Geometry: Mensuration, Trigonometry

Hand of minute moves 360° in 60 minutes
In 20 minutes it will move = (360° × 20) / 60 = 120°

We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Question: If a regular polygon has each of its interior angle equal to 175°, then find the number of sides of the polygon.

Solution:
Given that,
Each interior angle = 175°

We know,
Sum of interior angles = (n - 2) × 180
Each interior angle = Sum of interior angles ÷ n

Now,
175 = [(n - 2) × 180] ÷ n
⇒ 175 × n = (n - 2) × 180
⇒ 175n = 180n - 360
⇒ 180n - 175n = 360
⇒ 5n = 360
⇒ n = 360/5
∴ n = 72

So, the polygon has 72 sides.

Question: The ratio of length and breadth of a rectangular park is 7 : 5. A man runs along its boundary at 8 km/hr and takes 9 minutes for one round. Find its area in sq. meters.

Solution:
One round of the park is equal to the perimeter of the park.
So, by completing one round, the man covers a distance equal to the perimeter of the park.
Now,
Distance or perimeter = speed × time
= 8 × (9/60)
= 1.2 km
= 1200 meters

Let,
Length = 7x and breadth = 5x
So, Perimeter,
2(7x + 5x) = 1200
⇒ 24x = 1200
∴ x =1200/24 = 50 meters

So, Length = 7 × 50 = 350 meters
And, Breadth = 5 × 50 = 250 meters

Area = Length × Breadth
= 350 × 250
= 87500 sq. m.

Question: The perimeter of a rectangular field is 110 meters. If the length of the field is 5 meters less than three times the width, what is the area of that field in square meters?

Solution:
ধরি, আয়তাকার ক্ষেত্রটির প্রস্থ = x মিটার
সুতরাং, ক্ষেত্রটির দৈর্ঘ্য = 3x - 5 মিটার

আয়তক্ষেত্রের পরিসীমা = 2(দৈর্ঘ্য + প্রস্থ)
প্রশ্নমতে,
2((3x - 5) + x) = 110
⇒ 2(4x - 5) = 110
⇒ 4x - 5 = 55
⇒ 4x = 60
⇒ x = 15 মিটার

সুতরাং, প্রস্থ = 15 মিটার।
দৈর্ঘ্য = 3x - 5 = (3 × 15) - 5 
= 45 - 5 = 40 মিটার।

আয়তক্ষেত্রের ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ
ক্ষেত্রফল = 40 × 15 = 600 বর্গ মিটার।

সুতরাং, ক্ষেত্রটির ক্ষেত্রফল হলো 600 বর্গ মিটার।

মনে করি, AB = h
খুঁটিটি ভূমির সাথে 60° কোণ তৈরি করে BC = √3 মিটার ছায়া তৈরি করে
তাহলে খুঁটির উচ্চতা h = ?
প্রশ্নমতে, tan60° = AB / BC
⇒ √3 = h/√3
∴ h = √3.√3 = 3 মিটার 

Question: Which trigonometric ratio is undefined at 90°?

Solution:
sin90° = 1
cos90° = 0
cot90°= 0

∴ tan90° = sin90°/cos90°
= 1 / 0
= ∞ (Undefined)

Question: ∠A and ∠B are complementary to each other. If ∠A = 30° + 3x and ∠B = 5x, find the value of ∠B.

Solution:
Here,
∠A = 30° + 3x and ∠B = 5x

For complementary angles,
∠A + ∠B = 90°
⇒ (30° + 3x) + 5x = 90°
⇒ 30° + 8x = 90°
⇒ 8x = 90° - 30°
⇒ 8x = 60°
⇒ x = 60°/8 = 7.5°

∴ ∠B = 5x = 5 × 7.5° = 37.5°

Question: The graphs of the equations 7x + 11y = 3 and 8x + y = 15 intersect at the point P, which also lies on the graph of the equation.

Solution: 
Given equations
7x + 11y = 3 .....(1)
8x + y = 15 .......(2)

From (1) we get,
8x + y = 15
∴ y = 15 - 8x ......(3)

Substitute into (1) then we get, 
⇒ 7x + 11(15 - 8x) = 3
⇒ 7x + 165 - 88x = 3
⇒ - 81x = - 162
∴ x = 2

Now from (3),
⇒ y = 15 - 8x
⇒ y = 15 - 16
∴ y = - 1

Now check which line also passes through P(2, -1),
A. 2x - y = 1
2(2) - (-1) = 4 + 1 = 5 ≠ 1 ; [Not valid]

B. 3x + 2y = 3
3(2) + 2(-1) = 6 - 2 = 4 ≠ 3 ; [Not valid]

C. 2x + y = 2
2(2) + (-1) = 4 - 1 = 3 ≠ 2 ; [Not valid]

D. 3x + 5y = 1
3(2) + 5(-1) = 6 - 5 = 1 = 1 ; [valid]

So correct answer is D. 3x + 5y = 1

let, the height is x
By applying Pythagoras Theorem, x² + 2² = 4²
⇒ x² = 16 - 4
⇒ x = √12 =  √(4.3)
⇒ x = 2√3 cm
So, the height is 2√3 cm

Question: In a right-angled triangle, the length of the medians from the vertices of acute angles are 7 cm and 4√6cm. What is the length of the hypotenuse of the triangle (in cm)?

Solution: 

Given that, 
AD = 7 cm
CE = 4√6 cm

Since, 4(AD² + CE²) = 5AC²
⇒ 4{(7)² + (4√6)²} = 5AC²
⇒ 4(49 + 96) = 5AC²
⇒ 4 × 145 = 5AC²
⇒ AC² = (4 × 145)/5
⇒ AC² = 4 × 29
⇒ AC = √(4 × 29)
∴ AC = 2√29 cm

Question: Point A is 10 km west of point B. Point B is 30 km north of point C. Point C is 20 km east of point D. What is the distance between points A and D?

Solution: 


AD = √(30² + 10²)
= √1000 
= 10√10 km

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

Let AE and BC be the heights of trees.

AE = 28 m
BC = 20 m

Horizontal distance between trees AB = DC

In EDC, EC² = ED² + DC² (Pythagoras theorem)
DC² = EC² - ED²
= 16² - 8²
= 256 - 64
DC² = 192
DC =√192 m.

Question: Two similar triangles have areas in the ratio 4 : 9. If the perimeter of the smaller triangle is 20, what is the perimeter of the larger triangle?

Solution: 
Given that, 
Two similar triangles
Areas ratio = 4 : 9
Perimeter of smaller triangle = 20

For similar triangles, the ratio of areas = square of the ratio of corresponding sides.
(Area of larger/Area of smaller) ​=(side of larger/side of smaller​)²
⇒ (9/4) = (k/1)² ; [Let the ratio of sides = k : 1 (larger : smaller)]
⇒ k² = 9/4
∴ k = 3/2

∴ Perimeter of larger : Perimeter of smaller = 3 : 2

∴ Perimeter of larger = 20 × (3/2) = 30 ; [Perimeter of smaller = 20]

So the perimeter of the larger triangle is 30.

Let, the area of larger part = x hector
∴ area of smaller part = (700 - x) hectors
The difference between the areas of the two parts
= x - (700 - x)
= 2x - 700
One-fifth of the average of the two areas
(1/5) × (700/2) [total area = 700]
= 70
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
2x−700 = 70
⇒ 2x = 770
⇒ x = 385.

∴ Smaller part of the land is = (700 - 385) = 315 hectares.

Question: If, cosec A + cot A = 5/2, find the value of, cosec A - cot A.

Solution:
Given that,
cosec A + cot A = 5/2

We know,
cosec²A - cot²A = 1
⇒ (cosec A + cot A)(cosec A - cot A) = 1

⇒ (5/2)(cosec A - cot A) = 1
⇒ cosec A - cot A = 1 ÷ (5/2)

∴ cosec A - cot A = 2/5

We know,
⇒ tanΘ = perpendicular/base
⇒ tanΘ = AB/BC
⇒ tanΘ = 6/2√3
⇒ tanΘ = √3
⇒ tanΘ = tan60°
∴ Θ = 60°

Question: Find the volume of a cylinder that is 14 cm tall with a base diameter of 6 cm.

Solution:
দেওয়া আছে,
সিলিন্ডারের উচ্চতা, h = 14 সে.মি.
সিলিন্ডারের ভূমির ব্যাস = 6 সে.মি.
∴ সিলিন্ডারের ভূমির ব্যাসার্ধ, r = 6/2 = 3 সে.মি.

আমরা জানি,
সিলিন্ডারের আয়তন, V = πr²h
= π × 3² × 14
= π × 9 × 14
= 126π ঘন সে.মি.

অতএব, সিলিন্ডারটির আয়তন 126π ঘন সে.মি.

Question: How many revolutions per minute does a 140 cm diameter scooter wheel need 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.

Question: If sinA + sin²A = 1, then the value of the expression (cos²A + cos⁴A) is - 

Solution:
Given,
sinA + sin²A = 1
⇒ sinA = 1 - sin²A
⇒ sinA = cos²A
⇒ sin²A = cos⁴A
⇒ 1 - cos²A = cos⁴A
∴ cos²A + cos⁴A = 1 

Question: If the areas of a circle and a square are equal then the ratio of their perimeters is-
(যদি একটি বৃত্ত এবং একটি বর্গের ক্ষেত্রফল সমান হয়, তবে তাদের পরিসরের অনুপাত হবে -)

Solution:
ধরা যাক, বর্গের প্রতিটি দিকের দৈর্ঘ্য = a cm এবং বৃত্তের ব্যাসার্ধ = r cm.

প্রশ্নানুসারে,
বৃত্তের ক্ষেত্রফল = বর্গের ক্ষেত্রফল
⇒ a² = πr²
⇒ a  = r√π​

∴ প্রয়োজনীয় অনুপাত = 2πr​/4a
= ​​2πr​/4r√π
= √π/2
= √π ​: 2

Question: The areas of a square and a rhombus are equal. The diagonals of the rhombus are 6 meters and 8 meters, respectively. What is the length of one side of the square?

Solution:
The area of the rhombus = (1/2) × Product of the diagonals
= (1/2) × 6 × 8
= 24 square meters

The area of the square = 24 square meters.
∴ Length of one side of the square = √24 meters
= 2√6 meters

∴ the length of one side of the square is 2√6 meters.

Let, length = x and Width = 3x - 8
ATQ, 
2(x + 3x - 8) = 80
Or, 4x - 8 = 80/2 = 40
Or, 4x = 48
Or, x = 12
So, width = 3 × 12 - 8 = 28

Question: If the area of a square is 529 square meters, what is the perimeter of the square? 

Solution:
Given,
The area of the square = 529 square meters.

Therefore,
The length of one side of the square = √529 meters = 23 meters.

We know,
The perimeter of a square = 4 × length of one side
= 23 × 4 meters
= 92 meters

Thus, the perimeter of the square is 92 meters.

Question: sin²12° + sin²78° = ?

Solution: 
Given that,
sin²12° + sin²78° 
= sin²12° + sin²(90° - 12°)
= sin²12° + cos²12°
= 1

Question: A cubical container with a side of 8 meters is to be painted on the entire outer surface area. If the cost of painting is Tk. 25 per square meter, what will be the total cost of painting?

Solution:
দেওয়া আছে,
ঘনকাকৃতির পাত্রের বাহুর দৈর্ঘ্য, a = 8 m
যেহেতু সম্পূর্ণ বাইরের পৃষ্ঠতলে রং করতে হবে, তাই রং করার ক্ষেত্রফল ঘনকটির সমগ্র পৃষ্ঠতলের ক্ষেত্রফলের সমান হবে।

ঘনকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 6a²
∴ রং করার ক্ষেত্রফল = 6 × 8²
⇒ রং করার ক্ষেত্রফল = 6 × 64
⇒ রং করার ক্ষেত্রফল = 384 বর্গমিটার।

এখন, প্রতি বর্গমিটারে রং করার খরচ = 25 টাকা
সুতরাং, মোট খরচ = (রং করার ক্ষেত্রফল) × (প্রতি বর্গমিটারে খরচ)
⇒ মোট খরচ = 384 × 25
⇒ মোট খরচ = 9600 টাকা।

অতএব, রং করতে মোট খরচ হবে 9600 টাকা।

Question: A kite is flown with a thread of length 200 meter. The thread is fully stretched and makes an angle of 60° with the horizontal, find the approximate height of the kite above the ground.

Solution:

Let height of the kite above the ground be AC = h.
Length of thread, BC = 200 m

From ΔABC,
sin60° = AC/BC
⇒ √3/2 = h/200
⇒ h = (200 × √3)/2
∴ h = 173.20

∴ Height of the kite above the ground be 173 m (approximately)

Question:

Solution: 
দেয়া আছে,

• ট্রেস (Trace): Matrix এর Trace হলো একটি বর্গাকার ম্যাট্রিক্সের প্রধান কর্ণের উপাদানগুলোর যোগফল। 
• প্রধান কর্ণ হলো ম্যাট্রিক্সের উপরের বাম কোণ থেকে নিচের ডান কোণ পর্যন্ত বিস্তৃত উপাদানগুলো।
প্রদত্ত ম্যাট্রিক্স A এর প্রধান কর্ণের উপাদানগুলো হলো 5, 9 এবং 13
সুতরাং, প্রদত্ত ম্যাট্রিক্সের ট্রেস (Trace) B = 5 + 9 + 13 = 27

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

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

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

∴ cosecθ = অতিভুজ/লম্ব
= 5/3

Question: If the difference between the circumference and diameter of a circle is 150 cm, then the diameter of the circle is-

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

প্রশ্নমতে,
2πr - 2r = 150
⇒ 2r(π - 1) = 150
⇒ r = (150/2){(22/7) - 1}
⇒ r = 75/(22 - 7)/7
⇒ r = (75 × 7)/15
∴ r = 35

∴ বৃত্তের ব্যাস = 2r = 2 × 35 = 70 সে.মি.

Question: What is the slope of a line perpendicular to the line whose equation is 20x - 2y = 6? 

Solution:
দেওয়া আছে,
20x - 2y = 6
⇒ - 2y = - 20x + 6
⇒ y = {(- 20)/(- 2)}x + 6/(- 2)
⇒ y = 10x - 3

এখানে, প্রদত্ত রেখার ঢাল, m₁= 10

আমরা জানি, দুটি রেখা পরস্পর লম্ব (Perpendicular) হলে তাদের ঢালদ্বয়ের গুণফল  -1 হয়।

ধরি, লম্ব রেখার ঢাল m₂
অর্থাৎ, m₁ × m₂ = -1

তাহলে,
10 × m₂ = -1
⇒ m₂= -1/10
∴m₂ = (-1/10)

∴ লম্ব রেখার ঢাল (-1/10)।

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 = (-1/2)x + 4
∴ ঢাল, m = -1/2

(খ) 2x + 4y = 12
⇒ 4y = -2x + 12
⇒ y = (-2/4)x + (12/4)
⇒ y = (-1/2)x + 3
∴ ঢাল, m = -1/2

(গ) y = 2x + 5
∴ ঢাল, m = 2

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

সুতরাং, দেখা যাচ্ছে যে শুধুমাত্র অপশন (গ) এর রেখার ঢাল মূল রেখার ঢাল থেকে ভিন্ন।

The perimeter is = 25 + 40 + 55 = 120 ft
Distance between pillars is 5 feet.
Required pillars = 120/5 + 1 = 25 [If it was a straight single line]
But as it is a triangular plot, one pillar overlaps 
So required pillar number is 25 - 1 = 24

From the figure, ∠DOC = 180°
Or, ∠DOB + ∠BOC = 180°
Or, ∠BOC = 180° - 63° = 117°

Question: If the volume of a sphere is 972π cm³, what is the surface area of the sphere?

Solution:
দেওয়া আছে,
গোলকের আয়তন, V = 972π cm³

আমরা জানি,
গোলকের আয়তন, V = (4/3)πr³
⇒ (4/3)πr³ = 972π
⇒ r³ = (972π × 3)/(4π)
⇒ r³ = (972 × 3)/4 
⇒ r³ = 243 × 3
⇒ r³ = 729 
∴ r = 9 cm

গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 4πr²
= 4π(9)²
= 4π × 81
= 324π cm².

অতএব, গোলকের সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 324π cm²

Question: A circular garden has a radius of 10 feet. If the radius is increased by 10%, what is the new area of the garden?

Solution:
Given that,
Original radius, r = 10 ft
And Increase radius by 10%

∴ New radius, r' = 10 + (10% of 10) = 10 + 1 = 11 ft.

We know, Area of circle = πr²

∴ New area with radius 11 ft
A′ = π(11²) = π × 121 = 121π ft²

Question: If a trapezium has two parallel sides measuring 4 cm and 6 cm, and its area is 100 square centimeters, what is the perpendicular distance (height) between the parallel sides?

Solution:
Given,
Parallel sides of a trapezium = 4 cm, and 6 cm

We know,
Area of trapezium = (1/2)(sum of the parallel sides) × distance between the parallel sides
100 = (1/2)(4 + 6) × distance
⇒ 100 = 5 × distance
⇒ distance = 100/5
∴ distance = 20 cm

So, the distance between the parallel lines of trapezium = 20 cm.

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

Question: In the triangle ABC, ∠B is equal to ∠C and D is a point strictly between B and C on side BC (i.e., D ≠ B, C). Which of the following is true?

Solution:
In △ABC, ∠B = ∠C.
∴ The triangle is isosceles.
AB = AC and point D lies strictly between B and C on side BC.



Now, In △ABC, 
∠ADC = ∠ABD + ∠BAD
∴ ∠ADC > ∠ABD 
∠ADC > ∠ACD        [∵ ∠B = ∠C and D ≠ B, C]
∴ AC > AD

Consider the diagram is shown above where QR represents the tree and PQ represents its shadow.
We have, QR = PQ
Let, ∠QPR = θ
tanθ = QR/PQ
= QR/QR [since QR = PQ]
= 1
= tan 45°
⇒ θ = 45°
i.e., the required angle of elevation = 45°

Question: What is the slope of a line perpendicular to the line whose equation is 20x - 2y = 6?

Solution:
The general equation of a straight line is
y = mx + c ......(1) (Where, m = slope)

If the slope of a line is m, then the slope of the line perpendicular to it is,
m' = - (1/m)

Now,
20x - 2y = 6
⇒ 2y = 20x - 6
∴ y = 10x - 3
Comparing with equation (1), we get,
∴ m = 10

∴ The slope of the perpendicular line is, m' = - (1/10)

The SHORTER sides have integral lengths equal to x and x + 1

Let the longest side be 'a'

So, a + x + (x +1) = 13
Or, a + 2x = 12 .......(1)

We know that the sum of the lengths of the shorter sides has to be more than the length of the longer one
Looking at the options, we can't have 8 or 10 as values for 'a'

Similarly, we can't have 2 or 4 as values for 'a' as it wouldn't be the longest side then.

So, the correct length of other side is 6

Question: Find the midpoint of the line segment joining the points P₁ = (- 5, - 2) and P₂ = (1, 6).

Solution:
দেওয়া আছে, P₁ = (- 5, - 2) এবং P₂ = (1, 6)

আমরা জানি, দুটি বিন্দুর (x₁, y₁) এবং (x₂, y₂) সংযোগকারী রেখাংশের মধ্যবিন্দু নির্ণয়ের সূত্র হলো: 
মধ্যবিন্দু = {(x₁ + x₂)/2 , (y₁+y₂)/2}

∴ মধ্যবিন্দু = {(- 5 + 1)/2 , (- 2 + 6)/2}
= ( - 4/2 , 4/2 )
= (- 2, 2)

সুতরাং, নির্ণেয় মধ্যবিন্দুটি হলো (- 2, 2)।

Question: If p is the circumference of the circle Q and the area of the circle is 25π, what is the value of p?

Solution:
বৃত্তের ক্ষেত্রফল (A) = πr²

প্রশ্নানুসারে, বৃত্তের ক্ষেত্রফল 25π।
∴ πr² = 25π
⇒ r²= 25
⇒ r = √25
⇒ r = 5

সুতরাং, বৃত্তের ব্যাসার্ধ (r) হলো 5।
এখন, বৃত্তের পরিধি (p) = 2πr
∴ p = 2π(5)
⇒ p = 10π

সুতরাং p-এর মান 10π

Let original length = 10
original breadth = 10
Then, original area = 10 × 10 = 100
Length is halved
⇒ New length = 10/2
= 5
breadth is tripled.
⇒ New breadth = 10 × 3
= 30
New area = 5 × 30
= 150
Increase in area
= new area - original area
= 150 - 100
= 50
Percentage increase in area
= {(increase in area/original area) × 100}%
= {(50/100) × 100}%
= 50%.

Question: What is the slope of a line perpendicular to the line whose equation is 2x + 5y = 10?

Solution:
প্রদত্ত সরল রেখার সমীকরণ: 2x + 5y = 10

y = mx + c আকারে লিখি, যেখানে m হলো রেখার ঢাল।
5y = - 2x + 10
⇒ y = (- 2/5)x + 2

অতএব, মূল রেখার ঢাল (m) = - 2/5

আমরা জানি, কোনো রেখার উপর লম্ব রেখার ঢাল m = - 1/m
= - 1/(- 2/5)
= 5/2

∴ লম্ব রেখার ঢাল = 5/2

Question:

Solution: