Geometry: Mensuration, Trigonometry

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

Question: cos²11° + cos²79° = ?

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

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.

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 

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

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 

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.

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

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°

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°

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°.

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.

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

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

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

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

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

Question: Which trigonometric ratio is undefined in value?

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

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।

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

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

Therefore, the length of the ladder is 16 m.

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

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

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

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

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

From (1), x = 8√3/2 = 4√3

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.

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

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

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.

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

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

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।

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

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°

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

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

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

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

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.

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।

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