Geometry: Mensuration, Trigonometry

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°

Question: If 5 workers can harvest 60 kg of wheat in 3 days, how many kilograms of wheat will 8 workers harvest in 5 days?

Solution: 
5 workers 3 days harvest = 60 kg
1 worker 1 day harvest = (60/15) kg
8 workers 5 days harvest = ( 60 × 40 ) / 15 kg
= 160 kg

Question: If a circle has a radius of 5 units, what is its circumference?

Solution: 
circumference = 2πr
=  2π × 5
= 10π units

Let the numerator be x.
The denominator of a fraction is 2 more than the numerator. Therefore, denominator = x + 2
Now,
Numerator/Denominator = 8/10
x/(x + 2) = 8/10

The numerator and denominator are increased by 4.

Therefore,
(x + 4)/{(x + 2) + 4} = 8/10
⇒ 10 (x + 4) = 8 ( x + 6)
⇒ 10 x + 40 = 8x + 48
⇒ 2x = 8
⇒ x = 4

Hence, the fraction is = 4/6
= 2/3.

Let the radius of the smaller circle be r cm
Then, radius of bigger circle = (r + 14) cm
Given, π(r+14)² - π(r)² = 1056
⇒ π(r² + 28r + 196) – πr² = 1056
⇒ 28πr + 196π = 1056
⇒ r + 7 = 1056/28π = 12
∴ r = 5

Question: If θ = 60°, then sec²θ - tan²θ = ?

Solution: 
Given, θ = 60°

Now,
sec²θ - tan²θ
= (sec60°)² - (tan60°)²
= 2² - (√3)²
= 4 - 3
= 1

As we know that volume of wire is volume of cylinder.
And in given question silver wire is drawn from 66 cubic centimetres of silver.
So we can equivalent the volume of cylinder with 66 cubic centimetres.
we know that volume of cylinder is Π×r²×l. [where l = length and r = radius]
Given, r = 1/2mm.
i.e. r = 0.5mm.
i.e. r = 0.05cm.
According to question Π×r²×l = 66
i.e. 22/7 × (0.05)²× l = 66
by solving this question we get l = 8400cm.
So, length of wire in metres is 84m.

Question: The measurement of a rectangle is 8 feet by 6 feet. What is the area of the smallest circle that can cover this rectangle entirely (so that no part of the rectangle is outside the circle)?

Solution: 
Here,
Diameter of the circle = √(8² + 6²)
= √(64 +36)
= √100
= 10 feet
∴ Radius of the circle = 10/2 = 5 feet

∴ Area of the circle = π × 5² = π × 25 = 25π sq. feet

Question: Find an equation of the vertical line containing the point (8, - 4)

Solution:

Given,
The given point is (8, -4).

A key property of a vertical line is that the x-coordinate of every point on the line always remains the same.
Since the line passes through the point (8, -4), the value of x for every point on the line will be 8.

Therefore, the required equation is:
x = 8

Question: A 60-meter cable is attached from the top of a vertical pole down to the ground. If the cable makes an angle of 30 degrees with the ground, find the height of the pole.

 Solution:
 
Let,
Height, AB = h 

Given that,
AC = 60m
∠ACB = 30°

∴ sin30°= AB/AC
⇒ 1/2 = h/60 
⇒ h = 60 × 1/2 
∴ h = 30 m

Δ ABC-তে AB = BC এবং AC অতিভুজ 
∴ ∠C = ∠A
এখন, ∠A + ∠B + ∠C = 180°
⇒ ∠C + 90° + ∠C = 180° [AB = BC এবং AC অতিভুজ (hypotenuse) হলে ∠B = 90°]
⇒ 2∠C = 180° - 90° = 90°
∴ ∠C = 45°

We know, Area of trapezium = 1/2(a + b)h
let the length of parallel side = x and x + 1
ATQ, 1/2(x + x + 1)2 = 25
⇒ 2x = 25 - 1
⇒ x = 12
∴ length of the longer side = 12 + 1 = 13 feet

The volume of the fish tank = 25×9×2 = 450 m³
Let height of the cylindrical container is = h
So, πr²h = 450
⇒ h = 450/πr²
= 450/π5² [As, Diameter of the cylinder is 10 m, so its radius is 5 m]
= 18/π

Question: When a diagonal of a square is 24 cm then find out of its perimeter.

solution:
Given that,
Diagonal of the square = 24 cm

We know, 
Diagonal of the square, d = √2 × a
a = d/√2 = 24/√2 = 12√2 
∴ a = 12√2 

∴ Perimeter of square = 4a = 4 × 12√2 = 48√2 cm

In △PQR,
QR + 2 = 2PQ
QR = 2PQ - 2 ..........(1)
PR = PQ + 10 ......... (2)
Perimeter = 40 m
Or, PQ + QR + RP = 40
Putting the value of PQ and QR from equation (1) and (2),
PQ + 2PQ - 2 + PQ + 10 = 40
Or, 4PQ = 32
Or, PQ = 8 cm ,which is the smallest side of the triangle.

Question: A cube has a total surface area of 384 square units. What is the volume of the cube?

Solution:
Given, total surface area of the cube, S = 384 square units.
We know, surface area of a cube, S = 6a²

According to the question,
6a² = 384
⇒ a² = 384 / 6
⇒ a² = 64
⇒ a² = 8²
⇒ a = 8

Again, we know, volume of the cube, V = a³
= 8³
= 512

Therefore, the volume of the cube is 512 cubic units.

Question: If r sinθ = 3, r cosθ = 4, then find the value of (4 tanθ + 1).

Solution:
r sinθ = 3
r cosθ = 4

Now,
(r sinθ)/(r cosθ) = 3/4
⇒ sinθ/cosθ = 3/4
⇒ tanθ = 3/4
⇒ 4 tanθ = 4 × 3/4 = 3
⇒ 4 tanθ + 1 = 3 + 1
∴ 4 tanθ + 1 = 4

Question: One diagonal of a rhombus is three times the other diagonal. If its area is 54 sq. cm, find the sum of the diagonals.

Solution:
ধরি,
রম্বসের একটি কর্ণ = x সেমি
রম্বসের অপর কর্ণ = 3x সেমি

আমরা জানি,
রম্বসের ক্ষেত্রফল = (1/2) × কর্ণদ্বয়ের গুণফল

প্রশ্নমতে,
(1/2) . x . 3x = 54
⇒ 3x² = 54 × 2
⇒ 3x² = 108
⇒ x² = 108/3
⇒ x² = 36
⇒ x = √36
∴ x = 6

এখন,
একটি কর্ণ = 6 সেমি
∴ অপর কর্ণ = 3 × 6 = 18 সেমি

∴ কর্ণদ্বয়ের সমষ্টি = 6 + 18 = 24 cm

Question: If cosA sinA = 0,then (cosA + sinA)² =?

Solution:
(cosA + sinA)²
= cos²A + 2 cosA sinA + sin²A
= 1 + 2.0 [sin²A + cos²A = 1]
= 1 + 0
= 1

Let the side of the square be a cm.
Then, its diagonal = √2 a cm.
Now, √2 a = 12√2
⇒ a = 12 cm.
Perimeter of square = 4a = 48 cm.
Perimeter of equilateral triangle = 48 cm.
Each side of the triangle = 16 cm.
Area of the triangle = ((√3/4)×16×16) cm²
= 64√3 cm²

Question: If a right-angled isosceles triangle has base 4 cm, then height is:

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Solution:
(Right-angled isosceles triangle) সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি = 4 cm.
সমকোণী সমদ্বিবাহু ত্রিভুজ এর ভূমি ও উচ্চতা সমান।
ভূমি = উচ্চতা = 4 cm.
∴ উচ্চতা = 4 cm

Question: If cotθ = 5/12, then secθ = ?

Solution:
এখানে,
cotθ = 5/12 = ভূমি/লম্ব

∴ ভূমি = 5, লম্ব= 12
∴ অতিভুজ = √(5²+ 12²)
= √169 = 13

∴ secθ = অতিভুজ/ভূমি
= 13/5

Question: A hollow cylinder has an internal radius of 8 cm, external radius of 12 cm, and height 15 cm. What is the volume of the material used?

Solution: 
Let, 
Internal radius (r) = 8 cm
External radius (R) = 12 cm
Height (h) = 15 cm

Volume of the the material used, 
V = πh(R² - r²) 
= π × 15 (144 - 64)
= π × 15 × 80
= 1200π

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

Solution: 
Given that, 
tan(θ + 45°) = 1
We know that, tan 45° = 1

Therefore, 
tan(θ + 45°) = tan 45°
⇒ θ + 45° = 45°
⇒ θ = 45° - 45°
∴ θ = 0°

So, sinθ = sin0° = 0

Question: A photographer who is 1.6 meters tall is standing 20 meters away from a tower. If the angle of elevation from his eye to the top of the tower is 45°, what is the height of the tower?

Solution:

এখানে,
ফটোগ্রাফারের উচ্চতা, CD = 1.6 মিটার
এখানে, CD = EB
টাওয়ারের উচ্চতা, = AB

এখন,
tan∠C = AE/CE
⇒ tan45° = AE/20
⇒ 1 = AE/20
∴ AE = 20

∴ AB = AE + BE
= 20 + 1.6
= 21.6 m

Question: If sec²θ - tan²θ = 1 and tan²θ = 3, then the value of θ when 0° ≤ θ ≤ 90° is?

Solution:
Given,
tan²θ = 3
⇒ tanθ = √3
⇒ tanθ = tan60°

∴ θ = 60°

Question: The perimeter of a circle measures 28π cm. What is the area of the circle in sq. cm?

Solution:
মনে করি
বৃত্তের ব্যাসার্ধ = r
∴ বৃত্তের পরিধি = 2πr
এবং বৃত্তের ক্ষেত্রফল = πr²

প্রশ্নমতে,
2πr = 28π
⇒ 2r = 28
⇒ r = 14

∴ বৃত্তের ক্ষেত্রফল = πr²
= π(14)²
= 196π sq. cm

Question: A ladder 26 m long is placed against a wall of height 13 m such that it just touches the top of the wall. Find the angle of elevation made by the ladder with the ground.

Solution:

AC = 26 meter
AB = 13 meter
∠ACB = θ

∴ sin θ = AB/AC = 13/26 = 1/2
⇒ sin θ = sin 30°

∴ θ = 30°

Paint Area = Total area - Non-Paint area
Subtracting width of the border from all sides we get,
Length = 20 - 4 - 4 = 12
Breadth = 15 - 4 - 4 = 7

∴ Paint Area = (20 x 15) - (12 x 7) = 300 - 84 = 216

Total Cost of painting the border = Rate x Area
= 6 x 216
= Tk. 1296

Question: A circular garden with a diameter of 20 meters is surrounded by a walkway of width 1 meter. What is the area of the walkway?

Solution:
Given that,
Radius of the garden = 20/2 = 10 m
Width of walkway = 1 m
So, radius of the outer circle (garden + walkway) = 10 + 1 = 11 m

We know,
Area of circle = πr²

Now,
Area of outer circle = π × (11²) = π × 121 = 121π m²
And,
Area of inner circle (garden only) = π × (10²)= π × 100 = 100π m²

Now,
Area of walkway = Area of outer circle - Area of inner circle = (121π - 100π) m² = 21π m²

So the area of the walkway = 21π m²

Question: The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

Solution:
Let the length of the rectangle = L cm
Let the breadth of the rectangle = B cm

Given that,
Ratio of perimeter to breadth = 5 : 1
Area of the rectangle = 216 cm²

Perimeter of rectangle = 2(L + B)
So, according to the ratio,
⇒ 2(L + B)/B = 5/1
⇒ 2(L + B) = 5B
⇒ 2L + 2B = 5B
⇒ 2L = 5B − 2B
⇒ 2L = 3B
∴ L = (3/2)B .........(1)

Also given that, 
Area = L × B = 216
∴ L × B = 216 ......... (2)

Substitute L from equation (1) into equation (2). Then we get,
⇒ (3/2)B × B = 216
⇒ (3/2)B² = 216
⇒ B² = 216 × (2/3)
⇒ B² = 144 = 12²
∴ B = 12 cm
Now find length, L = (3/2) × 12 = 18 cm

So, the length of the rectangle is 18 cm.

Question: A 40-meter cable is attached from the top of a vertical pole down to the ground. If the cable makes an angle of 30 degrees with the ground, find the height of the pole.

 Solution:
 
ধরি, উচ্চতা(Height), AB = h
দেয়া আছে, AC = 40m
∠ACB = 30°

∴ sin30°= AB/AC
⇒ 1/2 = h/40
⇒ h = 40 × 1/2
∴ h = 20 m

Question: The ratio between the perimeter and the breadth of a rectangle is 3 : 1. If the area of the rectangle is 288 sq. cm, what is the length and breadth of the rectangle?

Solution: 
Let, Length = l
And, breadth = b
∴ Perimeter of a rectangle = 2(l + b)

Now,
2(l + b)/b = 3/1
⇒ 2l + 2b = 3b
⇒ b = 2l

∴ Area, b × l = 288
⇒ 2l × l = 288
⇒ l² = 144
⇒ l = 12
∴ b = 2 × 12 = 24

∴ The length & breadth of the rectangle is 12 cm & 24 cm respectively. 

Surface area of sphere = 4πr² 
Is the new radius is 10% increased, then new surface area will be = 4π(1.1)²  = 4.84πr² 

Surface area Increased in percentage = (4.84πr²/4πr² × 100) - 100 =  121 - 100 = 21%

Question: In the figure, AOC is the diameter of the circle and arc AXB = (1/2)arc BYC. Find ∠BOC = ?

Solution:
Given that,
arc AXB = (1/2) arc BYC
∴ ∠AOB = (1/2) ∠BOC

We know that,
 ∠AOB + ∠BOC = 180º

Therefore,
(1/2) ∠BOC + ∠BOC = 180º {linear pair since AOC is the diameter}
⇒ (3/2) ∠BOC 180º
⇒ ∠BOC = (2/3) × 180º = 120º
∴  ∠BOC = 120º

Question: The perimeter of an equilateral triangle is 84√3 cm. Find its height.

Solution:
Given,
The perimeter of the equilateral triangle = 84√3 cm.
∴ Each side of the equilateral triangle = (84√3/3) = 28√3 cm.

We know,
The height of the equilateral triangle will be = (√3/2) × (28√3) = 42 cm

Question: Choose the equation of a circle with radius 6 and center (3, -5).

Solution: 
দেওয়া আছে,
বৃত্তের কেন্দ্র = (3, -5)
ব্যাসার্ধ = 6

আমরা জানি,  
বৃত্তের আদর্শ সমীকরণ,
(x - h)² +(y - k)² = r²
⇒ (x - 3)² + {y - (- 5)}² = 6² ; [এখানে h = 3, k = - 5 এবং r = 6]
∴ (x - 3)² + (y + 5)² = 36

সুতরাং, বৃত্তের সমীকরণ (x - 3)² + (y + 5)² = 36

Question: What is the slope of a line perpendicular to the line whose equation is 6x + 4y = 18?

Solution:
সরল রেখার সাধারণ সমীকরণ,
y = mx + c ...... (1) (এখানে m = ঢাল)

যদি কোনো রেখার ঢাল হয় m, তবে তার লম্ব (perpendicular) রেখার ঢাল হবে,
 m' = - (1/m)

এখন, 6x + 4y = 18
⇒ 4y = - 6x + 18
⇒ y = - (6/4) × x + 18/4
⇒ y = - (3/2) × x + 9/2

(1) নং এর সাথে তুলনা করে পাই, m = - (3/2)

∴ লম্ব (perpendicular) রেখার ঢাল হবে, m' = - {1/- (3/2)}
= 2/3

∴ লম্ব রেখার ঢাল = 2/3

Given, BC = 20 m
∠ACB = 30°

The total height of the telegraph post is (AB + CA) = ?
In ABC, tan 30° = AB/BC
1/√3 = AB 20

∴ AB = 20/√3m

Now, cos 30° = BC/AC
√3/2 = 20/AC
∴ AC = 40/√3 m

So, AB + CA
= (20/√3) + (40/√3)
= (60/√3)
= 20√3 m

Question: Determine the value of the 4^th term of the sequence: sin⁡(nπ/6)

Solution:
এখানে,
sin(nπ/6) এর চতুর্থ পদ = {sin(4 × π)/6}
= {sin(4 × 180°)/6}
= sin120°
= sin(90° + 30°) 
= cos30°
= √3/2

Let, the side of the square = 11x 
and, diameter of the circle = 7x
ATQ,
2π(7x/2) = 4×11x
Or, 7πx = 44x
x omits from both side.
So, the radius of the circle can't be determined from the given information.

sin 30° = 1/2, sin 60° = √3/2, cos 30° = √3/2 and cos 60° = 1/2

Putting these values, we get:

(1/2 + 1/2) - (√3/2 + √3/2)
= 1 – [(2√3)/2]
= 1 – √3

দেয়া আছে,
1 + sinθ = x cosθ
⇒ (1 + sinθ)/cosθ = x cosθ/cosθ [উভয়পক্ষে cosθ দ্বারা ভাগ করে পাই]
⇒ (1/cosθ) + sinθ/cosθ = x
⇒ secθ + tanθ = x ..........(i)
আমরা জানি,
sec²θ - tan²θ = 1
⇒ (secθ + tanθ)(secθ - tanθ) = 1
⇒ x(secθ - tanθ) = 1
⇒ secθ - tanθ = 1/x .........(ii)
সমীকরণ (i) থেকে (ii) বিয়োগ করে পাই,
(secθ + tanθ) - (secθ - tanθ) = x - (1/x)
2 tanθ = (x² - 1)/x
tanθ = (x² - 1)/2x.

Solution: 
Diameter = 2 × 4 = 8 cm = diagonal of the square 

let, the side of square is x 

√2x = 8 
⇒ x = 8/√2

∴ Area of the square = x² 
= (8/√2)² 
= 64/2
= 32 sq. cm