Geometry: Mensuration, Trigonometry

PrepBank · পাতা ৬ / ২১ · ৫০১–৬০০ / ২,০৮৫

৫০১.

A storm breaks a tree. The broken part of tree bends so that the top of the tree touches the ground and makes an angle of 60° with the horizontal plane. If the distance between the base of the tree and the point where top of tree touches the ground is 10 m, find the height of the tree?

ক
37.3 m
খ
17.3 m
গ
27.3 m
ঘ
20.3 m

ব্যাখ্যা

Question: A storm breaks a tree. The broken part of tree bends so that the top of the tree touches the ground and makes an angle of 60° with the horizontal plane. If the distance between the base of the tree and the point where top of tree touches the ground is 10 m, find the height of the tree?

Solution:

PQ = 10 and let RQ be X.

RQ/PQ = tan60°
⇒ X/10 = √3
∴ X = 10√3

Now,
PR² = X² + (10)²
PR² = (10√3)² + (10)² = 300 + 100
PR² = 400
∴ PR = 20

Height of tree = RQ + PR
= X + 20
= 10√3 + 20
= 10 × 1.73 + 20
= 17.3 + 20
= 37.3 meter

৫০২.

The three sides of a triangle are 2x, 3x + 1, and 4x − 1 respectively, and the perimeter is 36 cm. What is the length of the longest side?

ক
18 cm
খ
21 cm
গ
15 cm
ঘ
20 cm

ব্যাখ্যা

Question: The three sides of a triangle are 2x, 3x + 1, and 4x − 1 respectively, and the perimeter is 36 cm. What is the length of the longest side?

Solution:
প্রশ্নমতে, ত্রিভুজের তিনটি বাহুর দৈর্ঘ্যের যোগফল তার পরিসীমার সমান।
2x + (3x + 1) + (4x − 1) = 36

সমীকরণটি সমাধান করে পাই,
(2x + 3x + 4x) + (1 − 1) = 36
9x = 36
x = 36 / 9
x = 4

এখন, x এর মান বসিয়ে বাহুগুলোর দৈর্ঘ্য নির্ণয় করি:
প্রথম বাহু = 2x = 2 × 4 = 8 সেমি
দ্বিতীয় বাহু = 3x + 1 = 3 × 4 + 1 = 13 সেমি
তৃতীয় বাহু = 4x − 1 = 4 × 4 − 1 = 15 সেমি

সুতরাং, সবচেয়ে বড় বাহুটি হলো 15 সেমি।

৫০৩.

A cube-shaped box with a side length of 9 meters contains another cube-shaped box inside it. The empty space accommodates 217,000 liters of water. What is the surface area of the smaller box?

ক
364m²
খ
334m²
গ
328m²
ঘ
384m²
ঙ
None of the above

ব্যাখ্যা

Question: A cube-shaped box with a side length of 9 meters contains another cube-shaped box inside it. The empty space accommodates 217,000 liters of water. What is the surface area of the smaller box?

Solution:
আমরা জানি,
১০০০ লিটার = ১ ঘন মি.
∴ ২১৭০০০ লিটার = (২১৭০০০/১০০০) = ২১৭ ঘন মি.

অর্থাৎ, খালি অংশের আয়তন = ২১৭ ঘন মি.

বড় বক্সের আয়তন = (৯)^৩ ঘন মি. = ৭২৯ ঘন মি.

∴ ছোট বক্সের আয়তন = ৭২৯ - ২১৭ = ৫১২ ঘন মি.

ছোট বক্সের এক বাহুর দৈর্ঘ্য = ^৩√৫১২ = ৮ মি.

∴ ছোট বক্সের সমগ্রতলের ক্ষেত্রফল = ৬ × (৮)^২ বর্গ মি.
= ৩৮৪ বর্গ মি.

৫০৪.

A right triangle with sides 6 cm, 8 cm, and 10 cm is rotated on the side of 6 cm to form a cone. The volume of the cone so formed is -

ক
ক) 48π cm³
খ
খ) 64π cm³
গ
গ) 72π cm³
ঘ
ঘ) 96π cm³

ব্যাখ্যা

Question: A right triangle with sides 6 cm, 8 cm, and 10 cm is rotated on the side of 6 cm to form a cone. The volume of the cone so formed is -

Solution:

We have,
r = 6cm
h = 8cm

∴ Volume = (1/3)πr²h
= (1/3) × π × 6² × 8 cm³
= 96π cm³

৫০৫.

If the area of a rhombus is 54 sq. cm and the length of one of the diagonals is 6 cm then the length of the other diagonal is–

ক
ক) 18
খ
খ) 12
গ
গ) 9
ঘ
ঘ) 6

ব্যাখ্যা

We know, Area of rhombus = 1/2 × x × y [Here, x and y are two diagonals of the rhombus]
Or, x = (54 × 2) / 6 = 18 cm

৫০৬.

A mirror is placed on the ground facing upwards. A man sees the top of a tower in the mirror which is at a distance of 100 m from the mirror. The man is 0.5 m away from the mirror, and his height is 1.5 m.

ক
300 m
খ
200 m
গ
50.5 m
ঘ
315 m

ব্যাখ্যা

Question: A mirror is placed on the ground facing upwards. A man sees the top of a tower in the mirror which is at a distance of 100 m from the mirror. The man is 0.5 m away from the mirror, and his height is 1.5 m.

Solution:

Given that,
Distance from the mirror to the tower = 100 m
Distance from the man to the mirror = 0.5 m
Height of the man = 1.5 m
Height of the tower, H = ?

Now,
⇒ Height of the man/Distance from man to mirror = Height of the tower/Distance from tower to mirror
⇒ 1.5/0.5 = H/100
⇒ 3 = H/100
⇒ H = 100 × 3 = 300 m

৫০৭.

Ratio of Volumes of cube and Sphere is 6/π. Find the ratio of side of cube and radius of sphere.

ক
2 : 1
খ
3 : 1
গ
4 : 1
ঘ
5 : 1
ঙ
1 : 2

ব্যাখ্যা

Question: Ratio of Volumes of cube and Sphere is 6/π. Find the ratio of side of cube and radius of sphere.

Solution:
Let the side of cube is 'a' and radii of sphere is 'r'.
Now Volume of cube= a³
Volume of sphere= (4/3)πr³

a³/{(4/3)πr³} = 6/π
⇒ a³/r³ = (6 × 4)/3
⇒ a³/r³ = 8/1
⇒ a/r = 2/1
Hence the answer is 2 : 1

৫০৮.

The curved surface area and the diameter of a right circular cylinder are 660 sq.cm and 21 cm respectively. Find its height (in cm).

ক
8
খ
9
গ
10
ঘ
12
ঙ
None of these

ব্যাখ্যা

Question: The curved surface area and the diameter of a right circular cylinder are 660 sq.cm and 21 cm respectively. Find its height (in cm).

Solution:
Diameter of cylinder = 21 cm
Radius of cylinder = 21/2 cm

The curved Surface area of cylinder = 2πrh,
Where,
r = radius,
h = height

According to the question
660 = 2 × (22/7) × (21/2) × h
⇒ 660 = 66 × h
∴ h = 10 cm

৫০৯.

A box is made in the form of a cube. If a second cubical box has inside dimensions three times those of the first box, how many times as much does the second box contain?

ক
ক) 12
খ
খ) 27
গ
গ) 9
ঘ
ঘ) 6

ব্যাখ্যা

If the second box has each dimension three times that of the first box, then its volume is 3 × 3 × 3 = 27 times.
So, the second box contains 27 times as much the first box.

৫১০.

If sinA + sin²A = 1. Then the value of the expression (cos²A + cos⁴A) is -

ক
ক) 1
খ
খ) 2
গ
গ) (1/2)
ঘ
ঘ) 3

ব্যাখ্যা

দেওয়া আছে,
sinA + sin²A = 1
⇒ sinA = 1 - sin²A
⇒ sinA = cos²A
এখন,
cos²A + Cos⁴A
= cos²A + cos²A.cos²A
= cos²A + sinA.sinA [sina = cos²A]
= Cos²A + sin²A
= 1[sin²A + cos²A = 1]

৫১১.

A rhombus has diagonals of 10 cm and 20 cm. Find the side length of a square that has the same area as the rhombus.

ক
16 cm
খ
12 cm
গ
10 cm
ঘ
8 cm
ঙ
None of these

ব্যাখ্যা

Question: A rhombus has diagonals of 10 cm and 20 cm. Find the side length of a square that has the same area as the rhombus.

Solution:
Given that,
d₁ = 10 cm and d₂ = 20 cm
where d₁ and d₂ are the lengths of the diagonals.

We know,
The area of a rhombus is = (d₁ × d₂)/2
= (10 × 20)/2
= 200/2
= 100 cm²
Now, let the side length of the square be a cm.
So the area of the square is a².
Since the square has the same area as the rhombus.
⇒ a² = 100
⇒ a = √100
∴ a = 10 cm

So the side length of the square is 10 cm.

৫১২.

The perimeter of one face of a cube is 28 cm. Its volume must be-

ক
225 cm³
খ
216 cm³
গ
343 cm³
ঘ
None of these

ব্যাখ্যা

Question: The perimeter of one face of a cube is 28 cm. Its volume must be-

Solution:
perimeter of one face is = 28 cm

let, length of one side is = a cm
∴ perimeter = 4a cm

ATQ,
⇒ 4a = 28
⇒ a = 28/4
= 7 cm

∴ volume = a³
= 7³
= 343 cm³

৫১৩.

ক
ক) 12°
খ
খ) 18°
গ
গ) 30°
ঘ
ঘ) 42°

ব্যাখ্যা

Question:

Solution:
Given,
sin(A + 18°) = 1/2
⇒ sin(A + 18°) = sin30°
⇒ A + 18° = 30°
⇒ A = 30° - 18°
∴ A = 12°

৫১৪.

If tanθ = 1 then sinθ - cos(- θ) = ?

ক
- 1
খ
0
গ
1
ঘ
2

ব্যাখ্যা

Question: If tanθ = 1 then sinθ - cos(- θ) = ?

Solution:
Given,
tanθ = 1
⇒ sinθ/cosθ = 1
⇒ sinθ = cosθ

∴ sinθ - cos(- θ)
= sinθ - cosθ
= cosθ - cosθ
= 0

৫১৫.

The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is -

ক
ক) 9.2 m
খ
খ) 8.7 m
গ
গ) 7.2 m
ঘ
ঘ) 6.5 m

ব্যাখ্যা

ধরি,
AB হচ্ছে দেয়াল এবং BC হচ্ছে মই।
এখানে ∠ACB = 60° এবং AC = 4.6 মিটার
তাহলে, AC/BC = cos 60° = 1/2 [ যেহেতু, cosθ = ভূমি/অতিভুজ]
⇒ BC = 2 × AC
∴ BC = 2 × 4.6
= 9.2m

৫১৬.

A lady grows cabbages in her garden that is in the shape of a square. Each cabbage takes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year?

ক
11236 cabbages
খ
11025 cabbages
গ
10582 cabbages
ঘ
10644 cabbages
ঙ
None

ব্যাখ্যা

Question: A lady grows cabbages in her garden that is in the shape of a square. Each cabbage takes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year?

Solution:
Let,
the side of the square area used for growing cabbages this year = X ft.
∴ the area of the ground used for cultivation this year = X² sq. ft.

and
the side of the square area used for growing cabbages last year be Y ft.
∴ the area of the ground used for cultivation last year = Y² sq. ft.

The cabbage field remained square-shaped in both years.
Given the increase of 211 cabbages (1 sq ft per cabbage),
∴ the area difference is: X² - Y² = 211
⇒ (X + Y)(X - Y) = 211

Since 211 is prime, the only factor pair is (211, 1), so:
X + Y = 211
X - Y = 1
→ Solving gives: X = 106 and Y = 105

This year's production = 106² = 11236 cabbages

৫১৭.

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.

ক
792π cm²
খ
925π cm³
গ
972π cm³
ঘ
520π cm²
ঙ
None of these

ব্যাখ্যা

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 surface volume would be 972π cm³

৫১৮.

What is the ratio of the areas of two squares if one has its diagonal double than the other?

ক
ক) 4 : 1
খ
খ) 2 : 1
গ
গ) 8 : 1
ঘ
ঘ) 16 : 1

ব্যাখ্যা

Question: What is the ratio of the areas of two squares if one has its diagonal double than the other?

Solution:
Let the diagonal of two squares be 2x and x respectively

length of first square = 2x/√2 = √2x
length of second square = x/√2 = x/√2

Ratio = (√2x)² : (x/√2)²
= 2x² : x²/2
= 2 : 1/2
= 4 : 1

৫১৯.

The area of a rectangle is 40cm² and one of its sides is 8cm long. What will be its perimeter?

ক
ক) 26 cm
খ
খ) 13 cm
গ
গ) 28 cm
ঘ
ঘ) 20 cm

ব্যাখ্যা

দেয়া আছে,
আয়তক্ষেত্রের ক্ষেত্রফল 40 বর্গ সে.মি.
আয়তক্ষেত্রের এক পাশের দৈর্ঘ্য 8 সে.মি.
আয়তক্ষেত্রের অপর পাশের দৈর্ঘ্য 40/8 সে.মি.
=5 সে.মি.

আয়তক্ষেত্রের পরিসীমা = 2 (8 + 5) সে.মি.
= 26 সে.মি.

৫২০.

If tanθ = 5/12, then cosecθ = ?

ক
17/13
খ
12/5
গ
12/13
ঘ
13/5

ব্যাখ্যা

Question: If tanθ = 5/12, then cosecθ = ?

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

∴ লম্ব = 5, ভূমি = 12

∴ অতিভুজ = √(5²+ 12²)
= √(25+144) = √169 = 13

∴ cosecθ
= অতিভুজ/লম্ব
= 13/5

৫২১.

There are two stations of length 162 meters and 120 meter respectively. A train takes 18 seconds to pass first station and 15 seconds to pass another station. Determine the length of the train.

ক
ক) 90 m
খ
খ) 70 m
গ
গ) 100 m
ঘ
ঘ) 95 m

ব্যাখ্যা

Let length of the train be x m
Speed of train,
(x+162) / 18 = (x+120) / 15
∴ x = 90 m

৫২২.

A right triangle has sides 9 cm, 12 cm, and 15 cm. What is its area?

ক
24 cm²
খ
34 cm²
গ
54 cm²
ঘ
32 cm²

ব্যাখ্যা

Question: A right triangle has sides 9 cm, 12 cm, and 15 cm. What is its area?

Solution:
A right triangle with sides 9 cm, 12 cm, and 15 cm.

We know,
Area = (1/2) × base × height
= (1/2) × 9 × 12
= 54 cm²

So, the area of the right triangle is 54 cm².

৫২৩.

A rectangle’s length exceeds its width by 24 meters. If the total perimeter is 208 meters, what is the area?

ক
2200 square meters
খ
2520 square meters
গ
2560 square meters
ঘ
2740 square meters

ব্যাখ্যা

Question: A rectangle’s length exceeds its width by 24 meters. If the total perimeter is 208 meters, what is the area?

Solution:
মনে করি,
আয়তক্ষেত্রটির দৈর্ঘ্য = x মিটার
প্রস্থ = (x - 24) মিটার

আমরা জানি,
আয়তক্ষেত্রের পরিসীমা = 2(দৈর্ঘ্য + প্রস্থ)

প্রশ্নমতে,
2(x + x - 24) = 208
⇒ 2(2x - 24) = 208
⇒ 2x - 24 = 208/2
⇒ 2x - 24 = 104
⇒ 2x = 104 + 24
⇒ 2x = 128
⇒ x = 128/2
⇒ x = 64

∴ প্রস্থ = (64 - 24) মিটার = 40 মিটার

অতএব, আয়তক্ষেত্রের ক্ষেত্রফল = (দৈর্ঘ্য × প্রস্থ) = (64 × 40) বর্গমিটার = 2560 বর্গমিটার

৫২৪.

In triangle △ABC, If AB = BC and ∠B = 70°, ∠A will be:

ক
ক) 70°
খ
খ) 10°
গ
গ) 55°
ঘ
ঘ) 130°

ব্যাখ্যা

If AB = BC, then ∠A = ∠C
As we know, ∠A + ∠B + ∠C = 180°
Or, A + 70° + ∠A = 180°
Or, 2∠A = 180° - 70° = 110°
So, ∠A = 55°

৫২৫.

PQR ত্রিভুজের ∠QPR = ৮০°, PQ = PR হলে ∠PQR = ?

ক
৫০°
খ
৮০°
গ
৯০°
ঘ
১২০°
ঙ
কোনটিই নয়

ব্যাখ্যা

প্রশ্ন: PQR ত্রিভুজের ∠QPR = ৮০°, PQ = PR হলে ∠PQR = ?

সমাধান:

যেহেতু, PQ = PR
∠PQR = ∠PRQ
কিন্তু, ∠QPR = ৮০°

∠PQR + ∠PRQ + ∠QPR = ১৮০°
⇒ ∠PQR + ∠PRQ = ১৮০° - ৮০°
⇒ ২∠PQR = ১০০°
∴ ∠PQR = ৫০°

৫২৬.

ABCD is a rectangle. If its length is decreased by 5 meter and the width is increased by 3 meter, the area decreases by 9 square meters. If its length is increased by 3 meter and the width is increased by 2 meter, the area increases by 67 square meters. What is the width of ABCD?

ক
ক) 9 meter
খ
খ) 17 meter
গ
গ) 18 meter
ঘ
ঘ) 22 meter

ব্যাখ্যা

Question: ABCD is a rectangle. If its length is decreased by 5 meter and the width is increased by 3 meter, the area decreases by 9 square meters. If its length is increased by 3 meter and the width is increased by 2 meter, the area increases by 67 square meters. What is the width of ABCD?

Solution:
ধরি
আয়তক্ষেত্রের দৈর্ঘ্য = x মিটার
আয়তক্ষেত্রের প্রস্থ = y মিটার

১ম শর্তমতে
(x - 5)(y + 3) = xy - 9
xy - 5y + 3x - 15= xy - 9
3x - 5y = 6 ......................(1)

২য় শর্তমতে
(x + 3)(y + 2) = xy + 67
xy + 3y + 2x + 6 = xy + 67
2x + 3y = 61 ........................(2)

(1) × 2 - (2) × 3 ⇒
6x - 10y - 6x - 9y = 12 - 183
- 19y = - 171
y = 171/19
y = 9

অতএব
আয়তক্ষেত্রের প্রস্থ = 9 মিটার

৫২৭.

If the diameter of a sphere is 6 m, its hemisphere will have a volume of?

ক
9π cubic meters
খ
18π cubic meters
গ
54 cubic meters
ঘ
27π cubic meters
ঙ
None of these

ব্যাখ্যা

Question: If the diameter of a sphere is 6 m, its hemisphere will have a volume of?

Solution:
Given that,
Diameter of the sphere = 6 m
Radius of the sphere = 6/2 = 3

We know that,
The volume of a hemisphere is, V = (2/3)πr³
= (2/3) × π × (3)³
= 18π cubic meters

৫২৮.

What is the perimeter of the figure above?

ক
380
খ
360
গ
330
ঘ
300
ঙ
230

ব্যাখ্যা

Question:

What is the perimeter of the figure above?

Solution:

Apply the Pythagorean theorem
We get: 60² + x² = 100²
⇒ 3600 + x² = 10000
⇒ x² = 6400
∴ x = 80

∴ Perimeter = 60 + 70 + 100 + 80 + 70 = 380

৫২৯.

The perimeter of a circle measures 16πcm, what is the area of the circle in sq.cm?

ক
32√2π
খ
64π
গ
256π
ঘ
128π

ব্যাখ্যা

ধরি, বৃত্তের ব্যাসার্ধ r cm
এর ক্ষেত্রফল πr² cm²

প্রশ্নমতে, 2πr = 16π
r = 16/2 =8

ক্ষেত্রফল = πr²cm²π × (8)² cm²=64π cm²

৫৩০.

The area of a square plot is twice that of another square plot. If the diagonal of the bigger plot is x, what is the diagonal of the smaller plot?

ক
x/2
খ
x/4
গ
2x/3
ঘ
√2x/2
ঙ
x

ব্যাখ্যা

Area of square = d²/2 (d for diagonal)
so, area of bigger square = x²/2
so area of smaller square = x²/4 (half of the bigger one)
one side of smaller square = √(x²/4)
= x/2
diagonal of smaller square d² = (x/2)² + (x/2)²
d = root 2 × x/2.

৫৩১.

If tanθ = 1 then sinθ - cos(- θ) = ?

ক
0
খ
1
গ
1/2
ঘ
- 1

ব্যাখ্যা

Question: If tanθ = 1 then sinθ - cos(- θ) = ?

Solution:
Given, tanθ = 1
⇒ sinθ/cosθ = 1
⇒ sinθ = cosθ

∴ sinθ - cos(- θ)
= sinθ - cosθ
= cosθ - cosθ
= 0

৫৩২.

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

ক
5/2
খ
- 2/5
গ
3/5
ঘ
1/2
ঙ
- 1/5

ব্যাখ্যা

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

Solution:
প্রদত্ত সরলরেখার সমীকরণটি হলো: 2x + 5y = 10
কোনো সরলরেখার ঢাল নির্ণয়ের জন্য সেটিকে y = mx + c আকারে আনতে হয়, যেখানে m হলো ঢাল।
2x + 5y = 10
⇒ 5y = - 2x + 10
⇒ y = (- 2/5)x + 2
অতএব, প্রদত্ত রেখার ঢাল (m₁) = - 2/5

আমরা জানি, দুটি সমান্তরাল রেখার ঢাল সমান হয় (m₁ = m₂)।
∴ সমান্তরাল রেখাটির ঢাল, m₂= - 2/5

৫৩৩.

The ratio of the angles of a triangle is 2 : 3 : 4. What is the largest angle in degrees?

ক
30°
খ
60°
গ
90°
ঘ
75°
ঙ
80°

ব্যাখ্যা

Question: The ratio of the angles of a triangle is 2 : 3 : 4. What is the largest angle in degrees?

Solution:
ত্রিভুজের কোণগুলোর অনুপাত = 2 : 3 : 4
ধরি
কোণগুলো = 2x , 3x 4x

প্রশ্নমতে,
2x + 3x + 4x = 180°
বা, 9x = 180°
বা, x = 180°/9
x = 20°

বৃহত্তম কোণ = 4 × 20° = 80°

৫৩৪.

The angle of elevation of the top of a building from a point on the ground, which is 30 m away from the foot of the building, is 30°. The height of the building is:

ক
ক) 10√3 m
খ
খ) 30√3 m
গ
গ) 1/√3 m
ঘ
ঘ) None of these

ব্যাখ্যা

Question: The angle of elevation of the top of a building from a point on the ground, which is 30 m away from the foot of the building, is 30°. The height of the building is:

Solution:

let, x is the height of the building.

Hence,
tan 30° = perpendicular/base = x/30
⇒ 1/√3 = x/30
⇒ x = 30/√3 m
⇒ x = (30√3)/(√3 × √3)
⇒ x = (30√3)/3
⇒ x = 10√3 m

৫৩৫.

The height of an equilateral triangle with a side of 6 cm is -

ক
ক) 2√3 cm
খ
খ) 5√3 cm
গ
গ) 3√3 cm
ঘ
ঘ) √3 cm

ব্যাখ্যা

প্রশ্ন : The height of an equilateral triangle with a side of 6 cm is -

সমাধান :

সমবাহু ত্রিভুজের বাহুর দৈর্ঘ্য ৬ সেমি হলে,
ক্ষেত্রফল = (√৩/৪) × ৬^২ --- --- --- (১)

আমরা জানি, ত্রিভুজের ক্ষেত্রফল = ১/২ × ভূমি × উচ্চতা
সমবাহু ত্রিভুজের যেকোন একটি বাহুকে ভূমি ধরলে,
ভূমি = ৬ সেমি
অতএব, ক্ষেত্রফল = ১/২ × ৬ × উচ্চতা --- --- --- (২)

অতএব, সমীকরণ (১) ও (২) থেকে পাই,
১/২ × ৬ × উচ্চতা = (√৩/৪) × ৬^২
⇒ উচ্চতা = (√৩/৪) × ৩৬/৩
∴ উচ্চতা = ৩√৩ সেমি

৫৩৬.

Here is a square and a circular segment in given figure. Find the Total area of the whole region.

ক
147.7 m²
খ
181.7 m²
গ
150 m²
ঘ
165.5 m²

ব্যাখ্যা

Question: Here is a square and a circular segment in given figure. Find the Total area of the whole region.

Solution:
Given,
The radius of the circular segment r = 12 m
The angle subtended at centre θ = 30°
∴ The area of circular segment = (θ/360) × π × r² m²
= (30/360) × π × 144 m²
= 12π m²
= 37.7 m²

The are of square is 12² = 144 m²

∴ The area of whole region = 37.7 + 144 m²
= 181.7 m²

৫৩৭.

∠P and ∠Q are complementary to each other. If angle P = 20° + 4x and angle Q = 6x, find the value of ∠Q.

ক
34°
খ
37.5°
গ
42°
ঘ
48°

ব্যাখ্যা

Question: ∠P and ∠Q are complementary to each other. If angle P = 20° + 4x and angle Q = 6x, find the value of ∠Q.

Solution:
Here, ∠P = 20° + 4x and ∠Q = 6x

For complementary angles,
∠P + ∠Q = 90°
⇒ (20° + 4x) + 6x = 90°
⇒ 20° + 10x = 90°
⇒ 10x = 90° - 20°
⇒ 10x = 70°
⇒ x = 70°/10 = 7°

∴ ∠Q = 6x = 6 × 7° = 42°

৫৩৮.

An accurate clock shows 8 o’clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 2 o’clock in the afternoon?

ক
ক) 144°
খ
খ) 150°
গ
গ) 168°
ঘ
ঘ) 180°

ব্যাখ্যা

সকাল আটটা থেকে দুপুর দুইটা পর্যন্ত ঘড়ির ঘণ্টার কাটা ছয় ঘর সামনে যায়, অর্থাৎ = 360/12 × 6 ° = 180° ঘুরে

৫৩৯.

An open circular tank is dug 20 m deep. The earth is taken out and has been spread all round it, to form a circular embankment. The width of an Embankment is 2 m,& diameter of circular tank is 18m. Find the height of circular embankment.

ক
36.5 m
খ
38.5 m
গ
40.5 m
ঘ
43.5 m

ব্যাখ্যা

Question: An open circular tank is dug 20 m deep. The earth is taken out and has been spread all round it, to form a circular embankment. The width of an Embankment is 2 m,& diameter of circular tank is 18m. Find the height of circular embankment.

Solution:
Volume of the Open Circular Tank = πr²h
= π × [(18)/2]² × 20
= π × 9 × 9 × 20 m³

Area of the earth dug out = π(9 + 2)² - π(9)²
= π[(11 + 9)(11 - 9)]
= π(20)(2)
= 40πm²
∴ Height of circular embankment = (= π × 9 × 9 × 20 )/( 40π )
= 40.5 m

৫৪০.

The length, breadth, and height of a rectangular box are in the ratio 3 : 2 : 1. If its total surface area is 352 cm², what is its volume in cm³?

ক
324
খ
384
গ
432
ঘ
480

ব্যাখ্যা

Question: The length, breadth, and height of a rectangular box are in the ratio 3 : 2 : 1. If its total surface area is 352 cm², what is its volume in cm³?

Solution:
দেওয়া আছে,
আয়তাকার বাক্সের দৈর্ঘ্য, প্রস্থ ও উচ্চতার অনুপাত 3 : 2 : 1।
সমগ্রতলের ক্ষেত্রফল 352 বর্গ সে. মি.

মনে করি,
আয়তাকার বাক্সের দৈর্ঘ্য, a = 3x সে. মি.
আয়তাকার বাক্সের প্রস্থ, b = 2x সে. মি.
আয়তাকার বাক্সের উচ্চতা, c = x সে. মি.

আমরা জানি,
আয়তাকার বাক্সের সমগ্রতলের ক্ষেত্রফল = 2(ab + bc + ca) বর্গ একক
= 2(3x × 2x + 2x × x + x × 3x) বর্গ একক
= 2(6x²+ 2x²+ 3x²) বর্গ একক
= 2 × 11x²
= 22x² বর্গ একক

প্রশ্নমতে,
22x²= 352
⇒x²= 352/22
⇒x²= 16
⇒x = √16
∴ x = 4

আয়তাকার বাক্সের আয়তন = abc ঘন একক
= 3x × 2x × x ঘন একক
= 6x³ ঘন একক
= 6 × 4³ ঘন সে. মি.
= 384 ঘন সে. মি.

∴ আয়তাকার বাক্সের আয়তন 384 ঘন সে. মি. ।

৫৪১.

A cone’s slant height is 21 cm, and the curved surface area is 264 cm². Determine the diameter of the cone’s base.

ক
4 cm
খ
6 cm
গ
8 cm
ঘ
9 cm

ব্যাখ্যা

Question: A cone’s slant height is 21 cm, and the curved surface area is 264 cm². Determine the diameter of the cone’s base.

Solution:
দেওয়া আছে,
তির্যক উচ্চতা, l = 21 সে.মি
বক্রপৃষ্ঠের ক্ষেত্রফল = 264 cm²

মনে করি,
কোণকটির ভূমির ব্যাসার্ধ = r

∴ কোণকটির বক্রপৃষ্ঠের ক্ষেত্রফল = πrl

প্রশ্নমতে,
πrl = 264
⇒ r × (22/7) × 21 = 264
⇒ 66r = 264
⇒ r = 264/66
⇒ r = 4

∴ কোণকটির ভূমির ব্যাস = 2r = 2 × 4 = 8 সে.মি

৫৪২.

A rectangular park is 500 meters long and 300 meters wide. There is a 10-meter wide road around the park. What is the area of the park excluding the road?

ক
137,600 square meters
খ
126,300 square meters
গ
134,400 square meters
ঘ
141,500 square meters
ঙ
128,200 square meters

ব্যাখ্যা

Question: A rectangular park is 500 meters long and 300 meters wide. There is a 10-meter wide road around the park. What is the area of the park excluding the road?

Solution:
Total park area = (500 × 300) = 150,000 square meters
Park length after removing road = 500 - (2 × 10) = 480 meters
Park width after removing road = 300 - (2 × 10) = 280 meters
Park area excluding road = (480 × 280) square meters
= 134,400 square meters

৫৪৩.

How many interior reflex angles are there in a pentagon?

ক
1
খ
2
গ
5
ঘ
None of these

ব্যাখ্যা

There is no interior reflex angle in a polygon( বহুভুজে কোন অন্তঃস্থ প্রবৃদ্ধ কোণ নেই)।
So, there is no interior reflex angle in a pentagon( তাই পঞ্চভুজে কোন অন্তঃস্থ প্রবৃদ্ধ কোণ নেই).

৫৪৪.

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2 cm, then the area of the triangle is-

ক
36√3 cm²
খ
49√2 cm²
গ
60√2 cm²
ঘ
64√3 cm²

ব্যাখ্যা

Question: A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2 cm, then the area of the triangle is-

Solution:
Let, the side of the square be a cm
Then, its diagonal √2a
√2a = 12√2
⇒ a = 12

∴ Perimeter of the square = 4a
= 4 × 12
= 48 cm

and also perimeter of the equilateral triangle = 48 cm

∴ Each side of the triangle = 48/3
= 16 cm

Area of the triangle = (√3/4) × (16)² cm²
= (√3/4) × 256 cm²
= 64√3 cm²

৫৪৫.

What is the lowest value of sinθ?

ক
1
খ
0.5
গ
- 1
ঘ
1/3

ব্যাখ্যা

Question: What is the lowest value of sinθ?

Solution:
θ কোণ নির্দেশ করলে
sinθ এর বৃহত্তম মান = 1
sinθ এর ক্ষুদ্রতম মান = -1

৫৪৬.

The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is-

ক
30 m
খ
15 m
গ
15√2 m
ঘ
30√2 m
ঙ
60 m

ব্যাখ্যা

Question: The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is-

Solution:
Length of room = 30 m
Breadth of room = 24 m
Height of room = 18 m

Length of the longest rod = Diagonal of the room = √(30² + 24² + 18²)
= √(900 + 576 + 324)
= √(1800)
= √(900 × 2)
= 30√2

৫৪৭.

The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower is:

ক
25√3 m
খ
50√3 m
গ
75√3 m
ঘ
190 m

ব্যাখ্যা

Question: The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower is:

Solution:
AB is a tower and AB = 75 m
From A, the angle of depression of a car C
on the ground is 30°

Let distance BC = x

Now, in ΔACB,
tanθ = AB/BC
⇒ tan30° = 75/x
⇒ 1/√3 = 75/x
⇒ x = 75√3
∴ BC = 75√3 m

৫৪৮.

A tank is 12 m long, 8 m wide and 5 m deep. The cost of plastering its walls and bottom at 75 paisa per sq. m is-

ক
222 Tk
খ
348 Tk
গ
480 Tk
ঘ
592 Tk

ব্যাখ্যা

Question: A tank is 12 m long, 8 m wide and 5 m deep. The cost of plastering its walls and bottom at 75 paisa per sq. m is-

Solution:
Let, l = 12 m, b = 8 m and, h = 5 m

∴ Area to be plastered = [2(l + b) × h] + (l × b)
= [2(12 + 8) × 5] + (12 × 8) sq. m
= (200 + 96) sq. m
= 296 sq. m

∴ Cost of plastering = 296 × (75/100) Tk
= 296 × (3/4) Tk
= (74 × 3) Tk
= 222 Tk

৫৪৯.

The total cost of flooring a room at Tk. 8.50 per square meter is Tk. 510. If the length of the room is 8m, what is its breadth?

ক
7.5 m
খ
8.5 m
গ
10.5 m
ঘ
12.5 m

ব্যাখ্যা

Question: The total cost of flooring a room at Tk. 8.50 per square meter is Tk. 510. If the length of the room is 8m, what is its breadth?

Solution:
ঘরের ক্ষেত্রফল = 510/8.50
= 60 বর্গমিটার

প্রশ্নমতে
8 × প্রস্থ = 60
বা, প্রস্থ = 60/8
∴ প্রস্থ =7.5

আয়তাকার ক্ষেত্রের প্রস্থ =7.5 মিটার

৫৫০.

A pole of 66 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.

ক
22 m
খ
30 m
গ
36 m
ঘ
44 m

ব্যাখ্যা

Question: A pole of 66 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/(66 - x)
⇒ 1/2 = x/(66 - x)
⇒ 66 - x = 2x
⇒ 3x = 66
⇒ x = 66/3 = 22

the length of the broken part of the pole = 66 - 22 = 44 m

৫৫১.

7 মিটার উঁচু খুঁটির ছায়ার দৈর্ঘ্য 7√3 হলে, সূর্যের উন্নতি কোণ কত?

ক
20°
খ
30°
গ
45°
ঘ
60°
ঙ
কোনোটিই নয়

ব্যাখ্যা

প্রশ্ন: 7 মিটার উঁচু খুঁটির ছায়ার দৈর্ঘ্য 7√3 হলে, সূর্যের উন্নতি কোণ কত?

সমাধান:

খুঁটির দৈর্ঘ্য AB = 7 মিটার
ছায়ার দৈর্ঘ্য BC = 7√3
সূর্যের উন্নতি কোণ ∠ACB = θ = ?

এখন,
ΔABC এ
tanθ = AB/BC
⇒ tanθ = 7/(7√3)
⇒ tanθ = 1/√3
⇒ tanθ = tan30°
⇒ θ = 30°

৫৫২.

If the radius of a circle is decreased by 20% then the area is decreased by:

ক
ক) 20%
খ
খ) 30%
গ
গ) 36%
ঘ
ঘ) 40%

ব্যাখ্যা

ধরি,
বৃত্তের ব্যাসার্ধ = r একক,
ক্ষেত্রফল = πr² বর্গ একক।

ব্যাসার্ধ 20% কমে = r - r এর 20% = 0.8r একক
তাহলে ক্ষেত্রফল = π (0.8r)² = 0.64πr² বর্গ একক

ক্ষেত্রফল কমবে = πr²– 0.64πr² = 0.36πr² বর্গ একক

ক্ষেত্রফল কমার হার = (0.36πr²/πr²) x 100 = 36%

শর্টকাট পদ্ধতি:
[- 20 - 20 + (20 × 20)/100]% = - 36%

সুতরাং বৃত্তের ক্ষেত্রফল ৩৬% হ্রাস পাবে।

৫৫৩.

The complementary angle of supplementary angle of 130°-

ক
50°
খ
30°
গ
40°
ঘ
60°
ঙ
70°

ব্যাখ্যা

Question: The complementary angle of supplementary angle of 130°-

Solution:
For supplementary angle: The sum of two angles is 180°.
For complementary angle: The sum of two angles is 90°.

The supplement angle of 130° = 180° - 130° = 50°
The complement angle of 50° = 90° - 50° = 40°

∴ The complement angle of the supplement angle of 130° is 40°

৫৫৪.

If the length of each side of an equilateral triangle is increased by 2 meters, the area is found to be increased by 3 + √3 square meters. The length of each side of the triangle is:

ক
3 meters
খ
3√2 meters
গ
√3 meters
ঘ
5√3 meters
ঙ
None

ব্যাখ্যা

Question: If the length of each side of an equilateral triangle is increased by 2 meters, the area is found to be increased by 3 + √3 square meters. The length of each side of the triangle is:

Solution:
Let,
the length of each side of the equilateral triangle = a meters
∴ Its area = √3/4 × a² sq. meter

The area of the triangle when the length of each side increases by 2 meters = √3/4 × (a + 2)² sq. metre

ATQ,

৫৫৫.

Determine sin(A + B) for A = π/2 and B = π/4.

ক
1/√3
খ
1/4
গ
1/√2
ঘ
1/2

ব্যাখ্যা

Question: Determine sin(A + B) for A = π/2 and B = π/4.

Solution:
A = π/2
B = π/4

sin(A + B) = sin(π/2 + π/4)
= sin(90° + 45°)
= cos 45°
= 1/√2

৫৫৬.

The length of the tangent drawn from a point P to a circle of radius 6 cm is 8 cm. What is the distance of P from the center of the circle?

ক
8 cm
খ
10 cm
গ
12 cm
ঘ
14 cm

ব্যাখ্যা

Question: The length of the tangent drawn from a point P to a circle of radius 6 cm is 8 cm. What is the distance of P from the center of the circle?

Solution:

O be the center of the circle.
P be the point from which the tangent is drawn.
PT be the tangent from P to the circle, where T is the point of tangency.

The radius OT = 6 cm.
The length of the tangent PT = 8 cm.
Since OT is the radius and PT is a tangent, OT ⊥ PT, forming a right triangle △OPT.
OP² = OT² + PT²
⇒ OP² = 6² + 8²
⇒ OP² = 36 + 64
⇒ OP² = 100
⇒ OP = 10 cm

৫৫৭.

The angle of depression of a point situated at a distance of 70 m from the base of a tower is 60°. The height of the tower is -

ক
65√3 meter
খ
55√3 meter
গ
60√3 meter
ঘ
70√3 meter

ব্যাখ্যা

Question: The angle of depression of a point situated at a distance of 70 m from the base of a tower is 60°. The height of the tower is -

Solution:

Let,
Height of the tower AB = h meter.
Now, ∠DAC = ∠ACB = 60° and BC = 70 meter.

In ΔABC
tan60° = AB/BC
⇒ √3 = h/70
∴ h = 70√3

Height of the tower 70√3 meter.

৫৫৮.

The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.5 m away from the wall. The length of the ladder is:

ক
4.5 m
খ
8 m
গ
9 m
ঘ
11 m

ব্যাখ্যা

Question: The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.5 m away from the wall. The length of the ladder is:

Solution:
Let,
AB be the wall and BC be the ladder.
Then, ∠ACB = 60° and AC = 4.5 m.

Here,
AC/BC = cos60°
⇒ AC/BC = 1/2
⇒ BC = 2 × AC
⇒ BC = 2 × 4.5
∴ BC = 9

∴ The length of the ladder is 9 m.

৫৫৯.

If θ = 45°, what is the value of sec²θ - tan²θ?

ক
- √2
খ
- 1
গ
√2
ঘ
1

ব্যাখ্যা

Question: If θ = 45°, what is the value of sec²θ - tan²θ?

Solution:
θ = 45°

given,
sec²θ - tan²θ
= (√2)² - (1)²
= 2 - 1
= 1

৫৬০.

The minimum value of 2sin²θ + 3cos²θ is ?

ক
0
খ
2
গ
3
ঘ
1

ব্যাখ্যা

Question: The minimum value of 2sin²θ + 3cos²θ is ?

Solution:
Let, x = 2sin²θ + 3cos²θ
⇒ x = 2sin²θ + 2cos²θ + cos²θ
⇒ x = 2(sin²θ + cos²θ) + cos²θ
⇒ x = 2 + cos²θ ;[since sin²θ + cos²θ = 1]

Therefore x will be the minimum when cosθ = 0. i.e. minimum value of x will 2

৫৬১.

In a garden, there are 10 rows and 12 columns of mango trees. The distance between two trees is 2 meters and a distance of one meter is left from all sides of the boundary of the garden. What is the length of the garden?

ক
ক) 20 m
খ
খ) 22 m
গ
গ) 24 m
ঘ
ঘ) 26 m
ঙ
ঙ) None

ব্যাখ্যা

Each row contains 12 plants.
There are 11 gapes between the two corner trees (11 x 2) metres and 1 metre on each side is left.
Therefore Length = (22 + 2) m = 24 m.

৫৬২.

The volume V of a right circular cylinder is V =πr²h where r is the radius of the base and h is the height of the cylinder. If the volume of a right circular cylinder is 64π and its height is 4, what is the circumference of its base?

ক
2π
খ
3π
গ
8π
ঘ
5π

ব্যাখ্যা

Question: The volume V of a right circular cylinder is V = πr²h where r is the radius of the base and h is the height of the cylinder. If the volume of a right circular cylinder is 64π and its height is 4, what is the circumference of its base?
(একটি সোজা বৃত্তাকার সিলিন্ডারের ভলিউম V হল V = πr²h, যেখানে r হল ভিত্তির ব্যাসার্ধ এবং h হল সিলিন্ডারের উচ্চতা। যদি সোজা বৃত্তাকার সিলিন্ডারের ভলিউম 64π হয় এবং এর উচ্চতা 4 হয়, তবে তার ভিত্তির পরিধি কত হবে?)

Solution:
একটি সিলিন্ডারের উচ্চতা h একক ও ব্যাসার্ধ r একক হলে,
উক্ত সিলিন্ডারের আয়তন = πr²h ঘন একক

প্রশ্নমতে,
πr² × h = 64π
বা, πr² × 4 = 64π
বা, r² = 16
বা, r = 4

সুতরাং বৃত্তের পরিধি =2πr = 2π × 4 = 8π

৫৬৩.

If cos(x - 15°) = √3/2, what is the value of tan(x + 15°)?

ক
√3
খ
1
গ
1/√3
ঘ
0

ব্যাখ্যা

Question: If cos(x - 15°) = √3/2, what is the value of tan(x + 15°)?

Solution:
cos(x - 15°) = √3/2
⇒ cos(x - 15°) = cos 30°
⇒ x - 15° = 30°
⇒ x = 30° + 15°
∴ x = 45°

Now,
tan(x + 15°)
= tan(45° + 15°)
= tan 60°
= √3

৫৬৪.

The two diagonals of a rhombus are respectively 10 and 20 m. If the area of a square is same as that of rhombus, find the perimeter of the square.

ক
ক) 20 m
খ
খ) 30 m
গ
গ) 40 m
ঘ
ঘ) 50 m

ব্যাখ্যা

Question: The two diagonals of a rhombus are respectively 10 and 20 m. If the area of a square is same as that of rhombus, find the perimeter of the square.

Solution:
area of rhombus = (1/2) × 10 × 20
= 100 m2

area of square = 100 m2
side of square = √100 m
= 10 m

∴ perimeter of the square = 4 × 10 m
= 40 m

৫৬৫.

If θ is a positive acute angle and 4cos²θ - 1 = 0, then the value of tan(θ - 30°) is equal to?

ক
1/√3
খ
√3
গ
1
ঘ
0

ব্যাখ্যা

Question: If θ is a positive acute angle and 4cos²θ - 1 = 0, then the value of tan(θ - 30°) is equal to?

Solution:
Given,
4cos²θ - 1 = 0
⇒ 4cos²θ = 1
⇒ cos²θ = 1/4
⇒ cosθ = 1/2
⇒ cosθ = cos60°
∴ θ = 60°

Now,
tan(θ - 30°) = tan(60° - 30°)
= tan 30°
= 1/√3

৫৬৬.

If a rectangle's length and width are both doubled, by what percent is the rectangle's area increased?

ক
ক) 50%
খ
খ) 100%
গ
গ) 200%
ঘ
ঘ) 300%

ব্যাখ্যা

Question: If a rectangle's length and width are both doubled, by what percent is the rectangle's area increased?

Solution:
ধরি,
আয়তক্ষেত্রের দৈর্ঘ্য ক একক
আয়তক্ষেত্রের প্রস্থ খ একক
∴ আয়তক্ষেত্রের ক্ষেত্রফল কখ বর্গএকক

দৈর্ঘ্য ও প্রস্থ দ্বিগুণ হলে ক্ষেত্রফল হবে = ২ক × ২খ বর্গএকক
= ৪কখ বর্গএকক

ক্ষেত্রফল বৃদ্ধি পায় = (৪কখ - কখ) বর্গএকক
= ৩কখ

কখ বর্গএকক এ বৃদ্ধি পায় ৩কখ বর্গএকক
∴ ১ বর্গএকক এ বৃদ্ধি পায় (৩কখ)/(কখ) = ৩ বর্গএকক
∴ ১০০ বর্গএকক এ বৃদ্ধি পায় ৩ × ১০০ বর্গএকক
= ৩০০ বর্গএকক

∴ আয়তক্ষেত্রের ক্ষেত্রফল ৩০০% বৃদ্ধি পাবে।

৫৬৭.

The area of a triangle is 216 cm² and sides are in the ratio 3 : 4 : 5. The perimeter of the triangle is-

ক
65 cm
খ
70 cm
গ
72 cm
ঘ
75 cm

ব্যাখ্যা

Question: The area of a triangle is 216 cm² and sides are in the ratio 3 : 4 : 5. The perimeter of the triangle is-

Solution:
3² + 4² = 5²

It is a right-angled triangle.
let, the sides 3x, 4x, 5x

(1/2) × 3x × 4x = 216
⇒ 12x² = 432
⇒ x² = 36
⇒ x = 6

perimeter = (3 × 6) + (4 × 6) + (4 × 6)
= 72 cm

৫৬৮.

If A = 45° then

ক
ক) 2
খ
খ) 0
গ
গ) 1
ঘ
ঘ) 1/2

ব্যাখ্যা

Question: If A = 45° then

Solution:
A = 45°

এখন
(1 - tan²A)/(1 + tan²A)
= (1 - tan²45°)/(1 + tan²45°)
= (1 - 1²)/(1 + 1²)
= (1 - 1)/(1 + 1)
= 0/2
= 0

৫৬৯.

The two angles of a quadrilateral are 76° and 68°. If the other two angles are in the ratio of 5 : 7, then find the measure of each of them.

ক
96°, 120°
খ
100°, 116°
গ
90°, 126°
ঘ
80°, 136°

ব্যাখ্যা

Question: The two angles of a quadrilateral are 76° and 68°. If the other two angles are in the ratio of 5 : 7, then find the measure of each of them.

Solution:
Given two angles are 76° and 68°.
Let the other two angles be 5x and 7x.
As we know, the sum of interior angles of a quadrilateral is 360°.

Therefore,
76° + 68° + 5x + 7x = 360°
⇒ 144° + 12x = 360°
⇒ 12x = 360° - 144°
⇒ 12x = 216°
⇒ x = 216°/12
∴ x = 18°

Hence, the other two angles are:
5x = 5(18)° = 90°
7x = 7(18°) = 126°.

৫৭০.

What is the value of sin60°

ক
√3/2
খ
1/2
গ
1/√2
ঘ
1/√3

ব্যাখ্যা

Question: What is the value of sin60°

Solution:
sin60° = √3/2
sin30° = 1/2
sin45° = 1/√2
tan30° = 1/√3

৫৭১.

Two sides of a triangle are 7 and 16. Which of the following is not the length of the third side ?

ক
ক) 22
খ
খ) 17
গ
গ) 12
ঘ
ঘ) 9

ব্যাখ্যা

Question: Two sides of a triangle are 7 and 16. Which of the following is not the length of the third side ?

Solution:
আমরা জানি
ত্রিভুজের যে কোন দুই বাহুর সমষ্টি এর তৃতীয় বাহু অপেক্ষা বৃহত্তর

অপশন ক) 22 + 7 = 29 > 16 [এটি ত্রিভুজের বাহু হতে পারে]
অপশন খ) 17 + 7 = 24 > 16 [এটি ত্রিভুজের বাহু হতে পারে]
অপশন গ) 12 + 7 = 19 > 16[এটি ত্রিভুজের বাহু হতে পারে]
অপশন ঘ) 9 + 7 = 16 = 16 [এটি ত্রিভুজের বাহু হতে পারে না]

৫৭২.

The ratio of length and breadth of a rectangular field is 5 : 3. A dog runs along the boundary of the field at a speed of 12 km/h, and takes 8 minutes to complete one full round. Find the area of the field in square meters.

ক
132000 sq. m
খ
141500 sq. m
গ
150000 sq. m
ঘ
155600 sq. m
ঙ
None

ব্যাখ্যা

Question: The ratio of length and breadth of a rectangular field is 5 : 3. A dog runs along the boundary of the field at a speed of 12 km/h, and takes 8 minutes to complete one full round. Find the area of the field in square meters.

Solution:
One round of the park is equal to the perimeter of the park.
So, by completing one round, the dog covers a distance equal to the perimeter of the park.
Now,
Distance or perimeter = speed × time
= 12 × (8/60)
= 8/5 km
= 1.6 km
= 1600 meters

Let
Length = 5x and breadth = 3x
So, Perimeter: 2(5x + 3x) = 1600
⇒ 2 × 8x = 1600
⇒ 16x = 1600
∴ x = 1600/16 = 100 meters

So, Length = 5 × 100 = 500 meters
And, Breadth = 3 × 100 = 300 meters

Area = Length × Breadth
= 500 × 300
= 150000 sq. m.

৫৭৩.

The area of a trapezium is 120 square cm. The length of one of the parallel sides is 10 cm, and the distance between the parallel sides is 15 cm. Find the length of the other parallel side.

ক
4 cm
খ
6 cm
গ
8 cm
ঘ
12 cm

ব্যাখ্যা

Question: The area of a trapezium is 120 square cm. The length of one of the parallel sides is 10 cm, and the distance between the parallel sides is 15 cm. Find the length of the other parallel side.

Solution:
দেওয়া আছে,
ট্রাপিজিয়ামের ক্ষেত্রফল = 120 সেমি²
একটি সমান্তরাল বাহু, a = 10 সেমি
সমান্তরাল বাহুদ্বয়ের মধ্যবর্তী দূরত্ব, h = 15 সেমি

ধরি, অপর সমান্তরাল বাহু = b সেমি

আমরা জানি,
ট্রাপিজিয়ামের ক্ষেত্রফল = (1/2) × (a + b) × h

প্রশ্নমতে,
120 = (1/2) × (10 + b) × 15
⇒ 120 × 2 = (10 + b) × 15
⇒ 240 = (10 + b) × 15
⇒ (10 + b) = 240 / 15
⇒ 10 + b = 16
⇒ b = 16 - 10
⇒ b = 6 সেমি

সুতরাং, অপর সমান্তরাল বাহুটির দৈর্ঘ্য 6 সেমি।

৫৭৪.

ABC is an isosceles triangle with AB = AC, A circle through B touching AC at the middle point intersects AB at P. What is the ratio of AB and AP?

ক
4 : 1
খ
4 : 9
গ
4 : 3
ঘ
4 : 5
ঙ
4 : 7

ব্যাখ্যা

Let AB = AC = 2p and AQ = CQ = P
AP × AB = AQ²
AP × 2p = p²
AP = p/2
AP/AB = (p/2)/(2p) = 1/4
AP : AB = 1 : 4
---------------------------------------------------------------------------------------------
Alternative way:

Given that, AB=AC and AQ=QC
In △ABQ and △AQP,
∠BAQ=∠QAP [ For Common angle]
∠ABQ=∠AQP
[Angle between a chord and tangent is equal to angle subtended by that chord in the alternate segment]

[N. B. - The alternate segment theorem (also known as the tangent-chord theorem) states that in any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.]

△ABQ∼△AQP
AB/AQ = AQ/AP
or, AQ² = AB × AP
or, (AB/2)² = AB × AP
or, AB²/4 = AB × AP
or, AB/4 = AP
or, AB/AP = 4
∴ AB : AP = 4 : 1

৫৭৫.

Diameter of a circle is 4r unit. Then the area of the circle-

ক
ক) πr² sq.unit
খ
খ) 2πr² sq.unit
গ
গ) 4πr² sq.unit
ঘ
ঘ) None of these

ব্যাখ্যা

Quesion: Diameter of a circle is 4r unit. Then the area of the circle-

Solution:
Diameter of a circle is 4r m
Radius = 4r/2
= 2r

∴ Area = π(2r)²
= 4πr² sq.unit

৫৭৬.

The height of a cone is 12 cm and the radius of its base is 5 cm. Find its total surface area.

ক
60π cm²
খ
65π cm²
গ
85π cm²
ঘ
90π cm²

ব্যাখ্যা

Question: The height of a cone is 12 cm and the radius of its base is 5 cm. Find its total surface area.

Solution:
Let,
The height of a cone, h = 12 cm
the radius, r = 5 cm

∴ the slant height of the cone, l = √(r² + h²)
⇒ l = √(5² + 12²)
⇒ l = √(25 + 144)
⇒ l = √169
∴ l = 13

∴ total surface area = (πrl + πr²)
= π × 5 × 13 + π × 5²
= 65π + 25π
= 90π cm²

৫৭৭.

The perimeter of a rectangular field is 80 meters. If the ratio of its length to its width is 5:3, what is the area of the field in square meters?

ক
300 square meters
খ
375 square meters
গ
400 square meters
ঘ
484 square meters

ব্যাখ্যা

Question: The perimeter of a rectangular field is 80 meters. If the ratio of its length to its width is 5:3, what is the area of the field in square meters?

Solution:
ধরি,
আয়তাকার ক্ষেত্রের দৈর্ঘ্য = 5x মিটার এবং প্রস্থ = 3x মিটার।
পরিসীমা, P = 80 মিটার

আমরা জানি,
আয়তক্ষেত্রের পরিসীমা, P = 2 × (দৈর্ঘ্য + প্রস্থ)
∴ 80 = 2 × (5x + 3x)
⇒ 80 = 2 × (8x)
⇒ 80 = 16x
⇒ x = 80/16
∴ x = 5

সুতরাং,
দৈর্ঘ্য = 5x = 5 × 5 = 25 মিটার
প্রস্থ = 3x = 3 × 5 = 15 মিটার

∴ ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ
= 25 × 15 বর্গ মিটার
= 375 বর্গ মিটার

অতএব, মাঠটির ক্ষেত্রফল = 375 বর্গ মিটার।

৫৭৮.

A hollow cylinder has an internal radius of 6 cm, an external radius of 10 cm, and a height of 7 cm. Find the volume of the material used to make the cylinder.

ক
120π cm³
খ
448π cm³
গ
1000π cm³
ঘ
800π cm³

ব্যাখ্যা

Question: A hollow cylinder has an internal radius of 6 cm, an external radius of 10 cm, and a height of 7 cm. Find the volume of the material used to make the cylinder.

Solution:
Let,
Internal radius (r) = 6 cm
External radius (R) = 10 cm
Height (h) = 7 cm

Volume of the the material used,
V = πh(R² - r²)
= π × 7 (100 - 36)
= π × 7 × 64
= 448π

৫৭৯.

sin(θ + 15°) = 3/√12 হলে √2cosθ = ?

ক
ক) 1/√2
খ
খ) 1/2
গ
গ) 2
ঘ
ঘ) 1

ব্যাখ্যা

Question: sin(θ + 15°) = 3/√12 হলে √2cosθ = ?

Solution:
sin(θ + 15°) = 3/√12
⇒ sin(θ + 15°) = 3/(2√3)
⇒ sin(θ + 15°) = (√3 . √3)/2√3
⇒ sin(θ + 15°) = √3/2
⇒ sin(θ + 15°) = sin60°
⇒ θ + 15° = 60°
⇒ θ = 45°

Now,
√2cosθ = √2(cos 45°)
= √2(1/√2)
= 1

৫৮০.

If secθ + tanθ = x, then 2tanθ is -

ক
ক) (x² - 1)/2x
খ
খ) (x² - 1)/x
গ
গ) x/(x² - 1)
ঘ
ঘ) (x² + 1)/x

ব্যাখ্যা

Question: If secθ + tanθ = x, then tanθ is -

Solution:
দেওয়া আছে,
secθ + tanθ = x ................. (1)

আমরা জানি,
sec²θ - tan²θ = 1
বা, (secθ + tanθ)(secθ - tanθ) = 1
বা, x(secθ - tanθ) = 1
বা, secθ - tanθ = 1/x ................ (2)

(1) - (2) হতে পাই,
(secθ + tanθ) - (secθ - tanθ) = x - (1/x)
বা, 2tanθ = (x² - 1)/x
∴ 2tanθ = (x² - 1)/x

৫৮১.

A rectangular garden is 30 meters long and 18 meters wide. A walkway, 2.5 meters wide, is made all around the inside of the garden. What are the new length and width of the garden area left after building the walkway?

ক
27.5 meters by 15.5 meters
খ
32.5 meters by 20.5 meters
গ
25 meters by 13 meters
ঘ
35 meters by 23 meters
ঙ
None

ব্যাখ্যা

Question: A rectangular garden is 30 meters long and 18 meters wide. A walkway, 2.5 meters wide, is made all around the inside of the garden. What are the new length and width of the garden area left after building the walkway?

Solution:

Given,
Total outer dimensions including the walkway:
Length = 30 meters
Width = 18 meters

Walkway is 2.5 meters wide on all sides, so:

Length of remaining garden = (30 - 2.5 - 2.5) meters
= (30 - 5) meters
=25 meters

Width of remaining garden = (18 - 2.5 - 2.5) meters
= (18 - 5) meters
=13 meters

Hence The dimensions of the remaining garden (excluding the walkway) are 25 meters by 13 meters

৫৮২.

The complement of an angle exceeds the angle by 60°. Then the angle is equal to -

ক
15°
খ
25°
গ
30°
ঘ
35°

ব্যাখ্যা

Question: The complement of an angle exceeds the angle by 60°. Then the angle is equal to -

Solution:
Let, the angle be x
complement of the angle 90 - x

ATQ,
90 - x = x + 60°
⇒ 2x = 90 - 60
⇒ x = 30/2 = 15°

৫৮৩.

The area of the largest rectangle that can be inscribed in a circle of radius r is:

ক
2r
খ
2r²
গ
√2r
ঘ
2/r

ব্যাখ্যা

Question: The area of the largest rectangle that can be inscribed in a circle of radius r is:

Solution:
The largest rectangle inscribed in a circle is a square.
Diagonal of the square = diameter of the circle = 2r.

Let, side of square = s
By Pythagoras:
s² + s² = (2r)²
⇒ 2s² = 4r²
⇒ s² = 2r²
⇒ s = r√2

∴ Area = s²= 2r²

৫৮৪.

tan15° + tan75° + tan105° + tan165° = ?

ক
6
খ
0
গ
13
ঘ
1

ব্যাখ্যা

Question: tan15° + tan75° + tan105° + tan165° = ?

Solution:
tan15° + tan75° + tan105° + tan165°
= tan15° + tan(90° - 15°) + tan(90° + 15°) + tan(2 × 90° - 15°)
= tan15° + cot15° - cot15° - tan15°
= 0

৫৮৫.

In a right triangle, the length of one of the legs is 3 and the length of the hypotenuse is 5. What is the perimeter of the triangle?

ক
10
খ
12
গ
14
ঘ
16

ব্যাখ্যা

Question: In a right triangle, the length of one of the legs is 3 and the length of the hypotenuse is 5. What is the perimeter of the triangle?

Solution:
সমকোণী ত্রিভুজের অতিভুজ = 5
সমকোণ সংলগ্ন এক বাহু = 3
সমকোণ সংলগ্ন অপর বাহু = a

প্রশ্নমতে
a² + 3² = 5²
⇒ a² + 9 = 25
⇒ a² = 25 - 9
⇒ a² = 16
⇒ a² = 4²
∴ a = 4

∴ ত্রিভুজটির পরিসীমা = (3 + 4 + 5) = 12

৫৮৬.

What is the value of (2√3 - cot 30°) = ?

ক
0
খ
3√3
গ
√3
ঘ
3

ব্যাখ্যা

Question: What is the value of (2√3 - cot 30°) = ?

Solution:
Given that,
= (2√3 - cot 30°)
= (2√3 - √3)
= √3

৫৮৭.

The diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. The curved surface area is:

ক
ক) 150 sq. cm
খ
খ) 165 sq. cm
গ
গ) 177 sq. cm
ঘ
ঘ) 10 sq.cm

ব্যাখ্যা

Given that.
Diameter of a right circular cone = 10.5 cm
Radius of a right circular cone = 10.5/2 = 5.25 cm
and slant height of a right circular cone (l) = 10 cm

∴ Lateral surface area of a cone = πrl
= 22/7 × 5.25 × 10
= 165 cm²

৫৮৮.

If the length of the face diagonal of a cube is 7√2 meters, what is the volume of the cube?

ক
256 m³
খ
343 m³
গ
294 m³
ঘ
392 m³

ব্যাখ্যা

Question: If the length of the face diagonal of a cube is 7√2 meters, what is the volume of the cube?

Solution:
Given that,
The length of the face diagonal of the cube is 7√2 meters.

Let the edge length of the cube be a meters.
∴ The face diagonal of the cube = a√2 meters

According to the given condition:
a√2 = 7√2
∴ a = 7
Therefore, the volume of the cube = a³= 7³ = 343 cubic meters

∴ The volume of the cube is 343 m³.

৫৮৯.

If the land of the isosceles triangle is 16cm and the other two sides are 10cm each, what is the area of the triangle?

ক
ক) 44 sq cm
খ
খ) 48 sq cm
গ
গ) 49 sq cm
ঘ
ঘ) 42 sq cm

ব্যাখ্যা

The area of the triangle
= (16/4)√(4 × 10² - 16²)
= 4√(400 - 256)
= 4 × √144
= 4 × 12
= 48 sq cm

৫৯০.

66 cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in meters will be-

ক
80 m
খ
84 m
গ
88 m
ঘ
92 m

ব্যাখ্যা

Question: 66 cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in meters will be-

Solution:
ব্যাসার্ধ r = 1 mm/2 = 0.5 mm = 0.5/10 cm = 0.05 cm
তারের দৈর্ঘ্য h

π r²h = 66
⇒ π (0.05)² h = 66
⇒ h = 66/π (0.05)²
= 8400 cm
= 8400/100 m
= 84 m

৫৯১.

The length of each side of square A is increased by 100 percent to make square B. Then each side of square B is increased by 50 percent to make square C. By what percent is the area of square C greater than the sum of the areas of square A and B?

ক
40%
খ
60%
গ
80%
ঘ
120%

ব্যাখ্যা

Question: The length of each side of square A is increased by 100 percent to make square B. Then each side of square B is increased by 50 percent to make square C. By what percent is the area of square C greater than the sum of the areas of square A and B?

Solution:
let, length of square A is x m
area of A = x² cm²

length of square B is x + x cm = 2x cm
area of B = (2x)² = 4x² cm²

length of square C = 2x + 2x × 0.5 = 3x
area of C = (3x)²
= 9x² cm²

sum of the areas of square A and B = x² + 4x² = 5x²

percent increase of area of C = (9x² - 5x²) × 100%/5x²
= 80%

৫৯২.

If secθ + tanθ = x, then cotθ is -

ক
ক) (x² - 1)/2x
খ
খ) 2x/(x² - 1)
গ
গ) x/(x² - 1)
ঘ
ঘ) 2x/(x² + 1)

ব্যাখ্যা

Question: If secθ + tanθ = x, then cotθ is -

Solution:
দেওয়া আছে,
secθ + tanθ = x ................. (1)

আমরা জানি,
sec²θ - tan²θ = 1
বা, (secθ + tanθ)(secθ - tanθ) = 1
বা, x(secθ - tanθ) = 1
বা, secθ - tanθ = 1/x ................ (2)

(1) - (2) হতে পাই,
(secθ + tanθ) - (secθ - tanθ) = x - (1/x)
বা, 2tanθ = (x² - 1)/x
বা, tanθ = (x² - 1)/2x
∴ cotθ = 2x/(x² - 1)

৫৯৩.

From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 200 m high, the distance of point P from the foot of the tower is:

ক
298 meters
খ
312 meters
গ
346 meters
ঘ
450 meters

ব্যাখ্যা

tan30° = height/base
1/√3 = 200/base
or, base = 200√3 = 346m(approx.)
Therefore, the distance of point P from the foot of the tower = 346m

৫৯৪.

A ladder 17 feet long leans against a wall. The base is 8 feet from the wall. How high up the wall does the ladder reach?

ক
23 feet
খ
18 feet
গ
12 feet
ঘ
15 feet

ব্যাখ্যা

Question: A ladder 17 feet long leans against a wall. The base is 8 feet from the wall. How high up the wall does the ladder reach?

Solution:
Length of the ladder (hypotenuse) = 17 feet
Distance from base of ladder to the wall (one leg) = 8 feet

Using the Pythagorean theorem,
(Base)² + (Height)² = (Hypotenuse)²
⇒ 8² + h² = 17²
⇒ 64 + h² = 289
⇒ h² = 289 - 64
⇒ h² = 225
⇒ h = √225 = 15
∴ h = 15

So the ladder reaches 15 feet up the wall.

৫৯৫.

The surface area of a cube is 96 square units. What is the length of the longest stick that can be placed inside the cube?

ক
8
খ
4√3
গ
4√2
ঘ
6√2

ব্যাখ্যা

Question: The surface area of a cube is 96 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 = 96 square units

We know,
Surface area of a cube, S = 6a²
⇒ 6a² = 96
⇒ a² = 96/6
⇒ a²= 16 = 4²
∴ a = 4

The longest stick that can fit inside the cube runs along the space diagonal.
So the space diagonal of a cube, d = a√3
= 4√3 ; [a = 4]

So the length of the longest stick that can be placed inside the cube is 4√3 units.

৫৯৬.

A hall, 20m long and 15m broad, is surrounded by a verandah of uniform width of 3.5m. The cost of flooring the verandah at Tk.2.50 per square meter is-

ক
ক) Tk. 600
খ
খ) Tk. 594
গ
গ) Tk. 735
ঘ
ঘ) Tk. 800

ব্যাখ্যা

বারান্দাসহ হল ঘরের দৈর্ঘ্য = {20 + (2 × 3.5)} মিটার
= 27 মিটার

বারান্দাসহ হল ঘরের প্রস্থ = {15 + (2 × 3.5)} মিটার
= (15 + 7) মিটার
= 22 মিটার
বারান্দাসহ হল ঘরের ক্ষেত্রফল = (27 × 22) বর্গমিটার
= 594 বর্গমিটার

বারান্দাবাদে হল ঘরের ক্ষেত্রফল = (20 × 15) বর্গমিটার
= 300 বর্গমিটার
বারান্দার ক্ষেত্রফল = (594- 300) বর্গমিটার
= 294 বর্গমিটার
মোট খরচ =(294 × 2.50) টাকা
= 735 টাকা

৫৯৭.

Find the side of a square whose area is equal to the area of a rectangle with sides 10 m. and 6.4m .

ক
ক) 4 m.
খ
খ) 6 m.
গ
গ) 8 m.
ঘ
ঘ) 10 m.

ব্যাখ্যা

Given that
Length of a rectangle = 10 m
Breadth of a rectangle = 6.4 m
Area of a rectangle = 6.4 × 10 = 64m²
According to the question
Area of square = Area of rectangle
Suppose the side of square be x m.
64 = (x)²
(8)²=(x)²

Thus, the side is 8 m.

৫৯৮.

কোনটি ত্রিভুজের ক্ষেত্রফল?

ক
(1/2) × ভূমি × উচ্চতা
খ
ভূমি × উচ্চতা
গ
লম্ব × ভূমি
ঘ
(1/2) × লম্ব × উচ্চতা

ব্যাখ্যা

প্রশ্ন: কোনটি ত্রিভুজের ক্ষেত্রফল?

সমাধান:
• ত্রিভুজের ক্ষেত্রফল = (1/2) × ভূমি × উচ্চতা

• সামন্তরিকের ক্ষেত্রফল = ভূমি × উচ্চতা

৫৯৯.

The sum of the interior angles of a regular polygon is 1260°. How many sides does the polygon have?

ক
10 sides
খ
8 sides
গ
9 sides
ঘ
6 sides

ব্যাখ্যা

Question: The sum of the interior angles of a regular polygon is 1260°. How many sides does the polygon have?

Solution:
We know, the sum of the interior angles of a polygon = (n - 2) × 180°

Given,
(n - 2) × 180 = 1260
⇒ n - 2 = 1260/180
⇒ n - 2 = 7
⇒ n = 7 + 2
n = 9

∴ The polygon has 9 sides.

৬০০.

The area of a rectangle that has length = 2a²b and breadth = 3ab² is:

ক
6a³b³
খ
a³b³
গ
2a³b³
ঘ
4a³b³

ব্যাখ্যা

Question: The area of a rectangle that has length = 2a²b and breadth = 3ab² is:

Solution:
দেওয়া আছে,
আয়তক্ষেত্রের দৈর্ঘ্য = 2a²b
এবং আয়তক্ষেত্রের প্রস্থ = 3ab²
আয়তক্ষেত্রের ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ
= 2a²b × 3ab²
= 6a³b³