Geometry: Mensuration, Trigonometry

PrepBank · পাতা ৯ / ২১ · ৮০১–৯০০ / ২,০৮৫

৮০১.

How many coins 3 mm thick and 1.2 cm in diameter should be melted in order to form a right circular cylinder, having base diameter 4 cm and height 27 cm?

ক
850
খ
950
গ
980
ঘ
1000
ঙ
900

সঠিক উত্তর: ঘ

1000

উত্তর

সঠিক উত্তর: ঘ

1000

ঘ

ব্যাখ্যা

Question: How many coins 3 mm thick and 1.2 cm in diameter should be melted in order to form a right circular cylinder, having base diameter 4 cm and height 27 cm?

Solution:
Let the number of coins be n.

We have
n × π × (1.2/2)² × 0.3 = π × (4/2)² × 27
⇒ n × 0.36 × 0.3 = 4 × 27
⇒ n = (4 × 27 × 100 × 10)/(36 × 3)
⇒ n = 1000

৮০২.

An observer 1.5 m tall stands 10√3 meters away from a flagpole. The angle of elevation from his eye to the top of the flagpole is 30°. What is the height of the flagpole?

ক
10 m
খ
6 m
গ
12.5 m
ঘ
15 m
ঙ
11.5 m

সঠিক উত্তর: ঙ

11.5 m

উত্তর

সঠিক উত্তর: ঙ

11.5 m

ঙ

ব্যাখ্যা

Question: An observer 1.5 m tall stands 10√3 meters away from a flagpole. The angle of elevation from his eye to the top of the flagpole is 30°. What is the height of the flagpole?

Solution:

Here,
Flagpole Height = AB

Now,
tan∠c = AE/CE
⇒ tan30° = AE/10√3
⇒ 1/√3 = AE/10√3
∴ AE = 10

∴ AB = AE + BE
= 10 + 1.5
= 11.5 m

৮০৩.

A and B are centers of two circles that touch each other externally, as shown in the figure. What is the area of the circle whose diameter is AB?

ক
49π/4 square cm
খ
49π square cm
গ
25π/4 square cm
ঘ
36π square cm

সঠিক উত্তর: ক

49π/4 square cm

উত্তর

সঠিক উত্তর: ক

49π/4 square cm

ক

ব্যাখ্যা

Question: A and B are centers of two circles that touch each other externally, as shown in the figure. What is the area of the circle whose diameter is AB?

Solution:
যেহেতু বৃত্ত দুটি পরস্পরকে বহিস্থভাবে স্পর্শ করে, তাই তাদের কেন্দ্রবিন্দুদ্বয়ের মধ্যবর্তী দূরত্ব (AB) হবে তাদের ব্যাসার্ধের যোগফলের সমান।

এখন, নতুন বৃত্তের ব্যাস, AB = (4 + 3) সেমি = 7 সেমি।
সুতরাং, নতুন বৃত্তের ব্যাসার্ধ, r = 7/2 সেমি।

∴নতুন বৃত্তের ক্ষেত্রফল = πr²
= π(7/2)²
= π(49/4)
= 49π/4 বর্গ সেমি।

সুতরাং, নতুন বৃত্তের ক্ষেত্রফল হবে 49π/4 বর্গ সেমি।

৮০৪.

The area of a square is 1024 sq.cm. What is the ratio of the length to the breadth of a rectangle whose length is twice the side of the square and breadth is 12 cm less than the side of this square?

ক
5 : 18
খ
14 : 5
গ
16 : 5
ঘ
32 : 5

সঠিক উত্তর: গ

16 : 5

উত্তর

সঠিক উত্তর: গ

16 : 5

গ

ব্যাখ্যা

Question: The area of a square is 1024 sq.cm. What is the ratio of the length to the breadth of a rectangle whose length is twice the side of the square and breadth is 12 cm less than the side of this square?

Solution:
Let,
Arm of the square be a cm.
Area of square = 1024 sq.cm.
∴ a² = 1024
⇒ a = √1024
∴ a = 32

Length of rectangle = 2a = (2 × 32) cm = 64 cm.
Breadth of rectangle = (32 - 12) cm = 20 cm.

∴ Required ratio = 64 : 20 = 16 : 5.

৮০৫.

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

ক
1
খ
1/√2
গ
0
ঘ
1/√3

সঠিক উত্তর: ঘ

1/√3

উত্তর

সঠিক উত্তর: ঘ

1/√3

ঘ

ব্যাখ্যা

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

solution:
7sin²θ + 3cos²θ = 4
⇒7sin²θ + 3(1 - sin²θ) = 4 [cos²θ = 1 - sin²θ]
⇒7sin²θ + 3 - 3sin²θ = 4
⇒ 4sin²θ = 1
⇒ sin²θ = 1/4
⇒ sinθ = 1/2
⇒ sinθ = sin30°
∴ θ = 30°

Now,
tanθ
= tan30^∘
= 1/√3

৮০৬.

If the diagonal of a square measures 16√2 cm, what is the area of the square in sq. cm?

ক
ক) 32√2
খ
খ) 64√2
গ
গ) 128
ঘ
ঘ) 256

সঠিক উত্তর: ঘ

ঘ) 256

উত্তর

সঠিক উত্তর: ঘ

ঘ) 256

ঘ

ব্যাখ্যা

Given, the diagonal of a square measures, a√2 = 16√2 cm
So, side of the square is = 16 cm
Area of the square = 16² = 256 cm²

৮০৭.

The length of a rope, to which a cow is tied assume that the cow is able to move on all sides with equal ease, is increased from 21 m to 28 m. How much additional ground will it be able to graze?

ক
995 sq m.
খ
1055 sq m.
গ
1078 sq m.
ঘ
1135 sq m.
ঙ
None

সঠিক উত্তর: গ

1078 sq m.

উত্তর

সঠিক উত্তর: গ

1078 sq m.

গ

ব্যাখ্যা

Question: The length of a rope, to which a cow is tied assume that the cow is able to move on all sides with equal ease, is increased from 21 m to 28 m. How much additional ground will it be able to graze?

Solution:
We know,
Area of a circle = π × (radius)²

Given,
The cow can graze the area covered by the circle of radius 21 m initially, because the length of the rope is 21 m.
Therefore, the initial area that the cow can graze = (22/7) × 21² sq m.
= 1386 sq m.

When the length of the rope is increased to 28 m, grazing area becomes = (22/7) × 28² sq m.
= 2464 sq m.

The additional area it could graze when length is increased from 21 m to 28 m = (2464 - 1386) sq m.
= 1078 sq m.

৮০৮.

If the length of each side of an equilateral triangle is decreased by 1 units, the area is found to be decreased by 2√3 square unit. The length of each side of the triangle is-

ক
6 units
খ
5.5 units
গ
3.75 units
ঘ
9 units
ঙ
4.5 units

সঠিক উত্তর: ঙ

4.5 units

উত্তর

সঠিক উত্তর: ঙ

4.5 units

ঙ

ব্যাখ্যা

Question: If the length of each side of an equilateral triangle is decreased by 1 units, the area is found to be decreased by 2√3 square unit. The length of each side of the triangle is-

Solution:
Let,
Original side length = x
New side length = x - 1
Decrease in area = 2√3

Now,
⇒ (√3/4)x² - (√3/4)(x - 1)² = 2√3
⇒ x² - (x - 1)² = 8
⇒ x² - (x² - 2x + 1) = 8
⇒ x² - x² + 2x - 1 = 8
⇒ 2x - 1 = 8
⇒ 2x = 9
⇒ x = 9/2
∴ x = 4.5 units

So the original length of each side of the equilateral triangle is 4.5 units.

৮০৯.

The radius of a circular plate is 20 cm. If the radius is decreased by 20%, what is the percentage decrease in its area?

ক
32%
খ
44%
গ
40%
ঘ
36%

সঠিক উত্তর: ঘ

36%

উত্তর

সঠিক উত্তর: ঘ

36%

ঘ

ব্যাখ্যা

Question: The radius of a circular plate is 20 cm. If the radius is decreased by 20%, what is the percentage decrease in its area?

Solution:
Given that,
Original radius = 20 cm
And decreased by 20%
∴ Decrease = 20% of 20 = 4 cm
∴ New radius = 20 - 4 = 16 cm

We know,
Area of circle = πr²
∴ Original area = π × (20)² = 400π cm²
∴ New area = π × (16)² = 256π cm²

∴ Decrease in area = original area - new area = 400π - 256π = 144π cm²

∴ Percentage decrease in area,
= (decrease/original area) × 100%
= (144π/400π) × 100%
= (144/400) × 100%
= 0.36 × 100%
= 36%

So the area decreases by 36%.

৮১০.

A ladder is leaning against a wall. It makes a 60° angle with the ground. If the distance between the foot of the ladder and the wall is 8 meters, what is the length of the ladder?

ক
12 meters
খ
8√2 meters
গ
16 meters
ঘ
20 meters

সঠিক উত্তর: গ

16 meters

উত্তর

সঠিক উত্তর: গ

16 meters

গ

ব্যাখ্যা

Question: A ladder is leaning against a wall. It makes a 60° angle with the ground. If the distance between the foot of the ladder and the wall is 8 meters, what is the length of the ladder?

Solution:

ধরি, দেয়ালটি হলো AB এবং মইটি হলো AC
মইটি ভূমির সাথে ∠ACB = 60° কোণ তৈরি করে।
মইয়ের গোড়া থেকে দেয়ালের দূরত্ব, BC = 8 মিটার।
মইয়ের দৈর্ঘ্য হলো AC

এখন, ΔABC -এ
cos 60° = BC/AC
⇒ 1/2 = 8/AC
⇒ AC = 8 × 2
⇒ AC = 16
∴ মইয়ের দৈর্ঘ্য 16 মিটার।

৮১১.

A wire can be bent in the form of a circle of radius 21 cm. If it is bent in the form of a square, then its area will be -

ক
10 cm²
খ
109 cm²
গ
189 cm²
ঘ
1089 cm²

সঠিক উত্তর: ঘ

1089 cm²

উত্তর

সঠিক উত্তর: ঘ

1089 cm²

ঘ

ব্যাখ্যা

Question: A wire can be bent in the form of a circle of radius 21 cm. If it is bent in the form of a square, then its area will be -

Solution:
Given,
radius of the circle r = 21 cm

Circumference of the circle = 2πr
= 2 × (22/7) × 21
= 2 × 22 × 3
= 132 cm

The length of one side of the square = 132/4 = 33 cm

Area of the square = (33)² cm²
= 1089 cm²

৮১২.

If cosθ + sinθ = 1, then θ = ?

ক
ক) 30°
খ
খ) 45°
গ
গ) 60°
ঘ
ঘ) 90°

সঠিক উত্তর: ঘ

ঘ) 90°

উত্তর

সঠিক উত্তর: ঘ

ঘ) 90°

ঘ

ব্যাখ্যা

Given, cosθ + sinθ = 1
Or, (cosθ + sin)² = 1²
Or, cos²θ + sin²θ + 2cosθsinθ = 1
Or, 2cosθsinθ = 0 [As, cos²θ + sin²θ = 1]
Or, cosθsinθ = 0

So, either cosθ = 0 = cos90° or, sinθ = 0 = sin0°
⇒ θ = 90° or, 0°

৮১৩.

What is the side length of a square whose area is four times the area of another square with a side of 5m?

ক
ক) 10m
খ
খ) 20m
গ
গ) 25m
ঘ
ঘ) 50m

সঠিক উত্তর: ক

ক) 10m

উত্তর

সঠিক উত্তর: ক

ক) 10m

ক

ব্যাখ্যা

Question: What is the side length of a square whose area is four times the area of another square with a side of 5m?

Solution:
Area of given square = 5² = 25 m²
Area of new square = 25 × 4 = 100 m²
Side of new square = √100 = 10 m

৮১৪.

The area of a trapezium is 96 square cm. The length of one of the parallel sides is 12 cm, and the distance between the parallel sides is 8 cm. Find the length of the other parallel side.

ক
12 cm
খ
14 cm
গ
15 cm
ঘ
16 cm

সঠিক উত্তর: ক

12 cm

উত্তর

সঠিক উত্তর: ক

12 cm

ক

ব্যাখ্যা

Question: The area of a trapezium is 96 square cm. The length of one of the parallel sides is 12 cm, and the distance between the parallel sides is 8 cm. Find the length of the other parallel side.

Solution:
Given,
Area of the trapezium = 96 cm²
One parallel side a = 12 cm
Distance between the parallel sides h = 8 cm

Let
the other parallel side = b cm

We know,
The area of a trapezium = (1/2) × (a + b) × h
⇒ 96 = (1/2) × (12 + b) × 8
⇒ 96 = (12 + b) × 4
⇒ (12 + b) = 96/4
⇒ 12 + b = 24
⇒ b = 24 - 12
∴ b = 12

∴ The other parallel side is 12 cm.

৮১৫.

The ratio between the perimeter and the length of a rectangle is 3 : 1. If the area of the rectangle is 50 sq. cm, what is the breadth of the rectangle?

ক
25 cm
খ
10 cm
গ
5 cm
ঘ
20 cm

সঠিক উত্তর: গ

5 cm

উত্তর

সঠিক উত্তর: গ

5 cm

গ

ব্যাখ্যা

Question: The ratio between the perimeter and the length of a rectangle is 3 : 1. If the area of the rectangle is 50 sq. cm, what is the breadth of the rectangle?

Solution:
Let the length and breadth be x and y respectively

So, 2(x + y) : x = 3 : 1
⇒ (2x + 2y)/x = 3/1
⇒ 2x + 2y = 3x
⇒ 2y = 3x - 2x
⇒ 2y = x
∴ x = 2y

Then,
Area = 50
⇒ x × y = 50
⇒ 2y × y = 50
⇒ 2y² = 50
⇒ y² = 25 = 5²
∴ y = 5

So the breadth of the rectangle is 5 cm.

৮১৬.

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

ক
ক) 16cm
খ
খ) 24cm
গ
গ) 20cm
ঘ
ঘ) 18cm

সঠিক উত্তর: ঘ

ঘ) 18cm

উত্তর

সঠিক উত্তর: ঘ

ঘ) 18cm

ঘ

ব্যাখ্যা

Let the length and breadth of the rectangle be 'l' and 'b' respectively.
According to the question,
2(l + b)/b = 5/1
⇒ 2l + 2b = 5b
⇒ 3b = 2l
Therefore, b = (2/3)l
Then, Area = 216 cm²
⇒ l × b = 216
⇒ l × (2/3) × l = 216
⇒ l² = 324
Therefore, l = 18cm

৮১৭.

একটি বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য ১০ সে.মি.। একদিকের দৈর্ঘ্য কত?

ক
ক) ৭.০৭ সে.মি.
খ
খ) ৮.০৭ সে.মি.
গ
গ) ৯.০৭ সে.মি
ঘ
ঘ) ৬.০৭ সে.মি.

সঠিক উত্তর: ক

ক) ৭.০৭ সে.মি.

উত্তর

সঠিক উত্তর: ক

ক) ৭.০৭ সে.মি.

ক

ব্যাখ্যা

প্রশ্ন: একটি বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য ১০ সে.মি.। একদিকের দৈর্ঘ্য কত?

সমাধান:
বর্গক্ষেত্রের একদিক বা বাহু a একক হলে এর কর্ণের দৈর্ঘ্য a√২ একক

শর্তমতে,
a√২ = ১০
বা, a = ১০/√২
∴ a = ৭.০৭

বর্গক্ষেত্রটির একদিকের দৈর্ঘ্য ৭.০৭ সে.মি.।

৮১৮.

If secA + tanA = 7/3, then what is the value of secA - tanA?

ক
2/7
খ
1/7
গ
5/7
ঘ
3/7

সঠিক উত্তর: ঘ

3/7

উত্তর

সঠিক উত্তর: ঘ

3/7

ঘ

ব্যাখ্যা

Question: If secA + tanA = 7/3, then what is the value of secA - tanA?

Solution:
দেয়া আছে,
secA + tanA = 7/3

আমরা জানি,
sec²A - tan²A = 1
⇒ (secA + tanA)(secA - tanA) = 1
⇒ 7/3 (secA - tanA) = 1
∴ secA - tanA = 3/7

৮১৯.

If the radius of a sphere is 3r, what is its volume?

ক
72. 75r³
খ
27πr³
গ
105.25r³
ঘ
36πr³

সঠিক উত্তর: ঘ

36πr³

উত্তর

সঠিক উত্তর: ঘ

36πr³

ঘ

ব্যাখ্যা

Question: If the radius of a sphere is 3r, what is its volume?

Solution:
Given that,
Radius of sphere = 3r

We know,
Volume of a sphere = (4/3) × πr³= (4/3) × π(3r)³= (4/3) × π × 27 × r³
= 36πr³

৮২০.

The volume of a sphere with radius r is (4/3)πr³ and the surface area is 4πr². If a spherical balloon has a volume of 972π cubic centimeters, what is the surface area of the balloon in square centimeters?

ক
324
খ
729
গ
243π
ঘ
324π
ঙ
729π

সঠিক উত্তর: ঘ

324π

উত্তর

সঠিক উত্তর: ঘ

324π

ঘ

ব্যাখ্যা

Question: The volume of a sphere with radius r is (4/3)πr³ and the surface area is 4πr². If a spherical balloon has a volume of 972π cubic centimeters, what is the surface area of the balloon in square centimeters?

Solution:
Volume = (4/3)πr³ = 972π
⇒ r³= (972 × 3)/4
⇒ r³ = 729
∴ r = 9

So, the surface area would be 4πr² = 4 × π × 81 = 324π

৮২১.

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

ক
ক) 120°
খ
খ) 150°
গ
গ) 180°
ঘ
ঘ) 140°

সঠিক উত্তর: গ

গ) 180°

উত্তর

সঠিক উত্তর: গ

গ) 180°

গ

ব্যাখ্যা

Question: An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

Solution:
From 8 o'clock to 20'clock total time is 6 hours.

Hours hand makes,
in 12 hours = 360° angle
in 6 hours = (360° × 6)/12
= 180°

৮২২.

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

ক
2520 m²
খ
2480 m²
গ
2420 m²
ঘ
1520 m²

সঠিক উত্তর: ক

2520 m²

উত্তর

সঠিক উত্তর: ক

2520 m²

ক

ব্যাখ্যা

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

Solution:
We have,
(l - b) = 23 and
2(l + b) = 206
⇒ (l + b) = 103

Solving the two equations, we get:
l = 63 and b = 40

∴ Area = (l x b) = (63 x 40) m² = 2520 m²

৮২৩.

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

ক
30500
খ
32500
গ
40000
ঘ
37500

সঠিক উত্তর: ঘ

37500

উত্তর

সঠিক উত্তর: ঘ

37500

ঘ

ব্যাখ্যা

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

৮২৪.

The midpoints of the sides of a square are connected to form a new inscribed square. How many times greater than the area of the inscribed square is the area of the original square?

ক
1/2
খ
4
গ
2
ঘ
√2

সঠিক উত্তর: গ

উত্তর

সঠিক উত্তর: গ

গ

ব্যাখ্যা

Question: The midpoints of the sides of a square are connected to form a new inscribed square. How many times greater than the area of the inscribed square is the area of the original square?

Solution:

ধরি, বর্গক্ষেত্রটির বাহুর দৈর্ঘ্য n মিটার
ক্ষেত্রফল = n² বর্গমিটার

অন্ত:স্থ বর্গক্ষেত্রের বাহুর দৈর্ঘ্য n/√2 মিটার
ক্ষেত্রফল = (n/√2)²
= n²/2

বর্গক্ষেত্রটি অন্ত:স্থ বর্গক্ষেত্রের = n²/n²/2
= 2 গুণ

৮২৫.

A hall is 20m long and 10m broad. If the sum of the areas of the floor and the ceiling is equal to of the areas of four walls, the volume of the hall is:

ক
ক) 1333.33 m³
খ
খ) 2633.33 m³
গ
গ) 1233.33 m³
ঘ
ঘ) 2733.33 m³

সঠিক উত্তর: ক

ক) 1333.33 m³

উত্তর

সঠিক উত্তর: ক

ক) 1333.33 m³

ক

ব্যাখ্যা

ধরি
হল ঘরের উচ্চতা = h
দেওয়া আছে,
ঘরের দৈর্ঘ্য 20 মি.
ঘরের প্রস্থ 10 মি.

প্রশ্নমতে,
2 ( 20 × 10 ) = 2 × ( 20 + 10 ) × h
বা, 2 × 30 × h = 2 ( 20 × 10 )
বা, h × 30 = 200
বা, h = 20/3
আমরা জানি,
আয়তন= 20 × 10 × 20/3
= 1333.33 ঘন মিটার

৮২৬.

The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot at Tk. 26.50 per meter is Tk. 5,300. What is the length of the plot in meters?

ক
ক) 60
খ
খ) 100
গ
গ) 75
ঘ
ঘ) 50

সঠিক উত্তর: ক

ক) 60

উত্তর

সঠিক উত্তর: ক

ক) 60

ক

ব্যাখ্যা

বাগানের পরিসীমা = 5300/26.5
= 200 মিটার
এখন ধরি, বাগানের প্রস্থ x মিটার
∴ বাগানের দৈর্ঘ্য (x + 20) মিটার
প্রশ্নমতে, 2(x + x + 20) = 200
⇒ 2(2x + 20) = 200
⇒ 2x + 20 = 100
⇒ 2x = 100 - 20
⇒ 2x = 80
⇒ x = 40
∴ বাগানের দৈর্ঘ্য = 40 + 20 = 60 মিটার

৮২৭.

a + b + c + d = ?

ক
160°
খ
360°
গ
320°
ঘ
540°

সঠিক উত্তর: গ

320°

উত্তর

সঠিক উত্তর: গ

320°

গ

ব্যাখ্যা

Question: a + b + c + d = ?

Solution:

We know that,
Total amount of internal angles of pentagon is 540°
∴ 180° - a + 140° + 180° - b + 180° - c + 180° - d = 540°
⇒ 860° - a - b - c - d = 540°
⇒ a + b + c + d = 860° - 540° = 320°

৮২৮.

In Figure, we have BX = (1/2)AB, BY = (1/2)AB and AB = BC, then-

ক
BX = BY
খ
BX ≠ BY
গ
AX = AC
ঘ
More than one of the above
ঙ
None of the above

সঠিক উত্তর: ক

BX = BY

উত্তর

সঠিক উত্তর: ক

BX = BY

ক

ব্যাখ্যা

Question: In Figure, we have BX = (1/2)AB, BY = (1/2)AB and AB = BC, then-

Solution:
4th axiom of Euclid which state that, “the things which coincide with one another will be equal to one another.

In the given Figure, we have
BX = (1/2)AB
BY = (1/2)AB
AB = BC

Here, both BX and BY are equal to half of the line segment AB. Thus, from Euclid's axiom, all these three parts are equal to each other.
BX = (1/2)AB = BY
∴ BX = BY

৮২৯.

The value of 1 + {(tan 30° - tan 45°)/(cot 45° - cot 60°)} is -

ক
-1
খ
0
গ
1
ঘ
2

সঠিক উত্তর: খ

উত্তর

সঠিক উত্তর: খ

খ

ব্যাখ্যা

Question: The value of 1 + {(tan 30° - tan 45°)/(cot 45° - cot 60°)} is -

Solution:
1 + (tan 30° - tan 45°)/(cot 45° - cot 60°)
= 1 + (tan 30° - tan 45°)/{cot (90° - 45°) - cot (90° - 60°)}
= 1 + (tan 30° - tan 45°)/(tan 45° - tan 30°)
= 1 + (tan 30° - tan 45°)/(-1)(tan 30° - tan 45°)
= 1 - 1
= 0

৮৩০.

The ratio of total surface area to curved surface area of a cone whose radius is 7cm and height 24 cm, is -

ক
25 : 32
খ
24 : 25
গ
32 : 25
ঘ
35 : 25

সঠিক উত্তর: গ

32 : 25

উত্তর

সঠিক উত্তর: গ

32 : 25

গ

ব্যাখ্যা

Question: The ratio of total surface area to curved surface area of a cone whose radius is 7cm and height 24 cm, is -

Solution:
the slant height of the cone is l = √(r² + h²)
l = √(7² + 24²)
l = 25
Total surface area : curved surface area
= (πrl + πr²) : πrl
= (l + r) : l
= (25 + 7) : 25
= 32 : 25

৮৩১.

P, Q and R are three points on the circle. If PQ = PR = 6√2 cm and ∠QPR = 90° then the radius is -

ক
7 cm
খ
8 cm
গ
7.5 cm
ঘ
6 cm

সঠিক উত্তর: ঘ

6 cm

উত্তর

সঠিক উত্তর: ঘ

6 cm

ঘ

ব্যাখ্যা

Question: P, Q and R are three points on the circle. If PQ = PR = 6√2 cm and ∠QPR = 90° then the radius is -

Solution:

Given that,

P, Q, R are points on the circle
PQ = PR = 6√2 cm
∠QPR = 90°

Using Pythagoras theorem in △PQR,
QR² = PQ² + PR² = (6√2)² + (6√2)² = 72 + 72
⇒ QR² = 144 = 12²
∴ QR = 12

∴ Radius r = QR/2 = 12/2 = 6 cm

৮৩২.

A rectangle has a diagonal length of 14 meters and a width of 12meters. What is the area of the rectangle in square meters?

ক
52 square meters
খ
13√2 square meters
গ
24√13 square meters
ঘ
None

সঠিক উত্তর: গ

24√13 square meters

উত্তর

সঠিক উত্তর: গ

24√13 square meters

গ

ব্যাখ্যা

Question: A rectangle has a diagonal length of 14 meters and a width of 12meters. What is the area of the rectangle in square meters?

Solution:
ধরি,

আয়তক্ষেত্র ABCD এর কর্ণের দৈর্ঘ্য AC = 14 মিটার এবং প্রস্থ AB = 12 মিটার
∴ দৈর্ঘ্য, BC = √(14² - 12²) মিটার
=√(196 - 144) মিটার
= √52 মিটার
= 2√13 মিটার

আয়তক্ষেত্রের ক্ষেত্রফল = (2√13 × 12) বর্গমিটার
= 24√13 বর্গমিটার

৮৩৩.

The radius of a car wheel is 28 cm. How many revolutions will it make to cover a distance of 13.2 kilometers?

ক
8200
খ
10200
গ
6750
ঘ
11320
ঙ
7500

সঠিক উত্তর: ঙ

7500

উত্তর

সঠিক উত্তর: ঙ

7500

ঙ

ব্যাখ্যা

Question: The radius of a car wheel is 28 cm. How many revolutions will it make to cover a distance of 13.2 kilometers?

Solution:
আমরা জানি,
চাকার পরিধি = 2πr
= 2 × (22/7) × 28
= 2 × 4 × 22
= 176 সে. মি.

মোট দূরত্ব = 13.2 কি. মি.
= 13.2 × 1000 × 100) সে. মি. [১ কি. মি. = 1000 মি., ১ মি = 100 সে. মি.]
= 1320000 সে. মি.

∴ ঘূর্ণন সংখ্যা = 1320000/176
= 7500 টি

৮৩৪.

Find the area of an isosceles triangle given the length of the base is 10 cm and height is 15 cm.

ক
ক) 65 cm²
খ
খ) 75 cm²
গ
গ) 85 cm²
ঘ
ঘ) 95 cm²

সঠিক উত্তর: খ

খ) 75 cm²

উত্তর

সঠিক উত্তর: খ

খ) 75 cm²

খ

ব্যাখ্যা

প্রশ্ন: Find the area of an isosceles triangle given the length of the base is 10 cm and height is 15 cm.

সমাধান:
Base of the triangle (b) = 10 cm
Height of the triangle (h) = 15 cm

Area of Isosceles Triangle = (1/2) × b × h
= (1/2) × 10 × 15
= 5 × 15
= 75 cm²

৮৩৫.

If each side of a rectangle is increased by 50%, its area will increase by -

ক
ক) 200%
খ
খ) 125%
গ
গ) 150%
ঘ
ঘ) 50%

সঠিক উত্তর: খ

খ) 125%

উত্তর

সঠিক উত্তর: খ

খ) 125%

খ

ব্যাখ্যা

ধরি, আয়তক্ষেত্রটির দৈর্ঘ্য x একক এবং প্রস্থ y একক
সুতরাং ক্ষেত্রফল = xy বর্গ একক
৫০% বৃদ্ধিতে নতুন দৈর্ঘ্য = x + (xএর৫০%) = x + x/2 = ৩x/2
৫০% বৃদ্ধিতে নতুন প্রস্থ = y + (yএর ৫০%) = y + y/2 = ৩y/2
নতুন ক্ষেত্রফল = (৩x/2) × (৩y/2) = ৯xy/৪
ক্ষেত্রফল বৃদ্ধি পেয়েছে = ৯xy/৪ - xy = ৫xy/৪
∴ ক্ষেত্রফল শতকরা বৃদ্ধি পেয়েছে = (৫xy × ১০০)/৪xy = ১২৫%

শর্টকাটঃ ক্ষেত্রফল বৃদ্ধি পাবে = [50 + 50 + (50×50 / 100)] % = 125%

৮৩৬.

A square room has a square carpet symmetrically placed in it. This leaves an uncovered area of 9 meter². The area of the whole room is 25 meter². What is the length of the one side of the carpet?

ক
2 meter
খ
4 meter
গ
6 meter
ঘ
8 meter

সঠিক উত্তর: খ

4 meter

উত্তর

সঠিক উত্তর: খ

4 meter

খ

ব্যাখ্যা

Question: A square room has a square carpet symmetrically placed in it. This leaves an uncovered area of 9 meter². The area of the whole room is 25 meter². What is the length of the one side of the carpet?

Solution:
মনে করি,
কার্পেটের এক বাহুর দৈর্ঘ্য x মিটার

প্রশ্নমতে,
25 - 9 = x²
⇒ 16 = x²
⇒ 4² = x²
∴ x = 4

৮৩৭.

One of the diagonals of a rhombus is double the other diagonal. Its area is 36 sq. cm. The sum of the diagonal is?

ক
15
খ
18
গ
20
ঘ
22

সঠিক উত্তর: খ

উত্তর

সঠিক উত্তর: খ

খ

ব্যাখ্যা

Question: One of the diagonals of a rhombus is double the other diagonal. Its area is 36 sq. cm. The sum of the diagonal is?

Solution:
let, one diagonal is x m
other is 2x

0.5 × x × 2x = 36
⇒ x² = 36
⇒ x = 6

sum of diagonals = x + 2x
= 3x
= 3 × 6
= 18

৮৩৮.

The sides of a triangular field are 13 m, 14 m, and 15 m respectively. What is its area?

ক
90 square meters
খ
72 square meters
গ
84 square meters
ঘ
96 square meters
ঙ
100 square meters

সঠিক উত্তর: গ

84 square meters

উত্তর

সঠিক উত্তর: গ

84 square meters

গ

ব্যাখ্যা

Question: The sides of a triangular field are 13 m, 14 m, and 15 m respectively. What is its area?

Solution:
Let the sides of the triangle are,
a = 13 m, b = 14 m, c = 15 m

We know,
Semi-perimeter, s = (a + b + c)/2
= (13 + 14 + 15)/2
= 42/2
= 21 m

We know the formula for the area of a triangle,
Area = √[s(s - a)(s - b)(s - c)]
= √[21(21 - 13)(21 - 14)(21 - 15)]
= √[21 × 8 × 7 × 6]
= √[21 × 8 × 42]
= √7056
= 84

Therefore, the area of the triangular field is 84 square meters.

৮৩৯.

If the volume of a sphere is 288π cm³, what is the surface area of the sphere?

ক
108π cm²
খ
144π cm²
গ
72π cm²
ঘ
36π cm²

সঠিক উত্তর: খ

144π cm²

উত্তর

সঠিক উত্তর: খ

144π cm²

খ

ব্যাখ্যা

Question: If the volume of a sphere is 288π cm³, what is the surface area of the sphere?

Solution:
Given that the volume, V = 288π cm³
or, (4/3)πr³ = 288π
or, r³ = 216
∴ r = 6 cm

Surface area of a sphere, A = 4πr²
= 4π(6)²
= 144π cm²

৮৪০.

A square field is surrounded by a path of uniform width 4 meters. If the area of the path is 192 square meters, find the side length of the field.

ক
6 meters
খ
8 meters
গ
10 meters
ঘ
12 meters

সঠিক উত্তর: খ

8 meters

উত্তর

সঠিক উত্তর: খ

8 meters

খ

ব্যাখ্যা

Question: A square field is surrounded by a path of uniform width 4 meters. If the area of the path is 192 square meters, find the side length of the field.

Solution:
Let the side of the field = x meters.
Then, the side of the field including the path = x + (2 × 4)
= x + 8 meters.

Area of path = Area of field with path - Area of field
⇒ 192 = (x + 8)² - x²
⇒ 192 = x² + 16x + 64 - x²
⇒ 192 = 16x + 64
⇒ 16x = 192 - 64
⇒ 16x = 128
⇒ x = 128/16
⇒ x = 8 meters

∴ Therefore, the side length of the field is 8 meters.

৮৪১.

The diagonal of a rectangle is √41 cm and its area is 20cm². What is the perimeter of the rectangle?

ক
ক) 16cm
খ
খ) 17cm
গ
গ) 20cm
ঘ
ঘ) 18cm

সঠিক উত্তর: ঘ

ঘ) 18cm

উত্তর

সঠিক উত্তর: ঘ

ঘ) 18cm

ঘ

ব্যাখ্যা

Let, Length = a, width = b
So, area, ab = 20

ATQ,
diagonal = √(a² + b²) = √41
⇒ a² + b²= 41
⇒ (a + b)² – 2ab = 41
⇒ (a + b)² – 2×20 = 41
⇒ (a + b)² = 41 + 40 = 81
⇒ a + b = 9
∴ Perimeter = 2(a + b) = 2×9 = 18 cm

৮৪২.

Three angles of a triangle are in proportion 5 : 6 : 7. Then what is the difference in degrees between he biggest and the smallest angles?

ক
ক) 20°
খ
খ) 10°
গ
গ) 30°
ঘ
ঘ) 25°

সঠিক উত্তর: ক

ক) 20°

উত্তর

সঠিক উত্তর: ক

ক) 20°

ক

ব্যাখ্যা

Question: Three angles of a triangle are in proportion 5 : 6 : 7. Then what is the difference in degrees between he biggest and the smallest angles?

Solution:
অনুপাতগুলোর যোগফল = 18
তিন কোণের সমষ্টি = 180°

∴ বৃহত্তম কোণের পরিমাণ = (7 × 180)/18
= 70°

ক্ষুদ্রতম কোণের পরিমাণ = (5 × 180)/18
= 50°

∴ বৃহত্তম কোণ ও ক্ষুদ্রতম কোণের পার্থক্য = (70° - 50°)
= 20°

৮৪৩.

what is the angle between the hour and minute hands of a clock when it is 3 : 20?

ক
25°
খ
35°
গ
20°
ঘ
30°

সঠিক উত্তর: গ

20°

উত্তর

সঠিক উত্তর: গ

20°

গ

ব্যাখ্যা

Question: what is the angle between the hour and minute hands of a clock when it is 3 : 20?

Solution:
3 টা 20 মিনিট = 3 + (20/60) ঘণ্টা
= 3 + 1/3 = 10/3 ঘণ্টা

আমরা জানি,
ঘণ্টার কাঁটা 12 ঘণ্টায় 360° ঘোরে
ঘণ্টার কাঁটা 1 ঘণ্টায় = 360°/12 = 30°
ঘণ্টার কাঁটা 10/3 ঘণ্টায় = 30° × 10/3 = 100° ঘোরে

আবার,
মিনিটের কাঁটা 60 মিনিটে ঘোরে 360°
মিনিটের কাঁটা 1 মিনিটে ঘোরে 360°/60 = 6°
মিনিটের কাঁটা 20 মিনিটে ঘোরে (6 × 20) = 120° ঘোরে

∴ ঘড়ির কাঁটা দুটির মধ্যবর্তী কোণ = |120° - 100°| = 20°

৮৪৪.

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

ক
ক) 18
খ
খ) 12
গ
গ) 9
ঘ
ঘ) 6

সঠিক উত্তর: খ

খ) 12

উত্তর

সঠিক উত্তর: খ

খ) 12

খ

ব্যাখ্যা

আমরা জানি,
রম্বসের ক্ষেত্রফল = 1/2 × কর্ণদ্বয়ের গুণফল ।
অর্থ্যাৎ 1/2 × 16 × অপর কর্ণ = 96
⇒ অপর কর্ণ = (96 × 2 )/16 c.m.
⇒ অপর কর্ণ = 12 c.m.
Answer: অপর কর্ণের দৈর্ঘ্য 12 c.m.

৮৪৫.

In the picture area of the square ABCD is 196 square miles, O is the center of the circle. What will be the area of the circle in square miles?

ক
ক) 100π
খ
খ) 49π
গ
গ) 144π
ঘ
ঘ) 64π

সঠিক উত্তর: খ

খ) 49π

উত্তর

সঠিক উত্তর: খ

খ) 49π

খ

ব্যাখ্যা

Question: In the picture area of the square ABCD is 196 square miles, O is the center of the circle. What will be the area of the circle in square miles?

Solution:
Area of the square ABCD is 196 square miles
Side of square = √196
= 14 miles

Diameter of circle = side of square
= 14 miles
Radius = 14/2 miles
= 7 miles

∴ Area of circle = π7²
= 49π square miles

৮৪৬.

In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is-

ক
70°
খ
85°
গ
55°
ঘ
75°

সঠিক উত্তর: ক

70°

উত্তর

সঠিক উত্তর: ক

70°

ক

ব্যাখ্যা

Question: In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is-

Solution:
According to the figure.
OM = OY = ON
∴ In ΔOMY
∠OMY = ∠OYM = 15°
∴ ∠MOY = 180° - 15° - 15°
∠MOY = 150°
In ΔONY
∠ONY = ∠OYN = 50°
∴ ∠NOY = 180° - 50° - 50°
∠NOY = 80°
∴ ∠MON = 150° - 80°
∠MON = 70°

৮৪৭.

What is the area of a triangle with base 5 meters and height 10 meters?

ক
20 square meters
খ
35 square meters
গ
40 square meters
ঘ
25 square meters

সঠিক উত্তর: ঘ

25 square meters

উত্তর

সঠিক উত্তর: ঘ

25 square meters

ঘ

ব্যাখ্যা

Question: What is the area of a triangle with base 5 meters and height 10 meters?

Solution:
Area of a triangle = (1/2) × base × height
So, the area = (1/2) × 5 × 10
= 25 square meters

৮৪৮.

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:

ক
ক) 12π cm³
খ
খ) 8π cm³
গ
গ) 16π cm³
ঘ
ঘ) 15π cm³

সঠিক উত্তর: ক

ক) 12π cm³

উত্তর

সঠিক উত্তর: ক

ক) 12π cm³

ক

ব্যাখ্যা

We have
r=3cm
h=4cm

∴Volume=(1/3)πr²h
(1/3) × π × 3²× 4)cm³= 12π cm³

৮৪৯.

An observer 1.6m tall is 25√3m 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
ক) 24.6 m
খ
খ) 25.6 m
গ
গ) 26.6 m
ঘ
ঘ) 27.6 m

সঠিক উত্তর: গ

গ) 26.6 m

উত্তর

সঠিক উত্তর: গ

গ) 26.6 m

গ

ব্যাখ্যা

Let AB be the observer and CD tower
Draw BE perpendicular to CD
Then CE = AB = 1.6 m

And BE = AC = 25√3 m
Now
∴tan 30°=DE/BE
1/√3 =DE/25√3
DE = 25

CD = CE + DE = 1.6 + 25= 26.6 m

৮৫০.

In a triangle the lengths of two sides are 5 and 9 and the length of the third side is represented by x. Which statement is always true?

ক
x > 5
খ
x < 9
গ
5 ≤ x ≤ 9
ঘ
4 < x < 14
ঙ
None of these

সঠিক উত্তর: ঘ

4 < x < 14

উত্তর

সঠিক উত্তর: ঘ

4 < x < 14

ঘ

ব্যাখ্যা

Question: In a triangle the lengths of two sides are 5 and 9 and the length of the third side is represented by x. Which statement is always true?

Solution:
In a triangle,
The sum of the lengths of any two sides must be greater than the length of the third side,
- And the difference between the lengths of any two sides must be less than the length of the third side.
This is known as the Triangle Inequality Theorem.

Given that two sides of the triangle have lengths of 5 and 9, and the third side is represented by x, the theorem gives us the following inequalities:
x + 5 > 9, which simplifies to x > 4
x + 9 > 5, which is always true since x > - 4
5 + 9 > x, which simplifies to x < 14

Thus, the third side x must satisfy the condition 4 < x < 14.

৮৫১.

If the area of a small pizza is 154 inch², what size pizza box would best fit the small pizza?

ক
ক) 16 inch
খ
খ) 14 inch
গ
গ) 12 inch
ঘ
ঘ) 10 inch

সঠিক উত্তর: খ

খ) 14 inch

উত্তর

সঠিক উত্তর: খ

খ) 14 inch

খ

ব্যাখ্যা

প্রশ্ন : If the area of a small pizza is 154 inch², what size pizza box would best fit the small pizza?

সমাধান :

Area of a pizza (circle) = πr² = 154
Or, r² =154/π = 154/(22/7) = 49
Or, r = 7
So, the size is 2r = 2 × 7 = 14 inch

৮৫২.

The diameters of two cones are equal, If their slant heights be in the ratio of 5 : 7 then find the ratio of their Curved surface areas.

ক
3 : 7
খ
5 : 2
গ
5 : 7
ঘ
5 : 1

সঠিক উত্তর: গ

5 : 7

উত্তর

সঠিক উত্তর: গ

5 : 7

গ

ব্যাখ্যা

Question: The diameters of two cones are equal, If their slant heights be in the ratio of 5 : 7 then find the ratio of their Curved surface areas.

SolutioN:
Given,
l₁/l₂ = 5/7

Now,
curved surface area of first cone
= πrl₁

curved surface area of second cone
= πrl₂

Therefore, Ratio
= πrl₁ / πrl₂
= l₁ / l₂
= 5 : 7

৮৫৩.

Find the value of sin²130° + cos²130°

ক
2
খ
1/2
গ
1/√2
ঘ
1

সঠিক উত্তর: ঘ

উত্তর

সঠিক উত্তর: ঘ

ঘ

ব্যাখ্যা

Question: Find the value of sin²130° + cos²130°

Solution:
We know, sin²θ + cos²θ = 1

∴ sin²130° + cos²130° = 1

৮৫৪.

A right triangle has sides in the ratio of 5 : 12 : 13. What is the measure of the smallest angle in the triangle, in degree?

ক
13.34
খ
22.62
গ
34.14
ঘ
42.71

সঠিক উত্তর: খ

22.62

উত্তর

সঠিক উত্তর: খ

22.62

খ

ব্যাখ্যা

Question: A right triangle has sides in the ratio of 5 : 12 : 13. What is the measure of the smallest angle in the triangle, in degree?

Solution:

We know that,
sinθ = AB/AC
⇒ sin∠ACB = 5/13
⇒ θ = sin^-1(5/13)
∴ θ = 22.62°

৮৫৫.

Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100m high, the distance between the two ships is :

ক
ক) 173m
খ
খ) 200m
গ
গ) 273m
ঘ
ঘ) 300m

সঠিক উত্তর: গ

গ) 273m

উত্তর

সঠিক উত্তর: গ

গ) 273m

গ

ব্যাখ্যা

Question: Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100m high, the distance between the two ships is :

Solution:

ধরি
বাতিঘরের উচ্চতা AB = 100 মিটার
C ও D হলো জাহাজের অবস্থান

ΔABC এ
tan∠ACB = AB/BC
tan 30° = 100/BC
1/√3 = 100/BC
BC = 100√3

ΔABC এ
tan∠ADB = AB/BD
tan45° = 100/BD
1 = 100/BD
BD = 100

CD = 100√3 + 100
= 173.205 + 100
= 273.205
≈ 273

৮৫৬.

The area of a parallelogram is 72 square centimetre and its altitude is twice the corresponding base. What is the length of the base?

ক
9 c.m.
খ
4 c.m.
গ
5 c.m.
ঘ
6 c.m.
ঙ
15 c.m.

সঠিক উত্তর: ঘ

6 c.m.

উত্তর

সঠিক উত্তর: ঘ

6 c.m.

ঘ

ব্যাখ্যা

Let, base = x
Then, height = 2x
Area = base × height
= x × 2x
= 2x²
Area is given as 72 cm²
2x² = 72 cm²
⇒ x² = 36 cm²
⇒ x = 6 cm
Hence, the length of the base is 6 cm.

৮৫৭.

Find the value of cos(5π/4)

ক
1/2
খ
- √3/2
গ
- 1/2
ঘ
- 1/√2

সঠিক উত্তর: ঘ

- 1/√2

উত্তর

সঠিক উত্তর: ঘ

- 1/√2

ঘ

ব্যাখ্যা

Question: Find the value of cos(5π/4)

Solution:
cos(5π/4)
= cos(π + π/4) [যেহেতু (π + π/4) তৃতীয় চতুর্ভাগে পড়ে এবং তৃতীয় চতুর্ভাগে cos ঋণাত্মক, তাই cos(π + θ) = -cosθ]
= - cos(π/4)
= - cos(45°)
= - 1/√2

৮৫৮.

If cosθ.cosec23° = 1, the value of θ is:

ক
37°
খ
47°
গ
57°
ঘ
67°

সঠিক উত্তর: ঘ

67°

উত্তর

সঠিক উত্তর: ঘ

67°

ঘ

ব্যাখ্যা

Question: If cosθ.cosec23° = 1, the value of θ is:

Solution:
Here,
cosθ.cosec23° = 1
⇒ 1/cos θ = secθ
⇒ cosec23° = cosec(90° – θ)
⇒ 23° = 90° – θ
⇒ θ = 90° – 23° = 67°

Alternative way,
cosθ.cosec23° = 1
If cosA.cosecB = 1

Then, A + B = 90°
So, θ + 23° = 90°
∴ θ = 67°

৮৫৯.

The total surface area of solid hemisphere of radius 14 cm, is -

ক
ক) 1650 cm²
খ
খ) 1728 cm²
গ
গ) 1848 cm²
ঘ
ঘ) 1935 cm²

সঠিক উত্তর: গ

গ) 1848 cm²

উত্তর

সঠিক উত্তর: গ

গ) 1848 cm²

গ

ব্যাখ্যা

Question: The total surface area of solid hemisphere of radius 14 cm, is -

Solution:
Total surface area = 3πR²
= {3 × (22/7) × 14 × 14} cm²
= 1848 cm²

৮৬০.

If the radius of a sphere is increased by 20%, how much will the surface area be increased in percentage?

ক
ক) 21%
খ
খ) 24%
গ
গ) 44%
ঘ
ঘ) 40%

সঠিক উত্তর: গ

গ) 44%

উত্তর

সঠিক উত্তর: গ

গ) 44%

গ

ব্যাখ্যা

Surface area of sphere = 4πr²
If the new radius is 20% increased, then new surface area will be = 4π(1.2)² = 5.76πr²

Surface area Increased in percentage = (5.76πr²/4πr² × 100) - 100 = 144 - 100 = 44%

৮৬১.

A square and a circle have the same perimeter. The side length of the square is 11 cm. What is the area of the circle?

ক
154 square cm
খ
231 square cm
গ
77 square cm
ঘ
616 square cm

সঠিক উত্তর: ক

154 square cm

উত্তর

সঠিক উত্তর: ক

154 square cm

ক

ব্যাখ্যা

Question: A square and a circle have the same perimeter. The side length of the square is 11 cm. What is the area of the circle?

Solution:
দেওয়া আছে,
বর্গক্ষেত্রের এক বাহুর দৈর্ঘ্য, a = 11 সে.মি.
∴ বর্গক্ষেত্রের পরিসীমা = 4 × a
= 4 × 11
= 44 সে.মি.

প্রশ্নমতে,
বৃত্তের পরিধি = বর্গক্ষেত্রের পরিসীমা
∴ 2πr = 44
⇒ 2 × (22/7) × r = 44
⇒ (44/7) × r = 44
⇒ r = 44 × (7/44)
∴ r = 7 সে.মি.

এখন,
বৃত্তের ক্ষেত্রফল = πr²
= (22/7) × 7²
= (22/7) × 49
= 22 × 7
= 154 বর্গ সে.মি.

অতএব, বৃত্তের ক্ষেত্রফল = 154 বর্গ সে.মি.

৮৬২.

A ladder leans against a vertical wall making an angle of 60° with the ground. If the foot of the ladder is 4.6 m away from the base of the wall, find the length of the ladder.

ক
8 m
খ
9.2 m
গ
10.5 m
ঘ
12 m

সঠিক উত্তর: খ

9.2 m

উত্তর

সঠিক উত্তর: খ

9.2 m

খ

ব্যাখ্যা

Question: A ladder leans against a vertical wall making an angle of 60° with the ground. If the foot of the ladder is 4.6 m away from the base of the wall, find the length of the ladder.

Solution:

Let AB be the wall and BC be the ladder.

Then, ∠ACB = 60°
and AC = 4.6 m

We know,
cos∠ACB = AC/BC
⇒ cos60° = AC/BC
⇒ AC/BC= 1/2
⇒ BC = 2 × AC
⇒ BC = 2 × 4.6
∴ BC = 9.2 m

৮৬৩.

The angles of a triangle are in the proportion of 1 : 2 : 3 and the length of the smallest side is 1 cm. what is the length of the longest side of the triangle?

ক
ক) 2 cm
খ
খ) 2.3 cm
গ
গ) 4 cm
ঘ
ঘ) 5 cm

সঠিক উত্তর: ক

ক) 2 cm

উত্তর

সঠিক উত্তর: ক

ক) 2 cm

ক

ব্যাখ্যা

Question: The angles of a triangle are in the proportion of 1 : 2 : 3 and the length of the smallest side is 1 cm. what is the length of the longest side of the triangle?

Solution:
ত্রিভুজের তিনটি কোণের অনুপাত = 1 : 2 : 3

ধরি,
ত্রিভুজের তিনটি কোণ যথাক্রমে x, 2x, 3x

x + 2x + 3x = 180°
6x = 180°
x = 30°

ত্রিভুজের তিনটি কোণ যথাক্রমে = 30°, 60°, 90°
ত্রিভুজটি সমকোণী ত্রিভুজ।

ΔABC এ
cos60° = BC/AC
1/2 = 1/AC
AC = 2

৮৬৪.

The volume of a right circular cylinder is 25π cubic units and its height is 4 units. What is the circumference of its base?

ক
5π
খ
10π
গ
20π
ঘ
10√2π

সঠিক উত্তর: ক

5π

উত্তর

সঠিক উত্তর: ক

5π

ক

ব্যাখ্যা

Question: The volume of a right circular cylinder is 25π cubic units and its height is 4 units. What is the circumference of its base?

Solution:
আমরা জানি, একটি সিলিন্ডারের আয়তন = πr²h
যেখানে, r হলো ভূমির ব্যাসার্ধ এবং h হলো উচ্চতা।

প্রশ্নমতে,
πr² × 4 = 25π
⇒ 4r² = 25
⇒ r² = 25/4
⇒ r = √(25/4)
⇒ r = 5/2 = 2.5 একক

সিলিন্ডারের ভূমির পরিধি = 2πr
= 2π × 2.5
= 5π একক

∴ সিলিন্ডারটির ভূমির পরিধি হলো 5π একক।

৮৬৫.

If 1 + sinθ = mcosθ than what is the value of cotθ?

ক
(m² - 1)/2m
খ
2m/(m² + 1)
গ
m/(m² - 1)
ঘ
2m/(m² - 1)

সঠিক উত্তর: ঘ

2m/(m² - 1)

উত্তর

সঠিক উত্তর: ঘ

2m/(m² - 1)

ঘ

ব্যাখ্যা

Question: If 1 + sinθ = mcosθ than what is the value of cotθ?

Solution:
Given that,
1+ sinθ = m cos θ
⇒ (1 + sinθ)/cosθ = m
⇒ (1/cosθ) + (sinθ/cosθ) = m
∴ secθ + tanθ = m ...............(i)

We know,
(secθ + tanθ) (secθ - tanθ) = 1
⇒ m(secθ - tanθ) = 1
⇒ secθ - tanθ = 1/m .................(ii)

Now, (i) - (ii) ⇒
secθ + tanθ - (secθ - tanθ) = m - (1/m)
⇒ secθ + tanθ - secθ + tanθ = (m² - 1)/m
⇒ 2tanθ = (m² - 1)/m
⇒ tanθ = (m² - 1)/2m
⇒ 1/cotθ = 1/{(m² - 1)/2m}
∴ cotθ = 2m/(m² - 1)

৮৬৬.

A 30-meter pole has fractured and bent over to make a 30° angle with the ground, remaining partially attached. How high from the base did it break?

ক
8 meters
খ
10 meters
গ
18√3 meters
ঘ
15 meters

সঠিক উত্তর: খ

10 meters

উত্তর

সঠিক উত্তর: খ

10 meters

খ

ব্যাখ্যা

Question: A 30-meter pole has fractured and bent over to make a 30° angle with the ground, remaining partially attached. How high from the base did it break?

Solution:

ধরি,
খুটিটি x মিটার উচুতে ভেঙ্গেছিল।
∴ অপর ভাঙ্গা অংশের দৈর্ঘ্য = (30 - x) মিটার

এখন,
sin θ = লম্ব/অতিভুজ
বা, sin θ = x/(30 - x)
বা, sin 30° = x/(30 - x)
বা, 1/2 = x/(30 - x)
বা, 2x = 30 - x
বা, 2x + x = 30
বা, 3x = 30
⇒ x = 10

∴ খুটিটি ভূমি থেকে 10 মিটার উচুতে ভেঙ্গেছিল।

৮৬৭.

Dhruvo has to divide his rectangular field into two parts from one corner to the other using fence. If the area of the field is 540m² and the length of the field is 36m then what will be the length of the fence needed?

ক
32 m
খ
35 m
গ
39 m
ঘ
40 m

সঠিক উত্তর: গ

39 m

উত্তর

সঠিক উত্তর: গ

39 m

গ

ব্যাখ্যা

Question: Dhruvo has to divide his rectangular field into two parts from one corner to the other using fence. If the area of the field is 540m² and the length of the field is 36m then what will be the length of the fence needed?

Solution:
Breadth of the field = 540/36 = 15 m

the length of the fence = diagonal of the field
= √(15² + 36²)
= √1521
= 39 m

৮৬৮.

In the figure, AD and BC are lines intersecting at O. What is the value of a?

ক
18
খ
15
গ
12
ঘ
None of the above

সঠিক উত্তর: ঘ

None of the above

উত্তর

সঠিক উত্তর: ঘ

None of the above

ঘ

ব্যাখ্যা

Question: In the figure, AD and BC are lines intersecting at O. What is the value of a?

Solution:
Here,
y = 3x + 30 ..........(1)

5y/3 = 5x + 5a
⇒ y/3 = x + a
⇒ y = 3x + 3a ...........(2)

From (1) and (2) we get,
3x + 3a = 3x + 30
⇒ 3a = 30
∴ a = 10

৮৬৯.

The cost of the paint is 36.50 Tk. per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint the outside of a cube having 8 feet on each side?

ক
678 Taka
খ
676 Taka
গ
786 Taka
ঘ
876 Taka

সঠিক উত্তর: ঘ

876 Taka

উত্তর

সঠিক উত্তর: ঘ

876 Taka

ঘ

ব্যাখ্যা

Question: The cost of the paint is 36.50 Tk. per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint the outside of a cube having 8 feet on each side?

Solution:
Total surface area = 6 × 8² square ft.
∴ total paint needed = (6 × 8²)/16 kg
= 24 kg

Total cost = (24 × 36.5) taka
= 876 Taka

৮৭০.

A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:

ক
ক) 90 cm
খ
খ) 1 dm
গ
গ) 1.1 cm
ঘ
ঘ) 2 m

সঠিক উত্তর: খ

খ) 1 dm

উত্তর

সঠিক উত্তর: খ

খ) 1 dm

খ

ব্যাখ্যা

Let the thickness of the bottom be x cm.
Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000
320 x 250 x (110 - x) = 8000 x 1000
(110 - x) =8000 x 1000/320 x 250= 100
x = 10 cm = 1 dm.

৮৭১.

The difference between the length and breadth of a rectangle is 23 m. if its area is 2520 m², then its perimeter is:

ক
206 m
খ
114 m
গ
103 m
ঘ
298 m

সঠিক উত্তর: ক

206 m

উত্তর

সঠিক উত্তর: ক

206 m

ক

ব্যাখ্যা

Question: The difference between the length and breadth of a rectangle is 23 m. if its area is 2520 m², then its perimeter is:

Solution:
Let,
The breadth is x m
The length is x + 23 m

ATQ,
x(x + 23) = 2520
⇒ x² + 23x - 2520 = 0
⇒ x² + 63x - 40x - 2520 = 0
⇒ x(x + 63) - 40(x + 63) = 0
⇒ (x + 63)(x - 40) = 0
∴ x = - 63 Or x = 40
We ignore the negative value.
so, x = 40

∴ The breadth is 40 m
∴ The length is 40 + 23 m = 63 m

∴ The perimeter is 2(63 + 40)m = 2 × 103 m = 206 m

৮৭২.

If the area of the trapezium, whose parallel sides are 8 cm and 12 cm is 40 sq. cm, what will be the distance between the parallel sides?

ক
2 cm
খ
4 cm
গ
5 cm
ঘ
8 cm

সঠিক উত্তর: খ

4 cm

উত্তর

সঠিক উত্তর: খ

4 cm

খ

ব্যাখ্যা

Question: If the area of the trapezium, whose parallel sides are 8 cm and 12 cm is 40 sq. cm, what will be the distance between the parallel sides?

Solution:
Parallel sides of a trapezium = 8 cm, and 12 cm
Area of trapezium = (1/2)(sum of the parallel sides) × distance between the parallel sides
40 = (1/2)(8 + 12) × distance
⇒ 40 = 10 × distance
⇒ distance = 40/10 = 4 cm

So, the distance between the parallel lines of trapezium = 4 cm.

৮৭৩.

The area of incircle of an equilateral triangle of side 42 cm is ___ cm²

ক
ক) 462
খ
খ) 452
গ
গ) 442
ঘ
ঘ) 432

সঠিক উত্তর: ক

ক) 462

উত্তর

সঠিক উত্তর: ক

ক) 462

ক

ব্যাখ্যা

Radius of incircle
=a/2√3
=42/ 2√3
=7√3
Area of incircle
=22/7×49×3
=462 cm²

৮৭৪.

The area of a right-angled tringle is 20 time its base. What is its height?

ক
ক) 40 cm
খ
খ) 20 cm
গ
গ) 30 cm
ঘ
ঘ) 10 cm

সঠিক উত্তর: ক

ক) 40 cm

উত্তর

সঠিক উত্তর: ক

ক) 40 cm

ক

ব্যাখ্যা

Let the base and height of a right angled triangle be b, h respectively
Given area of a right angled triangle is 20times its base
⇒ (1/2)b h= 20b
⇒ h = 40 cm
Height of a right angled triangle is 40 cm

৮৭৫.

AB and CD are two parallel chords on the opposite sides of the center of the circle. If AB = 10 cm , CD = 24 cm and the radius of the circle is 13 cm, the distance between the chords is-

ক
17 cm
খ
10 cm
গ
16 cm
ঘ
18 cm
ঙ
24 cm

সঠিক উত্তর: ক

17 cm

উত্তর

সঠিক উত্তর: ক

17 cm

ক

ব্যাখ্যা

Question: AB and CD are two parallel chords on the opposite sides of the center of the circle. If AB = 10 cm , CD = 24 cm and the radius of the circle is 13 cm, the distance between the chords is-

Solution:

From O draw OL ⊥ AB and OM ⊥ CD. Join OA and OC.
AL = AB/2 = 5cm , OA = 13 cm.
OL² = OA² - AL² = (13)² - 5² = (169 - 25) = 144
⇒ OL = √144 = 12 cm.

Now,
CM = CD/2 = 12 cm and OC = 13c m.
∴ OM² = OC² - CM² = (13)² - (12)² = (169 - 144) = 25
⇒ OM =√25 = 5 cm.

∴ ML = OM + OL = (5 +12 ) cm = 17cm.

৮৭৬.

tanA + sinA = m and tanA - sinA = n, then (m² - n²)/4 = ?

ক
√m
খ
mn
গ
√mn
ঘ
√mn/2

সঠিক উত্তর: গ

√mn

উত্তর

সঠিক উত্তর: গ

√mn

গ

ব্যাখ্যা

প্রশ্ন: tanA + sinA = m and tanA - sinA = n, then (m² - n²)/4 = ?

সমাধান:
(m² - n²)/4
= {(tanA + sinA)² - (tanA - sinA)²}/4
= (4tanA . sinA)/4 [(a + b)² - (a - b)² = 4ab]
= √(tan²A . sin²A)
= √{tan²A (1- cos²A)}
= √(tan²A - tan²A . cos²A)
= √(tan²A - (sin²A/cos²A) . cos²A)
= √(tan²A - sin²A)
= √{(tanA + sinA)(tanA - sinA)}
= √mn

৮৭৭.

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is-

ক
173 m
খ
200 m
গ
273 m
ঘ
300 m

সঠিক উত্তর: গ

273 m

উত্তর

সঠিক উত্তর: গ

273 m

গ

ব্যাখ্যা

Question: Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is-

Solution:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100 m, ACB = 30° and ADB = 45°.

AB/AC = tan 30° = 1/√3
⇒ AC = AB × √3 = 100√3 m.

AB/AD = tan 45° = 1
⇒ AD = AB = 100 m.

∴ CD = (AC + AD) = (100√3 + 100) m
= 100(√3 + 1) m
= 100 × 2.73 m
= 273 m

৮৭৮.

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is

ক
ক) 456
খ
খ) 458
গ
গ) 558
ঘ
ঘ) 568

সঠিক উত্তর: গ

গ) 558

উত্তর

সঠিক উত্তর: গ

গ) 558

গ

ব্যাখ্যা

Area to be plastered= [2(l + b) x h] + (l x b)
= {[2(25 + 12) x 6] + (25 x 12)} m²
= (444 + 300) m²= 744 m².
Cost of plastering = Rs.744 x (75/100)
= Rs. 558

৮৭৯.

The angle of elevation of the top of a tower of height x metre from a point on the ground is found to be 60°. By going y metre away from that point, it becomes 30°. Which one of the following relations is correct?

ক
x = y
খ
2x = √3y
গ
2x = 3y
ঘ
None of the above

সঠিক উত্তর: খ

2x = √3y

উত্তর

সঠিক উত্তর: খ

2x = √3y

খ

ব্যাখ্যা

Question: The angle of elevation of the top of a tower of height x metre from a point on the ground is found to be 60°. By going y metre away from that point, it becomes 30°. Which one of the following relations is correct?

Solution:
Given that,
The angle of elevation of the top of a tower of height x meter from a point on the ground is found to be 60°.
By going y metre away from that point, it becomes 30°.

According to the question,
tan60° = AB/BC
⇒ √3 = x/BC
⇒ BC = x/√3 .........(1)

Again,
tan30° = AB/BD
⇒ 1/√3 = AB/BD
⇒ 1/√3 = x/BD
⇒ BD = √3x .........(2)

Now,
BD = BC + CD
⇒ √3x = (x/√3) + y [From equation (1 and 2)]
⇒ √3x = (x + √3y)/√3
⇒ 3x = x + √3y
⇒ 3x - x = √3y
⇒ 2x = √3y

৮৮০.

The area of an isosceles triangle is 25√3 cm², and the measure of each of the equal sides is 10 cm, what is the angle between the equal sides?

ক
30°
খ
45°
গ
50°
ঘ
60°

সঠিক উত্তর: ঘ

60°

উত্তর

সঠিক উত্তর: ঘ

60°

ঘ

ব্যাখ্যা

Question: The area of an isosceles triangle is 25√3 cm², and the measure of each of the equal sides is 10 cm, what is the angle between the equal sides?

Solution:
Area of the traingle :
⇒ 1/2absinθ = 25√3
⇒ 10 × 10 sinθ = 50√3
⇒ sinθ = √3/2
∴ θ = 60°

৮৮১.

Find the value of cosec(- π/3)

ক
- 2/√3
খ
√3/2
গ
1
ঘ
1/√2

সঠিক উত্তর: ক

- 2/√3

উত্তর

সঠিক উত্তর: ক

- 2/√3

ক

ব্যাখ্যা

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

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

৮৮২.

How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 metres by 9 metres?

ক
50 metres
খ
100 metres
গ
200 metres
ঘ
250 metres

সঠিক উত্তর: গ

200 metres

উত্তর

সঠিক উত্তর: গ

200 metres

গ

ব্যাখ্যা

Question: How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 metres by 9 metres?

Solution:
Area of the floor = (14 × 9) m² = 126 m²∴ Length of the carpet = {(126/63) × 100}m = 200 metres

৮৮৩.

A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

ক
ক) 98
খ
খ) 88
গ
গ) 99
ঘ
ঘ) 89

সঠিক উত্তর: খ

খ) 88

উত্তর

সঠিক উত্তর: খ

খ) 88

খ

ব্যাখ্যা

Given that,
The area of the field = 680 sq. feet
⇒ lb = 680 sq. feet
Length(l) = 20 feet
⇒ 20 × b = 680
⇒ b = 680/20
= 34 feet

∴ Required length of the fencing = l + 2b
= 20 + (2 × 34)
= 88 feet

৮৮৪.

If cos⁴θ - sin⁴θ = 2/3 then the value of (1 - 2sin²θ) is -

ক
ক) 1/3
খ
খ) 1/2
গ
গ) 2/3
ঘ
ঘ) 1

সঠিক উত্তর: গ

গ) 2/3

উত্তর

সঠিক উত্তর: গ

গ) 2/3

গ

ব্যাখ্যা

Question: If cos⁴θ - sin⁴θ = 2/3 then the value of (1 - 2sin²θ) is -

Solution:
cos⁴θ - sin⁴θ = 2/3
⇒ (cos²θ - sin²θ) (cos²θ + sin²θ) = 2/3
⇒ cos²θ - sin²θ = 2/3
⇒ 1 - sin²θ - sin²θ = 2/3
∴ 1 - 2sin²θ = 2/3

৮৮৫.

A hall measures 40 m in length, 25 m in width, and 20 m in height. If each person needs 200 cubic meters of space, how many people can the hall accommodate?

ক
80
খ
90
গ
100
ঘ
110

সঠিক উত্তর: গ

100

উত্তর

সঠিক উত্তর: গ

100

গ

ব্যাখ্যা

Question: A hall measures 40 m in length, 25 m in width, and 20 m in height. If each person needs 200 cubic meters of space, how many people can the hall accommodate?

Solution:
Length of the hall = 40 m
Width of hall = 25 m
Height of hall = 20 m

∴ Volume of the hall
= 40 × 25 × 20
= 20000 m³
∴ Space occupied by each person = 200 m³
∴ Number of people that can be accommodated in the hall
= 20000/200
= 100

৮৮৬.

If sin θ = 3/5 and θ is an acute angle, find the value of tan θ.

ক
1/4
খ
3/4
গ
2/3
ঘ
0

সঠিক উত্তর: খ

3/4

উত্তর

সঠিক উত্তর: খ

3/4

খ

ব্যাখ্যা

Question: If sin θ = 3/5 and θ is an acute angle, find the value of tan θ.

Solution:
Given sin θ = 3/5 and θ is acute.

We know,
sin²θ + cos²θ = 1
cos²θ = 1 - (3/5)² = 1 - 9/25 = 16/25
∴ cos θ = √(16/25) = 4/5

Now, tan θ = sin θ/cos θ
= (3/5) ÷ (4/5)
= 3/4

∴ tan θ = 3/4.

৮৮৭.

A wall 8 m long, 6 m high and 22.5 cm thick is made up of bricks, each measuring 25 cm × 11.25 cm × 6 cm. The number of bricks required is

ক
7,200
খ
6,400
গ
6,000
ঘ
5,600

সঠিক উত্তর: খ

6,400

উত্তর

সঠিক উত্তর: খ

6,400

খ

ব্যাখ্যা

Question: A wall 8 m long, 6 m high and 22.5 cm thick is made up of bricks, each measuring 25 cm × 11.25 cm × 6 cm. The number of bricks required is

Solution:
দেয়ালের দৈর্ঘ্য l = 8m = 800cm
দেয়ালের উচ্চতা h = 6m = 600cm
দেয়ালের প্রস্থ b = 22.5cm

দেয়ালের আয়তন = lbh
=800 × 600 × 22.5
=10800000 cm3

ইটের দৈর্ঘ্য l = 25cm
ইটের প্রস্থ b = 11.25cm
ইটের উচ্চতা h = 6cm
ইটের আয়তন =lbh
=25 ×11.25 × 6
=1687.5cm3

ইটের সংখ্যা = 10800000/1687.5
= 6400 টি

৮৮৮.

At what angle the hands of a clock are inclined at 15 minutes past 5?

ক
ক) 67.5°
খ
খ) 72.5°
গ
গ) 58.5°
ঘ
ঘ) 69.5°

সঠিক উত্তর: ক

ক) 67.5°

উত্তর

সঠিক উত্তর: ক

ক) 67.5°

ক

ব্যাখ্যা

Question: At what angle the hands of a clock are inclined at 15 minutes past 5?

Solution:
Hours hand moves in 15 past.
5 from 12 p.m = (5 + 15/60) hours = 21/4 hours
Angle of hours hand = (360/12) × (21/4)
= 157.5°

Minutes hands makes angle of = (360/60) × 15
= 90°

Angle between hours and minutes hands = (157.5° - 90°)
= 67.5°

৮৮৯.

A rectangular water tank is 2 m high, 4m long and 2.5 m high wide. How many liters of water can it hold?

ক
ক) 15000 litre
খ
খ) 20000 litre
গ
গ) 22000 litre
ঘ
ঘ) 25000 litre

সঠিক উত্তর: খ

খ) 20000 litre

উত্তর

সঠিক উত্তর: খ

খ) 20000 litre

খ

ব্যাখ্যা

Question: A rectangular water tank is 2 m high, 4m long and 2.5 m high wide. How many liters of water can it hold?

Solution:
Volume = length × width × height
= 2 × 2.5 × 4 m³
= 20 m³

1 m³ = 1000 litre
⇒ 20 m³ = 20 × 1000 litre
= 20000 litre

৮৯০.

What is the volume of a cylindrical shape water container, that has a height of 7cm and diameter of 10cm?

ক
1100 cm³
খ
75.35 cm³
গ
550 cm³
ঘ
110 cm³

সঠিক উত্তর: গ

550 cm³

উত্তর

সঠিক উত্তর: গ

550 cm³

গ

ব্যাখ্যা

Question: What is the volume of a cylindrical shape water container, that has a height of 7cm and diameter of 10cm?

Solution:
Given,
Diameter of the container = 10cm
Thus, radius of the container = 10/2 = 5cm
Height of container = 7cm

As we know, from the formula,
Volume of a cylinder = πr²h cubic units.

Therefore, volume of given container, V = π × 5² × 7
V = π × 25 × 7 = (22/7) × 25 × 7 = 22 × 25
V = 550 cm³

৮৯১.

The radius of a wheel is 7 cm. How many revolutions will it make in travelling 88 kilometers?

ক
100000
খ
200000
গ
250000
ঘ
100200

সঠিক উত্তর: খ

200000

উত্তর

সঠিক উত্তর: খ

200000

খ

ব্যাখ্যা

Question: The radius of a wheel is 7 cm. How many revolutions will it make in travelling 88 kilometers?

Solution:
আমরা জানি,
চাকার পরিধি = 2πr = 2 × (22/7) × 7 = 44 সে. মি.

∴ মোট দূরত্ব = 88 কি. মি. = 88 × 1000 × 100 = 8800000 সে. মি.

∴ ঘূর্ণন সংখ্যা = 8800000/44 = 200000 টি

৮৯২.

What is the area of an isosceles triangle if two of its sides measure 6 cm and 12 cm?

ক
7√5 cm²
খ
9√15 cm²
গ
9√11 cm²
ঘ
12√5 cm²
ঙ
None

সঠিক উত্তর: খ

9√15 cm²

উত্তর

সঠিক উত্তর: খ

9√15 cm²

খ

ব্যাখ্যা

Question: What is the area of an isosceles triangle if two of its sides measure 6 cm and 12 cm?

Solution:
In an isosceles triangle, two sides are equal. The possible third side can be either 6 cm or 12 cm.

If the equal sides are 6, the sides become 6, 6, and 12 — which violates the triangle inequality rule.
If the equal sides are 12, the sides become 6, 12, and 12 — which satisfies the triangle inequality.
∴ The valid third side is 12 cm.

Now,
a = 6 cm, b = 12 cm, c = 12 cm

∴ Semi-perimeter s =(6 + 12 + 12)/2 =15 cm

We know
from "Heron’s formula"
Area of the triangle = √{s(s - a)(s - b)(s - c)}
= √{15(15 - 6)(15 - 12)(15 - 12)}
= √(15 × 9 × 3 × 3)
= 9√15 cm²

৮৯৩.

If a right-angled isosceles triangle has height 5 cm, then base is:

ক
5√2
খ
5 cm
গ
12 cm
ঘ
13 cm

সঠিক উত্তর: খ

5 cm

উত্তর

সঠিক উত্তর: খ

5 cm

খ

ব্যাখ্যা

Question: If a right-angled isosceles triangle has height 5 cm, then base is:

Solution:

আমরা জানি,
সমকোণী সমদ্বিবাহু ত্রিভুজে সমকোণ সংলগ্ন দুইটি বাহু সমান হয়।

দেওয়া আছে, উচ্চতা = 5 cm

যেহেতু ত্রিভুজটি সমকোণী সমদ্বিবাহু,
∴ উচ্চতা = ভূমি
∴ ভূমি = 5 cm

৮৯৪.

An observer 2m tall is 10√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
ক) 14 m
খ
খ) 12 m
গ
গ) 10 m
ঘ
ঘ) 15 m

সঠিক উত্তর: খ

খ) 12 m

উত্তর

সঠিক উত্তর: খ

খ) 12 m

খ

ব্যাখ্যা

SR = PQ = 2 m
PS = QR = 10√3
tan 30° = TS/PS
1/√3 = TS/10√3
TS = 10√3/√3
= 10
TR = TS + SR
= 10 + 2
= 12 m.

৮৯৫.

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

ক
- 10
খ
- 1/10
গ
10
ঘ
1/10

সঠিক উত্তর: খ

- 1/10

উত্তর

সঠিক উত্তর: খ

- 1/10

খ

ব্যাখ্যা

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

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

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

এখন,
20x - 2y = 6
2y = 20x - 6
y = 10x + (-3)
(1) নং এর সাথে তুলনা করে পাই,
m = 10

∴ লম্ব (perpendicular) রেখার ঢাল হবে, m' = - (1/10)

৮৯৬.

A wheel of a car of radius 35 cm is rotating at 500 RPM. What is the speed of the car in km/hr?

ক
50
খ
55 km/hr
গ
60 km/hr
ঘ
66 km/hr

সঠিক উত্তর: ঘ

66 km/hr

উত্তর

সঠিক উত্তর: ঘ

66 km/hr

ঘ

ব্যাখ্যা

Question: A wheel of a car of radius 35 cm is rotating at 500 RPM. What is the speed of the car in km/hr?

Solution:
The radius of the wheel measures 35 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 × 35 = 2 × (22/7) × 35 = 220 cm.

In a minute, the distance covered by the wheel = circumference of the wheel × rpm
∴ this wheel will cover a distance of 220 × 500 = 110000 cm in a minute.

In an hour, the wheel will cover a distance of 110000 × 60 = 6600000 cm.
Therefore, the speed of the car = 6600000 cm/hr = 66 km/hr

৮৯৭.

The area of a rhombus is 198 sq.cm and the length of one of the diagonals is 22 cm. The length of other diagonal is-

ক
15 cm
খ
16 cm
গ
18 cm
ঘ
19 cm

সঠিক উত্তর: গ

18 cm

উত্তর

সঠিক উত্তর: গ

18 cm

গ

ব্যাখ্যা

Question: The area of a rhombus is 198 sq.cm and the length of one of the diagonals is 22 cm. The length of other diagonal is-

Solution:
⇒ We have given area of rhombus = 96cm2 and d1=22cm.
⇒ Area of rhombus = (1/2) × d1×d2
⇒ 198 = (1/2) × 22 × d2.
⇒ 11 × d2 = 198
∴ d2 = 18 cm

৮৯৮.

What is the 4^th term of the sequence: sin⁡(nπ/6)

ক
1
খ
√3/2
গ
1/2
ঘ
√2/2
ঙ
None

সঠিক উত্তর: খ

√3/2

উত্তর

সঠিক উত্তর: খ

√3/2

খ

ব্যাখ্যা

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

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

৮৯৯.

A rhombus is a quadrilateral -

ক
Whose all sides are equal
খ
Whose any two opposite sides are parallel
গ
Whose all sides are equal and four angels are equal to 90°
ঘ
Both (a) and (b)

সঠিক উত্তর: ঘ

Both (a) and (b)

উত্তর

সঠিক উত্তর: ঘ

Both (a) and (b)

ঘ

ব্যাখ্যা

Question: A rhombus is a quadrilateral -

Solution:
রম্বস
- যে চতুর্ভুজের চারটি বাহু সমান ও সমান্তরাল কিন্তু কর্ণ দুইটি অসমান তথা কোণগুলো সমকোণ নয় তাকে রম্বস বলে।
- সামান্তরিকের সন্নিহিত বাহুদ্বয় সমান হলে তখন তা রম্বস হয়ে
- রম্বসের কর্ণদ্বয় পরস্পরকে সমকোণে সমদ্বিখণ্ডিত করে।
- রম্বসের বিপরীত কোণগুলো পরস্পর সমান।
- রম্বসের কর্ণদ্বয়ের অন্তর্ভুক্ত কোণ 90°

৯০০.

If sinθ = 5/13 , then secθ = ?

ক
5/13
খ
4/3
গ
13/12
ঘ
5/12

সঠিক উত্তর: গ

13/12

উত্তর

সঠিক উত্তর: গ

13/12

গ

ব্যাখ্যা

Question: If sinθ = 5/13 , then secθ = ?

Solution:
এখানে,
sinθ = 5/13
∴ লম্ব = 5, অতিভুজ = 13

∴ ভূমি = √(13² - 5²) = 12

∴ secθ = অতিভূজ/ভূমি
= 13/12