Geometry: Mensuration, Trigonometry

PrepBank · পাতা ৭ / ২১ · ৬০১–৭০০ / ২,০৮৫

৬০১.

The perimeter of a circle measures 16π cm. What is the area of the circle in sq. cm?

ক
128π
খ
64π
গ
32
ঘ
256π

ব্যাখ্যা

Question: The perimeter of a circle measures 16π cm. What is the area of the circle in sq. cm?

Solution:
ধরি, বৃত্তের ব্যাসার্ধ = r
বৃত্তের পরিধি = 2πr
বৃত্তের ক্ষেত্রফল = πr²

প্রশ্ন অনুসারে,
2πr = 16π
⇒ 2r = 16
⇒ r = 8

বৃত্তের ক্ষেত্রফল = πr²
= π × (8)²
= 64π

৬০২.

Calculate the surface area of a cylinder with radius 6 cm and height 15 cm.

ক
252π square cm
খ
792π square cm
গ
540π square cm
ঘ
None of these

ব্যাখ্যা

Question: Calculate the surface area of a cylinder with radius 6 cm and height 15 cm.

Solution:
Here,
radius r = 6 cm
height h = 15 cm
Surface Area = (2πrh + 2πr²)
Surface Area = 2 × π × 6 × 15 + 2 × π × 6² = 252π square cm

৬০৩.

cos²11° + cos²79° = ?

ক
0
খ
1
গ
- 2
ঘ
2

ব্যাখ্যা

Question: cos²11° + cos²79° = ?

Solution:
Given that,
cos²11° + cos²79°
= cos²11° + cos²(90° - 11°)
= cos²11° + sin²11°
= 1

৬০৪.

In sin(A - B) = 1/2 and cos(A + B) = 1/2 where A > B > 0 and (A + B) is an acute angle, then the value of B is?

ক
π/10 radian
খ
π/12 radian
গ
π/8 radian
ঘ
π/3 radian

ব্যাখ্যা

Question: In sin(A - B) = 1/2 and cos(A + B) = 1/2 where A > B > 0 and (A + B) is an acute angle, then the value of B is?

Solution:
Given,
sin(A - B) = 1/2 (A - B) = 30^∘
and cos(A + B) = 1/2 (A + B) = 60^∘

Adding both side
(A - B) + (A + B) = 30^∘ + 60^∘
⇒ 2A = 90^∘⇒ A = 45^∘

∵ A - B = 30^∘B = A - 30^∘⇒ B = 45^∘- 30^∘⇒ B = 15^∘⇒ B = 15 × (π/180)
∴ B = π/12 radian

৬০৫.

The perimeter of a rectangular garden is 54 yards and the width is 36 feet. What is the length of the garden?

ক
12 yards
খ
15 yards
গ
15 feet
ঘ
27 feet

ব্যাখ্যা

Question: The perimeter of a rectangular garden is 54 yards and the width is 36 feet. What is the length of the garden?

Solution:
Given that,
Perimeter of rectangle = 54 yards
54 × 3 = 162 feet ; [1 yard = 3 feet]
And Width = 36 feet

We know,
Perimeter = 2(length + width)
⇒ 162 = 2(L + 36)
⇒ 81 = L + 36
∴ L = 81 - 36 = 45 feet

∴ The length of the garden is 45 feet = 45/3 = 15 yards.

৬০৬.

What would be the measure of the perimeter of a square whose area is equal to 256 square cm?

ক
16 cm
খ
36 cm
গ
64 cm
ঘ
128 cm

ব্যাখ্যা

Question: What would be the measure of the perimeter of a square whose area is equal to 256 square cm?

Solution:
দেওয়া আছে
বর্গক্ষেত্রের ক্ষেত্রফল = 256
বর্গের এক বাহুর দৈর্ঘ্য = a

প্রশ্নমতে
a² = 256
a² = (16)²
a = 16

বর্গক্ষেত্রের পরিসীমা = 4a
= 4 × 16
= 64

৬০৭.

In a cyclic quad. ABCD, ∠A = 80°. Then ∠C = ?

ক
80°
খ
100°
গ
120°
ঘ
160°
ঙ
180°

ব্যাখ্যা

Question: In a cyclic quad. ABCD, ∠A = 80°. Then ∠C = ?

Solution:
Opposite angles of a cyclic quadrilateral are supplementary.
∴ ∠A + ∠C = 180°
⇒ 80° + ∠C =180°
⇒ ∠C = 100°.

৬০৮.

The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. Find the height of a cylindrical pillar.

ক
7 meters
খ
8 meters
গ
6 meters
ঘ
9 meters

ব্যাখ্যা

Question: The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. Find the height of a cylindrical pillar.

Solution:
Let the radius of the cylinder be r meters and the height be h meters.
Curved surface area = 2πrh
∴ 2πrh = 264 .......(1)

And Volume = πr2h
∴ πr²h = 924 ......(2)

Now, (2) ÷ (1),
πr²h/2πrh = 924/264
⇒ r/2 = 924/264
⇒ r = (924/264) × 2
∴ r = 7

From (1) we get,
h = 264/2πr = 264 × (7/22 × 2 × 7) = 6m
∴ h = 6m

So the height of the cylindrical pillar is 6 meters

৬০৯.

Ramesh and Suresh’s mud forts have heights 8cm and 15 cm. They are 24 cm apart. How far are the fort tops from each other?

ক
ক) 22 cm
খ
খ) 23 cm
গ
গ) 25 cm
ঘ
ঘ) 24.5 cm

৬১০.

Two small circular parks of diameters 6 m and 8 m are to be replaced by a bigger circular park. What would be the radius of this new park, in meter, if the new park occupies the same space as the two small parks (in meter)?

ক
5
খ
10
গ
15
ঘ
20
ঙ
25

ব্যাখ্যা

Question: Two small circular parks of diameters 6 m and 8 m are to be replaced by a bigger circular park. What would be the radius of this new park, in meter, if the new park occupies the same space as the two small parks (in meter)?

Solution:
Let,
The radious of the new circular park = R
Area of the new circular park = sum of the areas of the 2 smaller parks
⇒ π (6/2)² + π (8/2)²
= π (3)² + π (4)²
= π 9 + π 16
= π(9 + 16)
= 25π

⇒ 25 π = π R².
∴ R² = 25
⇒ R = 5 m

৬১১.

A cylinder and a cone have the same base radius and height. What is the ratio of their volumes?

ক
1 : 2
খ
2 : 3
গ
3 : 1
ঘ
3 : 2

ব্যাখ্যা

Question: A cylinder and a cone have the same base radius and height. What is the ratio of their volumes?

Solution:
The radius of the base of both the cylinder and the cone be r.
The height of both the cylinder and the cone be h.

The volume of a cylinder is = πr²h
The volume of a cone is = (1/3)πr²h

∴ Ratio = (πr²h)/{(1/3)πr²h} = 1/(1/3) = 3
The ratio of the volumes of the cylinder to the cone is 3 : 1

৬১২.

If secA + tanA = 5/2, then what is the value of secA - tanA?

ক
√3/2
খ
1
গ
1/√2
ঘ
2/5

ব্যাখ্যা

Question: If secA + tanA = 5/2, then what is the value of secA - tanA?

Solution:
দেয়া আছে,
secA + tanA = 5/2

আমরা জানি,
sec²A - tan²A = 1
⇒ (secA + tanA) (secA - tanA ) =1
⇒ 5/2(secA - tanA) = 1
∴ (secA - tanA) = 2/5

৬১৩.

The average length of the sides of ∆ABC is 12. What is the perimeter at ∆ABC?

ক
24
খ
36
গ
12
ঘ
18

ব্যাখ্যা

Question: The average length of the sides of ∆ABC is 12. What is the perimeter at ∆ABC?

Solution:
তিনটি বাহুর গড় = 12
তিন বাহুর সমষ্টি = 12 × 3 = 36

∴ ∆ABC এর পরিসীমা = 36

৬১৪.

A room has a length of 10 m, width of 6 m, and height of 4 m. What is the area of the four walls of the room?

ক
144 m²
খ
136 m²
গ
128 m²
ঘ
120 m²

ব্যাখ্যা

Question: A room has a length of 10 m, width of 6 m, and height of 4 m. What is the area of the four walls of the room?

Solution:
Given that,
Length of the room = 10 m
Width = 6 m
Height = 4 m

The area of the four walls = 2 × (length × height) + 2 × (width × height)
= 2 × (10 × 4) + 2 × (6 × 4)
= 2 × 40 + 2 × 24
= (80 + 48)
= 128 m²

So the area of the four walls is 128 m².

৬১৫.

If a, b, and c are the lengths of the three sides of a triangle, then which of the following is true?

ক
a + b < c
খ
a - b < c
গ
a + b = c
ঘ
a + b ≥ c

ব্যাখ্যা

Question: If a, b, and c are the lengths of the three sides of a triangle, then which of the following is true?

Solution:
ত্রিভুজ হওয়ার শর্ত:
• ত্রিভুজের যেকোনো দুই বাহুর সমষ্টি তার তৃতীয় বাহু অপেক্ষা বৃহত্তর।
• যেকোনো দুই বাহুর অন্তর তার তৃতীয় বাহু অপেক্ষা ক্ষুদ্রতর।

দেওয়া আছে
ত্রিভুজের তিনটি বাহু যথাক্রমে a, b এবং c
সুতরাং, a - b < c অবশ্যই সঠিক।

৬১৬.

If 0 ≤ θ ≤ π/2 and cos θ + √3sinθ = 2, then what is the value of θ?

ক
π/2
খ
π/3
গ
π/4
ঘ
π/6

ব্যাখ্যা

Question: If 0 ≤ θ ≤ π/2 and cos θ + √3sinθ = 2, then what is the value of θ?

Solution:
cos θ + √3sinθ = 2
⇒ (1/2) cos + (√3/2) sinθ = 1
⇒ sin30° cosθ + cos30° sinθ = 1
⇒ sin(30° + θ) = sin90°
⇒ 30° + θ = 90°
∴ θ = 60° = π/3

৬১৭.

The length of the sides of a triangle are in the ratio of 3 to 5 to 6. If the perimeter of the triangle is 70. what is the length of the longest side?

ক
ক) 26
খ
খ) 35
গ
গ) 25
ঘ
ঘ) 30

ব্যাখ্যা

ত্রিভুজের তিন বাহুর দৈর্ঘ্যের অনুপাত = 3 : 5 : 6
ধরি ত্রিভুজের বাহু তিনটির দৈর্ঘ্য যথাক্রমে 3x, 5x, 6x
প্রশ্নমতে,
3x + 5x + 6x = 70
=> 14x = 70
=> x = 5
অতএব, বৃহত্তম বাহুর দৈর্ঘ্য = 6x = 6 × 5 = 30

৬১৮.

The area of a circle of radius √2 is approximately-

ক
ক) 1.4142
খ
খ) 6.2832
গ
গ) 9.4298
ঘ
ঘ) 3.1416

ব্যাখ্যা

Area of the circle = πr²
= π(√2)²
= 2π
= 6.2832

৬১৯.

The ratio of the four angles of a quadrilateral is 3 : 4 : 5 : 6. What is the largest angle?

ক
ক) 100°
খ
খ) 105°
গ
গ) 120°
ঘ
ঘ) 135°

ব্যাখ্যা

Question: The ratio of the four angles of a quadrilateral is 3 : 4 : 5 : 6. What is the largest angle?

Solution:
Sum of the angles of a quadrilateral = 360°
Largest angle = 360° × (6/18) = 120°

৬২০.

If a cube has a length of 8 cm, what is its total surface area?

ক
ক) 64
খ
খ) 96
গ
গ) 384
ঘ
ঘ) 256

ব্যাখ্যা

প্রশ্ন: If a cube has a length of 8 cm, what is its total surface area?

সমাধান:
দেওয়া আছে,
ঘনকের এক বাহুর দৈর্ঘ্য, a = 8 cm

∴ ঘনকের সমগ্রতলের ক্ষেত্রফল = 6a²
= 6 × 8 × 8 cm²
= 384 cm²

৬২১.

A circus artist is climbing a 28 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 60°.

ক
32√2 m
খ
24√3 m
গ
11√5 m
ঘ
14√3 m

ব্যাখ্যা

Question: A circus artist is climbing a 28m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 60°.

Solution:

By observing the figure, AB is the pole.
In triangle ABC,
⇒ AB/AC = sin60°
⇒ AB/28 = √3/2
⇒ AB = 14√3

Therefore, the height of the pole is 14√3 m

৬২২.

If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is-

ক
50°
খ
75°
গ
67°
ঘ
56°

ব্যাখ্যা

Question: If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is-

Solution:
Since, ΔABC and ΔPQR are similar triangles.
And ∠A = 47°
then,
∠B = ∠Q = 83°

Thus, in ΔABC,
∠A + ∠B + ∠C = 180°
∠C = 180° - (∠A + ∠ B)
⇒ ∠C = 180° - (47° + 83°)
∴ ∠C = 50°

৬২৩.

From a point at some distance from the base of a tree, the angle of elevation to the top of the tree is 30°. If the tree is 12 meters tall, how far is that point from the tree?

ক
12√2 meters
খ
12√3 meters
গ
8 meters
ঘ
8√2 meters

ব্যাখ্যা

Question: From a point at some distance from the base of a tree, the angle of elevation to the top of the tree is 30°. If the tree is 12 meters tall, how far is that point from the tree?

Solution:

Let's assume the tree is located a meters away.

Thus, tan 30° = AB/AC
⇒ 1/√3 = 12/a
∴ a = 12√3 meters

৬২৪.

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
300 m³
খ
462 m³
গ
250 m³
ঘ
148 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?
(একটি বৃত্তাকার কুয়া যার ব্যাস ১৪ মিটার, সেটি ৩ মিটার গভীর পর্যন্ত খনন করা হয়েছে। খননের ফলে উত্তোলিত মাটির আয়তন কত?)

Solution:
কূপের ব্যাস = 14 m
ব্যাসার্ধ 14/2 = 7 m

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

৬২৫.

If the base of a parallelogram is 8 and the height is 7 and perimeter is 24, what is the area of the parallelogram?

ক
ক) 28
খ
খ) 56
গ
গ) 65
ঘ
ঘ) 82

ব্যাখ্যা

Question: If the base of a parallelogram is 8 and the height is 7 and perimeter is 24, what is the area of the parallelogram?

Solution:
দেয়া আছে,
সামান্তরিকের ভূমি ৪ মিটার
সামান্তরিকের উচ্চতা 7 মিটার এবং
সামান্তরিকের পরিসীমা 24 মিটার

আমরা জানি,
সামান্তরিকের ক্ষেত্রফল = (ভূমি × উচ্চতা) বর্গমিটার
= 8 × 7 = 56 বর্গমিটার

৬২৬.

cos(θ + 25°) = √3/2, then the value of θ?

ক
15°
খ
5°
গ
10°
ঘ
20°

ব্যাখ্যা

Question: cos(θ + 25°) = √3/2, then the value of θ?

Solution:
Given that,
cos(θ + 25°) = √3/2
⇒ cos(θ + 25°) = cos30°
⇒ θ + 25° = 30°
⇒ θ = 30° - 25°
∴ θ = 5°

৬২৭.

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
9.2 m
খ
9.3 m
গ
9.4 m
ঘ
9.5 m

ব্যাখ্যা

Let,
AB be the wall and BC be the ladder.
Then, ∠ACB = 60° = AC = 4.6m
AC/BC = cos60° = 1/2
⇒ BC = 2 × AC = 2 × 4.6 = 9.2m

৬২৮.

If the height of a vertical pole is √3 times the length of its shadow on the ground, then the angle of elevation of the sun at that time is -

ক
30°
খ
60°
গ
45°
ঘ
75°

ব্যাখ্যা

Let AB be a vertical pole and let its shadow be BC
Let BC = x m, then length of pole = √3x

Θ be the angle of elevation.
∴ tanΘ = AB/BC = √3x/x
= √3
= tan60°
Θ = 60°.

৬২৯.

The diameter of a wheel is 1.26m. How far will it travel in 600 revolutions?

ক
ক) 2676 m
খ
খ) 7623 m
গ
গ) 2637 m
ঘ
ঘ) 2376 m

ব্যাখ্যা

Diameter of the wheel =1.26m
Radius of the wheel r=1.26/2m
=0.63m

The distance traveled in one revolution = perimeter of the wheel
=2πr
= 2 × (22/7) × 0.63 m
= 99/25 m
distance traveled in 600 revolutions = (99/25) × 600
= 2376 m

৬৩০.

The surface area of a cube is 150 square units. What is the length of the longest stick that can be placed inside the cube?

ক
25√2
খ
5
গ
6√3
ঘ
25
ঙ
5√3

ব্যাখ্যা

Question: The surface area of a cube is 150 square units. What is the length of the longest stick that can be placed inside the cube?

Solution:
Given that,
Surface area of a cube = 150 square units

We know,
Surface area of a cube, S = 6a²
⇒ 6a² = 150
⇒ a² = 150/6
⇒ a²= 25 = 5²
∴ a = 5

The longest stick that can fit inside the cube runs along the space diagonal.
So the space diagonal of a cube, d = a√3
= 5√3 ; [a = 5]

So the length of the longest stick that can be placed inside the cube is 5√3 units.

৬৩১.

In a circle, O is the center, ∠AOB = 60°. If the radius of the circle is 4 cm, then what is the value of AB?

ক
ক) 2 cm
খ
খ) 4 cm
গ
গ) 4√2 cm
ঘ
ঘ) 6 cm

ব্যাখ্যা

Question: In a circle, O is the center, ∠AOB = 60°. If the radius of the circle is 4 cm, then what is the value of AB?

Solution:
O কেন্দবিশিষ্ট বৃত্তে, ∠AOB = 60°
ব্যাসার্ধ OA = ব্যাসার্ধ OB = 4cm
∠OAB = ∠OBA = 60°

অতএব, AOB একটি সমবাহু ত্রিভুজ।
OA = OB = AB = 4 cm

৬৩২.

ABC is a triangle. The bisectors of the internal angle ∠B and external angle ∠C intersect at D. If ∠BDC = 50°, then ∠A = ?

ক
50°
খ
30°
গ
100°
ঘ
120°

ব্যাখ্যা

According to the property,
∠A = 2∠D
= 2 × 50°
= 100°
Therefore, the angle, ∠BAC = ∠A = 100°

৬৩৩.

From the top of a lighthouse which is 90 m above the sea, the angle of depression of a ship is 60°. How far is the ship from the lighthouse?

ক
30√3 m
খ
30 m
গ
17.34 m
ঘ
20.5 m

ব্যাখ্যা

Question: From the top of a lighthouse which is 90 m above the sea, the angle of depression of a ship is 60°. How far is the ship from the lighthouse?

Solution:

Let the height of the lighthouse above sea be AC and it is given 90 m.
Ship is at point B so the distance between the base of lighthouse A and ship is AB.
In ΔABC,
tan60° = AC/AB
⇒ √3 = 90/AB
⇒ AB = 90/√3 = 30√3

৬৩৪.

A garden of 100 meter length and 60 meter width has a walkway of 2 meter width on every side. What is the area of the garden, in square meters, excluding the walkway?

ক
ক) 5684
খ
খ) 6000
গ
গ) 624
ঘ
ঘ) 5376
ঙ
ঙ) 312

ব্যাখ্যা

Question: A garden of 100 meter length and 60 meter width has a walkway of 2 meter width on every side. What is the area of the garden, in square meters, excluding the walkway?

Solution:
The length of garden excluding the walkway = 100 - 4 meter
= 96 meter

The width of garden excluding the walkway = 60 - 4 meter
= 56 meter

∴ The are of the garden excluding the walkway = 96 × 56 square meter
= 5376 square meter

৬৩৫.

If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?

ক
20°
খ
30°
গ
45°
ঘ
60°

ব্যাখ্যা

Question: If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?

Solution:

ধরি,
খুঁটির উচ্চতা AB = 20 m
খুঁটির ছায়ার দৈর্ঘ্য BC = 20√3 m

এখন,
ΔABC হতে পাই,
tanθ = লম্ব/ভূমি
⇒ tanθ = AB/BC
⇒ tanθ = 20/(20√3)
⇒ tanθ = 1/√3
⇒ tanθ = tan30°
∴ θ = 30°

৬৩৬.

A pole of 60 metre long breaks into two parts without complete separation and makes an angle 30° with the ground. Find the length of the broken part of the pole.

ক
40 m
খ
30 m
গ
45 m
ঘ
50 m

ব্যাখ্যা

Question: A pole of 60 metre long breaks into two parts without complete separation and makes an angle 30° with the ground. Find the length of the broken part of the pole.

Solution:

sin30° = x/(60 - x)
⇒ 1/2 = x/(60 - x)
⇒ 60 - x = 2x
⇒ 3x = 60
⇒ x = 60/3 = 20

∴ The length of the broken part of the pole = 60 - 20 = 40 m

৬৩৭.

If each edge of a cube is increased by 25%, then the percentage increase in its surface area is?

ক
ক) 30%
খ
খ) 43.75%
গ
গ) 56.25%
ঘ
ঘ) 61%

ব্যাখ্যা

Question: If each edge of a cube is increased by 25%, then the percentage increase in its surface area is?

Solution:
Let the original edge = a
Then, the surafce area = 6a²
New edge = 125a/100 = 5a/4
New surafce area = 6 × (5a/4)² = 75a²/8
Increase in surface area = (75a²/8 - 6a²) = 27a²/8
The percentage increase in its surface area = (27a² × 1/6a² × 100)% = 56.25%

৬৩৮.

Given that the diagonal of a square measures 10√2 units, find the area of the square in square units.

ক
80 square units
খ
100 square units
গ
144 square units
ঘ
169 square units

ব্যাখ্যা

Question: Given that the diagonal of a square measures 10√2 units, find the area of the square in square units.

Solution:
দেয়া আছে,
বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = 10√2 একক
আমরা জানি,
বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = √2 × বাহু

প্রশ্নমতে,
√2 × বাহু = 10√2
⇒ বাহু = 10√2/√2
∴ বাহু = 10 একক

এখন,
বর্গক্ষেত্রের ক্ষেত্রফল = (বাহুর দৈর্ঘ্য)²
= (10)²
= 100 বর্গ একক

৬৩৯.

If a 38 meter ladder is placed against a 19 meter wall such that it just reaches the top of the wall, the angle of elevation of the wall is-

ক
20º
খ
30º
গ
40º
ঘ
60º

ব্যাখ্যা

Question: If a 38 meter ladder is placed against a 19 meter wall such that it just reaches the top of the wall, the angle of elevation of the wall is-

Solution:
Given that,
Ladder's length = 38 m
Wall's height = 19 m

Perpendicular = Wall's height = 19 m
Hypotenuse = Ladder's length = 38 m

We know,
Sinθ = Perpendicular/Hypotenuse
⇒ Sinθ = 19/38
⇒ Sinθ = 1/2
⇒ Sinθ = sin30º
∴ θ = 30º

৬৪০.

ক
ক) 30
খ
খ) 32
গ
গ) 22
ঘ
ঘ) None of these

ব্যাখ্যা

(5 × 6 × 4)/10 = 12
and (6 × 7 × 5)/10 = 21
∴ (4 × 8 × 10)/10 = 32

৬৪১.

The perimeter of the base of a cube is 24 cm. What is its volume?

ক
124 cm³
খ
216 cm³
গ
100 cm³
ঘ
150 cm³

ব্যাখ্যা

Question: The perimeter of the base of a cube is 24 cm. What is its volume?

Solution:
Let the side length of the cube be x.
So, 4x = 24
∴ x = 6 cm
Volume = (6)³
= 216 cm³

৬৪২.

Find the value of

ক
ক) 1
খ
খ) 0
গ
গ) - 1
ঘ
ঘ) 1/2

ব্যাখ্যা

Question: Find the value of

Solution:

৬৪৩.

A farmer has two rectangular fields. The larger field has twice the length and four times the width of the smaller field. If the smaller field has area K, then the area of the larger field is greater than the area of the smaller field by what amount?

ক
ক) 2K
খ
খ) 5K
গ
গ) 6K
ঘ
ঘ) 7K

ব্যাখ্যা

Question: A farmer has two rectangular fields. The larger field has twice the length and four times the width of the smaller field. If the smaller field has area K, then the area of the larger field is greater than the area of the smaller field by what amount?

Solution:

ধরি,
ছোট মাঠের দৈর্ঘ্য এবং প্রস্থ যথাক্রমে l, b.
∴ ক্ষেত্রফল, k = l × b

বর মাঠের দৈর্ঘ্য ও প্রস্থ যথাক্রমে 2l, 4b
∴ ক্ষেত্রফল = 2l × 4b = 8(l × b) = 8k

∴ বড় মাঠের ক্ষেত্রফল বেশি = 8k - k = 7k

৬৪৪.

The angles of a triangle are in the proportion of 1 : 2 : 3 and the length of the smallest side is 1 cm. What is the length of the longest side of the triangle?

ক
2 cm
খ
3 cm
গ
4 cm
ঘ
4.5 cm

ব্যাখ্যা

Question: The angles of a triangle are in the proportion of 1 : 2 : 3 and the length of the smallest side is 1 cm. What is the length of the longest side of the triangle?

Solution:
ত্রিভুজের তিনটি কোণের অনুপাত = 1 : 2 : 3

ধরি,
ত্রিভুজের তিনটি কোণ যথাক্রমে x, 2x, 3x

x + 2x + 3x = 180°
6x = 180°
x = 30°

ত্রিভুজের তিনটি কোণ যথাক্রমে = 30°, 60°, 90°
ত্রিভুজটি সমকোণী ত্রিভুজ।

ΔABC এ
cos60° = BC/AC
1/2 = 1/AC
AC = 2

৬৪৫.

Which trigonometric ratio is undefined in value?

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

ব্যাখ্যা

Question: Which trigonometric ratio is undefined in value?

Solution:
sin 90° = 1
cos 90° = 0
cosec 90° = 1
cot 0° = ∞ (Undefined)

৬৪৬.

sin(A + 12°) = 1/2, find the value of A?

ক
ক) 33°
খ
খ) 48°
গ
গ) 18°
ঘ
ঘ) 78°

ব্যাখ্যা

Question: sin(A + 12°) = 1/2, find the value of A?

Solution:
sin(A + 12°) = 1/2
⇒ sin(A + 12°) = sin30°
⇒ A + 12° = 30°
⇒ A = 30° - 12°
∴ A = 18°

৬৪৭.

Find the volume of a cylindrical-shaped water container that has a height of 15 cm and a diameter of 14 cm.

ক
2430 cm³
খ
2310 cm³
গ
2700 cm³
ঘ
2160 cm³

ব্যাখ্যা

Question: Find the volume of a cylindrical-shaped water container that has a height of 15 cm and a diameter of 14 cm.

Solution:
দেওয়া আছে,
উচ্চতা (h) = 15 cm এবং ব্যাস (d) = 14 cm

আমরা জানি, ব্যাস = 2 × ব্যাসার্ধ
⇒ 2 × ব্যাসার্ধ = 14 cm
⇒ ব্যাসার্ধ (r) = 14/2 = 7 cm

আমরা জানি,
সিলিন্ডারের আয়তন, V = πr²h ঘন একক
= 22/7 × (7)² × 15
= 22/7 × 49 × 15
= 22 × 7 × 15
= 2310 ঘন সেমি।

সুতরাং, সিলিন্ডার আকৃতির পানির পাত্রটির আয়তন হল 2310 cm³।

৬৪৮.

What is the greatest possible area of a triangle with one side of length 6 and another side of length 9?

ক
54
খ
49
গ
63
ঘ
27

ব্যাখ্যা

Question: What is the greatest possible area of a triangle with one side of length 6 and another side of length 9?

Solution:
দেওয়া আছে,
ত্রিভুজে নির্দিষ্ট দুইটি বাহু a = 6 এবং b = 9

আমরা জানি,
দুইটি নির্দিষ্ট বাহু দেওয়া থাকলে,
ত্রিভুজের ক্ষেত্রফল = (1/2) a × b × sinθ
θ হল 6 ও 9-এর মধ্যবর্তী কোণ।
sinθ-এর সর্বোচ্চ মান = sin90° = 1
অর্থাৎ, সর্বোচ্চ ক্ষেত্রফল তখনই হবে যখন এই দুই বাহু পরস্পর লম্ব হয়।

সর্বোচ্চ ক্ষেত্রফল = = (1/2) × 6 × 9 × sin90°
= 27 × 1 [sin90° = 1]
= 27

৬৪৯.

Three circles are mutually tangent externally Their centers form a triangle whose sides are of lengths 3, 4, and 5. The total area of the circles (in square units) is-

ক
35π
খ
27π
গ
21π
ঘ
14π

ব্যাখ্যা

Question: Three circles are mutually tangent externally Their centers form a triangle whose sides are of lengths 3, 4, and 5. The total area of the circles (in square units) is-

Solution:
let, the radiuses r₁, r₂, r₃and let r₃>r₂>r₁

ATQ,
r₁ + r₂ = 3
r₁ + r₃ = 4
r₂ + r₃ = 5

r₃ = 5 - r₂

r₁ + 5 - r₂ = 4
⇒ r₂ = r₁ + 1

r₁ + r₁ + 1 = 3
⇒ 2r₁ = 2
∴ r₁ = 1

r₂ = 1 + 1 = 2
r₃ = 5 - 2 = 3

total areas = πr₁² + πr₂² + πr₃²
= π + 4π + 9π
= 14π

৬৫০.

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

ক
15 m
খ
18 m
গ
21 m
ঘ
12 m

ব্যাখ্যা

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

৬৫১.

A building is under construction. The top of the building forms 30° angle of elevation from a point on the adjoining plot that is 300 m. After a month, the angle of elevation formed by the top of the building from the same point increased to 60°. How much was the building constructed in this 1 month.

ক
200√3
খ
100√3
গ
200/√3
ঘ
300(1/√3)

ব্যাখ্যা

Original Building height = h = MQ
New building height = PQ

In △MQN, tan 30° = 1/√3 = MQ/NQ
MQ = 300/√3

In △PQN, tan 60° = √3 = PQ/NQ
PQ = 300√3
Building grew = PQ - MQ

∴ Building grew = 300√3 - (300/√3 )
= 3 × 300 × (2/√3)
= 200√3

৬৫২.

A ladder makes an angle of 60° with the ground, when placed along a wall. If the foot of ladder is 8 m away from the wall, the length of ladder is

ক
ক) 4 m
খ
খ) 8 m
গ
গ) 8√3 m
ঘ
ঘ) 16 m
ঙ
ঙ) 3√2 m

ব্যাখ্যা

Let AB be the wall, AC be the length of the ladder.

In right angled triangle ABC,
cos 60° = BC/AC
1/2 = 8/AC
AC = 8 × 2 = 16

Therefore, the length of the ladder is 16 m.

৬৫৩.

If rsinθ = 1, rcosθ = √3 then the value of √3tanθ + 1 = ?

ক
ক) √3
খ
খ) 1/√3
গ
গ) 1
ঘ
ঘ) 2

ব্যাখ্যা

Question: If rsinθ = 1, rcosθ = √3 then the value of √3tanθ + 1 = ?

Solution:
দেওয়া আছে,
rsinθ = 1 ......... (1)
rcosθ = √3 .............. (2)

(1) ÷ (2) হতে পাই
⇒ rsinθ/rcosθ =1/√3
⇒ tanθ = 1/√3
⇒ √3tanθ = 1

এখন, √3tanθ + 1 = 1 + 1
∴ √3tanθ + 1 = 2

৬৫৪.

Find the value of cos(7π/6).

ক
- 1/2
খ
1/√2
গ
1
ঘ
- √3/2

ব্যাখ্যা

Question: Find the value of cos(7π/6).

Solution:
cos(7π/6)
= cos(π + π/6) [যেহেতু (π + π/6) তৃতীয় চতুর্ভাগে পড়ে এবং তৃতীয় চতুর্ভাগে cos ঋণাত্মক, তাই cos(π + θ) = - cosθ]
= - cos(π/6)
= - cos(30°)
= - √3/2

৬৫৫.

3cotA = 4 হলে, sinA এর মান কত?

ক
5/3
খ
5/6
গ
3/5
ঘ
1/3

ব্যাখ্যা

প্রশ্ন : 3cotA = 4 হলে, sinA এর মান কত?

সমাধান :
দেওয়া আছে,
3 cotA = 4
⇒ cot = 4/3
⇒ cot²A = 16/9
⇒ cosec²A - 1 = 16/9
⇒ cosec²A = (16/9) + 1
⇒ cosec²A = (16 + 9)/9
⇒ cosec²A = 25/9
⇒ cosecA = 5/3
⇒ 1/sinA = 5/3
⇒ sinA = 3/5

৬৫৬.

A room 8 m long, 6 m high and 22.5 cm thick is made up of bricks, each measuring 25 cm × 11.25 cm × 6 cm. The number of bricks required is.

ক
ক) 7200
খ
খ) 6400
গ
গ) 6000
ঘ
ঘ) 5600

ব্যাখ্যা

Number of bricks = Volume of the wall/Volume of 1 brick
= (800×600×22.5)/(25×11.25×6)
= 6400

৬৫৭.

If 4sin²(2θ) +1 = 4, then θ = ?

ক
ক) 60°
খ
খ) 45°
গ
গ) 30°
ঘ
ঘ) 0°

ব্যাখ্যা

Question: If 4sin²(2θ) +1 = 4, then θ = ?

Solution:
4sin²(2θ) +1 = 4
বা, 4sin²(2θ) = 4 - 1
বা, sin²(2θ) = 3/4
বা, sin(2θ) = √3/2
বা, sin(2θ) = sin60°
বা, 2θ = 60°
∴ θ = 30°

৬৫৮.

A rectangular floor that measures 4 meters by 8 meters is to be covered with carpet that each measures 2 meters by 2 meters. If the carpet cost Tk. 14 a piece, what is the total cost to cover the floor?

ক
ক) Tk. 115
খ
খ) Tk. 112
গ
গ) Tk. 240
ঘ
ঘ) Tk. 118

ব্যাখ্যা

Area of the floor = 4 × 8 = 32 m²
Area of each carpet = 2 × 2 = 4 m²
Required number of carpet = 32/4 = 8
one carpet costs Tk. 14
8 carpet costs Tk. 14 × 8 = Tk. 112

৬৫৯.

A piece of wire 156 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 3 : 2, then what is the length of the base?

ক
39 cm
খ
43 cm
গ
35 cm
ঘ
41 cm

ব্যাখ্যা

Question: A piece of wire 156 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 3 : 2, then what is the length of the base?

Solution:
Given,
Ratio of one of the equal sides to the base is 3 : 2
Therefore, let base be 2x and equal sides be 3x
156 cm piece of wire is bent to form an isosceles triangle.
Thus perimeter of triangle is 156 cm.
Therefore,
2x + 3x + 3x = 156
⇒ 8x = 156
⇒ x = 19.5
Thus the length of the base = 2 × 19.5 = 39 cm.

৬৬০.

If tanθ = a/b, find the value of asinθ + bcosθ.

ক
1/(a² + b²)
খ
(a² + b²)
গ
1/√(a² + b²)
ঘ
√(a² + b²)

ব্যাখ্যা

Question: If tanθ = a/b, find the value of asinθ + bcosθ.

Solution:
tanθ = লম্ব/ভূমি = a/b

অতিভুজ = √(a² + b²)

asinθ + bcosθ
= a{a/√(a² + b²)} + b{b/√(a² + b²)}
= a²/√(a² + b²) + b²/√(a² + b²)
= (a² + b²)/√(a² + b²)
= √(a² + b²)

৬৬১.

In a right triangle, the length of one of the legs is 12 and the length of the hypotenuse is 13. What is the length of the other leg?

ক
17
খ
9
গ
8
ঘ
5

ব্যাখ্যা

Question: In a right triangle, the length of one of the legs is 12 and the length of the hypotenuse is 13. What is the length of the other leg?

Solution:
এখানে, সমকোণী ত্রিভুজের (right triangle) অতিভুজ (hypotenuse) = 13 একক সমকোণ সংলগ্ন এক বাহু = 12 একক
সমকোণ সংলগ্ন অপর বাহু = a একক

প্রশ্নমতে,
a² + 12² = 13²
⇒ a² + 144 = 169
⇒ a² = 169 - 144
⇒ a² = 25
⇒ a = √25
⇒ a = 5

∴ সমকোণ সংলগ্ন অপর বাহুর দৈর্ঘ্য = 5 একক

৬৬২.

The lengths of the sides of a triangle are in the ratio 2 : 4 : 5. If the perimeter of the triangle is 66 cm, what is the length of the longest side?

ক
12 cm
খ
24 cm
গ
30 cm
ঘ
36 cm

ব্যাখ্যা

Question: The lengths of the sides of a triangle are in the ratio 2 : 4 : 5. If the perimeter of the triangle is 66 cm, what is the length of the longest side?

Solution:
মনে করি,
ত্রিভুজের তিন বাহুর দৈর্ঘ্য যথাক্রমে 2x, 4x এবং 5x
∴ ত্রিভুজের পরিসীমা = (2x + 4x + 5x)
= 11x

প্রশ্নমতে,
⇒ 11x = 66
⇒ x = 66/11
⇒ x = 6

∴ দীর্ঘতম বাহুর দৈর্ঘ্য = (6 × 5) cm
= 30 cm

৬৬৩.

The line perpendicular to the tangent line is called -

ক
ক) Normal line
খ
খ) Secant line
গ
গ) Limit
ঘ
ঘ) Derivative

ব্যাখ্যা

The line perpendicular to the tangent line to a curve at the point of tangency is called the normal line to the curve at that point.

৬৬৪.

Rectangular Floors X and Y have equal area. If Floor X is 12 feet by 18 feet and Floor Y is 9 feet wide, what is the length of Floor Y, in feet?

ক
ক) 13.5
খ
খ) 18
গ
গ) 21
ঘ
ঘ) 24

ব্যাখ্যা

Given: area = 12×18 = 9×length [As, x floor area = y floor area]
∴ length = 12×18 / 9 = 24 feet

৬৬৫.

In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is-

ক
75 m³
খ
750 m³
গ
7500 m³
ঘ
75000 m³

ব্যাখ্যা

Question: In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is-

Solution:

1 hectare = 10,000 m²

So, Area = (1.5 × 10000) m² = 15000 m².

Depth = 5 cm = 5/100 m = 1/20 m.

Volume = (Area × Depth) = 15000 × 1/20 m³= 750 m³.

৬৬৬.

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where to top touches the ground is 12 m then the height of the tree is-

ক
14√2 m
খ
12√3 m
গ
24√2 m
ঘ
18√3 m

ব্যাখ্যা

Question: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where to top touches the ground is 12 m then the height of the tree is-

Solution:

Let the height of the tree be h. Let the part that is still standing on the ground be x, and that part which has fallen be y.
Hence, we have
sin θ = x/y
⇒ sin θ = 1/2 ;[ θ = 30°]
⇒ x/y = 1/2
⇒ y = 2x ...............(1)

Also, it is given that the distance of the tip of the fallen tree to that of the base of the tree is 12 m
⇒ cos θ = 12/y
⇒ cos 30° = 12/y
⇒ √3/2 = 12/y
⇒ y = 24/√3 = (8 × √3 × √3)/√3
⇒ y = 8√3

From (1), x = 8√3/2 = 4√3

∴ Height of tree, h = x + y = 4√3 + 8√3 = 12√3 m

৬৬৭.

The area of a rectangle is 40 cm² and one of its sides is 5 cm long. What will be its perimeter?

ক
13 cm
খ
20 cm
গ
26 cm
ঘ
33 cm

ব্যাখ্যা

Question: The area of a rectangle is 40 cm² and one of its sides is 5 cm long. What will be its perimeter?

Solution:
another side = 40/5
= 8 cm

perimeter = 2 (8 + 5)
= 2 × 13
= 26 cm

৬৬৮.

is equal to?

ক
ক) 2cosecθ
খ
খ) secθ
গ
গ) 1/2secθ
ঘ
ঘ) 2secθ

ব্যাখ্যা

Question: is equal to?

Solution:
2secθ

৬৬৯.

The perimeter of an isosceles triangle is 100 cm. If the base is 36 cm, find the length of the equal sides.

ক
64 cm
খ
24 cm
গ
36 cm
ঘ
32 cm

ব্যাখ্যা

Question: The perimeter of an isosceles triangle is 100 cm. If the base is 36 cm, find the length of the equal sides.

Solution:
Let the length of equal side = x.
∴ x + x + 36 = 100
⇒ 2x = 64
∴ x = 32cm.

৬৭০.

Maximum value of (2sinθ + 3cosθ) is?

ক
√13
খ
√5
গ
√13/2
ঘ
√7

ব্যাখ্যা

Maximum value of (2sinθ+3cosθ) = √(2² + 3²) = √13

৬৭১.

A square has an area of 25 m². If a circle has a radius equal to the length of the square’s diagonal, what is the area of the circle?

ক
75π sq. m.
খ
50π sq. m.
গ
100π sq. m.
ঘ
65π sq. m.

ব্যাখ্যা

Question: A square has an area of 25 m². If a circle has a radius equal to the length of the square’s diagonal, what is the area of the circle?

Solution:
Area of square = 25
Side of square = √25 = 5

Diagonal of square = 5√2
So, the radius of the circle is 5√2 m

Area of circle = πr²
= π(5√2)²
= 50π m²

∴ The area of the circle is 50π sq. m.

৬৭২.

The diameter of the base of a cone is 10.5 cm and its slant height is 10cm. The curved surface area is:

ক
ক) 134 cm²
খ
খ) 165 cm²
গ
গ) 154 cm²
ঘ
ঘ) 195 cm²

ব্যাখ্যা

Question: The diameter of the base of a cone is 10.5 cm and its slant height is 10cm. The curved surface area is:

Solution:
কোণকের ভূমির ব্যাস = 10.5 cm
কোণকের ভূমির ব্যাসার্ধ r = 10.5/2 = 5.25 cm
কোণকের তীর্যক উচ্চতা l = 10 cm

কোণকের বক্রতলের ক্ষেত্রফল = πrl = (22/7) × 5.25 × 10 = 165 cm²

৬৭৩.

What is the area of the largest sphere that can be placed inside a cube of volume 216 cm³?

ক
ক) 49π/3
খ
খ) 36π
গ
গ) 15√2π
ঘ
ঘ) 3√2π
ঙ
ঙ) None

ব্যাখ্যা

Let the length of each side of the cube = a
So, the volume of the cube is a³
∴ a = 6

We know, area of a sphere is 4/3πr³
= 4/3π3³
= 4/3π.27
= 36π

৬৭৪.

tanθ.√(1 - cos²θ) = ?

ক
sinθ
খ
cosθ
গ
sin²θ / cosθ
ঘ
tanθ.cosθ

ব্যাখ্যা

Question: tanθ.√(1 - cos²θ) = ?

Solution:
tanθ.√(1 - cos²θ)
= (sinθ/cosθ)√(sin²θ)
= (sinθ/cosθ)(sinθ)
= sin²θ/cosθ

৬৭৫.

A sphere of maximum volume is cut out from a solid hemisphere of radius r. The ratio of the volume of the hemisphere to that of the cut out sphere is-

ক
ক) 4 : 3
খ
খ) 3 : 2
গ
গ) 3 : 4
ঘ
ঘ) 4 : 1

ব্যাখ্যা

Question: A sphere of maximum volume is cut out from a solid hemisphere of radius r. The ratio of the volume of the hemisphere to that of the cut out sphere is-

Solution:
Volume of hemisphere = (2/3)πr³
Volume of biggest sphere = Volume of sphere with diameter r
= (4/3)π(r/2)³
= (1/6)πr³

∴ Required ratio = {(2/3)πr³}/{(1/6)πr³}
= 4/1
= 4 : 1

৬৭৬.

The base of an isosceles triangle is 6cm and one of the equal sides is 12cm. Find the radius of the circle through the vertices of the triangle?

ক
ক) 7√13/3
খ
খ) 7√5/6
গ
গ) 8√15/5
ঘ
ঘ) 5√5/3

ব্যাখ্যা

ধরি,
বৃত্তের ব্যাসার্ধ r সেন্টিমিটার।
ত্রিভুজের উচ্চতা h = (r + x) সেন্টিমিটার।

এখানে,
AB = 6 সেন্টিমিটার
AD = 6/2 = 3 সেন্টিমিটার।

সুতরাং, ΔACD ∠D = 90°

পিথাগোরাসের সূত্র প্রয়োগ করে পাইঃ
AC²= CD² + AD²12² = h² + 3².
h² = 135
h = √135

ΔADE
পিথাগোরাসের সূত্র প্রয়োগ করে পাই
AE²= AD² + DE²
r² = 3²+ x²
r² = 9 + (h - r)²
r² = 9 + h² - 2hr + r²
0 = 9 + h² - 2hr
2hr = 9 + h²
r = (9 + h²)/2h
r = 9 + 135/2√135
r = 144/2√135
r = 72/√135
r = 72√135/135
r = 72√(9 × 15)/135
r = (72× 3√15)/135
r = 8√15/5

৬৭৭.

If θ + ϕ = π/2, and sinθ = 1/2, then find out the value of sinϕ-

ক
√3
খ
√3/2
গ
1/√3
ঘ
2/√3

ব্যাখ্যা

Question: If θ + ϕ = π/2, and sinθ = 1/2, then find out the value of sinϕ-

Solution:
দেওয়া আছে,
θ+ϕ = π/2
⇒ θ = π/2 - ϕ
⇒ sinθ = sin(π/2 - ϕ)
⇒ 1/2 = cosϕ
⇒ cos²ϕ = 1/4
⇒ 1 - sin²ϕ = 1/4
⇒ 1 - 1/4 = sin²ϕ
⇒ sin²ϕ = 3/4
∴ sinϕ = √3/2

৬৭৮.

A cube has a total surface area of 486 square meters. What is the volume of the cube?

ক
343 cubic meters
খ
512 cubic meters
গ
729 cubic meters
ঘ
1000 cubic meters

ব্যাখ্যা

Question: A cube has a total surface area of 486 square meters. What is the volume of the cube?

Solution:
ধরি, ঘনকের বাহুর দৈর্ঘ্য = a মিটার।

আমরা জানি, ঘনকের সম্পূর্ণ পৃষ্ঠের ক্ষেত্রফল = 6a²

প্রশ্নমতে,
6a² = 486
⇒ a² = 486/6
⇒ a² = 81
∴ a = 9 মিটার

এখন, ঘনকের আয়তন = a³ = 9³ = 729 ঘন মিটার

অতএব, ঘনকটির আয়তন = 729 ঘন মিটার

৬৭৯.

What is the total surface area of a right circular cone with a base radius of 7 cm and a height of 24 cm?

ক
550 cm²
খ
704 cm²
গ
810 cm²
ঘ
840 cm²

ব্যাখ্যা

Question: What is the total surface area of a right circular cone with a base radius of 7 cm and a height of 24 cm?

Solution:
দেওয়া আছে,
ভূমির ব্যাসার্ধ r = 7 cm, উচ্চতা h = 24 cm
∴ হেলানো উচ্চতা, l = √(r² + h²)
= √(7² + 24²)
= √(49 + 576)
= √(625)
= 25 cm.

সমগ্র পৃষ্ঠতলের ক্ষেত্রফল (total surface area), = πr(l + r)
= (22/7) × 7 × (25 + 7)
= 22 × 32
= 704 cm².

∴ সমগ্র পৃষ্ঠতলের ক্ষেত্রফল (total surface area) = 704 cm².

৬৮০.

What is the volume of a cylinder with radius 6 and height of 7?

ক
ক) 284π
খ
খ) 252π
গ
গ) 181π
ঘ
ঘ) 294π

ব্যাখ্যা

সিলিন্ডারের ব্যাসার্ধ r = 6 একক এবং উচ্চতা h = 7 একক।

নির্ণেয় আয়তন =πr²h ঘন একক
= π × 6² × 7
= 252π ঘন একক

৬৮১.

Calculate the area of a rhombus if the length of its side is 4 cm and one of its angles A is 120 degrees.

ক
16√3 cm²
খ
16 cm²
গ
6√2 cm²
ঘ
8√3 cm²

ব্যাখ্যা

Question: Calculate the area of a rhombus if the length of its side is 4 cm and one of its angles A is 120 degrees.

Solution:
Given that,
Side of rhombus, a = 4 cm
And One angle, A = 120°

We know,
Area of a rhombus = a² × sin⁡A [Where a = side of rhombus, A = any interior angle.]
= 4² × sin⁡120°
= 16 × (√3/2)
= 8√3

So the area of the rhombus is 8√3 cm²

Note:
sin(180∘ - θ) = sinθ,
So sin120° = sin⁡(180° - 60°) = sin⁡60° = √3/2

৬৮২.

If tan(θ + 30°) = √3, then what is the value of cosθ?

ক
1
খ
1/2
গ
√3/2
ঘ
1/√2
ঙ
0

ব্যাখ্যা

Question: If tan(θ + 30°) = √3, then what is the value of cosθ?

Solution:
Given that,
tan(θ + 30°) = √3
⇒ tan(θ + 30°) = tan60°
⇒ (θ + 30°) = 60°
∴ θ = 30°

Now,
cosθ
= cos30°
= √3/2

৬৮৩.

A cuboid has dimensions in the ratio 1 : 2 : 4 and a total surface area of 112 cm². What is its volume?

ক
48 cm³
খ
84 cm³
গ
64 cm³
ঘ
72 cm³

ব্যাখ্যা

Question: A cuboid has dimensions in the ratio 1 : 2 : 4 and a total surface area of 112 cm². What is its volume?

Solution:
দেয়া আছে,
আয়তাকার ঘনবস্তুর মাত্রাগুলির অনুপাত = 1 : 2 : 4
এবং সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 112 cm²

ধরি , আয়তাকার ঘনবস্তুর মাত্রাগুলির অনুপাত যথাক্রমে x, 2x এবং 4x

আমরা জানি,
আয়তাকার ঘনবস্তুর সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 2(lb + bh + lh)
⇒ সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 2(x)(2x) + (2x)(4x) + (4x)(x)
⇒ 112 = 2(2x² + 8x² + 4x²)
⇒ 112 = 2(14x²)
⇒ 112 = 28x²⇒ x² = 112/28
⇒ x² = 4
⇒ x = 2

সুতরাং, ঘনবস্তুটির মাত্রাগুলি হল,
দৈর্ঘ্য (l) = x = 2 cm
প্রস্থ (b) = 2x = 2 × 2 = 4 cm
উচ্চতা (h) = 4x = 4 × 2 = 8 cm

এখন, আয়তাকার ঘনবস্তুর আয়তন, V = l × b × h
⇒ V = 2 × 4 × 8
⇒ V = 64 ঘন সেমি।

সুতরাং, নির্ণেয় আয়তন হল 64 cm³।

৬৮৪.

A tank is 25 m long, 12 m wide and 6 m deep. The core of plastering its walls and bottom at 75 paisa per m² is:

ক
ক) tk. 456
খ
খ) tk. 458
গ
গ) tk. 558
ঘ
ঘ) tk. 568

ব্যাখ্যা

Area to be plastered = [2(l + b)×h] + (l×b)
= [2(25 + 12)×6] + (25×12)
= 744 m²
Cost of plastering = 744 × 75/100 = 558 tk

৬৮৫.

The supplement of an angle exceeds twice the angle by 30°. Then the angle is equal to-

ক
60°
খ
45°
গ
50°
ঘ
35°

ব্যাখ্যা

Question: The supplement of an angle exceeds twice the angle by 30°. Then the angle is equal to-

Solution:
Let the angle be x
Then, its supplement = 180 - x

According to the question,
180 - x = 2x + 30
⇒ 180 - 30 = 3x
⇒ 150 = 3x
⇒ x = 50°

৬৮৬.

If the difference between the circumference and diameter of a circle is 60 cm, then the radius of the circle is

ক
7cm
খ
9cm
গ
10cm
ঘ
14cm

ব্যাখ্যা

Question: If the difference between the circumference and diameter of a circle is 60 cm, then the radius of the circle is

Solution:
ধরি,
বৃত্তের ব্যাসার্ধ = r
বৃত্তের ব্যাস = 2r
বৃত্তের পরিধি = 2πr

প্রশ্নমতে,
2πr - 2r = 60
⇒ 2r(π - 1) = 60
⇒ r = (60/2)/{(22/7) - 1}
⇒ r = 30/{(22 - 7)/7}
⇒ r = (30 × 7)/15
∴ r = 14

∴ বৃত্তের ব্যাসার্ধ = r = 14 সে.মি.

৬৮৭.

If cosA = 8/17 than, what is the value of tanA = ?

ক
15/17
খ
15/8
গ
17/8
ঘ
8/15

ব্যাখ্যা

Question: If cosA = 8/17 than, what is the value of tanA = ?

Solution:
Given that,
cosA = 8/17

We know,
sin²A = 1 - cos²A = 1 - (8/17)²
= 1 - (64/289)
= (289 - 64)/289
= 225/289
∴ sinA = √(225/289) = 15/17

Now,
tanA = sinA/cosA = (15/17)/(8/17) = 15/8
∴ tanA = 15/8

৬৮৮.

In a rainfall, 7 cm of rain falls. What is the volume of water that falls on 2 hectares of ground?

ক
700 cubic meters
খ
850 cubic meters
গ
1400 cubic meters
ঘ
1700 cubic meters

ব্যাখ্যা

Question: In a rainfall, 7 cm of rain falls. What is the volume of water that falls on 2 hectares of ground?

Solution:

1 hectare = 10,000 m²

So, Area = (2 × 10000) m² = 20000 m²

Depth = 7 cm = 7/100 m

Volume = (Area × Depth) = 20000 × (7/100) m³= 1400 m³∴ The volume of water that falls on 2 hectares of ground is 1,400 cubic meters.

৬৮৯.

Find the area of the circle whose circumference is equal to the perimeter of a square of side 22 cm.

ক
ক) 767 cm²
খ
খ) 156 cm²
গ
গ) 263 cm²
ঘ
ঘ) 616 cm²

ব্যাখ্যা

Given that
Side of the square = 22 cm
According to the question,
Circumference of the circle = Perimeter of the square
⇒ 2πr = 4a
⇒ 2πr = 4 × 22
⇒ 2 × (22 / 7) × r = 88
⇒ r = 14 cm

Area of the circle = πr² ⇒ (22/ 7) × 14 × 14 = 616 cm²
∴ Area of circle is 616 cm².

৬৯০.

What is the slope of the line perpendicular to the line y = -5x + 9?

ক
ক) 5
খ
খ) -5
গ
গ) 1/5
ঘ
ঘ) -1/5

ব্যাখ্যা

y = -5x + 9
⇒ y + 5x = 9 .....(i)
সুতরাং (i) নং রেখাটির লম্বরেখার সমীকরণ 5y - x = k
⇒ y = 1/5x + k
∴ লম্ব রেখাটির ঢাল = 1/5

৬৯১.

A rectangular plot measuring 90 metre by 50 metre needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metre apart. How many poles will be needed?

ক
30
খ
60
গ
44
ঘ
56

ব্যাখ্যা

Length of the wire fencing
= perimeter
= 2 (90+ 50) = 280 metre

Two poles are kept 5 metre apart. Note that the poles are placed along the perimeter of the rectangular plot, not in a single straight line.

Hence, number of poles required
= 280/5
= 56 metre

৬৯২.

If sin(θ + 15°) = 3/√12 then 2sin²θ = ?

ক
ক) 1/2
খ
খ) 2
গ
গ) 1
ঘ
ঘ) 1/4

ব্যাখ্যা

Question: If sin(θ + 15°) = 3/√12 then 2sin²θ = ?

Solution:
sin(θ + 15°) = 3/√12
⇒ sin(θ + 15°) = 3/(2√3)
⇒ sin(θ + 15°) = √3/2
⇒ sin(θ + 15°) = sin60°
⇒ θ + 15° = 60°
⇒ θ = 45°

Now,
2sin²θ = 2(sin 45°)²
= 2 . (1/√2)²
= 2 . 1/2
= 1

৬৯৩.

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 5.5 meters, find the length of the ladder.

ক
11 m
খ
10.5 m
গ
10 m
ঘ
9.5 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 5.5 meters, find the length of the ladder.

Solution:

Let BC be the wall and AC be the ladder.
∠BAC = 60° and AB = 5.5 meter
In ΔABC,
cos60° = AB/AC
⇒ 1/2 = 5.5/AC
⇒ AC = 5.5 × 2
∴ AC = 11

৬৯৪.

The distance from the point P to two vertices A and B of an equilateral triangle are ।PA।= 2 and ।PB।= 3. What is the greatest possible value of ।PC। = ?

ক
ক) 5
খ
খ) 9
গ
গ) 6
ঘ
ঘ) 12

ব্যাখ্যা

ধরি,
। PA।= a = 2, ।PB।= b = 3 এবং । PC। = x

অর্ধ পরিসীমা = S = (a + b + x)/2
= (2 + 3 + x)/2
=(5 + x)/2
ΔABC = (1/2)[(√3/4)(a² + b² + x²) + 3√{s(s - a)(s - b)(s - c)}]
= (1/2)[(√3/4)(2² + 3² + x²) + 3√{(5 + x)/2 ((5 + x)/2 - 2)((5 + x)/2 - 3)((5 + x)/2 - x)}]
= (1/2)[(√3/4)(13 + x²) + 3√{((5 + x)/2)((x + 1)/2)((x - 1)/2)((5 - x)/2)}]
= (1/2)[(√3/4)(13 + x²) + 3√{((5 + x)/2)((5 - x)/2) × ((x + 1)/2)((x - 1)/2)}]
= (1/2)[(√3/4)(13 + x²) + 3√{((25 - x²)/4)((x² - 1)/4)}]
= (1/2)[(√3/4)(13 + x²) + 3√{(25 - x²)(x² - 1)/16}]
= (1/2)[(√3/4)(13 + x²) + (3/4)√{(25 - x²)(x² - 1)

ΔABC এর মান পাওয়া যাবে যখন (25 - x²)(x² - 1) ≥ 0

25 - x² ≥ 0 এবং x² - 1 ≥ 0
25 - x² ≥ 0 সত্য হবে যদি যখন
x = ±1, ±2, ±3, ±4, ±5
- 5 < x < 5
x এর সর্বোচ্চ মান হবে 5
। PC। = x = 5

৬৯৫.

What is the area of the triangle BCD?

ক
52
খ
48
গ
54
ঘ
42
ঙ
46

ব্যাখ্যা

Question: What is the area of the triangle BCD?

Solution:
ΔBAC থেকে পিথাগোরাসের সূত্রানুযায়ী,
BC² = AB² - AC² [BCA এক সমকোণ বলে]
⇒ BC² = 10² - 6²
⇒ BC² = 64
∴ BC = 8

আবার, ΔDEB তে DEB এক সমকোণ বলে,
BE² = BD² - DE²
⇒ BE2 = 13²- 5²
⇒ BE² = 144
∴ BE = 12

DEBC এর ক্ষেত্রফল = (1/2) × (BC + ED) × BE [ট্রাপিজিয়াম DEBC তে BC এবং DE সমান্তরাল এবং তাদের মধ্যে লম্ব দূরত্ব EB]
= (1/2) × (8 + 5) × 12 = 78
ΔDEB এর ক্ষেত্রফল = (1/2) × 12 × 5 = 30
∴ ΔBCD এর ক্ষেত্রফল = 78 - 30 = 48

৬৯৬.

If 100 square marbles of equal size were required to pave a corridor of dimension 6m x 24m then the length of each marble is -

ক
120 cm
খ
144 cm
গ
250cm
ঘ
100 cm

ব্যাখ্যা

Dimension of the corridor = 6m x 24 m
Area of the corridor = 6 x 24 m².
It is given that 100 square marbles are needed to cover the corridor of area 6 x 24 m².
Area of each marble = 6 x 24 / 100 m²= 144 / 100 m²= 1.44 m²
Since the marbles are in square shape, the length of each marble = sqrt(1.44) m = 1.2 m
Hence the answer is 1.2m = 1.2 x 100 cm = 120 cm.

৬৯৭.

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

ক
ক) 30, 22
খ
খ) 28, 20
গ
গ) 34, 26
ঘ
ঘ) 56, 48

ব্যাখ্যা

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

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

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

blackboard এর প্রস্থ = 20 সে.মি.
blackboard এর দৈর্ঘ্য = 20 + 8 = 28 সে.মি.

৬৯৮.

A hall is 15m long and 12m broad. If the sum of the areas of the floor and the ceiling is equal to of the areas of four walls, the volume of the hall is:

ক
ক) 720 m³
খ
খ) 1,200 m³
গ
গ) 900 m³
ঘ
ঘ) 1,800 m³

ব্যাখ্যা

ধরি
হল ঘরের উচ্চতা = h
দেওয়া আছে,
ঘরের দৈর্ঘ্য ১৫ মি.
ঘরের প্রস্থ ১২ মি.

প্রশ্নমতে,
২ ( ১৫ × ১২ ) = ২ × ( ১৫ + ১২ ) × h
বা, ২ × ২৭ × h = ২ ( ১৫ × ১২ )
বা, h × ২৭ = ১৫ × ১২
বা, h = (১৫ × ১২)/২৭
h = ২০/৩
আমরা জানি,
আয়তন= (১৫ × ১২×২০)/৩
= ১২০০ ঘন মিটার

৬৯৯.

A cube has a total surface area of 294 square meters. What is the length of its diagonal?

ক
7√3 m
খ
7 m
গ
5√3 m
ঘ
6√2 m

ব্যাখ্যা

Question: A cube has a total surface area of 294 square meters. What is the length of its diagonal?

Solution:
আমরা জানি, একটি ঘনকের মোট পৃষ্ঠের ক্ষেত্রফল = 6a²
প্রশ্নমতে, 6a² = 294
⇒ a² = 294/6
⇒ a² = 49
⇒ a = √49
⇒ a = 7 মিটার।

আমরা জানি,
একটি ঘনকের কর্ণের দৈর্ঘ্য = a√3
এখানে, a = 7
সুতরাং, কর্ণের দৈর্ঘ্য = 7√3 মিটার।

সুতরাং, ঘনকটির কর্ণের দৈর্ঘ্য হলো 7√3 মিটার।

৭০০.

If tanθ = 3/4, then cosθ = ?

ক
5/3
খ
4/3
গ
4/5
ঘ
3/5

ব্যাখ্যা

Question: If tanθ = 3/4, then cosθ = ?

Solution:
এখানে,
tanθ = 3/4 = লম্ব/ভূমি

∴ লম্ব = 3, ভূমি = 4
∴ অতিভুজ = √(32+ 42)
= √25 = 5

∴ cosθ = ভূমি/অতিভুজ
= 4/5