Geometry: Mensuration, Trigonometry

PrepBank · পাতা ১০ / ২১ · ৯০১–১,০০০ / ২,০৮৫

৯০১.

The side of an equilateral triangle is 2m. What is the area of the triangle?

ক
√2 m²
খ
√5 m²
গ
3 m²
ঘ
√3 m²

ব্যাখ্যা

Question: The side of an equilateral triangle is 2m. What is the area of the triangle?

Solution:
Area = (√3/4)2²
= (√3/4)× 4 m²
= √3 m²

৯০২.

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

ক
30°
খ
45°
গ
60°
ঘ
90°

ব্যাখ্যা

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

Solution:

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

∴ θ = 60°

৯০৩.

The diagonals of a rhombus are 72 cm and 30 cm respectively. What is its perimeter?

ক
ক) 12 cm
খ
খ) 144 cm
গ
গ) 156 cm
ঘ
ঘ) 168 cm

ব্যাখ্যা

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

Perimeter = 2√(72² + 30²)
= 156 cm.

৯০৪.

A rectangular floor that measures 8 meters by 10 meters is to be covered with carpet that each measure 2 meters by 2 meters. If the carpet cost Tk. 12 a piece, what is the total cost to cover the floor?

ক
ক) Tk. 200
খ
খ) Tk. 240
গ
গ) Tk. 480
ঘ
ঘ) Tk. 960

ব্যাখ্যা

Area of the floor = 8 × 10 = 80 m²
Area of each carpet = 2 × 2 = 4 m²
Required number of carpet = 80/4 = 20
one carpet costs Tk. 12
20 carpet costs Tk. 12 × 20 = Tk. 240

৯০৫.

If each side of the square is increased by 20%, what will be the ratio between the new area and the original area of the square?

ক
9 : 5
খ
15 : 7
গ
27 : 13
ঘ
36 : 25

ব্যাখ্যা

Question: If each side of the square is increased by 20%, what will be the ratio between the new area and the original area of the square?

Solution:
Let,
The side of original square is x
∴ The area of original square is x²

The side of new square is x + 20% of x = x + (x/5) = 6x/5
∴ The area of new square is (36x²)/25
∴ The ratio between the new area and the original area of the square = (36x²)/25 : x²
= 36/25 : 1
= 36 : 25

৯০৬.

The angles of elevation of the top of a tower from the top and bottom of a tree of height 15 m are 30° and 60° respectively. Find the height of the tower?

ক
7.5 m
খ
22.5 m
গ
11.5 m
ঘ
20 m

ব্যাখ্যা

Question: The angles of elevation of the top of a tower from the top and bottom of a tree of height 15 m are 30° and 60° respectively. Find the height of the tower?

Solution:

Let the CE be h meter.
Height of tree be AD = 15m
BE is the height of tower = BC + CE = 15 + h
AB = CD, let it is = X m

From ΔCDE,
tan30° = EC/CD
⇒ 1/√3 = h/X
∴ X = √3h ..........(1)

From ΔABE,
tan60° = EB/AB
⇒ √3 = (h + 15)/X
∴ X = (h + 15)/√3 ..............(2)

From (1) and (2) we get,
√3h = (h + 15)/√3
⇒ 3h = h + 15
⇒ 2h = 15
∴ h = 7.5

∴ Height of tower = 15 + 7.5 = 22.5 m

৯০৭.

A ladder rests against a wall that is perpendicular to the ground. If the bottom of the ladder is 4m away from the bottom of the wall, while the top of the ladder is at a height of 3m. What is the length of the ladder?

ক
ক) 7m
খ
খ) 35m
গ
গ) 5m
ঘ
ঘ) 25m

ব্যাখ্যা

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

৯০৮.

The angle of elevation of an aeroplane from a point A on the ground is 60°. After a straight flight of the plane for 30 seconds, the angle of elevation becomes 30°. If the palne flies at a constant height of 3600√3 metre, what is the speed of plane?

ক
432 m/sec
খ
480 m/sec
গ
240 m/sec
ঘ
864 m/sec

ব্যাখ্যা

Question: The angle of elevation of an aeroplane from a point A on the ground is 60°. After a straight flight of the plane for 30 seconds, the angle of elevation becomes 30°. If the palne flies at a constant height of 3600√3 metre, what is the speed of plane?

Solution:

P and Q = Positions of plane
∠PAB = 60°, ∠QAB = 30°, PB = 3600√3 metre
In ∆ABP, tan 60° = BP/AB
⇒ √3 = 3600√3/AB
⇒ AB = 3600 metre

In ∆ACQ, tan 30° = CQ/AC
⇒ 1/√3 = 3600√3/AC
⇒ AC = 3600 × 3 = 10800 metre
∴ PQ = BC = AC – AB = 10800 – 3600 = 7200 metre
This distance is covered in 30 seconds.

∴ Speed of plane = 7200/30 = 240 m/sec

৯০৯.

Find the value of Cos(3π/4).

ক
- (1/√3)
খ
- (1/√2)
গ
- (1/√5)
ঘ
(1/√2)
ঙ
- (1/√7)

ব্যাখ্যা

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

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

৯১০.

A wheel of a car of radius 21 cm is rotating at 600 RPM. What is the speed of the car in km/hr?

ক
79.2 km/hr
খ
47.52 km/hr
গ
7.92 km/hr
ঘ
39.6 km/hr

ব্যাখ্যা

Question: A wheel of a car of radius 21 cm is rotating at 600 RPM. What is the speed of the car in km/hr?

Solution:
The radius of the wheel measures 21 cm.

In one rotation, the wheel will cover a distance which is equal to the circumference of the wheel.
∴ in one rotation this wheel will cover 2 × π × 21 = 2 × (22/7) × 21 = 132 cm.

In a minute, the distance covered by the wheel = circumference of the wheel × rpm
∴ this wheel will cover a distance of 132 × 600 = 79200 cm in a minute.

In an hour, the wheel will cover a distance of 79200 × 60 = 4752000 cm.

Therefore, the speed of the car = 4752000 cm/hr = 47.52 km/hr

৯১১.

The ratio of the length to the width of a rectangular advertising display is approximately 3.3 to 2. If the width of the display is 8 meters, what is the approximate length of the display, in meters?

ক
ক) 7
খ
খ) 11
গ
গ) 13
ঘ
ঘ) 16

ব্যাখ্যা

Length : width = 3.3 : 2
Length : 8 = 3.3 : 2
Length/8 = 3.3/2
2 × length = 26.4
Length = 26.4/2 = 13.2 ≈ 13 meters(approximate)

৯১২.

The base of a rectangle is three times as long as the height. If the perimeter is 112, what is the area of the rectangle?

ক
536 square unit
খ
546 square unit
গ
560 square unit
ঘ
588 square unit

ব্যাখ্যা

Question: The base of a rectangle is three times as long as the height. If the perimeter is 112, what is the area of the rectangle?

Solution:
মনে করি,
আয়তক্ষেত্রের উচ্চতা = x একক
আয়তক্ষেত্রের ভূমি = 3x একক
আয়তক্ষেত্রের পরিসীমা = 2(x + 3x) একক

প্রশ্নমতে,
2(x + 3x) = 112
⇒ 2 × 4x = 112
⇒ 8x = 112
∴ x = 14

আয়তক্ষেত্রের উচ্চতা = 14 একক
এবং আয়তক্ষেত্রের ভূমি = 3 × 14 = 42 একক

∴ আয়তক্ষেত্রের ক্ষেত্রফল = ভূমি × উচ্চতা
= 42 × 14
= 588 বর্গ একক

৯১৩.

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

ক
Tk. 380
খ
Tk. 441
গ
Tk. 320
ঘ
Tk. 210

ব্যাখ্যা

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

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

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

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

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

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

So the total expenditure is Tk. 210.

৯১৪.

The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, what is the area of the field?

ক
ক) 37500 sq.m
খ
খ) 40000 sq.m
গ
গ) 48000 sq.m
ঘ
ঘ) 18750 sq.m

ব্যাখ্যা

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

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

∴ Area = lb
= (250 × 150)
= 37500 m²

৯১৫.

What is the perimeter of a square (in meter) if its area is 100 Sq. meter?

ক
10
খ
40
গ
120
ঘ
100

ব্যাখ্যা

Question: What is the perimeter of a square (in meter) if its area is 100 Sq. meter?

Solution:
∴ বর্গের ক্ষেত্রফল = ১০০ বর্গমিটার
= ১০^২ বর্গমিটার

∴ বর্গের একবাহুর দৈর্ঘ্য ১০ মিটার

∴ বর্গের পরিসীমা ৪ × ১০ মিটার
= ৪০ মিটার

৯১৬.

The length of a rectangle is thrice its breath, and its perimeter is 112 meters. What is its area?

ক
507 sq. m.
খ
508 sq. m.
গ
588 sq. m.
ঘ
510 sq. m.

ব্যাখ্যা

Question: The length of a rectangle is thrice its breath, and its perimeter is 112 meters. What is its area?

Solution:
Let the breath = x
So, the Length = 3x

Perimeter of a rectangle = 2 (Length + Breadth)
So, 2(3x + x) = 112
⇒ 6x + 2x = 112
⇒ 8x = 112
∴ x = 112/8 = 14

Now, Breadth = 14, so, length = 14 × 3 = 42

So, its area = Length × Breadth
= 42 × 14 = 588 sq. m.

৯১৭.

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

ক
78°
খ
45°
গ
60°
ঘ
42°

ব্যাখ্যা

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

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

৯১৮.

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is:

ক
3.696 kg
খ
6.696 kg
গ
7 kg
ঘ
9.369 kg

ব্যাখ্যা

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

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

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

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

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

৯১৯.

In a circle, if the inscribed angle on an arc is 35°, what is the measure of the central angle subtended by the same arc?

ক
40°
খ
17.5°
গ
70°
ঘ
105°

ব্যাখ্যা

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

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

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

অতএব, কেন্দ্রস্থ কোণের পরিমাপ 70°।

৯২০.

In the figure AC and BC are radii of circles. The length of AB is 8. If AC = 4, what is BC? (BC is tangent to the circle with center A.)

ক
3√2
খ
4√3
গ
4√2
ঘ
2√2

ব্যাখ্যা

Question: In the figure AC and BC are radii of circles. The length of AB is 8. If AC = 4, what is BC? (BC is tangent to the circle with center A.)

Solution:
Since BC is tangent to circle with centre A
∴ BC is perpendicular to AC.
ΔABC is right angled triangle.
So,
BC = √(AB² - AC²)
= √(8² - 4²)
= √(64 - 16)
= √48
= √(16 × 3)
= 4√3

৯২১.

How many triangle are there in the figure below.

ক
ক) 13
খ
খ) 12
গ
গ) 11
ঘ
ঘ) 9

ব্যাখ্যা

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

৯২২.

A wire can be bent in the form of a circle of radius 28cm. If it is bent in the form of a square, then what will be its area?

ক
ক) 1396 cm²
খ
খ) 1936 cm²
গ
গ) 1236 cm²
ঘ
ঘ) 1536 cm²

ব্যাখ্যা

দেয়া আছে,
বৃত্তের ব্যাসার্ধ r = 28 cm
বৃত্তের পরিধি = 2πr
= 2 × (22/7) × 28
= 2 × 22 × 4
= 176 cm
বর্গের এক বাহুর দৈর্ঘ্য = 176/4 cm
= 44 cm
বর্গের ক্ষেত্রফল = (44)² cm²
= 1936 cm²

৯২৩.

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

ক
18°
খ
25°
গ
28°
ঘ
30°

ব্যাখ্যা

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

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

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

∴ The angle is 18°

৯২৪.

A rectangle has a perimeter of 64 cm, with the lengths of its adjacent sides in a 7 : 9 ratio. What are the measurements of these sides?

ক
10 cm, 8 cm
খ
14 cm, 18 cm
গ
12 cm, 14 cm
ঘ
20 cm, 15 cm

ব্যাখ্যা

Question: A rectangle has a perimeter of 64 cm, with the lengths of its adjacent sides in a 7 : 9 ratio. What are the measurements of these sides?

Solution:
Perimeter of a rectangle = 2(Length + Breadth)
Also Length : Breadth = 9 : 7
Let, the actual values are 9p and 7p

Hence,
2(9p + 7p) = 64
⇒ 16p = 32
∴ p = 2

∴ The sides will be 14 cm and 18 cm.

৯২৫.

The sum of all the interior angles of a hexagon is:

ক
ক) 540°
খ
খ) 360°
গ
গ) 180°
ঘ
ঘ) 270°
ঙ
ঙ) 720°

ব্যাখ্যা

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

৯২৬.

Find the perimeter (in cm) of a square having an area equal to the area of a rhombus, measures of whose diagonals are 8 cm and 16 cm.

ক
ক) 38 cm
খ
খ) 36 cm
গ
গ) 32 cm
ঘ
ঘ) 39 cm

ব্যাখ্যা

প্রশ্ন : Find the perimeter (in cm) of a square having an area equal to the area of a rhombus, measures of whose diagonals are 8 cm and 16 cm.

সমাধান :

Measures of the diagonals of the rhombus are 8 cm and 16 cm.
Area of the square = Area of the rhombus

Concept used:

Area of a rhombus = Product of two diagonals ÷ 2
Area of a square = side²

Perimeter of a square = 4 × side

Area of the rhombus = (8 × 16)/2 = 64 cm²

According to the question,

Area of the square = 64 cm²

The measure of each side of the square = √64 = 8 cm
Now, the perimeter of the square = 4 × 8 = 32 cm

∴ The perimeter of the square is 32 cm.

৯২৭.

The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be-

ক
ক) 9 cm
খ
খ) 12 cm
গ
গ) 30 cm
ঘ
ঘ) 18 cm

ব্যাখ্যা

Question: The diagonal of a rectangle is √41 cm and its area is 20 sq.cm . The perimeter of the rectangle must be-

Solution:
l²+ b²= (√41)²= 41
lb = 20

We know that,
(l + b)²= l²+ b²+ 2lb
(l + b)² = 41 + 2 × 20
(l + b)² = 41 + 40
(l + b)² = 81
l + b = √81
= 9

Perimeter= 2(l + b) = 2 × 9 = 18 cm

৯২৮.

What is the volume of a hemisphere having radius 3 cm?

ক
18π m³
খ
9π m³
গ
27π m³
ঘ
36π m³
ঙ
None of these

ব্যাখ্যা

Question: What is the volume of a hemisphere having radius 3 cm?

Solution:
The radius of a hemisphere is 3cm
A hemisphere's volume is equal to (2/3)πr³

Volume = (2/3)π.3³)
⇒ 2π × 9
⇒ 18π

∴ The volume of a hemisphere having a radius 3 cm is 18π cm³

৯২৯.

A rectangular carpet has an area of 120 sq. metres and a perimeter of 46 metres. What is the length of the diagonal?

ক
ক) 15
খ
খ) 16
গ
গ) 20
ঘ
ঘ) 17

ব্যাখ্যা

Question: A rectangular carpet has an area of 120 sq. metres and a perimeter of 46 metres. What is the length of the diagonal?

Solution:
Let the length and breadth of the rectangle be x and y metres.

given 2(x + y) = 46
x + y = 23
x = 23 - y

and,
xy = 120
(23 - y) y = 120
23y - y² = 120
y² - 23y + 120 = 0
y² - 15y - 8y + 120 = 0
y(y - 15) - 8(y - 15) = 0
(y - 15) (y - 8) = 0
y = 15 or, 8

∴ length, x = 15 and breadth, y = 8

so, the diagonal is = √{(15)² + (8)²}
= √289
= 17

৯৩০.

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

ক
∠ABC = 90°
খ
∠ACB = 30°
গ
tan(A) = √3
ঘ
A, B and C
ঙ
None of these

ব্যাখ্যা

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

Solution:

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

2² = a² + 1²
⇒ a² = 4 - 1
⇒ a = √3

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

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

৯৩১.

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

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

ব্যাখ্যা

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

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

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

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

৯৩২.

In the figure below, what is the value of x?

ক
55°
খ
60°
গ
70°
ঘ
90°

ব্যাখ্যা

Question: In the figure below, what is the value of x?

Solution:
AB = √(4² + 3²) = √(16 + 9) = √25 = 5
13² = AC² + 12²
⇒ AC2 = 13² - 12² = 25
AC = 5

∠B = ∠C = 55°

∠x = 180 - 55 - 55
= 180 - 110
= 70°

৯৩৩.

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

ক
12 cm
খ
16√2 cm
গ
24 cm
ঘ
32 cm

ব্যাখ্যা

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

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

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

অতএব, অপর কর্ণের দৈর্ঘ্য = 24 সে.মি.

৯৩৪.

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

ক
5x - 2y - 16 = 0
খ
5x - 2y - 4 = 0
গ
2x + 5y - 11 = 0
ঘ
5x + 2y + 4 = 0

ব্যাখ্যা

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

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

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

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

∴ the required line 5x - 2y - 16 = 0.

৯৩৫.

If sec²θ + tan²θ = 5/12 then, sec⁴θ - tan⁴θ = ?

ক
7/12
খ
5/12
গ
5/7
ঘ
3/8

ব্যাখ্যা

প্রশ্ন: If sec²θ + tan²θ = 5/12 then, sec⁴θ - tan⁴θ = ?

সমাধান:
sec⁴θ - tan⁴θ
= (sec²θ - tan²θ)(sec²θ + tan²θ)
= 1 × (sec²θ + tan²θ) [cause 1 + tan²θ = sec²θ]
= 1 × (7/12)
= 5/12

৯৩৬.

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
7.9 m
খ
8.4 m
গ
5.3 m
ঘ
8.2 m
ঙ
9.2 m

ব্যাখ্যা

Let AB be the wall and BC be the ladder

Then, ACB = 60°
and AC = 4.6 m.
AC/BC = cos 60° = 1/2
BC = 2 × AC = (2 × 4.6) m = 9.2 m.

৯৩৭.

The breadth of a room is twice its height, one half of its length and the volume of the room is 512cu.m. The length of the room :

ক
ক) 4 m
খ
খ) 8 m
গ
গ) 12 m
ঘ
ঘ) 16 m

ব্যাখ্যা

Let
breadth of a room (b) = x
Then height (h)= x/2
and length (l)= 2x

Now,
Volume of the room =2x × x × x/2 = x³

Volume of the room is given as 512 m³.
⇒ x³= 512
⇒ x³= 8³
⇒ x = 8
∴ Length = 2x = 2 × 8 = 16 m

৯৩৮.

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,
একটি ত্রিভুজের যেকোন দুই বাহুর সমষ্টি তৃতীয় বাহু অপেক্ষা বৃহত্তর হবে।
16 - 7 < তৃতীয় বাহু < 16 + 7
9 < তৃতীয় বাহু < 23

9 কখনো তৃতীয় বাহুর দৈর্ঘ্য হতে পারে না।

৯৩৯.

The slope of the line perpendicular to the line y = - 5x + 9 is-

ক
ক) 5
খ
খ) - 5
গ
গ) 1/5
ঘ
ঘ) - 1/5

ব্যাখ্যা

প্রশ্ন: The slope of the line perpendicular to the line y = - 5x + 9 is-

সমাধান:
y = -5x + 9
⇒ y + 5x = 9
সরল রেখার উপর লম্ব রেখার সমীকরণ হবে,
5y - x = k
⇒ 5y = x + k
∴ y = x/5 + k/5

লম্ব রেখাটির ঢাল = 1/5

৯৪০.

In triangle ABC, AB = AC . All of the following statements are true except

ক
ক) AB < AC + BC
খ
খ) AC < AB + BC
গ
গ) BC + AC > AB + BC
ঘ
ঘ) AC + BC = AB + BC

ব্যাখ্যা

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

৯৪১.

What is the perimeter of the rectangle shown at the right?

ক
ক) 16
খ
খ) 24
গ
গ) 30
ঘ
ঘ) 28

ব্যাখ্যা

প্রশ্ন: What is the perimeter of the rectangle shown at the right?

সমাধান:
আয়তক্ষেত্রে,
কর্ণ^২ = দৈর্ঘ্য^২ + প্রস্থ^২
⇒ 10² = দৈর্ঘ্য২ + 6²
⇒ দৈর্ঘ্য^২ = 10² - 6²
⇒ দৈর্ঘ্য^২ = 100 - 36
⇒ দৈর্ঘ্য^২ = 64
⇒ দৈর্ঘ্য = √64
∴ দৈর্ঘ্য = 8 মিটার

পরিসীমা = ২ (দৈর্ঘ্য + প্রস্থ)
= 2(6 + 8) মিটার
= (2 × 14) মিটার
= 28 মিটার

৯৪২.

The ratio between the perimeter and the length of a rectangle is 7 : 2. If the area of the rectangle is 0.12 sq. m, what is the breadth of the rectangle?

ক
30 cm
খ
10 cm
গ
12 cm
ঘ
15 cm
ঙ
18 cm

ব্যাখ্যা

2 ( length + breadth ) / length = 7/2
or, ( length + breadth ) / length = 7/4
or, 4 × breadth + 4 × length = 7 × length
∴ length = 4 × breadth / 3

Area = 0.12 square meter = 0.12 × 100 × 100 square centimeter
∴ length × breadth = 1200 square centimeter
or, breadth × 4 × breadth / 3 = 1200
or, breadth ² = 900 = 30² square centimeter
∴ breadth = 30cm

৯৪৩.

= ?

ক
secA
খ
cosecA
গ
1/cosecA
ঘ
1/tanA

ব্যাখ্যা

Question: = ?

Solution:
1/{tanA√(1 - sin²A)}
= 1/(tanA × √cos²A)
= 1/(tanA × cosA)
= 1/{(sinA/cosA) × cosA}
= 1/sinA
= cosecA

৯৪৪.

What is the value of sin45° ?

ক
1
খ
1/2
গ
1/√2
ঘ
1/√3

ব্যাখ্যা

Question: What is the value of sin45°?

Solution:
sin45° = 1/√2
sin30° = 1/2
sin90° = 1
sin60° = √3/2

৯৪৫.

Find the area of the trapezium if height is 5 cm and AB and CD are given as 10 and 6 cm respectively.

ক
40 cm²
খ
60 cm²
গ
90 cm²
ঘ
80 cm²

ব্যাখ্যা

Question: Find the area of the trapezium if height is 5 cm and AB and CD are given as 10 and 6 cm respectively.

Solution:
Given,
AB = 10cm, CD = 6cm, height = 5cm

According to the formulae,
Area of Trapezium = (1/2) h (AB+CD)
⇒ Area of Trapezium = 1/2 × 5 × (10 + 6)
⇒ Area of Trapezium = 40 cm²

৯৪৬.

একটি ত্রিভুজাকৃতি কক্ষের বাহুগুলোর দৈর্ঘ্য যথাক্রমে ৫ মি. ১২ মি. ১৩ মি. হলে এর ক্ষেত্রফল কত?

ক
৫০ বর্গমিটার
খ
৪৫ বর্গমিটার
গ
৬০ বর্গমিটার
ঘ
৩০ বর্গমিটার
ঙ
৬৫ বর্গমিটার

ব্যাখ্যা

প্রশ্ন: একটি ত্রিভুজাকৃতি কক্ষের বাহুগুলোর দৈর্ঘ্য যথাক্রমে ৫ মি. ১২ মি. ১৩ মি. হলে এর ক্ষেত্রফল কত?

সমাধান:
ধরি, বাহুগুলোর দৈর্ঘ্য a = ৫ মি. B = ১২ মি. C = ১৩ মি.
পরিসীমা, ২s = (৫ + ১২ + ১৩) মি.
বা, s = ৩০/২ মি. = ১৫ মি.

আমরা জানি,
ক্ষেত্রফল = √{s(s - a) (s - b) (s - c)}
= √{১৫ (১৫ - ৫) (১৫ - ১২) (১৫ - ১৩)} বর্গমিটার
= √{১৫ × ১০ × ৩ × ২} বর্গমিটার
= √{৯০০} বর্গমিটার
= ৩০ বর্গমিটার

৯৪৭.

If the difference between the circumference and diameter of a circle is 90 cm, then the diameter of the circle is -

ক
ক) 21 cm
খ
খ) 28 cm
গ
গ) 42 cm
ঘ
ঘ) 24 cm

ব্যাখ্যা

Question: If the difference between the circumference and diameter of a circle is 90 cm, then the diameter of the circle is -

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

প্রশ্নমতে,
2πr - 2r = 90
⇒ 2r(π - 1) = 90
⇒ r = (90/2){(22/7) - 1}
⇒ r = 45/(22 - 7)/7
⇒ r = (45 × 7)/15
∴ r = 21

∴ বৃত্তের ব্যাস = 2r = 2 × 21 = 42 সে.মি.

৯৪৮.

A pole is broken such that the broken part makes an angle of 45° with the ground and touches it at a point 10 meters from the base. What is the length of the broken part?

ক
10√2 m
খ
20 m
গ
5√2 m
ঘ
15 m
ঙ
None of these

ব্যাখ্যা

Question: A pole is broken such that the broken part makes an angle of 45° with the ground and touches it at a point 10 meters from the base. What is the length of the broken part?

Solution:

To find the length of the broken part of the pole, we can model this as a right-angled triangle:

The broken part of the pole is the hypotenuse.
The horizontal distance from the base of the pole to where the top touches the ground is 10 meters.
The angle between the broken part and the ground is 45°.
let the hypotenuse = x

Using the cosine function:
cos(45°) = Adjacent(base)/hypotenuse
Or, 1/√2 = 10/x
Or, x = 10√2

৯৪৯.

What is the maximum value of cos θ?

ক
1/2
খ
90
গ
1
ঘ
180

ব্যাখ্যা

Question: What is the maximum value of cos θ?

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

৯৫০.

If sinθ - cosθ = 0, then what is the value of sin⁴θ + cos⁴θ?

ক
1
খ
1/2
গ
1/4
ঘ
0

ব্যাখ্যা

Question: If sinθ - cosθ = 0, then what is the value of sin⁴θ + cos⁴θ?

Solution:
Given,
sinθ - cosθ = 0
⇒ sinθ = cosθ

Hence, θ = 45° [∵ Sin45° = cos45° = 1/√2]

Now,
sin⁴θ + cos⁴θ
= sin⁴45° + cos⁴45°
= 2 × (1/√2)⁴
= 2 × (1/4)
= 1/2

৯৫১.

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
620π cm³
খ
578π cm³
গ
516π cm³
ঘ
972π cm³

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 volume would be 972π cm³

৯৫২.

The length of two sides of the right angled triangle is 13cm and 5cm respectively. The length of the third side is -

ক
ক) greater than 15cm
খ
খ) less than 10 cm
গ
গ) equal to 3cm
ঘ
ঘ) equal to 12cm

ব্যাখ্যা

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

৯৫৩.

If the volume of a cube is 1728 cm³, then the surface area of the cube will be -

ক
ক) 144 cm²
খ
খ) 432 cm²
গ
গ) 512 cm²
ঘ
ঘ) 864 cm²

ব্যাখ্যা

Question: If the volume of a cube is 1728 cm³, then the surface area of the cube will be -

Solution:
দেওয়া আছে,
আয়তন, a³ = 1728
⇒ a³ = 12³
⇒ a = 12

পৃষ্ঠের ক্ষেত্রফল = 6a²
= 6 × 12²
= 6 × 144
= 864 cm²

৯৫৪.

Painting a wall costs BDT. 50 per sq. m. If the 20m long wall has a width of 10m, how much would it cost to paint 7/10th of the wall?

ক
ক) 5000
খ
খ) 10000
গ
গ) 8000
ঘ
ঘ) 7000

ব্যাখ্যা

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

৯৫৫.

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

ক
3/2
খ
1
গ
1/2
ঘ
2

ব্যাখ্যা

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

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

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

৯৫৬.

If A = 45°, What is the value of 3 - sec²A?

ক
- 1
খ
1
গ
2
ঘ
- 2

ব্যাখ্যা

Question: If A = 45°, What is the value of 3 - sec²A?

Solution:
A = 30°

3 - sec²A
= 3 - (sec45°)²
= 3 - 2
= 1

৯৫৭.

A room has dimensions 6 m × 4 m × 3 m (length × width × height). If only the walls and the ceiling are to be painted, what is the total area to be painted?

ক
74m³
খ
64m²
গ
60m³
ঘ
84m²

ব্যাখ্যা

Question: A room has dimensions 6 m × 4 m × 3 m (length × width × height). If only the walls and the ceiling are to be painted, what is the total area to be painted?

Solution:
Here,
Length, L = 6 m
Width, W = 4 m
Height, H = 3 m

We know,
Wall Area = 2 × (L + W) × H
= 2 × (6 + 4) × 3
= 2 × 10 × 3
= 60 m²

And,
Ceiling is just the top surface of the floor (rectangle).
Ceiling Area = L × W = 6 × 4 = 24 m²

∴ Total Area = Wall Area + Ceiling Area = (60 + 24) m² = 84m²

৯৫৮.

D is any point on side AC of ΔABC. If P, Q, X, Y are the mid-point of AB, BC, AD and DC respectively, then the ratio of PX and QY is

ক
2 : 1
খ
3 : 2
গ
1 : 1
ঘ
2 : 5

ব্যাখ্যা

Question: D is any point on side AC of ΔABC. If P, Q, X, Y are the mid-point of AB, BC, AD and DC respectively, then the ratio of PX and QY is

Solution:
ATQ,

PX || BD
∴ PX = (1/2)BD
Similarly, QY || BD
∴ QY = (1/2)BD

∴ PX : QY = (1/2)BD : (1/2)BD = 1 : 1

৯৫৯.

What is the maximum value of 3sec²A - 4tan²A?

ক
0
খ
6
গ
2
ঘ
3

ব্যাখ্যা

Question: What is the maximum value of 3sec²A - 4tan²A?

Solution:
3sec²A - 4tan²A
= 3sec²A - 3tan²A - tan²A
= 3(sec²A - tan²A) - tan²A
= 3 - tan²A
= 3 - 0 [the maximum value of the equation depends on the minimum value of tan²A]
= 3

৯৬০.

The base of a rectangle is three times as long as the height. If the perimeter is 64, what is the area of the rectangle?

ক
ক) 24
খ
খ) 64
গ
গ) 96
ঘ
ঘ) 192

ব্যাখ্যা

Question: The base of a rectangle is three times as long as the height. If the perimeter is 64, what is the area of the rectangle?

Solution:
: মনেকরি,
আয়তক্ষেত্রের উচ্চতা = x একক
আয়তক্ষেত্রের ভূমি = 3x একক
আয়তক্ষেত্রের পরিসীমা = 2(x + 3x) একক

প্রশ্নমতে,
2(x + 3x) = 64
2 × 4x = 64
8x = 64
x = 8

আয়তক্ষেত্রের উচ্চতা = ৪ একক
এবং আয়তক্ষেত্রের ভূমি = 3 × 8 = 24 একক
আয়তক্ষেত্রের ক্ষেত্রফল = ভূমি × উচ্চতা
= 24 × 8
= 192 বর্গ একক

৯৬১.

An observer 1.6 m tall is 20√3 m 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
18 m
খ
19.2 m
গ
21.6 m
ঘ
20 m

ব্যাখ্যা

Question: An observer 1.6 m tall is 20√3 m 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 of the observer AB = 1.6 m
Distance between observer and tower AC = 20√3 m
∴ CE = AB = 1.6 m
Height of the tower is CD = CE + DE.

From ΔBDE we get,
BE = AC = 20√3 m

Now,
tan30° = DE/BE
⇒ 1/√3 = DE/(20√3)
⇒ DE = (20√3)/√3
∴ DE = 20

∴ The height of the tower is = CE + DE = 1.6 + 20 m = 21.6 m

৯৬২.

Find the number of cubes of 2cm edge will be required to fill a cubical box of 2m edge.

ক
ক) 10⁶
খ
খ) 10⁵
গ
গ) 10⁴
ঘ
ঘ) 10³

ব্যাখ্যা

Question: Find the number of cubes of 2cm edge will be required to fill a cubical box of 2m edge.

Solution:
Let,
N = number of cubes

ATQ
Volume of the box = N × volume of the cubes
⇒ 2³ = N × (0.02)³
⇒ 8 = N × 0.000008
⇒ N = 8/0.000008
∴ N = 1000000 = 10⁶

৯৬৩.

The area of a rhombus is 96 sq.cm and the length of one of the diagonals is 16cm. The length of the other diagonal is

ক
ক) 18
খ
খ) 12
গ
গ) 9
ঘ
ঘ) 6

ব্যাখ্যা

প্রশ্ন: The area of a rhombus is 96 sq.cm and the length of one of the diagonals is 16cm. The length of the other diagonal is
সমাধান:
দেওয়া আছে,
রম্বসের ক্ষেত্রফল 96 বর্গ সে.মি.
রম্বসের একটি কর্ণ 16 সে.মি.

ধরি,
রম্বসটির অপর কর্ণের দৈর্ঘ্য d

আমরা জানি,
রম্বসের ক্ষেত্রফল = 1/2 × কর্ণদ্বয়ের গুণফল
96 = (1/2) × 16 × d
d = (96 × 2)/16
d = 12

৯৬৪.

If sec A + tan A = 7/9, then sec A - tan A = ?

ক
2/7
খ
9/7
গ
4/5
ঘ
115

ব্যাখ্যা

Question: If sec A + tan A = 7/9, then sec A - tan A = ?

Solution:
Here,
sec A + tan A = 7/9

Now,
sec²A - tan²A = 1
⇒ (sec A + tan A)(sec A - tan A) =1
⇒ 7/9(sec A - tan A) =1
⇒ (sec A - tan A) = 9/7

৯৬৫.

If tanθ = √3^{- 1} then what is the value of θ?

ক
ক) 60°
খ
খ) 30°
গ
গ) 45°
ঘ
ঘ) 90°

ব্যাখ্যা

Question: If tanθ = √3^{- 1} then what is the value of θ?

Solution:
দেওয়া আছে,
tanθ = √3^{- 1}
⇒ tanθ = √(1/3)
⇒ tanθ = 1/√3
⇒ tanθ = tan30°
∴ θ = 30°

৯৬৬.

Let O be the in-center of a triangle ABC and D be a point on the side BC of ΔABC, such that OD ⊥ BC. If ∠BOD = 60°, then ∠ABC = ?

ক
ক) 30°
খ
খ) 45°
গ
গ) 15°
ঘ
ঘ) 60°

ব্যাখ্যা

Question: Let O be the in-center of a triangle ABC and D be a point on the side BC of ΔABC, such that OD ⊥ BC. If ∠BOD = 60°, then ∠ABC = ?

Solution:

Given,
∠BOD = 60°

We know,
∴ ∠BDO + ∠DOB + ∠DBO = 180°
⇒ ∠DBO = 180° - ∠BDO + ∠DOB
⇒ ∠DBO = 180° - 90° - 60°
∴ ∠DBO = 30°

Now,
∠ABC = 2 × ∠DBO
⇒ ∠ABC = 2 × 30°
∴ ∠ABC = 60°

৯৬৭.

If length and width of a rectangular plot were each increased by 20%, what would be the percentage increase in the area of the plot?

ক
20%
খ
24%
গ
36%
ঘ
44%
ঙ
None of these

ব্যাখ্যা

Question: If length and width of a rectangular plot were each increased by 20%, what would be the percentage increase in the area of the plot?

Solution:
মনে করি,
দৈর্ঘ্য = x একক এবং প্রস্থ = y একক
∴ ক্ষেত্রফল = xy বর্গ একক

20% বৃদ্ধিতে
নতুন দৈর্ঘ্য = x + x এর 20%
= 12x/10 একক

20% বৃদ্ধিতে
প্রস্থ = y + y এর 20%
= 12y/10 একক
∴ নতুন ক্ষেত্রফল = (12x/10) ×( 12y/10) = 144xy/100 বর্গ একক

ক্ষেত্রফল বৃদ্ধি =(144xy/100) - xy
=(144xy - 100xy)/100
= 44xy/100

শতকরা ক্ষেত্রফল বৃদ্ধি = {(44xy/100) × (1/xy) × 100}% = 44%

৯৬৮.

A right cylindrical container with radius 2 meters and height 1 meter is filled to capacity with oil. How many empty right cylindrical cans, each with radius 1/2 meter and height 4 meters, can be filled to capacity with the oil in this container?

ক
1
খ
2
গ
4
ঘ
8

ব্যাখ্যা

Question: A right cylindrical container with radius 2 meters and height 1 meter is filled to capacity with oil. How many empty right cylindrical cans, each with radius 1/2 meter and height 4 meters, can be filled to capacity with the oil in this container?

Solution:
Volume of cylinder = π × r² × h

Volume₁ = π × 2² × 1 = 4π
Volume₂ = π × (1/2)² × 4 = π

Number of Cylinder = Volume₁ /Volume₂ = 4π/ π = 4

৯৬৯.

The four walls and ceiling of a room of length 25 ft., breadth 12 ft. and height 10 ft. are to be painted. Painter A can Paint 200 sqr.ft in 5 days, painter B can paint 250 sqr.ft in 2 days. If A & B work together, in how many days do they finish the job?

ক
ক) 4(9/11)
খ
খ) 5(8/13)
গ
গ) 5(11/12)
ঘ
ঘ) 6(10/33)
ঙ
ঙ) 7(6/11)

ব্যাখ্যা

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

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

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

৯৭০.

A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is:

ক
36 kg
খ
3.696 g
গ
3.696 kg
ঘ
6.96 g

ব্যাখ্যা

Question: A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm³, then the weight of the pipe is:

Solution:
Volume of iron = π (4² - 3²) 21
= π (16 - 9) 21
= π 7 × 21
= 462 cm³

weight of the pipe is = 462 × 8 g
= 3696 g
= 3.696 kg

৯৭১.

A rectangle is 14 cm long and 10 cm wide. If the length is reduced by x cm and its width is increased also by x cm so as to make it a square, then its area changes by:

ক
4
খ
144
গ
12
ঘ
2

ব্যাখ্যা

Question: A rectangle is 14 cm long and 10 cm wide. If the length is reduced by x cm and its width is increased also by x cm so as to make it a square, then its area changes by:

Solution:
ATQ,
14 - x = 10 + x
⇒ x + x = 14 - 10
⇒ 2x = 4
∴ x = 2

Area of rectangle = 14 × 10 cm²
= 140 cm²

Area of square = 12²
= 144 cm²

∴ Area changes by = 144 - 140 cm²
= 4 cm²

৯৭২.

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

ক
21m²
খ
9m²
গ
12m²
ঘ
108m²

ব্যাখ্যা

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

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

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

Hence, area of the field = lb = 108m²

৯৭৩.

The angle measure of base angles of an isosceles triangle are represented by x and the vertex angle is 3x+10. Find the measure of base angle.

ক
ক) 112°
খ
খ) 42.5°
গ
গ) 34°
ঘ
ঘ) 16°

ব্যাখ্যা

ATQ, x + x + 3x + 10 = 180°
Or, 5x = 170°
Or, x = 34°

৯৭৪.

If the ratio of radius of two spheres is 4 : 7, the ratio of their volume is-

ক
4 : 7
খ
64 : 343
গ
49 : 16
ঘ
16 : 49
ঙ
None of these

ব্যাখ্যা

Question: If the ratio of radius of two spheres is 4 : 7, the ratio of their volume is-

Solution:
Ratio of radii of 2 spheres is 4 : 7.
∴ratio of their volume = 4³ : 7³ = 64 : 343

৯৭৫.

In triangle ABC, AB = AC and ∠C = 30°. Find the measure of ∠A.

ক
180°
খ
90°
গ
120°
ঘ
60°

ব্যাখ্যা

Question: In triangle ABC, AB = AC and ∠C = 30°. Find the measure of ∠A.

Solution:

AB = AC
∴ ∠B = ∠C = 30°

∴ ∠A = 180° - ∠B - ∠C
= 180° - 30° - 30°
= 180° - 60°
= 120°

৯৭৬.

The perimeter of one face of a cube is 20 cm. Its volume must be-

ক
100 cm³
খ
115 cm³
গ
125 cm³
ঘ
150 cm³

ব্যাখ্যা

Question: The perimeter of one face of a cube is 20 cm. Its volume must be-

Solution:
perimeter of one face is 20 cm

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

⇒ 4a = 20
⇒ a = 20/4
= 5 cm

volume = a³
= 5³
= 125 cm³

৯৭৭.

The length of a tangent from point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

ক
3 cm
খ
4 cm
গ
5 cm
ঘ
6 cm

ব্যাখ্যা

Question: The length of a tangent from point A at a distance of 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution:
Let O be the centre of the circle and AB is the tangent.

OB ⊥ AB then ∆OAB is a right-angled triangle.

OA² = AB² + OB² (by Pythagoras Theorem)
⇒ 5² = 4² + OB²
⇒ OB² = 5² – 4²
⇒ OB² = 25 – 16
⇒ OB² = 9
⇒ OB = √9 = 3

Hence, the radius of the circle = 3 cm.

৯৭৮.

The one-sixth of the complementary angle to 60° is

ক
5°
খ
10°
গ
15°
ঘ
20°

ব্যাখ্যা

প্রশ্ন: The one-sixth of the complementary angle to 60° is

সমাধান:
60° এর পূরক কোণ = 90° - 60° = 30°
30° এর 1/6 = 5°

৯৭৯.

The total area of 64 small squares of a chessboard is 400 cm². There is a 3 cm wide border around the chessboard. What is the length of the side of the chessboard?

ক
20 cm
খ
23 cm
গ
26 cm
ঘ
28 cm

ব্যাখ্যা

Question: The total area of 64 small squares of a chessboard is 400 cm². There is a 3 cm wide border around the chessboard. What is the length of the side of the chessboard?

Solution:
Length of 64 small squares = √400
= 20 cm

The length of the side of the chess board is = 20 + 3 + 3
= 26 cm

৯৮০.

An angle which is less than 360° and larger than 180° is classified as:

ক
acute angle
খ
reflex angle
গ
obtuse angle
ঘ
adjacent angle

ব্যাখ্যা

Question: An angle which is less than 360° and larger than 180° is classified as:

Solution:
- ৯০° অপেক্ষা অপেক্ষা ছোট কোণকে সূক্ষ্মকোণ বলে।
- ৯০° অপেক্ষা বড় কিন্তু ১৮০° অপেক্ষা ছোট কোণকে স্থূলকোণ বলে।
- ১৮০° অপেক্ষা বড় কিন্তু ৩৬০° অপেক্ষা ছোট কোণকে প্রবৃদ্ধ কোণ বলে।
- একটি সরলরেখার উপর আরেকটি সরলরেখা লম্বভাবে দন্ডায়মান হলে যে দুইটি সন্নিহিত কোণ উৎপন্ন হয় এবং তাদের মান সমান হলে (৯০°) তাদের প্রত্যেককেটিকে সমকোণ বলে।

৯৮১.

A garden of 50 meter length and 40 meter width has a walkway of 2 meter width on every side. What is the area of the garden, in square meters, excluding the walkway?

ক
ক) 5376 sq m
খ
খ) 1656 sq m
গ
গ) 2556 sq m
ঘ
ঘ) 7874 sq m

ব্যাখ্যা

The area of the garden, excluding the walkway, is
= {(50 - 2×2) × (40 - 2×2)}
= 46×36
= 1656 m²

৯৮২.

sin(180° + θ) = ?

ক
- sinθ
খ
- cosθ
গ
sinθ
ঘ
cosθ

ব্যাখ্যা

Question: sin(180° + θ) = ?

Solution:

sin(180° + θ) হলে sin তৃতীয় ভাগে পরবে।
তাই sin এর মান ঋণাত্নক হবে।
∴ sin(180° + θ) = - sinθ

৯৮৩.

ক
2/5
খ
√3/2
গ
1/2
ঘ
4

ব্যাখ্যা

Question:

Solution:

৯৮৪.

What is the length of the diagonal of a square whose area is 4 times of another square with diagonal as 5√2 cm?

ক
ক) 10 cm
খ
খ) 10√2 cm
গ
গ) 20√2 cm
ঘ
ঘ) 20 cm

ব্যাখ্যা

Question: What is the length of the diagonal of a square whose area is 4 times of another square with diagonal as 5√2 cm?

Solution:
Area of square = (1/2) × (length of diagonal)²
Area of square₂ =(1/2) × (5√2)² =
Area of square₁= 4 × 25 = 100

Length of diagonal of square₁ = √(2 × area)
= √(2 × 100)
= 10√2 cm

৯৮৫.

If the angle of depression at a point on the ground 20 meters from the top of the house is 30°, then find the height of the house.

ক
8 m
খ
10 m
গ
12 m
ঘ
14 m

ব্যাখ্যা

Question: If the angle of depression at a point on the ground 20 meters from the top of the house is 30°, then find the height of the house.

Solution:

height of the house, = AB

sin30° = AB/20
⇒ 1/2 = AB/20
⇒ AB = 10 m

৯৮৬.

512 small sphere balls are formed from a big ball with a radius of 16cm. What will be the radius of a small ball?

ক
1cm
খ
2cm
গ
3cm
ঘ
4cm

ব্যাখ্যা

Question: 512 small sphere balls are formed from a big ball with a radius of 16cm. What will be the radius of a small ball?

Solution:
converting a big sphere ball to 512 small balls,
the volume will be the same for both cases.
let,
small ball radius = r
given, big ball radius, R = 16cm

∴ (4/3)πR³ = 512 × (4/3)πr³
or, R³ = 512 × r³
or, R = 8 × r
or, r = 16/8
r = 2cm

৯৮৭.

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

ক
x - 2y = 8
খ
3x - 6y = 12
গ
y = 3x + 5
ঘ
y = x/2 + 4

ব্যাখ্যা

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

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

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

এবার, প্রদত্ত অপশনগুলোর প্রত্যেকটির ঢাল নির্ণয় করি:

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

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

গ) y = 3x + 5
∴ ঢাল, m = 3

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

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

৯৮৮.

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

ক
Tk. 8500
খ
Tk. 9600
গ
Tk. 10000
ঘ
Tk. 9200

ব্যাখ্যা

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

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

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

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

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

৯৮৯.

If θ is an acute angle and 7sin²θ + 3cos²θ = 4, what is the value of tanθ?

ক
1/√3
খ
√3
গ
0
ঘ
1

ব্যাখ্যা

Question: If θ is an acute angle and 7sin²θ + 3cos²θ = 4, what is the value of tanθ?

Solution:
7sin²θ + 3cos²θ = 4
⇒ 7sin²θ + 3(1 − sin²θ) = 4
⇒ 7sin²θ + 3 − 3sin²θ = 4
⇒ 4sin²θ = 1
⇒ sin²θ = 1/4
⇒ sinθ = 1/2
⇒ sinθ = sin30°
⇒ θ = 30°

∴ tanθ = tan30° = 1/√3

৯৯০.

The surface area of a sphere is the 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
4 cm
খ
5 cm
গ
6 cm
ঘ
8 cm

ব্যাখ্যা

Question: The surface area of a sphere is the 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:
The surface area of sphere = 4πr₁²
The curved surface area of cylinder =2πr₂h
diameter = 12 cm
radius r₂ = 6 cm

⇒ 4πr₁² = 2πrh
⇒ r₁² = (6 × 12)/2
⇒ r₁² = 36
⇒ r₁ = 6

radius of the sphere 6 cm

৯৯১.

A rectangular field will be fenced on three side leaving one side of 20 feet uncovered. If the area of the field 680 square feet. How many feet of fencing is required?

ক
ক) 88
খ
খ) 34
গ
গ) 40
ঘ
ঘ) 68

ব্যাখ্যা

আয়তাকার মাঠের এক পাশের দৈর্ঘ্য = 20 ফুট
আয়তাকার মাঠের ক্ষেত্রফল = 680 বর্গ ফুট

আয়তাকার মাঠের অন্য পাশের দৈর্ঘ্য = 680/20 ফুট = 34 ফুট

বেড়ার দৈর্ঘ্য = (34 + 20 + 34) = 88 ফুট

৯৯২.

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

ক
164 cm
খ
112 cm
গ
144 cm
ঘ
None of the above

ব্যাখ্যা

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

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

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

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

৯৯৩.

A square park is surrounded by a path of uniform width 2 meters all around it. The area of the path is 288 sq. meters. Find the perimeter of the park with path.

ক
136 m
খ
1444 m
গ
152 m
ঘ
1156 m

ব্যাখ্যা

Question: A square park is surrounded by a path of uniform width 2 meters all around it. The area of the path is 288 sq. meters. Find the perimeter of the park with path.

Solution:
Let,
One side of the park is x meter.
So, one side of the park with path = x + (2 + 2) meter
= x + 4

We know,
Area of the park = x²
Area of the path, (x + 4)²- x² = 288
⇒ x² + 8x + 16 - x² = 288
⇒ 8x + 16 = 288
⇒ 8x = 288 - 16
⇒ 8x = 272
⇒ x = 272/8
∴ x = 34

One side of the square park = 34 m.
One side of the square park with path = 34 + 4 = 38 m.

So, perimeter of the square park with path = 4 × 38
= 152m

৯৯৪.

The area of a square inscribed in a circle is 140 cm². What is the area of the semi-circle?

ক
ক) 220 cm²
খ
খ) 200 cm²
গ
গ) 150 cm²
ঘ
ঘ) 110 cm²

ব্যাখ্যা

Question: The area of a square inscribed in a circle is 140 cm². What is the area of the semi-circle?

Solution:
The area of a square inscribed in a circle is 140 cm²
side of square = √140 cm
= 2√35 cm

diagonal of the square = √2 × 2√35
= 2√70 cm

diameter of circle = 2√70 cm
radius of the circle = √70 cm
∴ area of the circle = π (√70)2 cm²
= (22/7) × 70 cm²
= 220 cm²

area of semi-circle = 220/2
= 110 cm²

৯৯৫.

A square is inscribed in a circle of diameter 2a and another square is circumscribing circle. The difference between the areas of outer and inner squares is:

ক
ক) a²
খ
খ) 2a²
গ
গ) 3a²
ঘ
ঘ) 4a²

ব্যাখ্যা

Question: A square is inscribed in a circle of diameter 2a and another square is circumscribing circle. The difference between the areas of outer and inner squares is:

Solution:
বৃত্তটির ব্যাস হলো অন্তবর্গের কর্ণ

অন্তবর্গের এক বাহুর দৈর্ঘ্য x একক হলে
x√2 = 2a
বা, x = 2a/√2
∴ x = √2a
অন্তবর্গের ক্ষেত্রফল = (√2a)² = 2a²

বহিবর্গের এক বাহুর দৈর্ঘ্য = 2a
বহিবর্গের ক্ষেত্রফল = (2a)²
= 4a²
∴ বহিবর্গ ও অন্তবর্গের ক্ষেত্রফলের পার্থক্য = 4a² - 2a² = 2a²

৯৯৬.

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is -

ক
ক) 40%
খ
খ) 42%
গ
গ) 44%
ঘ
ঘ) 48%

ব্যাখ্যা

Question: The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is -

Solution:
ধরি,
চতুর্ভুজের দৈর্ঘ্য ও প্রস্থ যথাক্রমে ক ও খ
ক্ষেত্রফল = কখ

দৈর্ঘ্য ২০% বাড়লে নতুন দৈর্ঘ্য = ক + ক এর ২০%
= ক + ২০ক/১০০
= ১২/১০ক

প্রস্থ ২০% বাড়লে নতুন প্রস্থ = খ + খ এর ২০%
= খ + ২০খ/১০০
= ১২/১০খ

নতুন ক্ষেত্রফল = (১২ক/১০) × (১২খ/১০)
= ১৪৪কখ /১০০
ক্ষেত্রফল বৃদ্ধি = ১৪৪কখ/১০০ - কখ
= ৪৪কখ/১০০

শতকরা বৃদ্ধির হার = {(৪৪কখ/১০০) / কখ }১০০%
= ৪৪%

৯৯৭.

If the radius and height of a right circular cylinder are 14 cm and 21 cm respectively, then the total surface area of the cylinder is:

ক
ক) 890π cm²
খ
খ) 809π cm²
গ
গ) 908π cm²
ঘ
ঘ) 980π cm²

ব্যাখ্যা

Here, r = 14 cm and h= 21 cm.
Total surface area of cylinder = 2πr(h + r)
= 2π × 14 × (14 + 21)
= 2 × π × 14 × 35
= 980π cm²

৯৯৮.

If θ is an acute angle, and cosθ =15/17, then find the value of cot(90 - θ)?

ক
ক) 3/4
খ
খ) 7/19
গ
গ) 8/15
ঘ
ঘ) 11/17
ঙ
ঙ) 2/5

ব্যাখ্যা

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

৯৯৯.

A wire can be bent in the form of a circle of radius 7cm. If it is bent in the form of a square, then what will be its area?

ক
100 cm²
খ
121 cm²
গ
130 cm²
ঘ
144 cm²

ব্যাখ্যা

প্রশ্ন: A wire can be bent in the form of a circle of radius 7cm. If it is bent in the form of a square, then what will be its area?

সমাধান:
দেওয়া আছে,
বৃত্তের ব্যাসার্ধ r = 7 cm
বৃত্তের পরিধি = 2πr
= 2 × (22/7) × 7
= 2 × 22 × 1
= 44 cm

বর্গের এক বাহুর দৈর্ঘ্য = 44/4 cm
= 11 cm

∴ বর্গের ক্ষেত্রফল = (11)² cm²
= 121 cm²

১,০০০.

The wheel of scooter has diameter 140 cm. How many revolutions per minute must the wheel make so that the speed of the scooter is kept at 132 km per hour?

ক
1100
খ
500
গ
250
ঘ
1000

ব্যাখ্যা

Question: The wheel of scooter has diameter 140 cm. How many revolutions per minute must the wheel make so that the speed of the scooter is kept at 132 km per hour?

Solution:
Distance travelled by wheel in one revolution = circumference of wheel = (22/7) × 140 = 440 cm.

Speed of scooter = 132 km/hr
= (132 × 1000 × 100)/60 cm/min = 220,000 cm/min.

The wheel has therefore got to travel 220,000 cm in 1 min
It has to perform 220,000/440 = 500 revolutions in 1 min