Geometry: Mensuration, Trigonometry

PrepBank · পাতা ৪ / ২১ · ৩০১–৪০০ / ২,০৮৫

৩০১.

Which of these values cannot be the value of sinθ?

ক
√3/2
খ
1/2
গ
3/2
ঘ
-√2/2

ব্যাখ্যা

Question: Which of these values cannot be the value of sinθ?

Solution:
আমরা জানি,
sin⁡θ এর মান সবসময় -1 থেকে 1 এর মধ্যে হয়।
অর্থাৎ, - 1 ≤ sin⁡θ ≤ 1

এখন,
√3/2 = 0.866 → সত।
1/2 = 0.5 → সত্য।
3/2 = 1.5 → এই মান 1-এর চেয়েও বড় সত্য নয়।
-√2/2 = - 0.707 → সত্য।

৩০২.

If cotA = 3/4 then, sinA = ?

ক
3/5
খ
4/3
গ
3/4
ঘ
4/5

ব্যাখ্যা

Question: If cotA = 3/4 then, sinA = ?

Solution:
We know,
cosec²A = 1 + cot²A
⇒ cosec²A = 1 + (3/4)²
⇒ cosec²A = 1 + 9/16
⇒ cosec²A = (16 + 9)/16
⇒ cosec²A = 25/16
⇒ cosecA = √(25/16)
⇒ 1/sinA = 5/4
∴ sinA = 4/5

৩০৩.

If secθ + tanθ = 4/3, then secθ - tanθ = ?

ক
4/3
খ
3/4
গ
1/3
ঘ
1/4

ব্যাখ্যা

We know,
sec²θ - tan²θ = 1
⇒ (secθ + tanθ)(secθ - tanθ) = 1
⇒ 4/3(secθ + tanθ) = 1
⇒ (secθ - tanθ) = 3/4

৩০৪.

Parimal wants to purchase a cylindrical can with a radius equivalent to 5 inches. The can contains 1 gallon of oil. Find the height of the cylinder.

ক
2.54 inches
খ
3.64 inches
গ
3.44 inches
ঘ
2.94 inches

ব্যাখ্যা

Question: Parimal wants to purchase a cylindrical can with a radius equivalent to 5 inches. The can contains 1 gallon of oil. Find the height of the cylinder.

Solution:
Volume V is given by= 1 gallon
1 gallon= 231 cubic inches
Radius r = 5 inches

Volume of the cylinder is given by,
V = πr²h
⇒ 231 = (22/7) × (5)² × h
⇒ (231 × 7)/(22 × 25) = h
∴ h = 2.94 inches.

Therefore, the height is equivalent to 2.94 inches.

৩০৫.

A triangular piece of land has sides measuring 25 meters, 20 meters, and 15 meters respectively. If it costs 3.50 Taka per square meter to plant grass on the land, how much will it cost to plant grass on the entire land?

ক
Tk. 450
খ
Tk. 480
গ
Tk. 510
ঘ
Tk. 525
ঙ
None

ব্যাখ্যা

Question: A triangular piece of land has sides measuring 25 meters, 20 meters, and 15 meters respectively. If it costs 3.50 Taka per square meter to plant grass on the land, how much will it cost to plant grass on the entire land?

Solution:
Given,
a=25 m, b=20 m and c=15 m

Let
the semi-perimeter of the triangle = s
∴ s = (a + b + c)/2
= (25 + 20 + 15)/2
= 30 m

∴ The area of the triangle = √{s(s - a)(s - b)(s - c)}
= √{30 × (30 - 25) × (30 - 20) × (30 - 15)}
= √(30 × 5 × 10 × 15)
= √22500
= 150 square meters

Cost to plant grass per square meter = 3.50 Taka
∴ Cost to plant grass in 150 square meter = (3.50 × 150) Taka
= Tk. 525

৩০৬.

The flowers in a basket double every minute and the basket gets full in 1 hr. How much time does it take to fill the basket half?

ক
ক) 30 minutes
খ
খ) 45 minutes
গ
গ) 58 minutes
ঘ
ঘ) 59 minutes

ব্যাখ্যা

Question: The flowers in a basket double every minute and the basket gets full in 1 hr. How much time does it take to fill the basket half?

Solution:
In 60 minutes the basket will be full 1 part
So, in 59 minutes the basket will be full 1/2 part [Because, every minute it gets double]

৩০৭.

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

ক
ক) 2,520m²
খ
খ) 1,520m²
গ
গ) 2,420m²
ঘ
ঘ) 2,480m²

ব্যাখ্যা

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

৩০৮.

The width of a rectangle is equal to 3/4 of its length. What is the length of the rectangle if the length of the diagonal is 25 cm?

ক
ক) 15 cm
খ
খ) 18 cm
গ
গ) 20 cm
ঘ
ঘ) None of the above

ব্যাখ্যা

প্রশ্ন : The width of a rectangle is equal to 3/4 of its length. What is the length of the rectangle if the length of the diagonal is 25 cm?

সমাধান :
ধরি,
দৈর্ঘ্য = 4x সে.মি.
∴ প্রস্থ = 4x এর 3/4 সে.মি.
= 3x সে.মি.

ΔBCD সমকোণী ত্রিভুজে,
(3x)² + (4x)² = (25)²
বা, 25x²= 625
বা, x² = 25
বা, x = 5
∴ x = 5

∴ দৈর্ঘ্য = 4x সে.মি.
= (4×5) সে.মি.
= 20 সে.মি.

৩০৯.

If sinθ + cosecθ = 2 then sin⁵θ + cosec⁵θ =?

ক
10
খ
4
গ
2
ঘ
0

ব্যাখ্যা

Question: If sinθ + cosecθ = 2 then sin⁵θ + cosec⁵θ =?

Solution:
sinθ + cosecθ = 2
or, sinθ + (1/sinθ) = 2
or, sin²θ + 1 = 2sinθ
or, sin²θ - 2sinθ + 1 = 0
or, (sinθ - 1)² = 0
or, sinθ - 1 = 0
∴ sinθ = 1

cosecθ = 1/sinθ = 1/1 = 1

∴ sin⁵θ + cosec⁵θ = (1)⁵ + (1)⁵
= 2

৩১০.

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

বৃত্তটির ব্যাস হলো অন্তবর্গের কর্ণ

অন্তবর্গের এক বাহুর দৈর্ঘ্য x একক হলে
x√2 = 2a
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

৩১১.

If each edge of a cube is increased by 50%, find the percentage increase in its surface area.

ক
125%
খ
150%
গ
175%
ঘ
110%

ব্যাখ্যা

Question: If each edge of a cube is increased by 50%, find the percentage increase in its surface area.

Solution:
Let the edge = a cm
New measure of the edges after increase = a + 50% of a = a + a/2 = 3a/2

Total surface area of the original cube= 6a²

Total surafec area of the new cube = 6 × (3a/2)²
= 6 × (9a²/4)
= (9/4) × 6a²
= 2.25 × 6a²
= 2.25 × Total surface area of the original cube

Increase in the area
= New total surface area - orignal suface area
= (2.25 - 1) × 6a²
= 1.25 × 6a²

Total surface area of the original cube = 1.25 × Total surface area of the original cube.

∴ Percentage increase in surface area = 125%

৩১২.

Let A, B, C, D be the angles of a quadrilateral. If they are concyclic, then the value of cos A + cos B + cos C + cos D is ?

ক
0
খ
1
গ
2
ঘ
3

ব্যাখ্যা

Every angle = 90°
So, A = B = C = D = 90°
∴ cosA + cosB + cosC + cosD
= cos90° + cos90° + cos90° + cos90°
= 0 + 0 + 0 + 0
= 0

৩১৩.

If each side of a square is increased by 25%, find the percentage change in its area.

ক
ক) 24.25%
খ
খ) 43.25%
গ
গ) 56.25%
ঘ
ঘ) 66.25%

ব্যাখ্যা

Question: If each side of a square is increased by 25%, find the percentage change in its area.

Solution:
Let, each side of the square be a,
then area = a²
Given that The side is increased by 25%, then

New side = 125a/100 = 5a/4
New area = (5a/4)²

Increased area = (25a²/16) − a²
= (25a² - 16a²)/16
= 9a²/16

∴ Increase % = [9a²/16]/a² × 100 %
= 56.25%

৩১৪.

A piece of wire 91 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?

ক
ক) 24 cm
খ
খ) 21 cm
গ
গ) 18 cm
ঘ
ঘ) 14 cm

ব্যাখ্যা

Question: A piece of wire 91 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?

Solution:
Given,
Ratio of one of the equal sides to the base is 5 : 3
Therefore, the sides are 5x, 3x, 5x.

91 cm piece of wire is bent to form an isosceles triangle.
Thus perimeter of triangle is 91 cm.

ATQ,
∴ 13x = 91
⇒ x = 7

Thus the length of the base = 3 × 7 = 21 cm.

৩১৫.

The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

ক
72π m²
খ
90π m²
গ
60π m²
ঘ
66π m²

ব্যাখ্যা

Question: The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

Solution:
Here, l = 10m
h = 8m

So, r = √(l² - h²)
= √(10² - 8²)
=√(100 - 64)
= √36
= 6

∴Curved surface area = πrl
= (π × 6 × 10) m² = 60π m²

৩১৬.

The perimeter of a rectangle is 400 meters. The breadth is 3/7 part of the length. What is the length?

ক
148 m
খ
126 m
গ
160 m
ঘ
140 m
ঙ
None of the above

ব্যাখ্যা

Question: The perimeter of a rectangle is 400 meters. The breadth is 3/7 part of the length. What is the length?

Solution:
Let
Breadth of rectangle = 3p
Length of rectangle = 7p

Now,
2(3p + 7p) = 400
⇒ 2 × 10p = 400
⇒ 20p = 400
⇒ p = 20

∴ Length of rectangle = 7 × 20 = 140 m

৩১৭.

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

ক
225 sq. cm
খ
154 sq. cm
গ
194 sq. cm
ঘ
144 sq. cm

ব্যাখ্যা

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

Solution:
Perimeter of the square = 4 × 11
= 44 cm

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

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

৩১৮.

cos(24°) + cos(5°) + cos(175°) + cos(204°) + cos(300°) =

ক
1/√2
খ
1/2
গ
1/3
ঘ
- 1/2

ব্যাখ্যা

Question: cos(24°) + cos(5°) + cos(175°) + cos(204°) + cos(300°) =

Solution:
cos(175°) = cos(180° - 5°) = - cos(5°)
cos(204°) = cos(180° + 24°) = - cos(24°)
cos(300°) = cos(360° - 60°) = cos(60°)

∴ cos(24°) + cos(5°) + cos(175°) + cos(204°) + cos(300°)
= cos(24°) + cos(5°) - cos(5°) - cos(24°) + cos(60°)
= cos(60°)
= 1/2

৩১৯.

If θ is a positive acute angle and 4cos²θ - 1 = 0, then the value of tan(θ - 15°) is equal to?

ক
ক) 0
খ
খ) 1
গ
গ) 1/√3
ঘ
ঘ) √3

ব্যাখ্যা

Question: If θ is a positive acute angle and 4cos²θ - 1 = 0, then the value of tan(θ - 15°) is equal to?

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

Now,
tan(θ - 15°) = tan(60° - 15°)
= tan 45°
= 1

৩২০.

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

ক
Tk. 695
খ
Tk. 702
গ
Tk. 712
ঘ
Tk. 725

ব্যাখ্যা

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

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

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

৩২১.

A field has a length of 60 meters and a width of 50 meters. Two roads, each 4 meters wide, cross each other perpendicularly through the middle of the field. What is the total area of the two roads?

ক
320 sq meters
খ
424 sq meters
গ
496 sq meters
ঘ
520 sq meters
ঙ
None

ব্যাখ্যা

Question: A field has a length of 60 meters and a width of 50 meters. Two roads, each 4 meters wide, cross each other perpendicularly through the middle of the field. What is the total area of the two roads?

Solution:
দৈর্ঘ্য বরাবর রাস্তার ক্ষেত্রফল = (60 × 4) বর্গমিটার
= 240 বর্গমিটার

প্রস্থ বরাবর রাস্তার ক্ষেত্রফল = (50 - 4) × 4 বর্গমিটার
= 46 × 4 বর্গমিটার
= 184 বর্গমিটার

রাস্তা দুইটির মোট ক্ষেত্রফল = (240 + 184) বর্গমিটার
= 424 বর্গমিটার

৩২২.

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

ক
6 cm
খ
5 cm
গ
4 cm
ঘ
3 cm

ব্যাখ্যা

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

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

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

৩২৩.

If measures of the angles in a triangle are in the ratio of 1 : 3 : 5, then the degrees in the largest angle:

ক
60°
খ
90°
গ
120°
ঘ
100°

ব্যাখ্যা

Question: If measures of the angles in a triangle are in the ratio of 1 : 3 : 5, then the degrees in the largest angle:
(Officer Cash 2022 অনুযায়ী)

Solution:
Given that,
The angles of a triangle are in the ratio 1 : 3 : 5
Let,
x, 3x, 5x

We know that,
Sum of angles in a triangle = 180°

Now
x + 3x +5x = 180°
9x = 180°
x = 180°/9 = 20°
∴ x = 20°

∴ Largest angle = 5x = 5 × 20 = 100°

৩২৪.

On the same side of the tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is -

ক
ক) 63.5 m
খ
খ) 76.9 m
গ
গ) 86.7 m
ঘ
ঘ) 90 m

ব্যাখ্যা

Let AB be the tower and C and D be the objects.
Then, AB = 150 m,
∠ACB = 45° and
∠ADB = 60°

AB/AD = tan 60° = √3
AD = AB/√3
= 150/√3 m.

AB/AC = tan 45° = 1
AC = AB = 150 m.

∴CD = (AC - AD)
= {150 - (150/√3)} m
= [{150(√3 - 1)/√3} × {(√3)/(√3)}] m
= 50(3 - √3) m
= (50 × 1.27) m
= 63.5 m.

৩২৫.

If sin x = 3/4, then cos x = ?

ক
ক) 2/3
খ
খ) √2/3
গ
গ) √7/4
ঘ
ঘ) 1/2

ব্যাখ্যা

cosx
= √(1 - sin²x)
= √{1 - (3/4)²}
= √(7/16)
= √7/4

৩২৬.

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

ক
220 cm²
খ
100 cm²
গ
210 cm²
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

৩২৭.

secA - tanA = 3/5 , then secA + tanA - (5/3) =?

ক
5/3
খ
3
গ
1/2
ঘ
0

ব্যাখ্যা

প্রশ্ন: secA - tanA = 3/5 , then secA + tanA - (5/3) =?

সমাধান:
আমরা জানি,
sec²A - tan²A = 1
⇒ (secA - tanA ) (secA + tanA) = 1
⇒ (3/5) (secA + tanA) = 1
⇒ secA + tanA = 5/3

∴ secA + tanA - (5/3)
= (5/3) - (5/3)
= 0

৩২৮.

∠P এবং ∠Q পরস্পর পূরক কোণ। যদি ∠P = ২০° + ৪ক এবং ∠Q = ৬ক হয়, তবে ∠Q এর মান কত?

ক
৪০°
খ
৫০°
গ
৫৫°
ঘ
৪২°
ঙ
কোনোটিই নয়

ব্যাখ্যা

প্রশ্ন: ∠P এবং ∠Q পরস্পর পূরক কোণ। যদি ∠P = ২০° + ৪ক এবং ∠Q = ৬ক হয়, তবে ∠Q এর মান কত?

সমাধান:
এখানে,
∠P = ২০° + ৪ক এবং ∠Q = ৬ক

পূরক কোণের ক্ষেত্রে,
∠P + ∠Q = ৯০°
⇒ (২০° + ৪ক) + ৬ক = ৯০°
⇒ ২০° + ৪ক + ৬ক = ৯০°
⇒ ২০° + ১০ক = ৯০°
⇒ ১০ক = ৯০° − ২০°
⇒ ১০ক = ৭০°
∴ ক = ৭°

∴ ∠Q = ৬ × ৭° = ৪২°

৩২৯.

The 4th term of a geometric sequence is 81 and the 2nd term is 9. What is the common ratio?

ক
3
খ
5
গ
9
ঘ
6
ঙ
None of these

ব্যাখ্যা

Question: The 4th term of a geometric sequence is 81, and the 2nd term is 9. What is the common ratio?

Solution:
Given,
The 2nd term of a geometric sequence, a₂ = 9
The 4th term of the same sequence, a₄ = 81

In a geometric sequence we know,
a_n = ar^{n - 1}

So,
a₂= ar^{2 - 1}= ar .......(1)
And
a₄ = ar^{4 - 1}= ar³........ (2)

Now (2) ÷ (1),
ar³/ar = 81/9
⇒ r² = 9
⇒ r = √9
∴ r = 3

Since all terms are positive (2nd term is 9 and 4th is 81), the common ratio is 3.

৩৩০.

A rectangular tank with a length of 4m and a width of 2m can store 20000 liters. What is the height of the tank?

ক
2 m
খ
1.5 m
গ
3.5 m
ঘ
2.5 m

ব্যাখ্যা

প্রশ্ন: A rectangular tank with a length of 4m and a width of 2m can store 20000 liters. What is the height of the tank?

Solution:
দেয়া আছে,
ট্যাংকের দৈর্ঘ্য (l) = 4 m, প্রস্থ (b) = 2 m, এবং আয়তন (V) = 20000 লিটার।

ধরি, ট্যাংকটির উচ্চতা হল h মিটার।

আমরা জানি,
আয়তাকার ঘনবস্তুর আয়তন, V = l × b × h ঘন একক
= (4 × 2 × h) m³
= 8h m³

এখন,
1 m³ = 1000 লিটার।

প্রশ্নমতে,
(8h × 1000) = 20000
বা, 8h = 20000/1000
বা, 8h = 20
∴ h = 2.5

সুতরাং, ট্যাংকটির উচ্চতা হল 2.5 মিটার।

৩৩১.

The area of a rectangle R with width 4 ft is equal to the area of a square S, which has a perimeter of 24 ft, the perimeter of the rectangle R, in feet, is

ক
ক) 9
খ
খ) 16
গ
গ) 24
ঘ
ঘ) 26

ব্যাখ্যা

Given,
Perimeter of the square = 24 ft
Length of the side of the square = 24/4 =6 ft
So, its area = 6² = 36 ft²

ATQ,
Area of the rectangle is, length × width = 36 ft²
⇒ length = 36/4 = 9 [As the rectangle's width is 4 ft]

∴ Perimeter of the rectangle = 2(9 + 4) = 26 ft

৩৩২.

A cuboid has dimensions in the ratio 1:2:3 and a total surface area of 88 cm². What is its volume?

ক
24 cm³
খ
36 cm³
গ
48 cm³
ঘ
60 cm³

ব্যাখ্যা

Question: A cuboid has dimensions in the ratio 1:2:3 and a total surface area of 88 cm². What is its volume?

Solution:
Let the dimensions be x, 2x, and 3x.

Total surface area of the cube = 2(x.2x + 2x.3x + 3x.x)
= 22x²

According to the question,
22x² = 88
x² = 4
∴ x = 2

So, volume of the cuboid = 2 × 4 × 6
= 48 cm³

৩৩৩.

A square that has a side length of 4m is reduced by 50% of its area. The new side length will be -

ক
ক) 2m
খ
খ) 3m
গ
গ) 2√3m
ঘ
ঘ) 2√2m

ব্যাখ্যা

Question: A square that has a side length of 4m is reduced by 50% of its area. The new side length will be -

Solution:
here, the side length is = 4m
the area of the square = 4² =16m²

after reducing the area by 50% the remaining area is = 16 - (50% of 16)
= 16 - 8 = 8m²

the new side length is = √8 = 2√2 m

৩৩৪.

What is the value of cosA + secA is -

ক
ক) 2/5
খ
খ) 5/2
গ
গ) 1/2
ঘ
ঘ) 5/3

ব্যাখ্যা

Question:
What is the value of cosA + secA is -

Solution:
প্রদত্ত চিত্র হতে,
ΔABC সমকোণী ত্রিভুজ হতে,
অতিভূজ AC = 2 এবং ভূমি, BC = √3

আমরা জানি,
(লম্ব)² = (অতিভূজ)² - (লম্ব)²
বা, AC² = (2)² - (√3)²
বা, AC² = 4 - 3
∴ AC = 1

cosA = AB/AC = 1/2
secA = AC/AB = 2/1 = 2
∴ cosA + secA = 1/2 + 2 = 5/2

৩৩৫.

secA + tanA = 4/3, then find, secA - tanA = ?

ক
4/5
খ
3/4
গ
5/12
ঘ
3/2

ব্যাখ্যা

Question: secA + tanA = 4/3, then find, secA - tanA = ?

Solution:
Given that,
secA + tanA = 4/3

We know,
sec²A - tan²A = 1
⇒ (secA + tanA)(secA - tanA) = 1
⇒ (4/3)(secA - tanA) = 1
⇒ (secA - tanA) = 1/(4/3)
∴ secA - tanA = 3/4

৩৩৬.

Find the length of the altitude of an equilateral triangle of side 3√3 cm.

ক
6.5 cm
খ
5.5 cm
গ
4.5 cm
ঘ
7 cm

ব্যাখ্যা

Question: Find the length of the altitude of an equilateral triangle of side 3√3 cm.

Solution:
Given that,
Side of equilateral triangle = 3√3 cm

We know,
Altitude (height) of an equilateral triangle is,
h = (√3/2) × side
h = (√3/2) × 3√3
= (3√3 × √3)/2
= (3 × 3)/2
= 9/2
∴ h = 4.5 cm

So the length of the altitude is 4.5 cm or 9/2 cm.

৩৩৭.

In a right triangle, the length of one of the legs is 8 and the length of the hypotenuse is 17. What is the length of the other leg?

ক
10
খ
12
গ
15
ঘ
18

ব্যাখ্যা

Question: In a right triangle, the length of one of the legs is 8 and the length of the hypotenuse is 17. What is the length of the other leg?

Solution:
এখানে,
সমকোণী ত্রিভুজের (right triangle) অতিভুজ (hypotenuse)= 17 একক
সমকোণ সংলগ্ন এক বাহু = 8 একক
সমকোণ সংলগ্ন অপর বাহু = a একক

প্রশ্নমতে,
a² + 8² = 17²
⇒ a² + 64 = 289
⇒ a² = 289 - 64
⇒ a² = 225
⇒ a = √225
∴ a = 15

৩৩৮.

A trapezium ABCD has side AD parallel to BC, ∠BAD = 90°, BC = 3 cm and AD = 8 cm. If the perimeter of the trapezium is 36 cm, then its area is

ক
44 sq. cm.
খ
50 sq. cm.
গ
66 sq. cm.
ঘ
72 sq. cm.

ব্যাখ্যা

Question: A trapezium ABCD has side AD parallel to BC, ∠BAD = 90°, BC = 3 cm and AD = 8 cm. If the perimeter of the trapezium is 36 cm, then its area is

Solution:

AB + CD = 36 - 8 - 3 = 25
CD = 25 - AB = 25 - x

CD² = x² + 5²
⇒ (25 - x)² = x² + 25
⇒ 625 - 50x + x² = x² + 25
⇒ 50x = 625 - 25 - 600
⇒ x = 12

Area = (1/2) (8 + 3) 12
= 66 sq. cm.

৩৩৯.

The ratio between the length and breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park.

ক
ক) 153600 m²
খ
খ) 150035 m²
গ
গ) 133636 m²
ঘ
ঘ) 148600 m²

ব্যাখ্যা

Perimeter = Distance covered in 8 min.
Perimeter = (12000/60)×8 m
Perimeter = 1600 m

Let length = 3x metres and
breadth = 2x metres.
Then,
2(3x + 2x) = 1600
5x = 800
x = 160
Therefore Length = 480 m and Breadth = 320 m

Therefore Area = (480 x 320) m²
= 153600 m²

৩৪০.

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
12π cm³
খ
15π cm³
গ
16π cm³
ঘ
20π cm³

ব্যাখ্যা

Question: 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-

Solution:

Clearly, we have r = 3 cm and h = 4 cm.

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

৩৪১.

Find the ratio of the areas of the incircle and the circumcircle of a square.

ক
3 : 2
খ
1 : 2
গ
5 : 2
ঘ
1 : 1

ব্যাখ্যা

Question: Find the ratio of the areas of the incircle and circumcircle of a square.

Solution:
ধরি, বর্গক্ষেত্রের বাহুর দৈর্ঘ্য r

বর্গের অন্তর্বৃত্তের ব্যাস বর্গের বাহুর সমান।
অন্তর্বৃত্তের ক্ষেত্রফল = π(r/2)²
= πr²/4

বর্গের বহিঃবৃত্তের ব্যাস বর্গের কর্ণের সমান।
বহিঃবৃত্তের ক্ষেত্রফল = π(√2 r/2)²
= πr²/2

অনুপাত = πr²/4 : πr²/2
= (1/4) : (1/2)
= 1 : 2

৩৪২.

The total surface area of a cube is 96 cm². The volume of the cube is:

ক
ক) 8 cm³
খ
খ) 512 cm³
গ
গ) 64 cm³
ঘ
ঘ) 27 cm³
ঙ
ঙ) 32 cm³

ব্যাখ্যা

We know that the Total Surface Area of the cone = 6a².
6a² = 96 cm²
a² = 96/6 = 16
a = 4 cm

The volume of cone = a³ cubic units
V = 4³ = 64cm³.

৩৪৩.

If sin x + cos x = 1, then x = ?

ক
30°
খ
45°
গ
60°
ঘ
90°

ব্যাখ্যা

Question: If sin x + cos x = 1, then x = ?

Solution:
অপশন টেস্ট করে পাই,
x = 30° হলে,
sin 30° + cos 30°
= (1/2) + (√3/2)
= (1 + √3)/2 ;[যা সত্য নয়]

x = 45° হলে,
sin 45° + cos 45°
= (1/√2) + (1/√2)
= 2/√2
= √2 ;[যা সত্য নয়]

x = 60° হলে,
sin 60° + cos 60°
= (√3/2) + (1/2)
= (1 + √3)/2 ;[যা সত্য নয়]

x = 90° হলে,
sin 90° + cos 90°
= 1 + 0
= 1 ;[যা সত্য]

∴ x = 90° হলে, sin x + cos x = 1 হবে।

৩৪৪.

Find the equation of the line with x-intercept = -3 and y-intercept = 2.

ক
2x - 5y + 10 = 0
খ
3x - 2y - 6 = 0
গ
4x - 3y + 12 = 0
ঘ
2x - 3y + 6 = 0

ব্যাখ্যা

Question: Find the equation of the line with x-intercept = -3 and y-intercept = 2.

Solution:
Given, x-intercept = - 3,
So, the line passes through (-3, 0).
y-intercept = 2,
So, the line passes through (0, 2).

We know, the intercept form of a line is:
(x/a) + (y/b) = 1, where a = x-intercept, b = y-intercept.
⇒ x/(- 3) + (y/2) = 1
⇒ (- 2x + 3y)/6 = 1
⇒ - 2x + 3y = 6
⇒ 2x - 3y + 6 = 0

∴ The equation of the line is 2x - 3y + 6 = 0

৩৪৫.

A rectangular field is 3.2 yards long. A fence marking the boundary is 11.2 yards in length. What is the area of the field in square yards?

ক
4.68
খ
7.68
গ
9.28
ঘ
11.28

ব্যাখ্যা

Question: A rectangular field is 3.2 yards long. A fence marking the boundary Is 11.2 yards in length. What is the area of the field in square yards?

Solution:
দেওয়া আছে,
আয়তাকার মাঠের দৈর্ঘ্য = 3.2 গজ
মাঠের চতুর্দিকে বেড়ার দৈর্ঘ্য = মাঠের পরিসীমা = 2(দৈর্ঘ্য + প্রস্থ) = 11.2 গজ

ধরি,
মাঠের প্রস্থ = x গজ

প্রশ্নমতে,
2(3.2 + x) = 11.2
⇒ 6.4 + 2x = 11.2
⇒ 2x = 11.2 - 6.4
⇒ 2x = 4.8
⇒ x = 4.8/2
⇒ x = 2.4

∴ আয়তাকার মাঠের ক্ষেত্রফল = দৈর্ঘ্য × প্রস্থ = (3.2 × 2.4) বর্গগজ = 7.68 বর্গগজ

৩৪৬.

The total cost of flooring a room at Tk. 6.50 per square meter is Tk. 390. If the length of the room is 8m, what is its breadth?

ক
7.5 m
খ
8.5 m
গ
10.5 m
ঘ
12.5 m

ব্যাখ্যা

Question: The total cost of flooring a room at Tk. 6.50 per square meter is Tk. 390. If the length of the room is 8m, what is its breadth?

Solution:
ঘরের ক্ষেত্রফল = 390/6.50
= 60 বর্গমিটার

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

আয়তাকার ক্ষেত্রের প্রস্থ =7.5 মিটার

৩৪৭.

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

ক
2
খ
3
গ
4
ঘ
6

ব্যাখ্যা

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

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

৩৪৮.

Read the following questions carefully and choose the right answer. (33-56):
৩৩) The length of a rectangle is 20% more than its breadth. Find the ratio of the area of the rectangle to that of the square whose side is equal to the breadth of the rectangle.

ক
ক) 8:9
খ
খ) 3:2
গ
গ) 6:5
ঘ
ঘ) 2:1

ব্যাখ্যা

Let the breadth of the rectangle be x
∴ According to question,
Length = 1.20x
∴ Area of rectangle = 1.20x × x = 1.20x²
Area of square = x × x = x²
∴ Required ratio = 1.20x²/x²
= 12/10
= 6/5

৩৪৯.

If y = sinx then for which value of x, y has the highest value?

ক
45°
খ
90°
গ
60°
ঘ
0°

ব্যাখ্যা

Question: If y = sinx then for which value of x, y has the highest value?

Solution:
y = sinx
y এর মান সর্বোচ্চ হবে যদি sinx এর মান সর্বোচ্চ হয়।
sinx এর সর্বোচ্চ মান 1
তাহলে,
x = 90°

৩৫০.

Compute the surface area of a cuboid with length 10 cm, width 6 cm, and height 4 cm.

ক
332 square cm
খ
240 square cm
গ
166 square cm
ঘ
248 square cm

ব্যাখ্যা

Question: Compute the surface area of a cuboid with length 10 cm, width 6 cm, and height 4 cm.

Solution:
the surface area of a cuboid = 2(ab + bc + ca)

Surface Area= 2 × (10 × 6 + 10 × 4 + 6 × 4) = 248 square cm

৩৫১.

cot³30° - 2sin60° = ?

ক
√3
খ
3√3
গ
2√3
ঘ
1/√3

ব্যাখ্যা

Question: cot³30° - 2sin60° = ?

Solution:
Given that,
cot³30° - 2sin60°
= (√3)³ - 2(√3/2)
= 3√3 - √3
= 2√3

৩৫২.

The area of the right angle triangle is 184 square cm one of its leg is 16 cm long. Find the length of the other leg.

ক
23 cm
খ
22 cm
গ
24 cm
ঘ
20 cm
ঙ
18 cm

ব্যাখ্যা

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

৩৫৩.

A rectangular block 6 cm by 12 cm by 15 cm is cut up into an exact number of equal cubes. Find the least possible number of cubes.

ক
ক) 30
খ
খ) 35
গ
গ) 40
ঘ
ঘ) 45

ব্যাখ্যা

Volume of the block = (6 × 12 × 15) cm³
= 1080 cm³

Side of the largest cube = H.C.F of 6 cm, 12 cm, 15 cm
= 3 cm.

Volume of this cube = (3 × 3 × 3) cm³
= 27 cm³

Number of cubes = 1080/27
= 40.

৩৫৪.

Given that 1 cubic cm of marble weighs 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. What is the length of the block?

ক
ক) 26.5 cm
খ
খ) 32 cm
গ
গ) 36 cm
ঘ
ঘ) 37.25 cm

ব্যাখ্যা

Question: Given that 1 cubic cm of marble weighs 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. What is the length of the block?

Solution:
Let the length of the block = x cm
Then (x × 28 × 5 × (25/10000)) = 112
∴ x = (112 × (1/28) × (1/5) × (1000/25)) = 32 cm

৩৫৫.

If the ratio of the area of a sector to the area of the circle is 3:5, what is the ratio of the length of the are in the sector to the circumference of the circle?

ক
3/5
খ
2/7
গ
1/3
ঘ
1/2

ব্যাখ্যা

Question: If the ratio of the area of a sector to the area of the circle is 3 : 5, what is the ratio of the length of the are in the sector to the circumference of the circle?

Solution:
বৃত্তের ব্যাসার্ধ = r
বৃত্তের ক্ষেত্রফল = πr²
বৃত্তের পরিধি = 2πr
বৃত্তের চাপের দৈর্ঘ্য = θπr/180
বৃত্তকলার ক্ষেত্রফল = θπr²/360

প্রশ্নমতে
(θπr²/360) : πr² = 3 : 5
(θπr2/360)/πr2 = 3/5
θ/360 = 3/5

বৃত্তের চাপের দৈর্ঘ্য : বৃত্তের পরিধি = (θπr/180) : 2πr
= (θπr/180)/2πr
= (θπr/180) × 1/(2πr)
= θ/360
= 3/5

৩৫৬.

The ratio of the area of a square to that of the square drawn on its diagonal, is-

ক
ক) 1 : √2
খ
খ) 1 : 2
গ
গ) 1 : 4
ঘ
ঘ) √2 : 2

ব্যাখ্যা

Let the side of a square be a.
Area of square = a²
Length of the diagonal = √(a² + a²) =a√2
Area of a square formed on diagonal of first square is = (a√2)²=2a²

Now
a²/2a²=1/2
= 1 : 2

৩৫৭.

The area of the triangle with sides 5cm, 12cm and 13cm is

ক
30 cm²
খ
42 cm²
গ
84 cm²
ঘ
78 cm²

ব্যাখ্যা

The area of the triangle with sides 5cm, 12cm and 13cm
= 1/2 × 12 × 5
= 30 cm²
[ This triangle is a right angled triangle because of sides 5cm, 12cm and 13cm
5² + 12² = 13²]

৩৫৮.

If the area of a square garden is 50 sq. meters, what is the maximum distance between two points on its boundary?

ক
5√2 meters
খ
20 meters
গ
10√2 meters
ঘ
10 meters

ব্যাখ্যা

Question: If the area of a square garden is 50 sq. meters, what is the maximum distance between two points on its boundary?

Solution:
Given that,
Area of square = 50 m²
∴ Side of length = √Area = √50 =√(25 × 2)
= 5√2 meters

The maximum distance between any two points on the boundary of a square is the length of the diagonal.
∴ Diagonal of the square = side × √2
= 5√2 × √2
= 5 × 2
= 10 meters

So the maximum distance between two points on the boundary is 10 meters.

৩৫৯.

The minimum value of 2sin²θ + 3cos²θ is?

ক
0
খ
1
গ
2
ঘ
4

ব্যাখ্যা

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

Solution:
2 sin²θ + 3cos²θ
= 2 sin²θ + 2cos²θ + cos²θ
= 2 + cos²θ
= 2 + 0 [minimum value of cos²θ is 0]
= 2

৩৬০.

A metallic sphere of radius 6 cm is melted to make a cone with base of the same radius. What is the height of the cone ?

ক
ক) 12cm
খ
খ) 18cm
গ
গ) 24cm
ঘ
ঘ) 36cm

ব্যাখ্যা

Question: A metallic sphere of radius 6 cm is melted to make a cone with base of the same radius. What is the height of the cone ?

Solution:
Let
The height of the cone be h cm
Then,
(4/3)π × (6)³= (1/3)π × (6)² × h
⇒ 4 × 6 = h
⇒ h = 24cm

৩৬১.

Given that the diagonal of a square measures 12√2, find the area of the square in square units.

ক
144 square units
খ
184 square units
গ
192 square units
ঘ
282 square units

ব্যাখ্যা

Question: Given that the diagonal of a square measures 12√2, find the area of the square in square units.
(একটি বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য 12√2 হলে ঐ বর্গক্ষেত্রের ক্ষেত্রফল কত বর্গ একক?)

Solution:
আমরা জানি,
বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = √2 × বাহু

প্রশ্নমতে,
√2 × বাহু = 12√2
⇒ বাহু = 12√2/√2 = 12

∴ বর্গক্ষেত্রের ক্ষেত্রফল = 12²
= 144 বর্গ একক

৩৬২.

The value of {(tan 15° - tan 60°)/(cot 30° - cot 75°)} + 1 is -

ক
ক) - 1
খ
খ) 1
গ
গ) 2
ঘ
ঘ) 0

ব্যাখ্যা

Question: The value of {(tan 15° - tan 60°)/(cot 30° - cot 75°)} + 1 is -

Solution:
(tan 15° - tan 60°)/(cot 30° - cot 75°) + 1
= (tan 15° - tan 60°)/cot (90° - 60°) - cot (90° - 15°) + 1
= (tan 15°- tan 60°)/(tan 60° - tan 15°) + 1
= (tan 15°- tan 60°)/(- 1)(tan 15° - tan 60°) + 1
= - 1 + 1
= 0

৩৬৩.

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

ক
29.5%
খ
33.5%
গ
36.5%
ঘ
39.5%

ব্যাখ্যা

Question: A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

Solution:
Let the side of the square (ABCD) be x metres.
Then, AB + BC = 2x metres

AC = √2x = (1.41x) m
Saving on 2x metres = (0.59x) m

∴ Saving% = {(0.59x/2x).100%}
= 29.5%

৩৬৪.

The total surface area of a hemisphere of radius r is

ক
ক) 2πr²
খ
খ) 4πr²
গ
গ) πr²
ঘ
ঘ) 3πr²

ব্যাখ্যা

গোলকের ব্যাসার্ধ r হলে গোলকের ক্ষেত্রফল হবে = 4πr²
অর্ধগোলকের ক্ষেত্রফল = অর্ধগোলকের ক্ষেত্রফল + বৃত্তের ক্ষেত্রফল
= (4πr²)/2 + πr²
= 2πr² + πr²
= 3πr²

৩৬৫.

The numerical value of is?

ক
ক) 2
খ
খ) 3
গ
গ) 5
ঘ
ঘ) 0

ব্যাখ্যা

Question: The numerical value of is?

Solution:

৩৬৬.

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

ক
18π cm³
খ
52π cm³
গ
27π cm³
ঘ
36π cm³

ব্যাখ্যা

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

Solution:
Given that,
Radius of sphere = 3 cm

We know,
Volume of a sphere = (4/3) × πr³= (4/3) × π(3)³= (4/3) × π × 27
= 36π cm³

৩৬৭.

The perimeter of one face of a cube is 16 cm. Its volume must be-

ক
ক) 25 cm³
খ
খ) 64 cm³
গ
গ) 75 cm³
ঘ
ঘ) 125 cm³

ব্যাখ্যা

Question: The perimeter of one face of a cube is 16 cm. Its volume must be-

Solution:
the perimeter of one face is 16 cm

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

⇒ 4a = 16
⇒ a = 16/4
= 4 cm

volume = a³
= 4³
= 64 cm³

৩৬৮.

If the length of the shorter diagonal is four, what is the length of the longer diagonal of this kite?

ক
3√5
খ
4√5
গ
5√3
ঘ
5√4

ব্যাখ্যা

Question: If the length of the shorter diagonal is four, what is the length of the longer diagonal of this kite?

Solution:
We can find the longer diagonal by adding together the altitude of the top triangle and the altitude of the bottom triangle. To find these, use Pythagorean Theorem. We can use Pythagorean Theorem because one of the properties of a kite is that the two diagonals are perpendicular.

The top triangle has two sides of length 3 [labeled in the picture], and a base of 4 [provided in the written directions]. To figure out the altitude, split this triangle into 2 right triangles. The two legs are x [the altitude] and 2 [half of the base 4], and the hypotenuse is 3:
x² + 2² = 3²
⇒ x² + 4 = 9
⇒ x² = 5
∴ x = √5

We will do something similar for the bottom triangle. Consider one of the right triangles. It will have a hypotenuse of 7, one leg that we don't know, x [the altitude], and one leg 2 [half the shorter diagonal]. Set up the equation using the Pythagorean Theorem:
x² + 2² = 7²
⇒ x² + 4 = 49
⇒ x² = 45
∴ x = √45 = 3√5

∴ The length of the longer diagonal of this kite = 3√5 + √5 = 4√5

৩৬৯.

The angle of elevation of the sun, when the length of the shadow of a tree 1/√3 times the height of the tree, is:

ক
30°
খ
45°
গ
60°
ঘ
90°

ব্যাখ্যা

Question: The angle of elevation of the sun, when the length of the shadow of a tree 1/√3 times the height of the tree, is:

Solution:

Let,
Height of the tree BC = x
Length of the shadow AB = (1/√3)x
The angle of elevation of the sun ∠A = θ

Now,
tanθ = BC/AB
⇒ tanθ = x/{(1/√3)x}
⇒ tanθ = √3
⇒ tanθ = tan60°
∴ θ = 60°

৩৭০.

If tan(θ + 15°) = √3, what is the value of sinθ?

ক
0
খ
1/2
গ
1/√2
ঘ
√3/2

ব্যাখ্যা

প্রশ্ন: If tan(θ + 15°) = √3, what is the value of sinθ?

সমাধান:
দেওয়া আছে,
tan(θ + 15°) = √3
⇒ tan(θ + 15°) = tan 60°
⇒ θ + 15° = 60°
⇒ θ = 60° - 15°
⇒ θ = 45°

এখন,
sinθ
= sin45°
= 1/√2

৩৭১.

Given that the diagonal of a square measures 10√6 units, find the area of the square in units.

ক
600 square units
খ
300 square units
গ
100√3 square units
ঘ
500 square units

ব্যাখ্যা

Question: Given that the diagonal of a square measures 10√6 units, find the area of the square in units.

Solution:
দেয়া আছে,
বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = 10√6 একক

আমরা জানি,
বর্গক্ষেত্রের কর্ণের দৈর্ঘ্য = √2 × বাহু

প্রশ্নমতে,
√2 × বাহু = 10√6
⇒ বাহু = 10√6/√2
⇒ বাহু = 10√3 একক

এখন, বর্গক্ষেত্রের ক্ষেত্রফল = বাহু²
= (10√3)²
= 300 বর্গ একক

∴ বর্গক্ষেত্রের ক্ষেত্রফল 300 বর্গ একক

৩৭২.

If 3sec²θ - 2 = 2, find the value of θ

ক
30°
খ
0°
গ
45°
ঘ
60°

ব্যাখ্যা

Question: If 3sec²θ - 2 = 2, find the value of θ.

Solution:
3sec²θ - 2 = 2
3sec²θ = 4
sec²θ = 4/3
secθ = 2/√3
secθ = sec30°
∴ θ = 30°

৩৭৩.

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

ক
ক) 3√15 cm²
খ
খ) 2√14 cm²
গ
গ) 5√7 cm²
ঘ
ঘ) 7√11 cm²
ঙ
ঙ) 4√3 cm²

ব্যাখ্যা

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

৩৭৪.

A new square is formed by joining the midpoints of the sides of a square. The same process is repeated indefinitely. If the side of the first square is 4 cm, find the sum of the areas of all the squares

ক
16 cm2
খ
32 cm²
গ
48 cm²
ঘ
24 cm²
ঙ
None

ব্যাখ্যা

Question: A new square is formed by joining the midpoints of the sides of a square. The same process is repeated indefinitely. If the side of the first square is 4 cm, find the sum of the areas of all the squares

Solution:
Side of the first square is 4 cm.
side of second square = 2√2 cm.
Side of third square = 2 cm.
Side of fourth square = √2 cm.
...............................
...............................

Area of these squares will be = 16, 8, 4, 2, ........................
the sum of the areas of all the squares = 16 + 8 + 4 + 2, ........................
= 16/{1 - (1/2)}
= 16/(1/2)
= 32 cm²

৩৭৫.

A circle and a rectangle have the same perimeter. The sides of the rectangle are 36 cm and 52 cm. what is the area of the circle?

ক
2412 sq.cm
খ
2426 sq.cm
গ
2448 sq.cm
ঘ
2464 sq.cm

ব্যাখ্যা

Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 36 cm and 52 cm. what is the area of the circle?

Solution:
Perimeter of the rectangle =2(36 + 52)
= 176 cm

∴ Circumference of circle = 176 cm
⇒ 2πr = 176
⇒ r = 176/2π
⇒ r = (176 × 7)/(2 × 22)
∴ r = 28 cm

∴ Area of circle = πr²
= (22/7) × (28)²
= 2464 sq.cm

৩৭৬.

What is the value of 1 + {tan²A/(1 + secA)} ?

ক
cosecA
খ
cosA
গ
sinA
ঘ
secA

ব্যাখ্যা

Question: What is the value of 1 + {tan²A/(1 + secA)} ?

Solution:
1 + {tan²A/(1 + secA)}
= 1 + {(sce²A - 1)/(1 + secA)}
= {(1 + secA) + (sce²A - 1)}/(1 + secA)
= (1 + secA + sce²A - 1)/(1 + secA)
= (secA + sce²A)/(1 + secA)
= secA(1 + secA)/(1 + secA)
= secA

৩৭৭.

A dice is thrown. What is the probability that the number shown on the dice is even number ?

ক
ক) 1/4
খ
খ) 1/6
গ
গ) 1/3
ঘ
ঘ) 1/2

ব্যাখ্যা

Question: A dice is thrown. What is the probability that the number shown on the dice is even number ?

Solution:
When a dice is thrown once.
The total number of outcomes is 6 (1, 2, 3, 4, 5, and 6)
Even numbers = 3 (2, 4, 6)

Probability = No of Favorable Outcomes/Total no of Outcomes
P(Even numbers) = 3/6
= 1/2

∴ The required probability is 1/2

৩৭৮.

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
5 : 1
খ
2 : 7
গ
5 : 7
ঘ
5 : 3

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
and curved surface area of second cone
= πrl₂
Therefore, Ratio
= πrl₁ / πrl₂
= l₁ / l₂
= 5 : 7

৩৭৯.

A cylinder has a radius of 50 cm and a height of 100 cm. Find the volume of the cylinder.

ক
196250 cm³
খ
3140000 cm³
গ
785000 cm³
ঘ
7850 cm³
ঙ
None of these

ব্যাখ্যা

Question: A cylinder has a radius of 50 cm and a height of 100 cm. Find the volume of the cylinder.

Solution:
We know the volume of a cylinder is given by the formula = πr²h, where r is the radius of the cylinder and h is the height.

Therefore, putting the values, we get,
V = πr²h
= 3.14 × 50² × 100
= 785000 cm³

৩৮০.

The volume of a sphere is the same as the volume of a right circular cylinder whose radius is 4 cm and height is 18 cm. What is the radius of the sphere?

ক
6 cm
খ
8 cm
গ
10 cm
ঘ
14 cm

ব্যাখ্যা

Question: The volume of a sphere is the same as the volume of a right circular cylinder whose radius is 4 cm and height is 18 cm. What is the radius of the sphere?

Solution:
ধরি, গোলকের ব্যাসার্ধ = r₁
এবং বেলনের ব্যাসার্ধ = r₂

দেওয়া আছে,
বেলনের ব্যাসার্ধ, r₂ = 4 সেমি
বেলনের উচ্চতা, h = 18 সেমি

আমরা জানি,
গোলকের আয়তন = (4/3)πr₁³
বেলনের আয়তন = πr₂²h

প্রশ্নমতে,
গোলকের আয়তন = বেলনের আয়তন
(4/3)πr₁³ = πr₂²h
⇒ (4/3)r₁³ = (4)² × 18
⇒ (4/3)r₁³ = 16 × 18
⇒ 4r₁³ = 16 × 18 × 3
⇒ r₁³ = (16 × 18 × 3)/4
⇒ r₁³ = 4 × 18 × 3
⇒ r₁³ = 216
⇒ r₁ = 6

∴ গোলকের ব্যাসার্ধ = 6 সেমি

৩৮১.

A rectangular hall measures 10 meters in length and 6 meters in width. If carpeting costs Tk. 15 per square meter, what will be the total cost to carpet the entire hall?

ক
Tk. 700
খ
Tk. 900
গ
Tk. 1000
ঘ
Tk. 500

ব্যাখ্যা

Question: A rectangular hall measures 10 meters in length and 6 meters in width. If carpeting costs Tk. 15 per square meter, what will be the total cost to carpet the entire hall?

Solution:
Area of the hall = Length × Width
= 10 × 6 = 60 m²

Cost of carpet = Area × Cost per m²
= 60 × 15
= Tk. 900

৩৮২.

Find the value of
is -

ক
ক) 1/2
খ
খ) 1/3
গ
গ) 1/4
ঘ
ঘ) - 1/4

ব্যাখ্যা

Question: Find the value of
is -

Solution:
এখানে,

= 1/4

৩৮৩.

If the height of a tringle is increased by 60% and its base is decreased by 30%, what will be the effect on its area?

ক
12% Increase
খ
12% decrease
গ
112% increase
ঘ
No change

ব্যাখ্যা

Question: If the height of a tringle is increased by 60% and its base is decreased by 30%, what will be the effect on its area?

Solution:
ধরি,
ত্রিভুজের উচ্চতা = 10 একক
ভূমি = 10 একক
∴ ত্রিভুজের ক্ষেত্রফল = (1/2) × 10 × 10 = 50 বর্গ একক

আবার,
উচ্চতা 60% বৃদ্ধি করলে নতুন উচ্চতা = 10 + 10 এর 60% = 10 + 6 = 16 একক
ভূমি 30% হ্রাস করলে হ্রাসকৃত ভূমি = 10 - 10 এর 30% = 10 - 3 = 7 একক

∴ নতুন ত্রিভুজের ক্ষেত্রফল = (1/2) × 16 × 7 = 56 বর্গ একক

∴ ক্ষেত্রফল বৃদ্ধি পেয়েছে = (56 - 50) বর্গ একক = 6 বর্গ একক

এখন, ক্ষেত্রফল,
50 বর্গ এককে বৃদ্ধি পেয়েছে = 6 বর্গ একক
∴ বর্গ এককে বৃদ্ধি পেয়েছে = 6/50 বর্গ একক
∴ 100 বর্গ এককে বৃদ্ধি পেয়েছে = (6 × 100)/50 = 12 বর্গ একক

সুতরাং,
ক্ষেত্রফল বৃদ্ধি পেয়েছে = 12%

৩৮৪.

The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is:

ক
5 : 2
খ
3 : 2
গ
1 : 2
ঘ
1 : 1

ব্যাখ্যা

Question: The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is:

Solution:
Let the radius of the sphere be r
Then, the radius of the cylinder = r
Height of the cylinder = 2r
Surface area of sphere = 4πr²

Surface area of the cylinder = 2πr(2r) = 4πr²

∴ Required ratio
= 4πr² : 4πr²
= 1 : 1

৩৮৫.

If a rectangle has length m and the width is half of the length, the area of the rectangle is-

ক
ক) m²
খ
খ) 2m²
গ
গ) m²/2
ঘ
ঘ) m²/4

ব্যাখ্যা

Given that
length = m
Width = m/2

The area of the rectangle is = m(m/2) = m²/2

৩৮৬.

The circumference of the back-sided wheel of a vehicle is 1 m greater than that of the front-side wheel. To travel 600 m, the front wheel rotates 30 times more than the back wheel. The circumference of the back-sided wheel is:

ক
4 m
খ
5 m
গ
6 m
ঘ
7 m

ব্যাখ্যা

Question: The circumference of the back-sided wheel of a vehicle is 1 m greater than that of the front-side wheel. To travel 600 m, the front wheel rotates 30 times more than the back wheel. The circumference of the back-sided wheel is:

Solution:
ধরি, সামনের চাকার পরিধি x m
পেছনের চাকার পরিধি = x + 1 m

600 m যেতে সামনের চাকা ঘোরে 600/x বার
600 m যেতে পেছনের চাকা ঘোরে 600/x + 1 বার

(600/x) - (600/x + 1) = 30
⇒ (x + 1 - x)/x (x + 1) = 30 / 600
⇒ 1/x² + x = 1/20
⇒ x² + x = 20
⇒ x² + x - 20 = 0
⇒ x² + 5x - 4x - 20 = 0
⇒ x (x + 5) - 4 (x + 5) = 0
⇒ (x + 5) (x - 4) = 0

x এর ঋণাত্মক মান গ্রহণযোগ্য নয়।
∴ x - 4 = 0
x = 4

পেছনের চাকার পরিধি = 4 + 1 = 5 m

৩৮৭.

A parallelogram has a base of 12 meters and a height of 8 meters. If the area of the parallelogram is increased by 50%, what will be the new height if the base remains the same?

ক
10 meters
খ
12 meters
গ
14 meters
ঘ
16 meters

ব্যাখ্যা

Question: A parallelogram has a base of 12 meters and a height of 8 meters. If the area of the parallelogram is increased by 50%, what will be the new height if the base remains the same?

Solution:
The area of a parallelogram is given by the formula:
Area = Base × Height
Original Area = 12 meters × 8 meters = 96 square meters

New area after a 50% increase:
New Area = 96 square meters × 1.5 = 144 square meters

New Height = New Area/Base = 144 square meters/12 meters = 12 meters

৩৮৮.

The surface area of a cube is 150 cm². Its volume must be-

ক
ক) 50 cm³
খ
খ) 95 cm³
গ
গ) 100 cm³
ঘ
ঘ) 125 cm³

ব্যাখ্যা

Question: The surface area of a cube is 150 cm². Its volume must be-

Solution:
let, length of one side is a cm
surface area = 6a² cm

⇒ 6a² = 150
⇒ a² = 150/6
⇒ a² = 25
⇒ a = 5 cm

volume = a³
= 5³
= 125 cm³

৩৮৯.

If a pole 12 m high casts a shadow 4√3 m long on the ground, then the elevation of the sun is -

ক
30°
খ
60°
গ
45°
ঘ
90°

ব্যাখ্যা

Question: If a pole 12 m high casts a shadow 4√3 m long on the ground, then the elevation of the sun is -

Solution:

ধরি,
AB = 12, BC = 4√3

ABC সমকোণী ত্রিভুজ হতে পাই,
tanθ = AB/BC
⇒ tanθ = 12/4√3
⇒ tanθ = 3/√3
⇒ tanθ = (√3 × √3)/√3
⇒ tanθ = √3
⇒ tanθ = tan60°
∴ θ = 60°

So the elevation of the sun is 60°.

৩৯০.

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

ক
6 cm
খ
3.5 cm
গ
4 cm
ঘ
3 cm

ব্যাখ্যা

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

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

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

৩৯১.

In a right triangle, the length of one of the legs is 5 and the length of the hypotenuse is 13. What is the length of the other leg?

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

ব্যাখ্যা

Question: In a right triangle, the length of one of the legs is 5 cm and the length of the hypotenuse is 13cm . What is the length of the other leg?

Solution:
সমকোণী ত্রিভুজের অতিভুজ = 13
সমকোণ সংলগ্ন এক বাহু = 5
সমকোণ সংলগ্ন অপর বাহু = a

প্রশ্নমতে,
a² + 5² = 13²
⇒ a² + 25 = 169
⇒ a² = 169 - 25
⇒ a² = 144
⇒ a² = 12²
∴ a = 12

∴ সমকোণ সংলগ্ন অপর বাহু = 12 সে.মি.

৩৯২.

Which set of three sides cannot form a triangle?

ক
7cm, 10cm, 12cm
খ
6cm, 9cm, 16cm
গ
5cm, 12cm, 13cm
ঘ
8cm, 15cm, 20cm

ব্যাখ্যা

Question: Which set of three sides cannot form a triangle?

Solution:
আমরা জানি,
ত্রিভুজের যেকোনো দুই বাহুর সমষ্টি তৃতীয় বাহু অপেক্ষা বৃহত্তর হতে হবে।

এখানে, আমরা প্রত্যেকটি ত্রিভুজের ক্ষুদ্রতম দুইটি বাহুর যোগফলকে তৃতীয় (বৃহত্তম) বাহুর সাথে তুলনা করে পাই:

ক) 7 + 10 = 17 > 12; ∴ ত্রিভুজ আঁকা সম্ভব।
খ) 6 + 9 = 15 < 16; ∴ ত্রিভুজ আঁকা সম্ভব নয়।
গ) 5 + 12 = 17 > 13; ∴ ত্রিভুজ আঁকা সম্ভব।
ঘ) 8 + 15 = 23 > 20; ∴ ত্রিভুজ আঁকা সম্ভব।

৩৯৩.

If the radius of a circle is increased by 100%, by what % is the area of the circle increased?

ক
300%
খ
100%
গ
200%
ঘ
250%

ব্যাখ্যা

Question: If the radius of a circle is increased by 100%, by what % is the area of the circle increased?
(যদি একটি বৃত্তের ব্যাসার্ধ ১০০% বৃদ্ধি পায়, তবে বৃত্তের ক্ষেত্রফল কত শতাংশ বৃদ্ধি পাবে?)

Solution:
ধরা যাক,
বৃত্তের ব্যাসার্ধ, r = 10

∴ বৃত্তের ক্ষেত্রফল = πr²
= π (10)²
= 100π

আবার,
বৃত্তের ব্যাসার্ধ 100% বৃদ্ধিতে,
বৃত্তের নতুন ব্যাসার্ধ = 10 + 10 এর 100%
= 10 + 10 এর 100/100
= 20

∴ বৃত্তের নতুন ক্ষেত্রফল = πr²
= π (20)²
= 400π

∴ ক্ষেত্রফল বৃদ্ধি পায় = 400π - 100π
= 300π

100π থেকে ক্ষেত্রফল বৃদ্ধি পায় = 300π
1 থেকে ক্ষেত্রফল বৃদ্ধি পায় = 300π/100π
∴ 100 থেকে ক্ষেত্রফল বৃদ্ধি পায় = (300π × 100)/100π
= 300%

∴ বৃত্তের ক্ষেত্রফল শতকরা 300% বৃদ্ধি পায়।

৩৯৪.

What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Tk. 10 per square meter?

ক
ক) Tk. 3430
খ
খ) Tk. 3440
গ
গ) Tk. 3450
ঘ
ঘ) Tk. 3460

ব্যাখ্যা

In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So, 2(l+b) = 340
As we have to make 1 meter boundary around this,
so Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344
So required cost will be = 344 × 10 = 3440

৩৯৫.

The perimeter of a rectangular field is 104 meters. If the length of the field is 10 meters more than twice the width, what is the area of that field in square meters?

ক
530
খ
532
গ
580
ঘ
588
ঙ
None

ব্যাখ্যা

Question: The perimeter of a rectangular field is 104 meters. If the length of the field is 10 meters more than twice the width, what is the area of that field in square meters?

Solution:
Let,
The width of the rectangular field is x meter
∴ The length of the rectangular field is 2x + 10 meter

ATQ,
2(2x + 10 + x) = 104
⇒ 3x + 10 = 52
⇒ 3x = 42
∴ x = 14

∴ The area of that field is = (2x + 10) × x = (2 × 14 + 10) × 14 = (28 + 10) × 14 square meters
= 38 × 14 square meters
= 532 square meters

৩৯৬.

A rope makes 70 rounds of the circumference of a cylinder whose radius of the base is 14 cm. How many times can it go round a cylinder with radius 20 cm?

ক
ক) 40
খ
খ) 49
গ
গ) 70
ঘ
ঘ) 100
ঙ
ঙ) None of these

ব্যাখ্যা

Let the required number of rounds be x
More radius, Less rounds (Indirect proportion)
∴ 20:14::70:x
⇔ (20×x) = (14×70)
⇔ x = (14×70)/20
⇔ x = 49

৩৯৭.

tanA = 3/4 হলে sinA এর মান কত?

ক
1/4
খ
3/5
গ
2/5
ঘ
4/5
ঙ
কোনটিই নয়

ব্যাখ্যা

প্রশ্ন: tanA = 3/4 হলে sinA এর মান কত?

সমাধান:
sec²A = 1 + tan²A
⇒ sec²A = 1 + (3/4)²
⇒ sec²A = 1 + (9/16)
⇒ sec²A = 25/16
⇒ secA = 5/4
⇒ 1/cosA = 5/4
∴ cosA = 4/5

আমরা জানি,
sin²A = 1 - cos²A
= 1 - (4/5)²
= 1 - (16/25)
= (25 - 16)/25
∴ sin²A = 9/25
∴ sinA = 3/5

৩৯৮.

A, B and C are three points on the circle. If AB = AC = 7√2 cm and ∠BAC = 90° then the radius is equal to:

ক
7 cm
খ
7√2 cm
গ
14 cm
ঘ
14√2 cm
ঙ
None of these

ব্যাখ্যা

Question: A, B and C are three points on the circle. If AB = AC = 7√2 cm and ∠BAC = 90° then the radius is equal to:

Solution:

Radius = hypotenuse/2
= 14/2
= 7

৩৯৯.

A piece of wire 91 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?

ক
ক) 14 cm
খ
খ) 18 cm
গ
গ) 21 cm
ঘ
ঘ) 24 cm

ব্যাখ্যা

Question: A piece of wire 91 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, then what is the length of the base?

Solution:
Given,
Ratio of one of the equal sides to the base is 5 : 3
Therefore, the sides are 5x, 3x, 5x.

91 cm piece of wire is bent to form an isosceles triangle.
Thus perimeter of triangle is 91 cm.

ATQ,
∴ 13x = 91
⇒ x = 7
Thus the length of the base = 3 × 7 = 21 cm.

৪০০.

If sec²θ + tan²θ = 7, then the value of θ when 0° ≤ θ ≤ 90° is?

ক
30°
খ
45°
গ
60°
ঘ
90°

ব্যাখ্যা

Question: If sec²θ + tan²θ = 7, then the value of θ when 0° ≤ θ ≤ 90° is?

Solution:
sec²θ + tan²θ = 7
⇒ 1 + tan²θ + tan²θ = 7 - 1
⇒ 2tan²θ = 6
⇒ tan²θ = 3
⇒ tanθ = √3
⇒ tanθ = tan60°
∴ θ = 60°