Geometry: Mensuration, Trigonometry

Question: The angle of elevation of the sun, when the height of a tower is √3 times the length of its shadow, is-

Solution:

Let, ∠ACB = θ
Then, AB/AC = √3
⇒ tan θ = √3 = tan60°

∴ θ = 60°

In Rhombus
Let,
a = length of each side
b = base
h = height
d₁,d₂ are diagonals
Then Perimeter = 4a
= 2√(d₁² + d₂²)

Perimeter = 2√(72² + 30²)
= 156 cm.

Length of the ladder = √(3² + 4²) = 5m

Question: Find the value of Cos(3π/4).

Solution:
Cos(3π/4)
= Cos{π - (π/4)}
= - Cos(π/4) [(π - θ) দ্বিতীয় চতুর্ভাগে পড়ে এবং দ্বিতীয় চতুর্ভাগে Cos ঋণাত্মক, তাই Cos(π - θ) = - Cosθ ]
= - Cos(45°)
= - (1/√2)

Question: The diagonal of a rectangular field is 15 m and its area is 108 sq. m. What will be the total expenditure in fencing the field at the rate of Tk. 5 per metre?

Solution:
Let the length and breadth of the rectangular field be x metres and y metres respectively.

Given that,
Diagonal, √(x² + y²) = 15
⇒ x² + y² = 225
And area, xy = 108 m²

We know, 
(x + y)² = x² + y² + 2xy
⇒ (x + y)² = 225 + 2 × 108
⇒ (x + y)² = 225 + 216
⇒ (x + y)² = 441
⇒ x + y = √441 = 21
∴ x + y = 21

∴ Perimeter of the field = 2(x + y) = 2(21) = 42 m

∴ Total expenditure for fencing = Perimeter × Rate
= 42 m × Tk. 5 per metre
= Tk. 210

So the total expenditure is Tk. 210.

Let, the length of the rectangle is l and breadth is b.
Where the breadth of the rectangular field is 60% of its length.
∴ b = 60l/100
      = 3l/5

Given that, Perimeter of the field = 800 m
⇒ 2(l + b) = 800
⇒ 2{l + (3l/5)} = 800
⇒ l + (3l/5) = 400
⇒ 8l/5 = 400
⇒ l = 250 m.
∴ b = 3l/5
= (3 × 250)/5
= 150 m

∴ Area = lb
= (250 × 150)
= 37500 m²

Question: sin(A + 18°) = √3/2, find the value of A.

Solution:
sin(A + 18°) = √3/2
⇒ sin(A + 18°) = sin60°
⇒ A + 18° = 60°
⇒ A = 60° - 18°
∴ A = 42°

Question: A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is:
(একটি ফাঁপা লোহার পাইপের দৈর্ঘ্য ২১ সেন্টিমিটার এবং এর বাইরের ব্যাস ৮ সেন্টিমিটার। যদি পাইপের পুরুত্ব ১ সেন্টিমিটার হয় এবং লোহার ঘনত্ব ৮ গ্রাম/সেন্টিমিটার^৩ হয়, তাহলে পাইপের ওজন কত হবে?)

Solution: 
পাইপটির দৈর্ঘ্য (L) = 21 cm
বাহ্যিক ব্যাস (D) = 8 cm
পাইপের প্রস্থ (t) = 1 cm

বাহ্যিক ব্যাসার্ধ R_external​ হল বাহ্যিক ব্যাসের অর্ধেক: 8/2 = 4 cm
অভ্যন্তরীণ ব্যাসার্ধ R_internal​ হল বাহ্যিক ব্যাসার্ধ থেকে প্রস্থ বিয়োগ:
R_internal = R_{external - t} = 4 - 1= 3 cm

লোহার আয়তন = π (4² - 3²) 21 [πr²h সূত্রানুসারে]
=  π (16 - 9) 21
= π 7 × 21
= 462 cm³

পাইপের ওজন = 462 × 8 g
= 3696 g
= 3.696 kg

Question: In a circle, if the inscribed angle on an arc is 35°, what is the measure of the central angle subtended by the same arc?
(কোন বৃত্তের একই চাপের উপর দণ্ডায়মান বৃত্তস্থ কোণ 35° হলে, কেন্দ্রস্থ কোণের পরিমাণ কত?)

Solution:
দেয়া আছে,
একই চাপের উপর দণ্ডায়মান বৃত্তস্থ কোণ = 35°
আমরা জানি,
কোন বৃত্তের একই চাপের উপর দণ্ডায়মান কেন্দ্রস্থ কোণ বৃত্তস্থ কোণের দ্বিগুণ।

∴ কেন্দ্রস্থ কোণ = 2 × বৃত্তস্থ কোণ
=2 × 35°
=70°

অতএব, কেন্দ্রস্থ কোণের পরিমাপ 70°।

There are 12 triangles in the figure.
These are: ABF, AEF, BCF, DEF, CDF, ACF, ADF, ACE, ABD, BCD, CDE, ACD

Question: The complement of an angle is 4 times the angle. Find the angle.

Solution:
Let the angle be x degrees.
The complement of an angle = 90° - x

According to the question,
90° - x = 4x
⇒ 90° = 4x + x
⇒ 90° = 5x
⇒ x = 90°/5
∴ x = 18°

∴ The angle is 18°

Number of sides in hexagon, n = 6
Sum of interior angles = (n - 2) x 180°
= (6 – 2) x 180°
= 720°

Question: ΔABC is a right triangle. In ΔABC, hypotenuse AC = 2, normal AB = 1; then which of the following is correct? 

Solution:



Given,
In right triangle ABC, hypotenuse AC = 2, normal AB = 1
Let, base BC = a

2² = a² + 1²
⇒ a² = 4 - 1
⇒ a = √3

sin(C) = AB/AC = 1/2 = sin 30° 
⇒ C = 30°
∴ ∠ACB = 30° 

Again, for ∠BAC, hypotenuse AC = 2, base AB = 1 and normal BC = √3
tan(A) = BC/AB = √3/1 = √3
∴ tan(A) = √3

Question: In the figure below, AB is perpendicular to BC and DB = DC. If AD = √10 and AC = 4 cm, what is the value of BC?

Solution:
ΔABD-এ,
BD² + AB² = AD² 
⇒ BD² + AB² = (√10)²
⇒ BD² + AB² = 10

 আবার, ΔABC-এ,
BC² + AB² = AC²
⇒ (BD + DC)² + AB² = 4²
⇒ BD² + DC² + 2BD.DC + AB² = 16
⇒ DC² + 2BD.DC = 6 (যেহেতু BD² + AB² = 10)
⇒ DC² + 2DC² = 6 (যেহেতু BD = DC)
⇒ 3DC² = 6
⇒ DC² = 2
⇒ DC = √2

অতএব, BC = 2DC (যেহেতু BD = DC)
= 2√2 cm

Question: A rhombus has one diagonal of 16 centimeters and an area of 192 square centimeters. What is the length of the second diagonal?

solution:
দেওয়া আছে,
রম্বসের ক্ষেত্রফল = 192 বর্গ সে.মি.
একটি কর্ণের দৈর্ঘ্য, d₁ = 16 সে.মি.
ধরি, অপর কর্ণের দৈর্ঘ্য = d₂ সে.মি.

আমরা জানি,
রম্বসের ক্ষেত্রফল = (1/2) × (কর্ণদ্বয়ের গুণফল)
∴ 192 = 1/2 × d₁ × d₂
⇒ 192 = 1/2 × 16 × d₂
⇒ 192 = 8 × d₂
⇒ d₂ = 192/8
∴ d₂ = 24 সে.মি.

অতএব, অপর কর্ণের দৈর্ঘ্য = 24 সে.মি.

Question: Find the equation of the line passing through (2, -3) and parallel to 5x - 2y + 6 = 0.

Solution:
Slope of given line: 
5x - 2y + 6 = 0 ---------(1)
⇒ y = (5/2)x + 3
∴ slope, m = 5/2

If a line has slope = m and passes through a point (x₁ , y₁).
Then the equation of the line is -
∴ y – y₁ = m(x – x₁)

Now, parallel to equation (1), has the same slope and pass through (2, -3).
So, the required line,
y + 3 = (5/2)(x - 2)
⇒ 2y + 6 = 5x - 10
⇒ 5x - 2y - 16 = 0

∴ the required line 5x - 2y - 16 = 0.

BC + AC > AB + BC can’t be true because AB = AC,
So, BC + AC > AC + BC
or, BC > BC, which is impossible

Question: What is the maximum value of cos θ?

Solution:
cosθ এর সর্বনিম্ন মান −1 এবং সর্বোচ্চ মান 1
∴ cosθ এর সর্বোচ্চ মান = 1

Question: The volume of a sphere with radius r is (4/3)πr³ and the surface area is 4πr². If a spherical ball has a surface area of 324π square centimeters. Find its volume.

Solution:
surface area = 4πr²

ATQ,
⇒ 4πr² = 324π
⇒ r² = 324π/4π
⇒ r² = 81 = 9²
∴ r = 9

Now,
volume = (4/3)πr³ 
= (4/3)π × (9)³
= (4/3)π × 729
= 972π

So, the volume would be 972π cm³

c²  = a²  + b²
⇒ 13²  = 5²  + b²
⇒ 169 = 25 + b²
⇒ b²  = 144
⇒ b = 12
So, Length of the third side is 12 cm

Walls area = 20 × 10 = 200 sq. m
∴ Cost of painting of 7/10th of the wall = (200 × 7/10 × 50) = 7000 taka

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

Solution:
Given that,
cos²θ + cos⁴θ = 1 ..........(1)
⇒ cos⁴θ = 1 - cos²θ
⇒ cos⁴θ = sin²θ
⇒ cos²θ.cos²θ = sin²θ
⇒ cos²θ = sin²θ/cos²θ
⇒ cos²θ = tan²θ

Now, putting cos²θ = tan²θ in equation (1) than we get,
⇒ tan²θ + tan⁴θ = 1

Total area to be painted = 25×12 +2(10×12 + 10×25) = 1040 sqr.ft

A paints = 200/5 = 40 sqr.ft per day
B paints = 250/2 = 125 sqr.ft per day

A + B = 40 + 125 = 165 sqr.ft
Number of days = 1040/165 = 6(10/33)

Question: The diagonal of a rectangular field is 15 metres and the difference between its length its length and width is 3 metres. The area of the rectangular field is-

Solution:
Let l and b be the length and breadth of the rectangle respectively.
Then,
⇒ √(l² + b²) = 15
⇒ (l² + b²) = (15)2
⇒ l² + b² = 225

And,
∴  l - b = 3
⇒ (l - b)² = 9
⇒ l² + b² - 2lb = 9
⇒ 225 - 2lb = 9
⇒ 2lb = 216
∴ lb = 108

Hence, area of the field = lb = 108m²

Question:

Solution:

Question: The slope of the line 4x - 8y = 16 is not the same as the slope of which one of the following lines?

Solution:
প্রথমে, প্রদত্ত রেখাটির ঢাল নির্ণয় করতে হবে। রেখাটির সমীকরণকে y = mx + c আকারে রূপান্তর করতে হবে। এখানে 'm' হলো ঢাল (Slope)।

প্রদত্ত রেখার সমীকরণ:
4x - 8y = 16
⇒ - 8y = - 4x + 16
 ⇒ y = (- 4/- 8)x + (16/- 8)
⇒ y = (1/2)x - 2
∴ এই রেখাটির ঢাল (m) হলো 1/2.

এবার, প্রদত্ত অপশনগুলোর প্রত্যেকটির ঢাল নির্ণয় করি:

ক) x - 2y = 8
⇒ - 2y = - x + 8
⇒ y = (- x/- 2) + (8/- 2)
⇒ y = (1/2)x - 4
∴ ঢাল, m = 1/2

খ) 3x - 6y = 12
⇒ - 6y = - 3x + 12
⇒ y = (- 3/- 6)x + (12/- 6)
⇒ y = (1/2)x - 2
∴ ঢাল, m = 1/2

গ) y = 3x + 5
∴ ঢাল, m = 3

ঘ) y = x/2 + 4
⇒ y = (1/2)x + 4
∴ ঢাল, m = 1/2

সুতরাং, দেখা যাচ্ছে যে শুধুমাত্র অপশন (গ) এর রেখার ঢাল মূল রেখার ঢাল থেকে ভিন্ন।
∴ সঠিক উত্তর: গ) y = 3x + 5

Question: A rectangular hall is 12.5 meters long and 6.4 meters wide. The cost of installing wooden flooring is Tk. 120 per square meter. What is the total cost of flooring the hall?

Solution: 
Given that,
Length of the hall = 12.5 meters
Width of the hall = 6.4 meters
Cost of wooden flooring = Tk. 120 per square meter

We know,
Area = Length × Width
= 12.5 m × 6.4 m
= 80 m²

∴ Total cost = Area × Cost per square meter
= 80 × 120 
= Tk. 9600 

Therefore, the total cost of installing wooden flooring in the hall is Tk. 9600.

Question: If the difference between the circumference and diameter of a circle is 240 cm, then the diameter of the circle is:

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

প্রশ্নমতে,
2πr - 2r = 240
⇒ 2r(π - 1) = 240
⇒ r = (240/2){(22/7) - 1}
⇒ r = 120/(22 - 7)/7
⇒ r = (120 × 7)/15
∴ r = 56

∴ বৃত্তের ব্যাস = 2r = 2 × 56 = 112 সে.মি.

Given that,
cos θ = 15/17
⇒ secθ = 17/15
⇒ sec²θ = 289/225
⇒ 1 + tan²θ = 289/225
⇒ tan²θ = 289/225 - 1
⇒ tan²θ = 64/225
⇒ tanθ = 8/15
⇒ cot(90 - θ) = 8/15