + ৪° = ৯০°
⇒ ২ক = ৯০° - ৪°
⇒ ক = ৮৬°/২
∴ ক = ৪৩°

১,১২৭.

A cube with side length 6 cm fits perfectly inside a hollow spherical ball. What is the total surface area of the sphere?

ক
112π cm²
খ
92π cm²
গ
120π cm²
ঘ
108π cm²

ব্যাখ্যা

Question: A cube with side length 6 cm fits perfectly inside a hollow spherical ball. What is the total surface area of the sphere?

Solution:
If a cube fits perfectly inside a sphere, then the diameter of the sphere = space diagonal of the cube

Diagonal of the cube = √3a = 6√3
So, diameter of sphere = 6√3
Radius = 6√3/2 = 3√3

Surface area of the sphere = 4πr²
= 4π(3√3)²
= 108π cm²

১,১২৮.

If a 1.5 m tall boy stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is

ক
2.5 meters
খ
4.5 meters
গ
3.5 meters
ঘ
1.5 meters

ব্যাখ্যা

Let AB be boy
Therefore, AB = 1.5m
CD be lamp
Let CD be 'h' meter.
'EB' be shadow cast by boy
Therefore, EB = 4.5meter.
BD be distance between boy and lamp
Therefore, BD = 3meters.

Therefore
In ΔAEB,
=> tanθ = AB/EB
In ΔCED,
=> tanθ = CD/ED = CD/(EB+BD)
∴ AB/EB = CD/(EB+BD)
=> 1.5/4.5 = CD/(4.5+3)
=> CD = 7.5/3 = 2.5m

১,১২৯.

In the xy-plane, a triangle has vertices (0, 0), (k, 0), and (k, - 5k), where k > 0. If the area of the region enclosed by the triangle is 40, what is the value of k?

ক
2
খ
4
গ
5
ঘ
6

ব্যাখ্যা

Question: In the xy-plane, a triangle has vertices (0, 0), (k, 0) and (k, - 5k), where k > 0. If the area of the region enclosed by the triangle is 40, what is the value of k?

Solution:

প্রদত্ত ত্রিভুজটির শীর্ষবিন্দুগুলো হলো (0, 0), (k, 0) এবং (k, - 5k)।

যেহেতু B এবং C বিন্দুর x-স্থানাঙ্ক একই (k), তাই BC রেখাটি y-অক্ষের সমান্তরাল।
যেহেতু A এবং B বিন্দুর y-স্থানাঙ্ক একই (0), তাই AB রেখাটি x-অক্ষের সমান্তরাল।
সুতরাং, ত্রিভুজটি B বিন্দুতে একটি সমকোণী ত্রিভুজ।

দুটি বিন্দুর স্থানাঙ্ক (x₁, y₁) এবং (x₂, y₂) হলে তাদের মধ্যবর্তী দূরত্ব = √{(x₂ - x₁)² + (y₂ - y₁)²}

∴ ভূমি = (0, 0) এবং (k, 0) বিন্দুর মধ্যবর্তী দূরত্ব = k (যেহেতু k > 0)।
∴ উচ্চতা = (k, 0) এবং (k, - 5k) বিন্দুর মধ্যবর্তী দূরত্ব = 5k (যেহেতু k > 0)।

ত্রিভুজের ক্ষেত্রফল = (1/2) × ভূমি × উচ্চতা

প্রশ্নমতে,
(1/2) × k × 5k = 40
⇒ (5/2)k²= 40
⇒ 5k² = 80
⇒ k² = 80/5
⇒ k² = 16
⇒ k = √16
⇒ k = ±4

যেহেতু প্রশ্নে দেওয়া আছে k > 0, তাই k এর মান হবে 4।
∴ k এর মান 4।

১,১৩০.

If the areas of a circle and a square are equal then the ratio of their perimeters is-

ক
π : 2
খ
√π : 1
গ
√π : 3
ঘ
√π : 2

ব্যাখ্যা

Question: If the areas of a circle and a square are equal then the ratio of their perimeters is-

Solution:
Let the length of each side of the square = a cm and the radius of the circle = r cm.
Given that
area of square = area of circle
⇒ a² = πr²
⇒ a = r√π

∴ Required ratio = 2πr/4a
= 2πr/4r√π
= √π/2
= √π : 2

১,১৩১.

If 0 ≤ θ ≤ 90° and 4 cos²θ - 4√3 cosθ + 3 = 0 then the value of θ is

ক
90°
খ
30°
গ
45°
ঘ
60°

ব্যাখ্যা

Question: If 0 ≤ θ ≤ 90° and 4 cos²θ - 4√3 cosθ + 3 = 0 then the value of θ is

Solution:
4 cos²θ – 4√3 cosθ + 3 = 0
⇒ (2cosθ)² – 2 · 2 cosθ · √3 + (√3)² = 0
⇒ (2cosθ – √3)² = 0
⇒ 2 cosθ – √3 = 0
⇒ 2 cosθ = √3
⇒ cosθ = √3/2
⇒ cosθ = cos 30°
∴ θ = 30°

১,১৩২.

An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:

ক
51.6 m
খ
41.6 m
গ
31.6 m
ঘ
21.6 m

ব্যাখ্যা

Question: An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The height of the tower is:

Solution:

Let AB be the observer and CD tower
Draw BE perpendicular to CD

Then CE = AB = 1.6 m
And BE = AC = 20√3 m

Then right angle triangle DEB
∴ tan30° = DE/BE
⇒ 1/√3 = DE/20√3
⇒ DE = 20√3m

Then CD = CE + DE = 1.6 + 20 = 21.6 m

১,১৩৩.

A parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Its area is-

ক
500√15 m²
খ
600√15 m²
গ
400√15 m²
ঘ
450√15 m²

ব্যাখ্যা

Question: A parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Its area is-

Solution:

১,১৩৪.

If the areas of a right triangle is 6 square metres and the hypotenuse is 5 meters. What is the perimeter of the triangle?

ক
ক) 10 meters
খ
খ) 12 meters
গ
গ) 15 meters
ঘ
ঘ) 15 meters
ঙ
ঙ) None

ব্যাখ্যা

Let the base of the triangle is x and height of the triangle is y
Area of right triangle = 1/2 × xy = 6
∴ xy = 12

According to the Pythagorean Law,
x² + y² = 5²
(x + y)² - 2xy = 25
(x + y)² = 25 + 24 = 49
(x + y) = 7

So, the perimeter = 7 + 5 = 12 meters

১,১৩৫.

A rectangular fish tank has internal dimensions 80 cm × 60 cm × 15 cm can be filled with tap A at the rate of 680 cm³ every 2 minutes, and can be emptied with tap B at a rate of 140 cm³ per minute. If Jamal leaves both taps are opened at 2:30 pm, when will the tank be filled to 1/3^rd its capacity?

ক
3.45 pm
খ
4.15 pm
গ
4.30 pm
ঘ
5.15 pm
ঙ
5.30 pm

ব্যাখ্যা

Question: A rectangular fish tank has internal dimensions 80 cm × 60 cm × 15 cm can be filled with tap A at the rate of 680 cm³ every 2 minutes, and can be emptied with tap B at a rate of 140 cm³ per minute. If Jamal leaves both taps are opened at 2 : 30 pm, when will the tank be filled to 1/3^rd its capacity?

Solution:
ট্যাংকের মোট আয়তন = 80 × 60 × 15 = 72000 cm³

তাহলে,
1/3 অংশ = 72000 ÷ 3 = 24000 cm³

এখন,
ট্যাপ A, প্রতি মিনিটে পানি ভর্তি করে = 680 ÷ 2 = 340 cm³

এবং,
ট্যাপ B, প্রতি মিনিটে পানি ফেলে দেয় = 140 cm³

∴ পানি জমার হার প্রতি মিনিটে = 340 - 140 = 200 cm³
∴ 24000 cm³ পানি জমতে কত সময় লাগবে = 24000/200 = 120 মিনিট = 2 ঘণ্টা [1 ঘণ্টা = 60মিনিট]

∴ সময় = ২ : ৩০ PM + 2 ঘণ্টা = 4 : 30 PM

ট্যাংকের 1/3 অংশ পূর্ণ হবে 4 : 30 PM-এ।

১,১৩৬.

A tiles of length 0.5 metre cost Tk.25. The total cost to overlay a square field of length 10 metres with tiles is -

ক
ক) 1000Tk.
খ
খ) 100000Tk.
গ
গ) 10000Tk.
ঘ
ঘ) 1000000Tk.

ব্যাখ্যা

Question: A tiles of length 0.5 metre cost Tk.25. The total cost to overlay a square field of length 10 metres with tiles is -

Solution:
The area of the field is = 102 = 100m2

The area of one tiles = 0.52 = 0.25m2

so, the total cost = (100/0.25) × 25 = 10000Tk.

১,১৩৭.

The sides of a triangle are consecutive integers. The perimeter of the triangle is 141 cm. Find the length of the greatest side:

ক
44 cm
খ
52 cm
গ
36 cm
ঘ
48 cm

ব্যাখ্যা

Question: The sides of a triangle are consecutive integers. The perimeter of the triangle is 141 cm. Find the length of the greatest side:

Solution:
Let,
the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively.

Then,
x + (x + 1) + (x + 2) = 141
⇒ 3x + 3 = 141
⇒ 3x = 138
∴ x = 46

∴ Length of the greatest side = (46 + 2) cm
= 48 cm

১,১৩৮.

A square park is surrounded by a path of uniform width 2 meters all around it. The area of the path is 288 sq. meters. Find the perimeter of the park.

ক
34 m
খ
1156 m
গ
136 m
ঘ
Cannot be determined

ব্যাখ্যা

Question: A square park is surrounded by a path of uniform width 2 meters all around it. The area of the path is 288 sq. meters. Find the perimeter of the park.

Solution:
Let, one side of the park is x meter.
So, one side of the park with path = x + (2 + 2)
= x + 4

We know,
Area of the park = x²
Area of the path, (x + 4)² - x² = 288
⇒ x² + 8x + 16 - x² = 288
⇒ 8x + 16 = 288
⇒ 8x = 288 - 16
⇒ 8x = 272
⇒ x = 272/8
∴ x = 34

One side of the square = 34 m.
So, perimeter of the square =4 × 34
= 136 m

১,১৩৯.

The wheel of a scooter has a diameter of 140 cm. How many revolutions per minute must the wheel make to maintain a speed of 132 km/h?

ক
250
খ
500
গ
1000
ঘ
850

ব্যাখ্যা

Question: The wheel of a scooter has a diameter of 140 cm. How many revolutions per minute must the wheel make to maintain a speed of 132 km/h?

Solution:
Distance travelled by wheel in one revolution = circumference of wheel
= (22/7) × 140 = 440 cm.

And
Speed of scooter = 132 km/hr = (132 × 1000 × 100)/60 cm/min = 220000 cm/min.

∴ Revolutions per minute = Distance covered per minute/Distance per revolution
= 220000/440 = 500

So the answer is indeed 500 revolutions per minute.

১,১৪০.

The altitude of an equilateral triangle of side 2√3 m is-

ক
1 m
খ
3 m
গ
5 m
ঘ
7 m

ব্যাখ্যা

Question: The altitude of an equilateral triangle of side 2√3 m is-

Solution:

Altitude AD = 2√3 × sin60
= 2√3 ×√3/2
= 3 m²

১,১৪১.

A ladder 25 m long leans against a wall. The foot of the ladder is 7 m from the wall. How high up the wall does the ladder reach?

ক
24 m
খ
32 m
গ
22 m
ঘ
26 m

ব্যাখ্যা

Question: A ladder 25 m long leans against a wall. The foot of the ladder is 7 m from the wall. How high up the wall does the ladder reach?

Solution:
Given that,
The ladder is the hypotenuse = 25 m
The distance from the foot of the ladder to the wall is one leg = 7 m

Applying the Pythagorean theorem,
Hypotenuse² = a² + h²
⇒ 25² = 7² + h²
⇒ h² = 625 - 49
⇒ h²= 576
⇒ h = √576
∴ h = 24

So the ladder reaches 24 m high up the wall.

১,১৪২.

The area of a circle whose radius is the diagonal of a square whose area is 16 square cm is -

ক
ক) 64π square cm
খ
খ) 128π square cm
গ
গ) 32π square cm
ঘ
ঘ) 256π square cm

ব্যাখ্যা

Question: The area of a circle whose radius is the diagonal of a square whose area is 16 square cm is -

Solution:
Area of square = 16
Side of square = √16 = 4

Diagonal of square = 4√2

So, the radius of the circle is 4√2 cm

Area of circle = πr²
= π(4√2)²
= 32π

১,১৪৩.

If the perimeter of a square becomes twice, its area becomes :

ক
2 times
খ
3 times
গ
4 times
ঘ
8 times

ব্যাখ্যা

Question: If the perimeter of a square becomes twice, its area becomes :

Solution:
Let, the perimeter of the Square is X
∴ the length of a single side is X/4
area = X²/16

new perimeter = 2X
new length = 2X/4
= X/2
area = X²/4

area becomes = ( X²/4 )/( X²/16 )
= 4 times

১,১৪৪.

When folded into two equal halves a rectangular sheet had a perimeter of 48cm for each part folded along one set of sides and the same is 66cm when folded along the other set of sides. Find the area of the sheet.

ক
1584 cm²
খ
1120 cm²
গ
792 cm²
ঘ
1320 cm²

ব্যাখ্যা

Let the sheet be folded along its breadth and its perimeter = 48cm

Therefore, (l/2 + b) = 48 ......(i)
Now, let the sheet be folded along its length, and the perimeter = 66cm
(l + b/2)= 66 ........(ii)
Solving (i) and (ii), we get,
l = 56 cm, b = 20 cm
Area = l × b = 56 cm × 20 cm
⇒ Area = 1120 cm²

১,১৪৫.

What is the value of θ when Cosθ.Cosec30° = 1 ?

ক
ক) 60°
খ
খ) 75°
গ
গ) 70°
ঘ
ঘ) 58.5°

ব্যাখ্যা

Question: Cosθ.Cosec30° = 1, the value of θ is?

Solution:
if, CosA.CosecB = 1
then, A + B = 90°

ATQ,
θ + 30°= 90°
θ = 60°

১,১৪৬.

∠M and ∠N are complementary to each other. If ∠M = 20° + 4x and ∠N = 6x, find the value of ∠N.

ক
75°
খ
42°
গ
38°
ঘ
52°

ব্যাখ্যা

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

Solution:
Here,
∠M = 20° + 4x and ∠N = 6x

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

So, ∠N = 6 × 7° = 42°

১,১৪৭.

Find the value of sin⁡60° × cos⁡60° × tan⁡60° = ?

ক
√3/2
খ
1/2
গ
3/4
ঘ
1

ব্যাখ্যা

Question: Find the value of sin⁡60° × cos⁡60° × tan⁡60° = ?

Solution:
We know,
sin60° = √3/2, cos60° = 1/2 and tan60° = √3

Given that,
sin⁡60° × cos⁡60° × tan⁡60°
= (√3/2) × (1/2) × √3
= 3/4

১,১৪৮.

If sinθ = 3/5 then tanθ =?

ক
3/4
খ
4/3
গ
5/3
ঘ
5/4

ব্যাখ্যা

Question: If sinθ = 3/5 then tanθ =?

Solution:
sinθ = 3/5

We know,
cosθ = √(1 - sin²θ)
= √(1 - 9/25)
=√(16/25)
= 4/5

∴ tanθ = sinθ/cosθ
= (3/5)/(4/5)
= (3/5) × (5/4)
= 3/4

১,১৪৯.

The length of two chords AB and AC of a circle are 8 cm and 6 cm and ∠BAC = 90º, then the radius of circle is-

ক
5 cm
খ
12 cm
গ
6 cm
ঘ
10 cm

ব্যাখ্যা

Question: The length of two chords AB and AC of a circle are 8 cm and 6 cm and ∠BAC = 90º, then the radius of circle is-

Solution:
Given that,
Chord AB = 8 cm
Chord AC = 6 cm
∠BAC = 90°
Since ∠BAC = 90°, triangle ABC is right-angled at A. The hypotenuse BC is the diameter of the circle.

We know,
BC = √(AB² + AC²)
⇒ BC = √(8² + 6²) = √(64 + 36) = √100
∴ BC = 10 cm

Radius = half of diameter
∴ r = BC/2 = 10/2 = 5 cm

১,১৫০.

When the shadow of a pole h metres high is √3h metres long, the angle of elevation of the Sun is-

ক
20°
খ
30°
গ
60°
ঘ
45°
ঙ
15°

ব্যাখ্যা

Let AB be the pole and BC be its shadow.
Consider θ is the angle of elevation of the Sun.

In right triangle ABC,

tan θ = AB/BC = h/√3h = 1/√3
tan θ = tan 30°
θ = 30°

১,১৫১.

The area of a trapezium is 1500 sq. m. If the parallel sides are in the ratio of 12: 18 and the distance between the parallel sides is 32 meters, what is the length of the smaller parallel side?

ক
3.3 m
খ
4.5 m
গ
5.3 m
ঘ
6.5 m

ব্যাখ্যা

The trapezium is a quadrilateral with one pair of parallel sides.
Let the length of the smaller side = 12x
Length of the bigger parallel side = 18x

Distance between parallel sides = 32 meters

Area of a trapezium = 1/2 × (sum of parallel sides) × distance between them.

So, as per the question
1500 = 1/2 × (12x +18x) × 30
⇒ 1500 = 1/2 × 30x × 30
⇒ 1500 = 30x × 15
⇒ 30x = 1500 /15
⇒ 30x = 100
⇒ x = 100 /30
= 3.3 meters.

১,১৫২.

A circular well with a diameter of 14 meters, is dug to a depth of 3 meters. What is the volume of the earth dug out?

ক
ক) 415 m³
খ
খ) 450 m³
গ
গ) 462 m³
ঘ
ঘ) 473 m³

ব্যাখ্যা

Question: A circular well with a diameter of 14 meters, is dug to a depth of 3 meters. What is the volume of the earth dug out?

Solution:
Diameter of the well is 14 meters
Radius of the well is 14/2 = 7 meters

Volume = πr²h
= (22/7) × 7 × 7 × 3
= 462 m³

১,১৫৩.

What is the distance between the points A(- 1, 3) and B(5, - 5)?

ক
8 units
খ
10 units
গ
12 units
ঘ
15 units
ঙ
9 units

ব্যাখ্যা

Question: What is the distance between the points A(- 1, 3) and B(5, - 5)?

Solution:

১,১৫৪.

If the ratio between the areas of two circles is 4 : 1 then the ratio between their radii will be-

ক
ক) 1 : 4
খ
খ) 1 : 1
গ
গ) 2 : 1
ঘ
ঘ) 2 : 2

ব্যাখ্যা

⇒ πr²₁/πr²₂= 4/1
⇒ r²₁/r²₂ = 4/1
⇒ (r₁/r₂)² = (2/1)²
⇒r₁/r₂ = 2/1
r₁ : r₂= 2 : 1

১,১৫৫.

The area of a circle is 36π cm². The circumference is equal to

ক
6π cm
খ
18π cm
গ
12√π cm
ঘ
12π cm

ব্যাখ্যা

Question: The area of a circle is 36π cm². The circumference is equal to

Solution:
Given that,
Area of a circle is = 36π cm²

We know that,
Area = πr²

ATQ,
⇒ πr² = 36π
⇒ r² = 36
∴ r = 6

∴ Circumference = 2πr = 2π × 6 = 12π cm

১,১৫৬.

In the triangle ABC if AB > AC then which of the following is true?

ক
ক) ∠ABC > ∠ACB
খ
খ) ∠ABC<∠BAC
গ
গ) ∠ACB > ∠BAC
ঘ
ঘ) ∠ACB > ∠ABC

ব্যাখ্যা

আমরা জানি,
ত্রিভুজের বৃহত্তম বাহুর বিপরীত কোণ ক্ষুদ্রতম বাহুর বিপরীত কোণ অপেক্ষা বৃহত্তর হবে।
তাই, AB > AC হলে অবশ্যই ∠ACB > ∠ABC

১,১৫৭.

Find the cost of a cylinder of radius 7 m and hight 3.5 m when the cost of its metal is Tk. 50 per cubicmettre?

ক
Tk. 16,950
খ
Tk. 17,250
গ
Tk. 21,580
ঘ
Tk. 26,950

ব্যাখ্যা

Question: Find the cost of a cylinder of radius 7 m and hight 3.5 m when the cost of its metal is Tk. 50 per cubicmettre?

Solution:
Given that,
Radius of the cylinder, r = 7 m
Height of the cylinder, h = 3.5 m
Cost per cubic meter = Tk. 50

We know,
The volume of a cylinder is,
V = πr²h = (22/7) × (7)² × 3.5 = 22 × 7 × 3.5 = 539 cubic meters

∴ Total Cost = Volume × Cost per cubic meter = 539 × 50 = 26,950

∴ The cost of the cylinder is Tk. 26,950

১,১৫৮.

If sinx = 1/2, then sin2x = ?

ক
1/2
খ
√3/2
গ
1
ঘ
0

ব্যাখ্যা

Question: If sinx = 1/2, then sin2x = ?

Solution:
দেওয়া আছে,
sin x = 1/2
বা, sin x = sin 30° [ Since sin 30° = 1/2 ]
বা, x = 30°

সুতরাং,
sin 2x = sin (2 × 30°) = sin 60° = √3/2

১,১৫৯.

What is the percentage increase in the area of a rectangle if each of its sides is increased by 20%?

ক
ক) 40%
খ
খ) 42%
গ
গ) 44%
ঘ
ঘ) 22%

ব্যাখ্যা

Let original length = 10
original breadth = 10
Then, original area
= 10 × 10
= 100
Length is increased by 20%
⇒ New length = 10 + 2 (2 is 20% of 10)
= 12
Breadth is increased by 20%
⇒ New breadth = 10 + 2 (2 is 20% of 10)
= 12
New area = 12 × 12
= 144
Increase in are =
= new area - original area
= 144 - 100
= 44
Percentage increase in area
= (increase in area/original area) × 100
= {(44/100) × 100}%
= 44%

১,১৬০.

যদি sec(x − 30°) = 2 হয় , তাহলে cot x = ?

ক
1/2
খ
Undefined
গ
0
ঘ
2/3

ব্যাখ্যা

sec (x − 30°) = 2
⇒ sec (x - 30°) = sec 60°
⇒ x - 30° = 60°
⇒ x = 90°
∴ cot 90° = 0

১,১৬১.

If C is the midpoint of the points A(2, 3) and B(8, 11), find the length of AC.

ক
5
খ
7.5
গ
6
ঘ
10.5

ব্যাখ্যা

Question: If C is the midpoint of the points A(2, 3) and B(8, 11), find the length of AC.

Solution:
দেওয়া আছে, A(2, 3) এবং B(8, 11), এবং C হলো AB-এর মধ্যবিন্দু।

দূরত্বের সূত্র ব্যবহার করে AB-এর দৈর্ঘ্য নির্ণয় করি।
AB = √{(x₂ - x₁)² + (y₂ - y₁)²}
AB = √{(8 - 2)² + (11 - 3)²}
AB = √(6² + 8²)
AB = √(36 + 64)
AB = √100
AB = 10

যেহেতু C হলো AB-এর মধ্যবিন্দু, তাই AC হবে AB-এর অর্ধেক।
∴ AC = AB/2
= 10/2
= 5

১,১৬২.

The base of a right prism is a trapezium whose lengths of two parallels sides are 10 cm and 6 cm and distance between them is 5 cm. If the height of the prism is 8 cm, its volume is -

ক
240 cm³
খ
320 cm³
গ
350 cm³
ঘ
380 cm³

ব্যাখ্যা

Question: The base of a right prism is a trapezium whose lengths of two parallels sides are 10 cm and 6 cm and distance between them is 5 cm. If the height of the prism is 8 cm, its volume is -

Solution:
Length of the parallel sides of the prism = 10 cm and 6 cm
Height of prism = 8 cm
∴ Volume of prism = (1/2){(10 + 6) × 5 × 8}
= (1/2) × 16 × 5 × 8
= 320 cm³

১,১৬৩.

In the figure below, the value of y is -

ক
12
খ
24
গ
42
ঘ
36

ব্যাখ্যা

Question: In the figure below, the value of y is -

Solution:
Here,
(2x)° = (y + 30)° [opposite angle]

And,
(2x)° + (3x)° = 180° [The sum of angles of a linear pair is always equal to 180°.]
⇒ (5x)° = 180°
∴ x = 36°

Now,
(y + 30)° = (2x)°
⇒ (y + 30)° = (2 × 36)°
⇒ y = 72° - 30°
∴ y = 42°

১,১৬৪.

The area of a square inscribed in a circle is 196 cm². What is the area of the circle?

ক
308 cm²
খ
268 cm²
গ
354 cm²
ঘ
412 cm²
ঙ
None

ব্যাখ্যা

Question: The area of a square inscribed in a circle is 196 cm². What is the area of the circle?

Solution:
Side of square = √196 cm = 14 cm
Diagonal of the square = 14 × √2 cm = 14√2 cm

diameter of circle = 14√2 cm
Radius of circle = 14√2/2= 7√2 cm

∴ Area = πr²
= (22/7) × (7√2)²
= (22/7) × 98
= 308 cm²

১,১৬৫.

If tan (x − 30°) = 1/√3, cosx = ?

ক
ক) 1/2
খ
খ) 0
গ
গ) 1/√2
ঘ
ঘ) 1

ব্যাখ্যা

tan (x - 30°) = 1/√3 = tan 30°
Or, x - 30° = 30°
Or, x = 60°

∴ cos 60° = 1/2

১,১৬৬.

Find the cost of a cylinder of radius 14 m and height 3.5 m when the cost of its metal is Tk. 50 per cubic meter-

ক
Tk. 107800
খ
Tk. 10800
গ
Tk. 109800
ঘ
Tk. 108700

ব্যাখ্যা

Question: Find the cost of a cylinder of radius 14 m and height 3.5 m when the cost of its metal is Tk. 50 per cubic meter-

Solution:
We know,
The volume of the cylinder = πr²h
= (22/7) × 14 × 14 × 3.5
= 2156 m³

Cost of the cylinder = 2156 × 50
= Tk. 107800

১,১৬৭.

A cube of dimension a fits perfectly in a hollow spherical ball. What is the total surface area of that ball?

ক
4πa²
খ
3πa²
গ
5πa²
ঘ
6πa²

ব্যাখ্যা

Question: A cube of dimension a fits perfectly in a hollow spherical ball. What is the total surface area of that ball?

Solution:
here,
the lenght of the cube is = a
∴ Diagonal of the cube is = √3a

as the diagonal of the cube is equal to the diameter of the ball,
2r = √3a
r = √3a/2

∴ the total surface area of the ball is = 4πr²
= 4π(√3a/2)²
= 3πa²

১,১৬৮.

Calculate sin(- 585°).

ক
1/3
খ
√2/3
গ
1/2
ঘ
1/√2

ব্যাখ্যা

Question: Calculate sin(- 585°).

Solution:
sin(- 585°)
= - sin(585°)
= - sin(360° + 225°)
= - sin(2π + 225°)
= - sin225°
= - sin(180° + 45°)
= - sin(π + 45°)
= sin45°
= 1/√2

১,১৬৯.

The angle of elevation of the top of a tower at a point on the ground 37 m away from the foot of the tower is 45°. What is the height of the tower?

ক
42 m
খ
36 m
গ
38 m
ঘ
37 m

ব্যাখ্যা

Question: The angle of elevation of the top of a tower at a point on the ground 37 m away from the foot of the tower is 45°. What is the height of the tower?

Solution:

Let AB be tower and C is a point on the ground 37 m away
From the foot of tower B
The angle of elevation is 45°

Let h be the height of the tower.
∴ tanθ = AB/BC
⇒ tan45^∘= AB/37
⇒ 1 = AB/37
∴ AB = 37 m

১,১৭০.

A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl?

ক
40
খ
48
গ
54
ঘ
60

ব্যাখ্যা

Question: A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl?

Solution:
Volume of hemisphere = (1/2) (4/3) πr3
= (1/2) (4/3) π9³
= (2/3) 729π

Volume of cylinder = π(3/2)² 4
= 9π

bottles will be needed to empty the bowl = (2 × 729π)/(3 × 9π)
= 54

১,১৭১.

A rectangle and a square have the same area. The square has a perimeter of 32 meters and the length of the rectangle is 4 meters. What is the width of the rectangle (in meters)?

ক
24
খ
16
গ
12
ঘ
8

ব্যাখ্যা

Question: A rectangle and a square have the same area. The square has a perimeter of 32 meters and the length of the rectangle is 4 meters. What is the width of the rectangle (in meters)?

Solution:
Given that,
Perimeter of square = 32 m
Area of rectangle = Area of square
And length of rectangle = 4 m

Now,
Perimeter of square,
4s = 32
⇒ s = 32/4 = 8
∴ s = 8 m
∴ Area of square = s² = 8² = 64 m²

According to the Question,
Area of rectangle = Area of square
∴ Area of rectangle = 64 m²

∴ Area of rectangle = length × width
64 = 4 × w
⇒ w = 64/4
∴ w = 16 m

So the width of the rectangle = 16 meters

১,১৭২.

A rectangular plot of land has a fence along three of its four sides, the unfenced side and the side opposite the unfenced side have a length that is three times the length of the other two sides. If the area of the plot is 675 square feet, what is the total length of the fence in feet?

ক
50 feet
খ
62 feet
গ
70 feet
ঘ
75 feet
ঙ
None

ব্যাখ্যা

Question: A rectangular plot of land has a fence along three of its four sides, the unfenced side and the side opposite the unfenced side have a length that is three times the length of the other two sides. If the area of the plot is 675 square feet, what is the total length of the fence in feet?

Solution:
Let,
Width of the rectangular land = x
∴ Length of the rectangular land = 3x

ATQ,
3x × x = 675
⇒ 3x² = 675
⇒ x² = 225
∴ x = 15

Total length of the fence = (x + 3x + x) feet
= 5x feet
= 5 × 15 feet
= 75 feet

১,১৭৩.

The volume of a sphere is 36π cm³ . What is the radius of the sphere?

ক
ক) 6 cm
খ
খ) 5 cm
গ
গ) 3 cm
ঘ
ঘ) 2 cm

ব্যাখ্যা

Volume of sphere = 36π cm³
Volume of sphere = (4/3) × πr³
⇒ (4/3) × πr³ = 36π
⇒ 36π = (4/3) × π × r³
⇒ 9 = (1/3) × r³
⇒ 27 = r³
⇒ r = 3

∴ The radius of sphere is 3 cm.

১,১৭৪.

A solid cylindrical block has a radius of 7 meters and a height of 10 meters. If the material of the cylinder costs Tk. 20 per cubic meter, find the total cost of the material required.

ক
Tk. 32,800
খ
Tk. 20,800
গ
Tk. 30,800
ঘ
Tk. 30,200
ঙ
None

ব্যাখ্যা

Question: A solid cylindrical block has a radius of 7 meters and a height of 10 meters. If the material of the cylinder costs Tk. 20 per cubic meter, find the total cost of the material required.

Solution:
Given,
Radius of the cylinder, r = 7 m
Height of the cylinder, h = 10 m
Cost per cubic meter = Tk. 20

The volume of the cylinder:
V = πr²h
= (22/7) × (7)² × 10
= (22/7) × 49 × 10
= 22 × 7 × 10
= 1540 cubic metres

Total cost = Volume × Cost per cubic metre
= 1540 × 20
= Tk. 30,800

∴ The total cost of the material is Tk. 30,800.

১,১৭৫.

What is the area of the hexagonal region shown in the figure?

ক
216
খ
108√3
গ
54√3
ঘ
216√3

ব্যাখ্যা

Question: What is the area of the hexagonal region shown in the figure?

Solution:
দেওয়া আছে,
সুষম ষড়ভুজের বাহুর দৈর্ঘ্য, a = 6 একক
বাহুর সংখ্যা, n = 6

আমরা জানি,
সুষম বহুভুজের ক্ষেত্রফল = {(n × a²)/4} × cot(180°/n) বর্গ একক
∴ সুষম ষড়ভুজের ক্ষেত্রফল = {(6 × 6²)/4} × cot(180°/6) বর্গ একক
= 54 × cot30° বর্গ একক
= 54√3 বর্গ একক

∴ নির্ণেয় ক্ষেত্রফল 54√3 বর্গ একক

১,১৭৬.

Find the value of sin(5π/6).

ক
- 1/2
খ
√3/2
গ
1/2
ঘ
- 1/√2

ব্যাখ্যা

Question: Find the value of sin(5π/6).

Solution:
sin(5π/6)
= sin(π - π/6) [যেহেতু (π - θ) দ্বিতীয় চতুর্ভাগে পড়ে এবং দ্বিতীয় চতুর্ভাগে sin ধনাত্মক, তাই sin(π - θ) = sin θ]
= sin(π/6)
= sin(30°)
= 1/2

১,১৭৭.

In triangle ABC, sides AB = AC, and angle ∠C = 35∘. Find the measure of angle ∠A.

ক
110°
খ
90°
গ
70°
ঘ
120°

ব্যাখ্যা

Question: In triangle ABC, sides AB = AC, and angle ∠C = 35^∘. Find the measure of angle ∠A.

Solution:
Since AB = AC, triangle ABC is isosceles with the equal sides being AB and AC. In an isosceles triangle, the angles opposite the equal sides are equal. Therefore ∠B =∠C
Given that ∠C = 35^∘, it follows that, ∠B = 35^∘
The sum of the interior angles in any triangle is 180^∘. Therefore,
⇒ ∠A + ∠B + ∠C = 180^∘
⇒ ∠A + 35^∘ + 35^∘ = 180^∘
⇒ ∠A = (180 - 70)°
∴ ∠A = 110°

১,১৭৮.

Between a square of perimeter 44 cm and a circle of circumference 44 cm, which figure has a larger area and by how much?

ক
ক) Both are equal
খ
খ) circle, 44 cm²
গ
গ) square, 33 cm²
ঘ
ঘ) circle, 33 cm²

ব্যাখ্যা

Perimeter of Square = 44 cm
⇒ Side of square is 11 cm
Area of Square = 11²= 121 cm²

Circumference of Circle = 44 cm
⇒ 2πr = 44
⇒ 2r(22/7) = 44
⇒ r = 7
⇒ Radius of circle is 7 cm
Area of circle =πr²=(22/7) × 7²= 154 cm²

Circle has larger area by 33 cm² than the area of square.

১,১৭৯.

∆ABC is right angled at B. If cosA = 8/17, then what is the value of cotC?

ক
15/8
খ
15/17
গ
8/17
ঘ
17/15

ব্যাখ্যা

Question: ∆ABC is right angled at B. If cosA = 8/17, then what is the value of cotC?

Solution:

cosA = 8/17 = AB/AC
∴ BC = √(AC² - AB²) = √(17² - 8²) = √(289 - 64)
= √225
= 15

cotC = BC/AB = 15/8

১,১৮০.

The difference between the length and the breadth of a table is 8 cm. If the breadth is decreased by 4 cm and the length is increased by 7 cm, the area remains the same. Find the area of the table?

ক
ক) 240 sq.cm
খ
খ) 540 sq.cm
গ
গ) 560 sq.cm
ঘ
ঘ) 660 sq.cm

ব্যাখ্যা

Question: The difference between the length and the breadth of a table is 8 cm. If the breadth is decreased by 4 cm and the length is increased by 7 cm, the area remains the same. Find the area of the table?

Solution:
টেবিলের প্রস্থ = x সে.মি.
টেবিলের দৈর্ঘ্য = x + 8 সে.মি.
টেবিলের ক্ষেত্রফল = x(x + 8) বর্গ সে.মি.
= x² + 8x বর্গ সে.মি.

প্রশ্নমতে
(x - 4)(x + 8 + 7) = x² + 8x
বা, (x - 4)(x + 15) = x² + 8x
বা, x² + 15x - 4x - 60 = x² + 8x
বা, x² - x² + 11x - 8x = 60
বা, 3x = 60
∴ x = 20

টেবিলের প্রস্থ = 20 সে.মি.
টেবিলের দৈর্ঘ্য = 20 + 8 = 28 সে.মি.

টেবিলের ক্ষেত্রফল = 20 × 28 = 560 বর্গসে.মি.

১,১৮১.

The length of a rectangular floor is twice its breadth. If the total cost to concrete the floor is Tk 256 at the rate of Tk 2 per sq. meter, what is the length of the floor?

ক
6 meter
খ
8 meter
গ
12 meter
ঘ
14 meter
ঙ
16 meter

ব্যাখ্যা

Question: The length of a rectangular floor is twice its breadth. If the total cost to concrete the floor is Tk 256 at the rate of Tk 2 per sq. meter, what is the length of the floor?

Solution:
Area of the floor = 256/2 = 128 sq. meter
Let breadth = x meter and length = 2x meter

∴ x × 2x = 128
⇒ 2x² = 128
⇒ x² = 64
⇒ x = 8

∴ Length of the floor = (2 × 8) meter
= 16 meter

১,১৮২.

The reflex angle between the hands of a clock at 10.25 is:

ক
395°/2
খ
397°/2
গ
399°/2
ঘ
393°/2

ব্যাখ্যা

Required angle
= । (11 M - 60 H) / 2 ।°
= । (11 × 25 - 60 × 10) / 2।°
= । (275 - 600) / 2।°
= । -325/2।°
= । 360 -325/2।°
= 395°/2

Angle traced minutes hand in 25 minutes = {(60 × 25)/360}°
=150°

The hour hand traces half a degree in one minute. Hence, angle traced by hour hand at 10 hour and 25 minutes = {(10 × 30) + 25/2}°
=312.5°
Thus, angle between two hands = 312.5 − 150 = 162.5°
Hence, the reflex angle is = (360−162.5)° = 197.5°

১,১৮৩.

A driving wheel makes 250 revolutions per minute to keep a speed of 66 km per hour. What is the diameter of the driving wheel?

ক
ক) 155 cm
খ
খ) 150 cm
গ
গ) 145 cm
ঘ
ঘ) 140 cm

ব্যাখ্যা

Question: A driving wheel makes 250 revolutions per minute to keep a speed of 66 km per hour. What is the diameter of the driving wheel?

Solution:
Total distance of one hour = 66 km = 66 × 1000 m = 66000 × 100 cm = 6600000 cm.
∴ Distance of 1 minute = 6600000/60 cm
= 110000 cm

∴ Distance for 1 revolution = 110000/250 cm
= 440 cm

Let,
The radius of the wheel = r cm
∴ The diameter of the wheel, d = 2r cm

Now,
2πr = 440
⇒ 2r = 440/π
⇒ d = 440 × (7/22)
∴ d = 140

∴ The length of the diameter of the driving wheel is 140 cm

১,১৮৪.

125 small sphere balls are formed from a big ball with radius 25cm. What will be the surface area of a small ball?

ক
334.16 cm²
খ
324.16 cm²
গ
318.16 cm²
ঘ
314.16 cm²

ব্যাখ্যা

Question: 125 small sphere balls are formed from a big ball with radius 25cm. What will be the surface area of a small ball?

Solution:
converting a big sphere ball to 125 small balls,
the volume will be the same for both case.
let,
small ball radius = r
given, big ball radius, R = 25cm

∴ (4/3)πR³ = 125 × (4/3)πr³
⇒ R³ = 125 × r³
⇒ R = 5 × r
⇒ r = 25/5
r = 5cm

∴ total surface area of a small ball is
a = 4πr²
= 4 × 3.1416 × (5)²
= 314.16 cm²

১,১৮৫.

The ratio between the perimeter and the length of a rectangle is 5 : 2. If the area of the rectangle is 484 sq. cm, what is the length of the rectangle?

ক
22 cm
খ
33 cm
গ
44 cm
ঘ
55 cm
ঙ
66 cm

ব্যাখ্যা

Question: The ratio between the perimeter and the length of a rectangle is 5 : 2. If the area of the rectangle is 484 sq. cm, what is the length of the rectangle?

Solution:
Let Length = l
& Breadth = b
Perimeter of a rectangle = 2(l + b)

Now,
2(l + b)/l = 5/2
⇒ 4(l + b) = 5l
⇒ l = 4b

Area, l × b = 484
⇒ 4b × b = 484
⇒ b² = 121 = 11²
⇒ b = 11

∴ l = 4 × 11 = 44

So the length of the rectangle is 44 cm.

১,১৮৬.

If tan(θ - 45°) = 1, then what is the value of sinθ?

ক
1/2
খ
√2/2
গ
√3/2
ঘ
1
ঙ
None

ব্যাখ্যা

Question: If tan(θ - 45°) = 1, then what is the value of sinθ?

Solution:
Given that,
tan(θ - 45°) = 1

We know,
tan45° = 1

So,
tan(θ - 45°) = tan45°
⇒ (θ - 45°) = 45°
⇒ θ = 90°

Now,
∴ sinθ = sin90° = 1

১,১৮৭.

If 3 sides of a triangle are 6 cm, 8 cm, and 10 cm, then the altitude of the triangle, using the largest side as its base, will be -

ক
4.8 cm
খ
4.4 cm
গ
6 cm
ঘ
8 cm

ব্যাখ্যা

Question: If 3 sides of a triangle are 6 cm, 8 cm, and 10 cm, then the altitude of the triangle, using the largest side as its base, will be -

Solution:
Semi perimeter of the triangle is, S = (6 + 8 + 10)/2
= 12 cm

Area of the triangle is = √{s(s - a)(s - b)(s - c)}
= √{12(12 - 6) (12 - 8) (12 - 10)
= √(12 × 6 × 4 × 2)
= √576
= 24 sq. cm

Area of the triangle = (1/2) × base × height
⇒ 24 = (1/2) × b × h
⇒ b × h = 48
⇒ h = 48/b
⇒ h = 48/10
∴ h = 4.8 cm

১,১৮৮.

The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is-

ক
30 m
খ
15√2 m
গ
30√2 m
ঘ
60 m

ব্যাখ্যা

Question: The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is-

Solution:
Length of room = 30 m
Breadth of room = 24 m
Height of room = 18 m

Length of the longest rod = Diagonal of the room = √(30² + 24² + 18²)
= √(900 + 576 + 324)
= √(1800)
= √(900 × 2)
= 30√2

১,১৮৯.

Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9), and D(5, 4). What is the shape of the quadrilateral?

ক
Square
খ
Rectangle but not a square
গ
Rhombus
ঘ
Parallelogram but not a rhombus

ব্যাখ্যা

Question: Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9), and D(5, 4). What is the shape of the quadrilateral?

Solution:

∴ The shape of the quadrilateral is Rhombus.

১,১৯০.

The difference between the circumference and the radius of a circle is 74 cm. Find the radius of a circle is -

ক
ক) 7 cm
খ
খ) 14 cm
গ
গ) 21 cm
ঘ
ঘ) 28 cm

ব্যাখ্যা

Let r be the radius of circle

Given that,
2πr - r = 74
⇒ r(2π - 1) = 74
⇒ r{(44/7) - 1} = 74
⇒ r (44 - 7)/7 }=74
⇒ r(37/7) = 74
⇒ r = 74 (7/37)
r = 14

১,১৯১.

A 6 m long and 4 m wide cistern contains water up to a breadth of 1 m 25 cm. Find the total surface area of the surface immersed in water?

ক
42 m²
খ
49 m²
গ
52 m²
ঘ
64 m²

ব্যাখ্যা

Question: A 6 m long and 4 m wide cistern contains water up to a breadth of 1 m 25 cm. Find the total surface area of the surface immersed in water?

Solution:
The area of the wet surface = Area of the base + area of the two walls 2 × (4 ×1.25) + area of the other two walls 2 × (6 × 1.25) cm
= 6×4 + 2 (4×1.25) + 2 (6×1.25)
= 24 + 10 + 15
= 49 m²

১,১৯২.

A solid metal sphere of radius 6 cm is melted and recast into solid cones of radius 4 cm and height 9 cm. How many cones can be made?

ক
6
খ
12
গ
10
ঘ
11

ব্যাখ্যা

Question: A solid metal sphere of radius 6 cm is melted and recast into solid cones of radius 4 cm and height 9 cm. How many cones can be made?

Solution:
The volume of a sphere = (4/3)πr³
= (4/3)π(6)³= (864/3) π
= 288π cm³

The volume of a cone = (1/3)πr²h
= (1/3)π(4)²(9)
= (144/3)π cm^3
=48π cm³

Number of cones = 288π/48π
= 288/48
= 6

১,১৯৩.

What is the area of a circle whose radius is the diagonal of a square whose area is 4?

ক
ক) 14π
খ
খ) 12π
গ
গ) 10π
ঘ
ঘ) 8π

ব্যাখ্যা

Area of square = 4
Side of square = √4 = 2
Diagonal of square = 2√2

Area of circle = πr²
= π(2√2)²
=8π

১,১৯৪.

The area of a square inscribed in a circle is 140 cm². What is the area of the circle?

ক
200 cm²
খ
220 cm²
গ
250 cm²
ঘ
230 cm²

ব্যাখ্যা

Question: The area of a square inscribed in a circle is 140 cm². What is the area of the circle?

Solution:
The area of a square inscribed in a circle is 140 cm²
side of square = √140 cm = 2√35 cm
diagonal of the square = √2 × 2√35
= 2√70 cm

diameter of circle = 2√70 cm
radius of the circle = √70 cm
∴ area of the circle = π (√70)2 cm²
= (22/7) × 70 cm²
= 220 cm²

১,১৯৫.

The area of the base of a cylinder is 100π m². The volume of the cylinder is 900π m³. What is the height of the cylinder?

ক
ক) 5 m
খ
খ) 9 m
গ
গ) 7 m
ঘ
ঘ) 4 m
ঙ
ঙ) 6 m

ব্যাখ্যা

Area of the base of a cylinder, πr² = 100π
The volume of the cylinder, πr²h = 900π
∴ h = πr²h/πr²
= 900/100
= 9 m

১,১৯৬.

The angle of elevation of a ladder leaning against a wall is 60º and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:

ক
ক) 2.5 m
খ
খ) 4.9 m
গ
গ) 5.7 m
ঘ
ঘ) 9.2 m

ব্যাখ্যা

Let AB be the wall and BC be the ladder.
Then, ∠ACB = 60° = AC = 4.6m
AC/BC=cos⁡60∘=1/2
⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

১,১৯৭.

Find an equation for the line with x-intercept = 2, y-intercept = - 1.

ক
2x - y = 1
খ
x - 2y = 2
গ
y = - 1
ঘ
x = 2

ব্যাখ্যা

Question: Find an equation for the line with x-intercept = 2, y-intercept = - 1.

Solution:
দেওয়া আছে,
রেখাটি x-অক্ষকে ছেদ করে (x₁, y₁) = (2, 0) বিন্দুতে
এবং রেখাটি y-অক্ষকে ছেদ করে (x₂, y₂) = (0, - 1) বিন্দুতে

আমরা জানি,
ঢাল m=(y₂ - y₁)/(x₂ - x₁)
=(-1- 0)/(0 - 2)
=1/2

এখানে,
m =1/2
c = y এর ছেদক = - 1

∴ সরলরেখার ঢালের সমীকরণ হতে পাই,
y = mx + c
⇒ y = (1/2)x+(-1)
⇒ y = (x - 2)/2
⇒ 2y = x - 2
∴ x - 2y = 2

১,১৯৮.

What is the perimeter of a rectangle that is 21 meter wide and has the same area as another rectangle that is 66 meter long and 42 meter wide?

ক
ক) 304 m
খ
খ) 302 m
গ
গ) 310 m
ঘ
ঘ) 306 m

ব্যাখ্যা

Question: What is the perimeter of a rectangle that is 21 meter wide and has the same area as another rectangle that is 66 meter long and 42 meter wide?

Solution:
ধরি, আয়তক্ষেত্রের দৈর্ঘ্য = x

প্রশ্নমতে,
x × 21 = 66 × 42
x = (66 × 42)/21
x = 132

অতএব
পরিসীমা = 2(132 + 21) মিটার = 306 মিটার

১,১৯৯.

If sin A + cos A = a and sec A+ cosecA = b then, b(a² - 1) = ?

ক
2a
খ
1/a
গ
√a
ঘ
1/√a

ব্যাখ্যা

Question: If sin A + cos A = a and sec A+ cosecA = b then, b(a² - 1) = ?

Solution:

১,২০০.

∠B is the right angle of a right angles triangle ABC. If tanA = 1, then 4sinACosA = ?

ক
ক) 1
খ
খ) 2
গ
গ) 4
ঘ
ঘ) 1/2

ব্যাখ্যা

Question: ∠B is the right angle of a right angles triangle ABC. If tanA = 1, then 4sinACosA = ?

Solution:

দেওয়া আছে,
tanA = 1
ধরি,
বিপরীত বাহু = সন্নিহিত বাহু = a
অতিভুজ = √(a² + a²) = √2 a

∴ sinA = a/√2a = 1/√2
cosA = a/√2 a = 1/√2

∴ 4sinACosA = 4 × (1/√2) × (1/√2)
= 4 × 1/2
= 2