Geometry: Mensuration, Trigonometry

PrepBank · পাতা ১৭ / ২১ · ১,৬০১–১,৭০০ / ২,০৮৫

১,৬০১.

A room 6.2m × 8m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Tk. 15 per sq. meter, the cost of carpeting the room will be:

ক
Tk. 650
খ
Tk. 702
গ
Tk. 799
ঘ
Tk. 852

ব্যাখ্যা

Question: A room 6.2m × 8m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Tk. 15 per sq. meter, the cost of carpeting the room will be:

Solution:
Area of the carpet :
= [(6.20 - 0.20) × (8 - 0.20)] m²
= (6 × 7.8) m²
= 46.8 m²

∴ Cost of carpeting :
= Tk. (46.8 × 15)
= Tk. 702

১,৬০২.

A man walking at the speed of 6 kmph crosses a square field diagonally in 4 minutes. What is the area of the field?

ক
ক) 40000 Square Meters
খ
খ) 50000 Square Meters
গ
গ) 60000 Square Meters
ঘ
ঘ) 80000 Square Meters

ব্যাখ্যা

Question: A man walking at the speed of 6 kmph crosses a square field diagonally in 4 minutes. What is the area of the field?

Solution:
In 60 minutes he goes 6000 m
In 4 minutes he goes (6000 × 4)/60 m
= 400 m

Now,
√2 × length = 400
⇒ length = 400/√2
⇒ (length)² = (400/√2)²
⇒ Area = 160000/2
⇒ Area = 80000 Square Meters

১,৬০৩.

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

ক
ক) a + b < c
খ
খ) a - b < c
গ
গ) a + b = ৫
ঘ
ঘ) a + b ≥ c

ব্যাখ্যা

আমরা জানি,
ত্রিভুজের যে কোন দুই বাহুর সমষ্টি তার তার তৃতীয় বাহু অপেক্ষা বৃহত্তর।
আবার, ত্রিভুজের যে কোন দুই বাহুর অন্তর বা ব্যবধান তৃতীয় বাহু অপেক্ষা ক্ষুদ্রতর।
২য় অনুসিদ্ধান্ত অনুসারে,
অপশন b তে প্রদত্ত a - b < c সঠিক।
অর্থাৎ দুটি বাহুর অন্তর তৃতীয় বাহু থেকে ছােট।

১,৬০৪.

If cosB + cos²B = 1, then sin²B + sin⁴B =?

ক
0
খ
1
গ
2
ঘ
3

ব্যাখ্যা

Question: If cosB + cos²B = 1, then sin²B + sin⁴B =?

Solution:
cosB + cos²B = 1
⇒ cosB = 1 - cos²B
⇒ cosB = sin²B

sin²B + sin⁴B
= sin²B + (sin²B)²
= sin²B + cos²B
= 1

১,৬০৫.

If 0 ≤ x ≤ 5. What is the maximum value of cos(x/3)?

ক
ক) 1
খ
খ) 0
গ
গ) 1/2
ঘ
ঘ) √3/2

ব্যাখ্যা

x= 0 হলে
cos(x/3) = cos(0/3)
= cos0
= 1
অন্য যেকোনো মানের জন্য cos এর মান 1 অপেক্ষা ছোট হবে।

১,৬০৬.

There are two poles, one on each side of the road. The higher pole is 54 m high. From the top of this pole, the angle of depression of the top and bottom of the shorter pole is 30° and 60° respectively. Find the height of the shorter pole.

ক
40 m
খ
32 m
গ
36 m
ঘ
35 m

ব্যাখ্যা

Question: There are two poles, one on each side of the road. The higher pole is 54 m high. From the top of this pole, the angle of depression of the top and bottom of the shorter pole is 30° and 60° respectively. Find the height of the shorter pole.

Solution:

Let AB and CD be the two poles.
Let AC = x m
CD = h m

Now, in triangle ABC,
tan60° = AB/AC
⇒ √3 = 54/AC
∴ AC = 18√3 m

Clearly, AC = DE = 18√3 m

In triangle BED,
tan30° = BE/DE
⇒ BE = DE tan 30
⇒ BE = 18 √3 / √3 m
⇒ BE = 18 m
⇒ CD = AE = AB - BE
⇒ CD = 54 - 18 = 36 m

Therefore, the height of the shorter pole = 36 m.

১,৬০৭.

A school room is be built to accommodate 70 children so as to allow 2.2 m² of floor and 11 m³ of space for each child. If the room be 14 metres long, what must be its breadth and height?

ক
10 m, 5 m
খ
11 m, 5 m
গ
11.5 m, 5.5 m
ঘ
9 m, 2 m

ব্যাখ্যা

Question: A school room is be built to accommodate 70 children so as to allow 2.2 m² of floor and 11 m³ of space for each child. If the room be 14 metres long, what must be its breadth and height ?
(৭০ জন ছাত্র-ছাত্রী ধারণ করার জন্য একটি বিদ্যালয় কক্ষ তৈরি করা হবে, যেখানে প্রতিটি শিশুর জন্য ২.২ বর্গমিটার মেঝে এবং ১১ ঘনমিটার স্থান থাকবে। কক্ষটির দৈর্ঘ্য ১৪ মিটার হলে, তার প্রস্থ এবং উচ্চতা কত হবে?)

Solution:
দেওয়া আছে,
ছাত্র সংখ্যা = ৭০
প্রতিটি ছাত্রের জন্য:
মেঝের ক্ষেত্রফল = ২.২ বর্গ মিটার
স্থান (ঘন মিটার): ১১ ঘন মিটার
রুমটির দৈর্ঘ্য = ১৪ মিটার

প্রতিটি ছাত্রের জন্য ২.২ বর্গ মিটার মেঝে স্থান প্রয়োজন। তাহলে, ৭০ জন ছাত্রের জন্য মোট মেঝে এলাকা হবে:
মোট মেঝের ক্ষেত্রফল = ৭০ × ২.২ = ১৫৪ বর্গ মিটার

প্রতিটি ছাত্রের জন্য ১১ ঘন মিটার স্থান প্রয়োজন। তাহলে, ৭০ জন ছাত্রের জন্য মোট স্থান হবে:
মোট স্থান = ৭০ × ১১ = ৭৭০ ঘন মিটার

রুমটির দৈর্ঘ্য দেওয়া আছে L = ১৪ মিটার, প্রস্থ b এবং উচ্চতা h বের করতে হবে।

মেঝের ক্ষেত্রফল হলো দৈর্ঘ্য এবং প্রস্থের গুণফল:
মেঝের ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ = ১৪ × b
১৪ × b = ১৫৪
এটি থেকে b বের করি:
b = ১৫৪ / ১৪ = ১১ মিটার

তাহলে, রুমটির প্রস্থ ১১ মিটার।

রুমটির মোট স্থান হলো দৈর্ঘ্য, প্রস্থ এবং উচ্চতার গুণফল:
স্থান = দৈর্ঘ্য × প্রস্থ × উচ্চতা = ১৪ × ১১ × h
১৪ × ১১ × h = ৭৭০

এটি থেকে h বের করি:
১৫৪ × h = ৭৭০
h = ৭৭০ / ১৫৪ = ৫ মিটার

তাহলে, রুমটির উচ্চতা ৫ মিটার।

চূড়ান্ত উত্তর:
রুমটির প্রস্থ ১১ মিটার।
রুমটির উচ্চতা ৫ মিটার।

১,৬০৮.

What is the volume of a cube whose surface area is 150?

ক
125
খ
118
গ
112
ঘ
108

ব্যাখ্যা

Question: What is the volume of a cube whose surface area is 150?

Solution:
Let,
One side of the cube = x

ATQ,
6x² = 150
⇒ x² = 25
∴ x = 5

∴ the volume of the cube = 5³ = 125

১,৬০৯.

If x is the length of a median of an equilateral triangle, then the area is

ক
ক) √(x² +2)
খ
খ) x - 2
গ
গ) x²√3/3
ঘ
ঘ) x/2

ব্যাখ্যা

Consider an equilateral triangle ABC having sides a and a median AD of length x unit'.
In an equilateral triangle, the median is always the perpendicular bisector of the triangle.
So, BD=a/2
In triangle ABD, by pythagoras theorem, we have
AB²=AD²+BD²⟹a²=x²+(a/2)²⟹a²=x²+ a²/4
⟹3a²/4=x²or,a²=4x²/3

Now, area of equilateral triangle =√3a²/4
=(√3/4) × (4x²/3)
=√3x²/3

১,৬১০.

A sphere is created with half the radius of the original sphere. What is the ratio of the volume of the original sphere to the volume of the new sphere?

ক
ক) 1 : 8
খ
খ) 8 : 1
গ
গ) 2 : 1
ঘ
ঘ) 1 : 2

ব্যাখ্যা

Question: A sphere is created with half the radius of the original sphere. What is the ratio of the volume of the original sphere to the volume of the new sphere?

Solution:
ধরি,
মূল গোলকের ব্যাসার্ধ r = 2x একক
নতুন গোলকের ব্যাসার্ধ r₁ = x একক

মূল গোলকের আয়তন = (4/3)πr³
= (4/3)π(2x)³
= (4/3)π × 8x³

নতুন গোলকের আয়তন = (4/3)πr₁³
= (4/3)πx³
মূল গোলক : নতুন গোলক = (4/3)π × 8x³ : (4/3)πx³
= 8 : 1

১,৬১১.

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is ____ metre square.

ক
ক) 1520
খ
খ) 2420
গ
গ) 2480
ঘ
ঘ) 2520

ব্যাখ্যা

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103
Solving the two equations, we get:
l = 63 and b = 40
∴ Area = (l x b) = (63 x 40) m² = 2520 m²

১,৬১২.

If θ = 30°, then sec²θ − tan²θ = ?

ক
1/√2
খ
1/2
গ
1
ঘ
√3/2

ব্যাখ্যা

Question: If θ = 30°, then sec²θ − tan²θ = ?

Solution:
Given, θ = 30°

Now,
sec²θ - tan²θ
= (sec30°)² - (tan30°)²
= (2/√3)² - (1/√3)²
= 4/3 - 1/3
= 3/3
= 1

১,৬১৩.

The area of the right triangle is 184cm^2.One of its leg is 16cm long. Find the length of the other leg.

ক
ক) 23cm
খ
খ) 22cm
গ
গ) 24cm
ঘ
ঘ) 20cm

ব্যাখ্যা

Area of the triangle = (1/2) × base × height
⇒ 184 = (1/2) × 16 × other leg
So,
other leg = (184 × 2)/16
= 23 cm

১,৬১৪.

Which one of the following is true for 0° < θ < 90° ?

ক
cosθ ≤ cos²θ
খ
cosθ < cos²θ
গ
cosθ > cos²θ
ঘ
cosθ ≥ cos²θ

ব্যাখ্যা

Let, θ = 60^∘
⇒ cosθ = cos60^∘ = 1/2 = 0.5
⇒ cos²θ= cos²60^∘ = (1/2)² = 1/4 = 0.25
0.5 > 0.25
∴ cosθ > cos²θ

১,৬১৫.

How many bricks, each measuring 25 cm × 12 cm × 6 cm, are required to build a wall measuring 5 m × 3 m × 12 cm?

ক
2,000
খ
1,000
গ
3,000
ঘ
2,400
ঙ
4,000

ব্যাখ্যা

Question: How many bricks, each measuring 25 cm × 12 cm × 6 cm, are required to build a wall measuring 5 m × 3 m × 12 cm?

Solution:
Wall Dimensions,
Length = 5 m = 500 cm
Width = 3 m = 300 cm
Height = 12 cm

Brick Dimensions,
Length = 25 cm
Width = 12 cm
Height = 6 cm

Volume of the wall = Length × Width × Height
= 500 × 300 × 12
= 1,800,000 cm³

Volume of one brick = Length × Width × Height
= 25 × 12 × 6
= 1,800 cm³

∴ Number of bricks = Volume of the wall ÷ Volume of one brick
= 1,800,000 ÷ 1,800
= 1,000

∴ The number of bricks needed to construct the wall is 1,000.

১,৬১৬.

The supplement of an angle is twice the angle. Find the angle.

ক
50°
খ
70°
গ
80°
ঘ
60°

ব্যাখ্যা

Question: The supplement of an angle is twice the angle. Find the angle.

Solution:
Let the angle be x degrees.
The supplement of an angle = 180° - x

According to the question,
180° - x = 2x
⇒ 180° = 2x + x
⇒ 180° = 3x
⇒ x = 180°/3
∴ x = 60°

∴ The angle is 60°

১,৬১৭.

A solid iron cube of 4m length is converted into a wire of 100m length. the circumference of the wire is-

ক
2.835m
খ
3.935m
গ
3.835m
ঘ
2.535m

ব্যাখ্যা

Question: A solid iron cube of 4m length is converted into a wire of 100m length. the circumference of the wire is-

Solution:
the volume of the cube is = 4³ m³
= 64 m³

the volume of the wire will be same as the cube.
let,
the radius of the wire is = r
∴ πr² × l = 64
⇒ r² = 64/(3.1416 × 100)
⇒ r² = 0.2037
⇒ r = 0.4513

∴ circumference = 2πr
= 2 × 3.1416 × 0.4513
= 2.835m

১,৬১৮.

{(1 - tan²60°)/(1 + tan²60°)} + sin²60° = ?

ক
√3/2
খ
1/2
গ
√3/4
ঘ
1/4

ব্যাখ্যা

Question: {(1 - tan²60°)/(1 + tan²60°)} + sin²60° = ?

Solution:
(1-tan²60°)/(1+tan²60°) + sin²60°
= (1-(√3)²)/(1+(√3)²) + (√3/2)²
= (1 - 3)/(1+3) + 3/4
= -2/4 + 3/4
= 1/4

১,৬১৯.

ক
3
খ
1/2
গ
4
ঘ
5/8

ব্যাখ্যা

Question:

Solution:

১,৬২০.

A circle and a rectangle have the same perimeter. The sides of the rectangle are 9 cm and 13 cm. what is the area of the circle?

ক
625 sq. cm
খ
308 sq. cm
গ
154 sq. cm
ঘ
77 sq. cm

ব্যাখ্যা

Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 9 cm and 13 cm. what is the area of the circle?

Solution:
Perimeter of the rectangle = 2(9 + 13)
= 44 cm

∴ Circumference of circle = 44 cm
⇒ 2πr = 44
⇒ r = 44/2π
⇒ r = (44 × 7)/(2 × 22)
∴ r = 7 cm

∴ Area of circle = πr²
= (22/7) × (7)²
= 154 sq. cm

১,৬২১.

Find the volume of the cylinder whose radius 14 m and height 1.5 m.

ক
ক) 780 m³
খ
খ) 820 m³
গ
গ) 924 m³
ঘ
ঘ) 960 m³

ব্যাখ্যা

Question: Find the volume of the cylinder whose radius 14 m and height 1.5 m.

Solution:
We know that,
Volume of a cylinder = πr²h
= ((22/7) × 14 × 14 × 1.5) m³
= 924 m³

১,৬২২.

The area of a rectangle and square are equal. The side of the square is 12 cm and the smaller side of the rectangle is one-third that of the square. The length of the other side of the rectangle would be-

ক
54 cm
খ
48 cm
গ
36 cm
ঘ
72 cm

ব্যাখ্যা

Question: The area of a rectangle and square are equal. The side of the square is 12 cm and the smaller side of the rectangle is one-third that of the square. The length of the other side of the rectangle would be-

Solution:
given that,
Side of the square = 12 cm
Smaller side of the rectangle = one-third of the square’s side = 12/3 = 4 cm
And The area of a rectangle and a square are equal.

Now,
Area of the square = 12² = 144 cm²

∴ Area of rectangle = 144cm² [The area of a rectangle and a square are equal]

Let the other side of rectangle = L
Now,
4 × L = 144
⇒ L = 144/4
∴ L = 36 cm

So the other side of the rectangle is 36 cm.

১,৬২৩.

If a ladder touches the roof of a wall and makes an angle of 45° with the 15 metre long wall, then the length of the ladder is-

ক
ক) 30m
খ
খ) 15m
গ
গ) 45m
ঘ
ঘ) 15√2m

ব্যাখ্যা

In right angled triangle ABC,
cos 45° = BC/AC
Or, 1/√2 = 15/AC [As, opposite angle of AB and BC is equal, length of both line is also equal]
∴ AC = 15√2

Therefore, the length of the ladder is 15√2 m.

১,৬২৪.

The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

ক
ক) 4 cm
খ
খ) 6 cm
গ
গ) 8 cm
ঘ
ঘ) 12 cm

ব্যাখ্যা

Question: The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:

Solution:
surface area of sphere = 4πr²
Curved Surface area of cylinder =2πr₁h
diameter = 12 cm
radius, r₁ = 6 cm

⇒ 4πr²=2πr₁h
⇒ r²= (6×12)/2
⇒ r²= 36
⇒ r = 6

radius of sphere 6 cm.

১,৬২৫.

A rectangular sheet of paper, 10cm long and 8cm wide has squares of side 2cm cut from each of its corner. The sheet is then folded to form a tray of depth 2cm. What is the volume of this tray?

ক
48 cm³
খ
56 cm³
গ
24 cm³
ঘ
36 cm³

ব্যাখ্যা

Question: A rectangular sheet of paper, 10cm long and 8cm wide has squares of side 2cm cut from each of its corner. The sheet is then folded to form a tray of depth 2cm. What is the volume of this tray?

Solution:
Length of tray = 10 - (2 × 2) = 10 - 4 = 6 cm.
Breadth of tray = 8 - (2 × 2) = 4 cm.
Depth of tray = 2 cm.

∴ Volume of tray = 6 × 4 × 2 = 48 cm³

১,৬২৬.

Due to sun, a 6ft man casts a shadow of 4ft, whereas a pole next to the man casts a shadow of 36ft. What is the height of the pole?

ক
ক) 23 ft
খ
খ) 42 ft
গ
গ) 54 ft
ঘ
ঘ) 68 ft

১,৬২৭.

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?

ক
16 cm
খ
18 cm
গ
15 cm
ঘ
24 cm

ব্যাখ্যা

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:

2(l + b)/b = 5/1

⇒ 2l + 2b = 5b

⇒ 3b = 2l

∴ b = (2/3)l

Then,
Area = 216 cm²
⇒ l × b = 216
⇒ l × (2/3)l = 216
⇒ l² = 324
∴ l = 18 cm

১,৬২৮.

Calculate the area of the triangle with a perimeter of 180cm and sides in the ratio 2 : 3 : 4.

ক
300 cm²
খ
900√15 cm²
গ
3000√15 cm²
ঘ
300√15 cm²

ব্যাখ্যা

Question: Calculate the area of the triangle with a perimeter of 180cm and sides in the ratio 2 : 3 : 4.

Solution:
Let,
The lengths of the triangle's sides be 2x, 3x, and 4x.
Triangle's perimeter = 180cm

Now,
2x + 3x + 4x = 180
⇒ 9x = 180
⇒ x = 20 cm

Then the sides of the triangle are 40 cm, 60 cm, and 80 cm.
Let a = 40 cm, b = 60 cm, c = 80 cm and s = 180/2 = 90 cm

∴ Area of triangle = √[s(s - a)(s - b)(s - c)]
= √[90 × 50 × 30 × 10]
= √1350000
= √(90000 × 15)
= 300√15 cm²

১,৬২৯.

The perimeter of a rectangle and a square are 160 m each. The area of the rectangle is less than that of the square by 100 sq m. The length of the rectangle is -

ক
ক) 30 m
খ
খ) 40 m
গ
গ) 50 m
ঘ
ঘ) 60 m

ব্যাখ্যা

The perimeter of the square = 160 m.
Side of square = (160/4) m
= 40 m.
Area of square = (40 × 40) m² = 1600 m²
Area of rectangle = (1600 - 100) m² = 1500 m²
Let the length and breadth of the rectangle be 'l' and 'b' respectively.
Then, 2(l + b) = 160
⇒ (l + b) = 80
⇒ b = 80 - l.

∴ lb = 1500
⇒ l(80 - l) = 1500
⇒ 80l - l² = 1500
⇒ l² - 80 l + 1500 = 0
⇒ (l - 50)(l - 30) = 0
⇒ l = 50. or l = 30

Hence, length = 50 m, breadth = 30 m.

১,৬৩০.

A person 1.5 meters tall sees the top of a building in a small mirror placed on the ground. The mirror is 2 meters away from the person's feet and 80 meters away from the base of the building. What is the height of the building?

ক
45 meters
খ
54.5 meters
গ
60 meters
ঘ
80 meters

ব্যাখ্যা

Question: A person 1.5 meters tall sees the top of a building in a small mirror placed on the ground. The mirror is 2 meters away from the person's feet and 80 meters away from the base of the building. What is the height of the building?

Solution:

ধরি, মানুষের উচ্চতা, AB = 1.5 m
মানুষ এবং আয়নার দূরত্ব, BC = 2 m
ভবনের উচ্চতা, ED = h
ভবন এবং আয়নার দূরত্ব, CD = 80 m

আলোর প্রতিফলনের সূত্র অনুসারে, ∠ACB = ∠ECD (আপতন কোণ = প্রতিফলন কোণ)।
∴ ΔABC এবং ΔEDC সদৃশ।

সদৃশ ত্রিভুজের ধর্ম অনুসারে:
AB/ED = BC/CD
⇒ 1.5/h = 2/80
⇒ h × 2 = 1.5 × 80
⇒ 2h = 120
∴ h = 60 m

অতএব, ভবনটির উচ্চতা = 60 meters

১,৬৩১.

What is the parameter of a rectangle that is 24 meter wide and has the same area as another rectangle that is 64 meter long and 48 meter wide?

ক
ক) 112 meter
খ
খ) 152 meter
গ
গ) 224 meter
ঘ
ঘ) 256 meter
ঙ
ঙ) 304 meter

ব্যাখ্যা

Question: What is the parameter of a rectangle that is 24 meter wide and has the same area as another rectangle that is 64 meter long and 48 meter wide?

Solution:
ধরি
আয়তক্ষেত্রের দৈর্ঘ্য = x

প্রশ্নমতে
x × 24 = 64 × 48
x = (64 × 48)/24
x = 128

অতএব
পরিসমা = 2(128 + 24) মিটার = 304 মিটার

১,৬৩২.

The surface area of a sphere is 144π cm². Find the radius of the sphere.

ক
12 cm
খ
10 cm
গ
6 cm
ঘ
11 cm

ব্যাখ্যা

Question: The surface area of a sphere is 144π cm². Find the radius of the sphere.

Solution:
Let, r is radius of the sphere

A.T.Q,
4πr² = 144π
r² = 36
∴ r = 6

The radius of the sphere is 6 cm.

১,৬৩৩.

A cylindrical rod of iron, whose height is equal to its radius, is melted and cast into spherical balls whose radius is 1/4 the radius of the rod. Find the number of balls?

ক
6
খ
48
গ
24
ঘ
36

ব্যাখ্যা

Question: A cylindrical rod of iron, whose height is equal to its radius, is melted and cast into spherical balls whose radius is 1/4 the radius of the rod. Find the number of balls?

Solution:
ধরি, সিলিন্ডার আকৃতির রডের ব্যাসার্ধ r এবং উচ্চতা h
দেওয়া আছে, রডের ব্যাসার্ধ এবং উচ্চতা সমান ।
∴ রডের আয়তন = πr²h = πr² × r = πr³ ঘন একক

আবার ধরি,
গোলকের ব্যাসার্ধ, r' = r/4

∴ প্রতিটি গোলকের আয়তন = (4/3) × πr'³ = (4/3) × π × (r/4)³
= πr³/48 ঘন একক

∴ গোলকের সংখ্যা = রডের আয়তন/প্রতিটি গোলকের আয়তন
= πr³/(πr³/48)
= (πr³/πr³) × 48
= 48

১,৬৩৪.

ABCD is a rhombus. AB = 3x - 2, AC = 4x + 4, and BD = 2x. Find x.

ক
5
খ
3
গ
2
ঘ
1.5

ব্যাখ্যা

Question: ABCD is a rhombus. AB = 3x - 2, AC = 4x + 4, and BD = 2x. Find x.

Solution:
A rhombus is a quadrilateral with four sides of equal length. Rhombuses have diagonals that bisect each other at right angles.

Thus, we can consider the right triangle AED and use the Pythagorean Theorem to solve for x. From the problem:
AD = 3x - 2
BD = 2x
AC = 4x + 4

Because the diagonals bisect each other, we know:
ED = x
AE = 2x + 2

Using the Pythagorean Theorem,
a² + b² = c²
∴ AE² + ED² = AD²
⇒ (2x + 2)² + (x)² = (3x - 2)²
⇒ 4x² + 8x + 4 + x² = 9x² - 12x + 4
⇒ 0 = 4x² - 20x

Factoring,
0 = x and 0 = 4x - 20
The first solution is nonsensical for this problem.
0 = 4x - 20
∴ x = 5

১,৬৩৫.

The top and bottom of a flag on a building subtend angles of 60° and 30° respectively at a point B which is 48 meter away from the building. Find the height of the flag?

ক
32 m
খ
32√3 m
গ
18.49 m
ঘ
16 m

ব্যাখ্যা

Question: The top and bottom of a flag on a building subtend angles of 60° and 30° respectively at a point B which is 48 meter away from the building. Find the height of the flag?

Solution:

Let height of building be AC = X and height of flag be CD = h.

In ΔDAB
tan60° = (X + h)/48
⇒ √3 = (X + h)/48
⇒ X + h = 48√3
∴ h = 48√3 - X ..................(1)

In ΔCAB
tan30° = X/48
⇒ 1/√3 = X/48
∴ X = 48/√3

From (1) we get,
h = 48√3 - 48/√3
= (48 × 3 - 48)/√3
= (144 - 48)/√3
= 96/√3
= (32 × 3)/√3
= 32√3

১,৬৩৬.

Find the area of a triangle with side lengths of 5 feet, 12 feet, and 13 feet.

ক
17 square feet
খ
27 square feet
গ
30 square feet
ঘ
37 square feet

ব্যাখ্যা

Question: Find the area of a triangle with side lengths of 5 feet, 12 feet, and 13 feet.

Solution:
ধরি,
তিনবাহুর দৈর্ঘ্য a = 5 ফুট, b = 12 ফুট ও c = 13 ফুট

∴ ত্রিভুজটির অর্ধ-পরিসীমা, s = (5 + 12 + 13)/2 = 15 ফুট

∴ ক্ষেত্রফল = √{s(s - a) (s - b) (s - c)} বর্গ ফুট
= √{15(15 - 5) (15 - 12) (15 - 13)} বর্গ ফুট
= √(15 × 10 × 3 × 2)
= √900 বর্গ ফুট
= 30 বর্গ ফুট

১,৬৩৭.

If ABC and PQR are similar triangles in which ∠A = 46° and ∠B = 82°, then ∠C is:

ক
50°
খ
52°
গ
56°
ঘ
60°

ব্যাখ্যা

Question: If ABC and PQR are similar triangles in which ∠A = 46° and ∠B = 82°, then ∠C is:

Solution:
Since ΔABC and ΔPQR are similar triangles.
Then, ∠B = ∠Q = 82° [জ্যামিতির নিয়ম অনুযায়ী, দুটি ত্রিভুজ সদৃশ হলে তাদের অনুরূপ কোণগুলো সমান হয়।]

আমরা জানি, কোনো ত্রিভুজের তিনটি কোণের সমষ্টি 180°। অর্থাৎ, ΔABC-এর ক্ষেত্রে, ∠A + ∠B + ∠C = 180°।

Thus, in ∆ABC,
∠C= 180° - (∠A + ∠ B)
or, ∠C= 180° - (46° + 82°)
∴ ∠C = 52°

১,৬৩৮.

The side length of a square inscribed in a circle is 2. What is the area of the circle shown at the right?

ক
ক) 2π
খ
খ) π
গ
গ) 2πr²
ঘ
ঘ) 2√2

ব্যাখ্যা

প্রশ্ন: The side length of a square inscribed in a circle is 2. What is the area of the circle shown at the right?

সমাধান:
বর্গক্ষেত্রের কর্ণ = √2 × এক বাহুর দৈর্ঘ্য
= 2√2 একক

চিত্রে,
বর্গক্ষেত্রটির কর্ণ = বৃত্তের ব্যস

বৃত্তের ব্যস = 2√2 একক
বৃত্তের ব্যাসার্ধ = 2√2/2 একক
= √2 একক

∴ বৃত্তের ক্ষেত্রফল = π ব্যাসার্ধ^২
= π (√2)²
= 2π বর্গএকক

১,৬৩৯.

What is the weight of 1 cubic meter of water?

ক
10 Kg
খ
100 Kg
গ
789 Kg
ঘ
1000 Kg
ঙ
10000 Kg

ব্যাখ্যা

Weight of 1 cubic meter of water is = 1000 Kg

১,৬৪০.

A cube has a total surface area of 150 square units. What is the volume of the cube?

ক
25
খ
64
গ
100
ঘ
125

ব্যাখ্যা

Question: A cube has a total surface area of 150 square units. What is the volume of the cube?

Solution:
Given that,
Total surface area of a cube, S = 150 square units.

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

And we know,
Volume of a cube, V = a³
= 5³
= 125

So the volume of the cube is 125 cubic units.

১,৬৪১.

ক
30°
খ
45°
গ
60°
ঘ
90°

ব্যাখ্যা

Question: If

Solution:

১,৬৪২.

ক
1/2
খ
1/3
গ
- 1/2
ঘ
2/3

ব্যাখ্যা

Question:

Solution:

১,৬৪৩.

An observer 2.8 m tall is 14√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is-

ক
25m
খ
16.8 m
গ
21.8 m
ঘ
18.5 m

ব্যাখ্যা

Question: An observer 2.8 m tall is 14√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is-

Solution:

Here, Height = AB
Now,
tan∠c = AE/CE
⇒ tan30° = AE/14√3
⇒ 1/√3 = AE/14√3
∴ AE = 14

∴ AB = AE + BE = 14 + 2.8 = 16.8m

১,৬৪৪.

The 2^nd angle of a right triangle is 30 degrees. Then how many degrees is the 3rd angle?

ক
150°
খ
120°
গ
60°
ঘ
30°
ঙ
None

ব্যাখ্যা

Question: The 2^nd angle of a right triangle is 30 degrees. Then how many degrees is the 3rd angle?

Solution:
আমরা জানি,
সমকোণী ত্রিভুজের একটি কোণ 90°
এখন, ২য় কোণ 30° হলে,

তিন কোণের সমষ্টি = 180°
বা, 90° + 30° + ৩য় কোণ = 180°
বা, ৩য় কোণ = 180° - 120°
∴ ৩য় কোণ = 60°

১,৬৪৫.

How many cubes of 8 cm edge can be put in a cubical box of 80 cm edge?

ক
700 cubes
খ
500 cubes
গ
2000 cubes
ঘ
1000 cubes

ব্যাখ্যা

Question: How many cubes of 8 cm edge can be put in a cubical box of 80 cm edge?

Solution:
Number of cubes = (80 × 80 × 80)/(8 × 8 × 8)
= 512000/512
= 1000 cubes

[Note: 1 m = 100 cm, here all dimensions are in cm]

১,৬৪৬.

If tanA = 3/4, then secA = ?

ক
3/4
খ
5/4
গ
5/6
ঘ
1/8

ব্যাখ্যা

Question: If tanA = 3/4, then secA =?

Solution:
tanA = 3/4
লম্ব/ভূমি = 3/4
অতিভুজ = √{(4)² + (3)²} = 5

secA = অতিভুজ/ভূমি
= 5/4

১,৬৪৭.

In the figure, DE is parallel to BC. If the area of a triangle ADE is half that of trapezium DECB, what is the ratio of ΔADE to ΔABC?

ক
2 : 1
খ
1 : 2
গ
2 : 3
ঘ
1 : 3

ব্যাখ্যা

Question: In the figure, DE is parallel to BC. If the area of a triangle ADE is half that of trapezium DECB, what is the ratio of ΔADE to ΔABC?

Solution:
দেওয়া আছে,
ADE ত্রিভুজের ক্ষেত্রফল = (1/2) × DECB ট্রাপিজিয়ামের ক্ষেত্রফল
∴ DECB ট্রাপিজিয়ামের ক্ষেত্রফল = 2 × ADE ত্রিভুজের ক্ষেত্রফল ........................(১)

এখানে,
ABC ত্রিভুজের ক্ষেত্রফল = ADE ত্রিভুজের ক্ষেত্রফল + DECB ট্রাপিজিয়ামের ক্ষেত্রফল
⇒ ABC ত্রিভুজের ক্ষেত্রফল = ADE ত্রিভুজের ক্ষেত্রফল + 2 × ADE ত্রিভুজের ক্ষেত্রফল [(১) নং এর সাহায্যে]
∴ ABC ত্রিভুজের ক্ষেত্রফল = 3 × ADE ত্রিভুজের ক্ষেত্রফল

অতএব,
ADE ত্রিভুজের ক্ষেত্রফল/ABC ত্রিভুজের ক্ষেত্রফল = 1/3
∴ ΔADE : ΔABC = 1 : 3

১,৬৪৮.

The value of the following is:

ক
0
খ
2
গ
7
ঘ
1

ব্যাখ্যা

Question: The value of the following is:

Solution:

১,৬৪৯.

Which of the following angle can be constructed with the help of a ruler and a pair of compasses?

ক
ক) 35°
খ
খ) 40°
গ
গ) 37.5°
ঘ
ঘ) 47.5°

ব্যাখ্যা

কম্পাস এবং রুলার ব্যবহার করে 15° দ্বারা ভাগ করা যায় এমন কোণ এবং ওই সব কোণের অর্ধেক অঙ্কন করা যায়।
একটি কম্পাসের সাহায্যে 60° এবং 90° কোণ অঙ্কন করার পর ওই দুই কোণের ছেদবিন্দু থেকে 75° কোণ অঙ্কন করা যায়।
এরপর 0° এবং 75° এর ছেদবিন্দু ব্যবহার করে 37.5° কোণ অঙ্কন করা যাবে।

১,৬৫০.

Find the area of a triangle whose sides are 12 cm, 5 cm and 13 cm-

ক
45 cm²
খ
50 cm²
গ
30 cm²
ঘ
25 cm²

ব্যাখ্যা

Question: Find the area of a triangle whose sides are 12 cm, 5 cm and 13 cm-

Solution:
ত্রিভুজের তিন বাহু যথাক্রমে,
a = 12 cm, b = 5 cm, c=13 cm
অর্ধপরিধি, S = (a + b + c)/2
= (12 + 5 + 13)/2
= 30/2 = 15 cm

আমরা জানি,
ত্রিভুজের ক্ষেত্রফল = √{s(s - a)(s - b)(s - c)}
= √{15(15 - 12)(15 - 5)(15 - 13)}
= √(15 × 3 × 10 × 2)
= √900
= 30 cm²

অতএব, ত্রিভুজের ক্ষেত্রফল 30 cm².

১,৬৫১.

The area of a circle is increased by 22 sq. cm if its radius is increased by 1 cm. The original radius of the circle is-

ক
3 cm
খ
4 cm
গ
6 cm
ঘ
7 cm
ঙ
None

ব্যাখ্যা

Question: The area of a circle is increased by 22 sq. cm if its radius is increased by 1 cm. The original radius of the circle is-

Solution:
Let,
the original radius of the circle = r cm

ATQ,
π(r + 1)² - πr² = 22
⇒ π[(r + 1)² - r²] = 22
⇒ r² + 2r + 1 - r² = 22/π
⇒ 2r + 1 = 22/(22/7)
⇒ 2r + 1 = 7
⇒ 2r = 7 - 1
⇒ 2r = 6
∴ r = 3

∴ the original radius of the circle = 3 cm

১,৬৫২.

What is the area of a parallelogram with a base of 23 cm and a height of 8 cm?

ক
84 cm²
খ
122 cm²
গ
148 cm²
ঘ
184 cm²

ব্যাখ্যা

Question: What is the area of a parallelogram with a base of 23 cm and a height of 8 cm?

Solution:
আমরা জানি,
সামন্তরিকের ক্ষেত্রফল = ভূমি × উচ্চতা
= ২৩ × ৮
= ১৮৪ বর্গ সে.মি.

১,৬৫৩.

The value of tan 90° is-

ক
0
খ
∞
গ
1
ঘ
undefined

ব্যাখ্যা

Question: The value of tan 90° is-

Solution:
We know that,
tan⁡θ = sin⁡θ/cos⁡θ
⇒ tan⁡θ = sin⁡90°/cos⁡90°
⇒ tan⁡θ = 1/0 ; [sin⁡90° = 1 ; cos⁡90° = 0 and Division by zero is undefined]
∴ tan⁡θ = undefined

১,৬৫৪.

If tanθ = 3/4, what is value of cotθ?

ক
3/4
খ
5/4
গ
4/3
ঘ
4/5

ব্যাখ্যা

Question: If tanθ = 3/4, what is value of cotθ?

Solution:
cotθ = 1/tanθ
= 1/3/4
= 4/3

১,৬৫৫.

The difference between the circumference and the radius of a circle is 185 cm. Find the radius of a circle is

ক
25 cm
খ
30 cm
গ
35 cm
ঘ
40 cm

ব্যাখ্যা

Question: The difference between the circumference and the radius of a circle is 185 cm. Find the radius of a circle is -

Solution:
Let r be the radius of circle

Given that,
2πr - r = 185
⇒ r(2π - 1) = 185
⇒ r{(44/7) - 1} = 185
⇒ r (44 - 7)/7 }= 185
⇒ r(37/7) = 185
⇒ r = 185 (7/37)
∴ r = 35

∴ The radius of the circle is 35 cm.

১,৬৫৬.

Find the value of sin(7π/6)

ক
√3/2
খ
- 1/2
গ
1/2
ঘ
- 1/√2

ব্যাখ্যা

Question: Find the value of sin(7π/6)

Solution:
sin(7π/6) = sin(π + π/6)
= - sin(π/6) [যেহেতু (π + π/6) তৃতীয় চতুর্ভাগে পড়ে এবং তৃতীয় চতুর্ভাগে sin ঋণাত্মক, তাই sin(π + θ) = - sinθ]
= - sin(30°)
= - 1/2

১,৬৫৭.

What is the value of cos150°?

ক
√3/2
খ
- √2/3
গ
- √3/2
ঘ
√2/3

ব্যাখ্যা

Question: What is the value of cos150°?

Solution:
cos150°
= cos(90° + 60°)
= - sin60°
= - √3/2

১,৬৫৮.

If the volume and surface area of a sphere are numerically the same, then its radius is:

ক
4 units
খ
3 units
গ
2 units
ঘ
1 unit

ব্যাখ্যা

Question: If the volume and surface area of a sphere are numerically the same, then its radius is:

Solution:
Let, the volume of a sphere = (4/3)πr³the surface area of a sphere = 4πr²

∴ (4/3)πr³= 4πr²
⇒ r = 3 units

১,৬৫৯.

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

ক
2 cm
খ
4 cm
গ
6 cm
ঘ
8 cm

ব্যাখ্যা

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

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

১,৬৬০.

cos²θ + (1/cosec²θ) + 17 = x. What is the value of x²

ক
324
খ
576
গ
144
ঘ
196

ব্যাখ্যা

Question: (cos²θ + 1/cosec²θ) + 17 = x. What is the value of x²?

Solution:
We know,
sin²θ + cos²θ = 1

Given that,
cos²θ + (1/cosec²θ) + 17 = x
⇒ cos²θ + sin²θ + 17 = x ; [1/cosecθ = sinθ]
⇒ 1 + 17 = x
⇒ x = 18
⇒ x² = 18²
⇒ x² = 324

∴ The value of x² is 324.

১,৬৬১.

If the diagonal of a rectangle is 10 cm long and its perimeter is 32 cm. Find the area of the rectangle.

ক
ক) 58 sq.cm
খ
খ) 68 sq.cm
গ
গ) 78 sq.cm
ঘ
ঘ) 88 sq.cm

ব্যাখ্যা

Question: If the diagonal of a rectangle is 10 cm long and its perimeter is 32 cm. Find the area of the rectangle.

Solution:
let length = x and breadth = y
2(x + y) = 32
⇒ x + y = 16

x² + y² = 10²
= 100

now (x + y)² = 16²
⇒ x² + y² + 2xy = 256
⇒ 100 + 2xy = 256
⇒ 2xy = 256 - 100 = 156
⇒ xy = 78

∴ area = xy = 78 sq.cm

১,৬৬২.

Find the distance between the points (2, - 4) and (- 4, 3).

ক
85
খ
√110
গ
√85
ঘ
110
ঙ
√87

ব্যাখ্যা

Question: Find the distance between the points (2, - 4) and (- 4, 3).

Solution:
আমরা জানি,
দুটি বিন্দুর মধ্যবর্তী দূরত্ব নির্ণয়ের সূত্র:
দূরত্ব, d = √{(x₂ - x₁)² + (y₂ - y₁)²}

দেওয়া আছে,
দুটি বিন্দু যথাক্রমে (2, - 4) and (- 4, 3)

∴ মধ্যবর্তী দূরত্ব, d = √{(- 4 - 2)² + (3 + 4)²}
= √{(- 6)² + (7)²}
= √(36 + 49)
= √85

১,৬৬৩.

The perimeter of a square is equal to the perimeter of a rectangle whose length and width are 6m and 4m respectively. The side of the square is

ক
ক) 3m
খ
খ) 4m
গ
গ) 5m
ঘ
ঘ) 6m

ব্যাখ্যা

Question: The perimeter of a square is equal to the perimeter of a rectangle whose length and width are 6m and 4m respectively. The side of the square is-

Solution:
আয়তক্ষেত্রের পরিসীমা = 2(6 + 4) = 20 মিটার
বর্গক্ষেত্রের পরিসীমা = 20 মিটার
বর্গক্ষেত্রের এক বাহুর দৈর্ঘ্য = 20 /4 = 5 মিটার

১,৬৬৪.

A ladder is placed in such a way that its foot is 15 m away from a wall and its top reaches a window 20 m above the ground. The length of the ladder is-

ক
35 m
খ
17.5 m
গ
25 m
ঘ
18 m
ঙ
None of these

ব্যাখ্যা

Question: A ladder is placed in such a way that its foot is 15 m away from a wall and its top reaches a window 20 m above the ground. The length of the ladder is-

Solution:

Let BC be the wall and AC be the ladder.
Then , BC = 20 m and AB =15m
∴ AC² = BC² + AB² = (20)² + (15)² = (400 + 225) = 625
⇒ AC = √625 = 25m.

১,৬৬৫.

If sinC = 3/4, then cosC = ?

ক
ক) 4/√7
খ
খ) √3/4
গ
গ) √7/4
ঘ
ঘ) 4/7

ব্যাখ্যা

Question: If sinC = 3/4, then cosC = ?

Solution:

আমরা জানি,
sinx = লম্ব/অতিভুজ = 3/4
আমরা জানি,
ভূমি = √(অতিভুজ)² - (লম্ব)²
বা, ভূমি = √(4)² - (3)²
বা, ভূমি = √(16 - 9)
∴ ভূমি = √7

আবার,
cosx = ভূমি/অতিভুজ
= √7/4

১,৬৬৬.

If sinA + cosA = 1 , then A = ?

ক
30°, 60°
খ
0°, 90°
গ
45°, 90°
ঘ
0°, 45°

ব্যাখ্যা

Question: If sinA + cosA = 1 , then A = ?

Solution:
sinA + cosA = 1
⇒ (sinA + cosA)² = 1²
⇒ sin²A + cos²A + 2sinAcosA = 1
⇒ 1 + 2sinAcosA = 1
⇒ 2sinAcosA = 1 - 1
⇒ 2sinAcosA = 0
∴ sinAcosA = 0

Here,
sinA = 0
⇒ sinA = sin0°
∴ A = 0°

Or,
cosA = 0
⇒ cosA = cos90°
∴ A = 90°

A = 0°, 90°

১,৬৬৭.

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 9√2 cm, then what is the area of the triangle?

ক
ক) 72√3 cm²
খ
খ) 84√3 cm²
গ
গ) 36√3 cm²
ঘ
ঘ) 144√3 cm²

ব্যাখ্যা

Question: A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 9√2 cm, then what is the area of the triangle?

Solution:
Let the side of the square be a cm
Then, its diagonal
√2a = 9√2
⇒ a = 9

Perimeter of the square
= 4a
= 4 × 9
= 36 cm

and also perimeter of the equilateral triangle = 36 cm

Each side of the triangle
= 36/3
= 12

Area of the triangle
= (√3/4) × (12)²
= 36√3 cm²

১,৬৬৮.

Given that the diagonal of a square measures 8√2 units, find the perimeter of the square in units.

ক
24 units
খ
32 units
গ
36 units
ঘ
64 units
ঙ
54 units

ব্যাখ্যা

Question: Given that the diagonal of a square measures 8√2 units, find the perimeter of the square in units.

Solution:
দেয়া আছে, বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = 8√2 একক

আমরা জানি, বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = √2 × বাহু

প্রশ্নমতে,
√2 × বাহু = 8√2
⇒ বাহু = 8√2/√2
⇒ বাহু = 8 একক

এখন, বর্গক্ষেত্রের পরিসীমা = 4 × বাহুর দৈর্ঘ্য
= 4 × 8
= 32 একক

∴ বর্গক্ষেত্রের পরিসীমা 32 একক।

১,৬৬৯.

What is the area of an isosceles triangle if two of its sides measure 10 and 8?

ক
9√13
খ
6√17
গ
7√3
ঘ
8√21

ব্যাখ্যা

Question: What is the area of an isosceles triangle if two of its sides measure 10 and 8?

Solution:
The given triangle is an Isosceles triangle and hence, two of the three sides of the triangle are equal.
Hence, the third side of the triangle can either be 10 or be 8.

If the two equal sides of the triangle measure 10, the sides of the triangle become 8, 10, and 10.
The sum of the smaller two sides is greater than the third side, and hence, this is a valid configuration.

If the two equal sides of the triangle measure 8, the sides of the triangle become 8, 8, and 10.
However, the sum of the two smaller sides (8 + 8 = 16) is not greater than the third side (10).
∴ 8 is not a possible value of the third side.
Let, a = 10, b = 8
∴ Area = (b/4) × √(4a² - b²)
= (8/4) × √{4 × (10)² - (8)²}
= 2 × √{4 × 100 - 64}
= 2 × √{400 - 64}
= 2 × √336
= 2 × √(16 × 21)
= 2 × 4 × √21
= 8√21

১,৬৭০.

Consider ΔABD such that angle ADB = 20° and C is a point on BD such that AB = AC and CD = CA. Then the measure of angle ABC is ____ .

ক
ক) 30°
খ
খ) 40°
গ
গ) 45°
ঘ
ঘ) 60°

ব্যাখ্যা

এখানে, ΔACD তে CD = CA বলে ∠CDA = ∠CAD = 20°
ΔACD এ
∠CDA + ∠CAD + ∠ACD = 180°
20° + 20° + ∠ACD = 180°
∠ACD =140°

এখানে
∠BCD = 180°
∠ACD + ∠ACB = 180°
140° + ∠ACB = 180°
∠ACB = 40°

AC = BC হলে
∠ACB = ∠ ABC = 40°

১,৬৭১.

Enayet has a triangle in mind. Its longest side has a length of 20 cm and another of its sides has a length of 10 cm. Its area is 80 cm². What is the exact length of its third side?

ক
√65 cm
খ
2√65 cm
গ
5√65 cm
ঘ
64 cm

ব্যাখ্যা

Question: Enayet has a triangle in mind. Its longest side has a length of 20 cm and another of its sides has a length of 10 cm. Its area is 80 cm². What is the exact length of its third side?

Solution:

(1/2) × 20 × h = 80
h = 8

BD = √(10² - 8²)
= √36 = 6
CD = 20 - 6 = 14

x = √(14² + 8²)
= √(196 + 64)
= √260
= 2√65 cm

১,৬৭২.

If the circle above has center O and circumference 18π, then the perimeter of sector RSTO is-

ক
3π + 9
খ
3π + 18
গ
6π + 9
ঘ
6π + 18
ঙ
6π + 24

ব্যাখ্যা

Question:

If the circle above has center O and circumference 18π, then the perimeter of sector RSTO is-

Solution:
Circumference = 2πr = 18π
∴ r = 18/2 = 9

Perimeter of sector RSTO = Arc Length + r + r

Arc Length = 2πr(C/360) [C = 60, is central angle of the sector]
Arc Length = 2πr(60/360) = 2 × 9 × π × (1/6) = 3π

Therefore Perimeter of sector RSTO = 3π + 9 + 9 = 3π + 18

১,৬৭৩.

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

ক
80 cubic units
খ
210 cubic units
গ
112 cubic units
ঘ
512 cubic units

ব্যাখ্যা

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
⇒ 6a² = 384
⇒ a² = 384 / 6
⇒ a² = 64
∴ a = 8

Volume of a cube, V = a³
= 8³
= 512

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

১,৬৭৪.

If A = π/4, what is the value of sin2A?

ক
8
খ
1
গ
2
ঘ
3

ব্যাখ্যা

Question: If A = π/4, what is the value of sin2A?

Solution:
given A = π/4

sin2A = 2sinAcosA
= 2sin(π/4)cos(π/4)
= 2sin(45°)cos(45°)
= 2 × 1/√2 × 1/√2
= 2/2
= 1

১,৬৭৫.

The distance between the points (4, 3) and (1, 7) is -

ক
4
খ
6
গ
5
ঘ
12

ব্যাখ্যা

Question: The distance between the points (4, 3) and (1, 7) is -

Solution:
Given points (4, 3) and (1, 7).
Now the formula for the distance between (x₁, y₁) and (x_2,y₂) is
= √[(x₂ - x₁)² + (y₂ - y₁)²]

∴ The distance between the points (4, 3) and (1, 7) is
= √[(1 - 4)² + (7 - 3)²]
= √[9 + 16]
= √25
= 5

১,৬৭৬.

Two solid balls are formed by melting a solid cylinder with a radius of 3 cm and height of 7 cm. Determine the volume of each ball.

ক
199 cm³
খ
99 cm³
গ
98 cm³
ঘ
198 cm³

ব্যাখ্যা

Question: Two solid balls are formed by melting a solid cylinder with a radius of 3 cm and height of 7 cm. Determine the volume of each ball.

Solution:
the volume of solid cylinder = π × r² × h
= (22/7) × 3² × 7 cm³
= 198 cm³

∴ the volume of each ball = 198/2 cm³
= 99 cm³

১,৬৭৭.

The area of a circle is increased by 22 square cm if its radius is increased by 1 cm. The original radius of the circle is -

ক
ক) 3 cm
খ
খ) 4 cm
গ
গ) 5 cm
ঘ
ঘ) 6 cm

ব্যাখ্যা

Question: The area of a circle is increased by 22 square cm if its radius is increased by 1 cm. The original radius of the circle is -

Solution:
Let the original radius of the circle be r cm.

ATQ,
π(r + 1)²- πr² = 22
⇒ π{(r + 1)² - r²} = 22
⇒ π(r² + 2r + 1 -r²) = 22
⇒ 2r + 1 = 22/π
⇒ 2r + 1 = (22 × 7)/22
⇒ 2r + 1 = 7
⇒ 2r = 6
⇒ r = 3 cm

১,৬৭৮.

What is the radius of a circle if its perimeter is numerically equal to thrice its area?

ক
2
খ
3
গ
2/3
ঘ
4

ব্যাখ্যা

Question: What is the radius of a circle if its perimeter is numerically equal to thrice its area?

Solution:
Let,
The radius be r

ATQ,
2πr = 3πr²
⇒ 3r = 2
⇒ r = 2/3

১,৬৭৯.

The difference between the circumference and the radius of a circle is 185 cm. Find the radius of a circle is -

ক
ক) 40 cm
খ
খ) 35 cm
গ
গ) 30 cm
ঘ
ঘ) 25 cm

ব্যাখ্যা

১,৬৮০.

The ratio of total surface area to lateral surface area of a cylinder whose radius is 25 cm and height 100 cm, is -

ক
5 : 3
খ
1 : 4
গ
5 : 4
ঘ
5 : 7

ব্যাখ্যা

Question: The ratio of total surface area to lateral surface area of a cylinder whose radius is 25 cm and height 100 cm, is -

Solution:
Total surface area : Lateral surface area
= (2πrh + 2πr²⁾ : 2πrh
= (h +r) : h
= (100 + 25) : 100
= 125 : 100
= 5 : 4

১,৬৮১.

The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 4400 cm³. Then its radius will be-

ক
ক) 4 cm
খ
খ) 7 cm
গ
গ) 10 cm
ঘ
ঘ) 13 cm

ব্যাখ্যা

Question: The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 4400 cm³. Then its radius will be-

Solution:
Let the radius and height of the cylinder be 5x and 7x cm respectively.
Then, volume = πr²h
= (22/7) × (5x)² × 7x
= 550x³ATQ,
550x³ = 4400
x³ = 4400/550 = 8
x = ∛8 = 2

Hence, the radius = (5 × 2) = 10 cm

১,৬৮২.

Find the area of an isosceles triangle whose sides are 10 cm, 6 cm and 6 cm.

ক
√11 sq. cm.
খ
25√11 sq. cm.
গ
5 sq. cm.
ঘ
11 sq. cm.
ঙ
5√11 sq. cm.

ব্যাখ্যা

Question: Find the area of an isosceles triangle whose sides are 10 cm, 6 cm and 6 cm.

Solution:
Semi perimeter of triangle s = (10 + 6 + 6)/2 = 11.
Area of triangle = √{11 × (11 - 10) × (11 - 6) × (11 - 6)}
= √(11 × 5 × 5)
= 5√11 cm²

১,৬৮৩.

The lengths of the two sides of a triangle are 9 cm and 10 cm respectively and the angle included between them is 30°. Find the area.

ক
45 cm²
খ
22.5 cm²
গ
50 cm²
ঘ
25.5 cm²

ব্যাখ্যা

Question: The lengths of the two sides of a triangle are 9 cm and 10 cm respectively and the angle included between them is 30°. Find the area.

Solution:
Given,
the sides of triangle are a = 9 cm. and b = 10 cm. respectively.
Their included angle θ = 30°

We know,
Area of the tringle = (1/2) × ab × sinθ
= (1/2) × 9 × 10 × sin30°
= (1/2) × 9 × 10 × (1/2)
= 22.5 cm²

১,৬৮৪.

Express the angle in degree included between hands of hour and minute of the clock at 4 : 30?

ক
ক) 40°
খ
খ) 45°
গ
গ) 50°
ঘ
ঘ) 55°

ব্যাখ্যা

Question: Express the angle in degree included between hands of hour and minute of the clock at 4 : 30?

Solution:
আমরা জানি,
মধ্যবর্তী কোণ, θ = Ι(11M - 60H)/2Ι°
= Ι(11 × 30) - (60 × 4)/2Ι°
= Ι(330 - 240)/2Ι°
= 45°

১,৬৮৫.

If the curved surface area of a sphere is same as the curved surface area of a hemisphere, find the radius of the hemisphere.

ক
Same as that of the sphere.
খ
√2 times that of the sphere.
গ
√3 times that of the sphere.
ঘ
2 times that of the sphere.

ব্যাখ্যা

Question: If the curved surface area of a sphere is same as the curved surface area of a hemisphere, find the radius of the hemisphere.

Solution:
Let,
R as the radius of the sphere.
r as the radius of the hemisphere.

Curved surface area of a sphere = Curved surface area of a hemisphere
4πR² = 2πr²
⇒ 2R² = r²
⇒ r = √2R

∴ The radius of the hemisphere = √2 times that of the sphere.

১,৬৮৬.

How many degrees are between the hands of a clock at 10:05?

ক
92.5°
খ
82.5°
গ
72.5°
ঘ
87.5°

ব্যাখ্যা

Required degrees = । (11M - 60H)/2 ।°
= । (11 × 5 - 60 × 10)/2 ।°
= । - 545/2 ।°
= । - 272.5 ।°
= 272.5°
Now, 360° - 272.5° = 87.5°

১,৬৮৭.

If sin A + sin²A = 1, then the value of the expression (cos²A + cos⁴A) is –

ক
ক) 1
খ
খ) 1/2
গ
গ) 2
ঘ
ঘ) 3

ব্যাখ্যা

Given, sin A + sin²A = 1
⇒ sinA = 1 - sin²A
⇒ sinA = cos²A
⇒ sin²A = cos⁴A
⇒ 1 - cos²A = cos⁴A
∴ cos²A + cos⁴A = 1

১,৬৮৮.

If cosA sinA = 1,then (cosA + sinA)² =?

ক
1
খ
2
গ
3
ঘ
4

ব্যাখ্যা

Question: If cosA sinA = 1,then (cosA + sinA)² =?

Solution:
(cosA + sinA)²
= cos²A + 2 cosA sinA + sin²A
= 1 + 2.1 [sin²A + cos²A = 1]
= 1 + 2
= 3

১,৬৮৯.

Find the value of cosec(π/3)

ক
√3
খ
2/√3
গ
1
ঘ
√3/2

ব্যাখ্যা

Question: Find the value of cosec(π/3)

Solution:
cosec(π/3)
= cosec(π/3)
= 1/sin(π/3)
= 1/sin60°
= 1/(√3/2)
= 2/√3

১,৬৯০.

ক
30°
খ
60°
গ
45°
ঘ
90°

ব্যাখ্যা

Question:

Solution:

১,৬৯১.

If two angles are said to be complementary angles and one angle is 52° then the other angle is of:

ক
100°
খ
68°
গ
128°
ঘ
38°

ব্যাখ্যা

Question: If two angles are said to be complementary angles and one angle is 52° then the other angle is of:

Solution:
দুটি কোণের সমষ্টি ৯০° হলে, একটিকে অপরটির পূরক কোণ বলে।

অপর কোণ = 90° - 52°
= 38°

১,৬৯২.

A rectangular floor is covered by a rug except fo a strip p meters along each of the four edges. If the floor is m meters by n meteres, What is the area of the rug in square meters?

ক
mn - p(m + n)
খ
mn - 2p(m + n)
গ
(m - p)(n - p)
ঘ
(m - 2p)(n - 2p)

ব্যাখ্যা

Question: A rectangular floor is covered by a rug except fo a strip p meters along each of the four edges. If the floor is m meters by n meteres, What is the area of the rug in square meters?

Solution:

The length of the rug = m - 2 × p = m - 2p
The width of the rug = n - 2 × p = n - 2p
∴ The area of the rectangular rug equals = (m - 2p)(n - 2p)

১,৬৯৩.

If tanθ = 9/40, then secθ = ?

ক
23/40
খ
41/10
গ
21/40
ঘ
41/40

ব্যাখ্যা

Question: If tanθ = 9/40, then secθ = ?

Solution:
এখানে,
tanθ = 9/40 = লম্ব/ভূমি

∴ লম্ব = 9, ভূমি = 40
∴ অতিভুজ = √(9² + 40²)
= √(81 + 1600)
= √1681 = 41

∴ secθ = 1/cosθ = অতিভুজ/ভূমি
= 41/40

১,৬৯৪.

Calculate the area of a rhombus if the length of its side is 2 cm and one of its angles A is 30 degrees.

ক
5 cm²
খ
4 cm²
গ
3 cm²
ঘ
2 cm²

ব্যাখ্যা

Question: Calculate the area of a rhombus if the length of its side is 2 cm and one of its angles A is 30 degrees.

Solution:
Given,
Side = s = 2 cm
Angle A = 30 degrees
Square of side = 2 × 2 = 4

Area, A = s² × sin (30°)
⇒ A = 4 × (1/2)
∴ A = 2 cm²

১,৬৯৫.

The area of a triangle with sides 3 cm, 5 cm and 6 cm is -

ক
ক) 2√3 cm²
খ
খ) 2√14 cm²
গ
গ) 5√12 cm²
ঘ
ঘ) 4√14 cm²

ব্যাখ্যা

অর্ধপরিসীমা, s = (3 + 5 + 6)/2 = 7 সে.মি
∴ ক্ষেত্রফল = √{s(s - a)(s - b)(s - c)} বর্গএকক
= √ {7 (7 - 3) (7 - 5) (7 - 6)} বর্গসে.মি
= √ (7 × 4 × 2 × 1)
= 2√14 বর্গসে.মি

১,৬৯৬.

A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m³) is:

ক
ক) 4830
খ
খ) 5120
গ
গ) 6420
ঘ
ঘ) 8960

ব্যাখ্যা

Clearly, l = (48 - 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m³ = 5120 m³.

১,৬৯৭.

The perimeter of a square is equal to the perimeter of a rectangle. The length of the rectangle is three times its width, and the area of the rectangle is 1,728 square metres. Find the perimeter of the square.

ক
192 m
খ
168 m
গ
144 m
ঘ
216 m

ব্যাখ্যা

Question: The perimeter of a square is equal to the perimeter of a rectangle. The length of the rectangle is three times its width, and the area of the rectangle is 1,728 square metres. Find the perimeter of the square.

Solution:
Let the width of the rectangle = x metres
Then the length of the rectangle = 3x metres

We know,
Area of the rectangle = length × width
⇒ 3x × x = 1728
⇒ 3x² = 1728
⇒ x² = 1728/3
⇒ x² = 576
⇒ x = √576
⇒ x = 24

Thus,
Width of rectangle = 24 m
Length of rectangle = 3 × 24 = 72 m

∴ Perimeter of the rectangle = 2(length + width)
= 2(72 + 24)
= 2 × 96
= 192 metres

Since the perimeter of the square = perimeter of the rectangle,
∴ Perimeter of the square = 192 metres.

১,৬৯৮.

Surface area of hollow cylinder with radius 'r' and height 'h' is measured by

ক
ক) 2πr - h
খ
খ) 2πrh
গ
গ) 2πh
ঘ
ঘ) πr²

ব্যাখ্যা

Question: Surface area of hollow cylinder with radius 'r' and height 'h' is measured by

Solution:
ফাঁপা সিলিন্ডারের বক্রতলের ক্ষেত্রফল = ভূমির পরিধি × উচ্চতা
=2πr × h
=2πrh

১,৬৯৯.

What is the slope of the line perpendicular to the line given by the equation y = 3/4x - 2?

ক
3/4
খ
- 3/4
গ
4/3
ঘ
- 4/3

ব্যাখ্যা

Question: What is the slope of the line perpendicular to the line given by the equation y = 3/4x - 2?

Solution:
The equation of the line is y = 3/4x - 2

This is in the slope-intercept form y = mx + c

So, Slope(m) = 3/4

For two lines to be perpendicular, the product of their slopes must equal -1.
∴ m₁ . m₂ = - 1

Here, m₁ = 3/4
∴ m₂ = -1/(3/4)
= - 4/3

১,৭০০.

The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? [Assume that the cow is able to move on all sides with equal ease.]

ক
1696 sq. m
খ
1694 sq. m
গ
1594 sq. m
ঘ
1756 sq. m

ব্যাখ্যা

Question: The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? [Assume that the cow is able to move on all sides with equal ease.]

Solution:
The cow can graze the area covered by the circle of radius 19 m initially, because the length of the rope is 19 m.
Area of a circle = π × (radius)²
Therefore, the initial area that the cow can graze = (22/7) × 19² sq. m.
When the length of the rope is increased to 30 m, grazing area becomes = (22/7) × 30² sq. m.
The additional area it could graze when length is increased from 19 m to 30 m = (22/7) × (30² - 19²) sq. m.
= (22/7) × (30 + 19)(30 - 19) = (22/7) × 49 × 11 = 1694 sq. m.