Geometry: Mensuration, Trigonometry

PrepBank · পাতা ১৬ / ২১ · ১,৫০১–১,৬০০ / ২,০৮৫

১,৫০১.

A square is inscribed in a circle of diameter 2a and another square is a 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

Area or outer space = 2a x 2a = 4a²
From ABCD rectangle, In ΔBAD,
BD² = (AD)² + (AB)²
⇒ (2a)²= x² + x²
⇒ 4a²= 2x²
⇒ 2a²= x²
∴ x = √2a

Area of inner square = √2a × √2a = 2a²

∴ The difference between the areas of outer and inner squares is = 4a² - 2a² = 2a²

১,৫০২.

The length of a rectangle is thrice its breath, and its perimeter is 120 meters. What is its area?

ক
688 sq. m.
খ
720 sq. m.
গ
580 sq. m.
ঘ
None of these

ব্যাখ্যা

Question: The length of a rectangle is thrice its breath, and its perimeter is 120 meters. What is its area?

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

Perimeter of a rectangle = 2 (Length + Breadth)
So, 2(3x + x) = 120
⇒ 6x + 2x = 120
⇒ 8x = 120
∴ x = 120/8 = 15

Now, Breadth = 15
so, length = 15 × 3 = 45

So, its area = Length × Breadth = 45 × 15 = 675 sq. m.

১,৫০৩.

How much will it cost to fence in a field that is 3000 cm long and 600 cm wide with fence that costs Tk. 50 a meter?

ক
ক) 720 Tk.
খ
খ) 1800 Tk.
গ
গ) 9000 Tk.
ঘ
ঘ) 3600 Tk.

ব্যাখ্যা

Question: How much will it cost to fence in a field that is 3000 cm long and 600 cm wide with fence that costs Tk. 50 a meter?

Solution:
The perimeter of the field is 2(3000 + 600) cm
= 2 × 3600 cm
= 7200 cm
= 72 m

Total cost = 72 × 50 = 3600 Tk.

১,৫০৪.

If tan 6A = √3, then find A.

ক
15°
খ
10°
গ
25°
ঘ
30°

ব্যাখ্যা

Question: If tan 6A = √3, then find A.

Solution:
tan 6A = √3
⇒ tan 6A = tan60°
⇒ 6A = 60°
⇒ A = 60°/6
∴ A = 10°

১,৫০৫.

A cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is.

ক
49 m²
খ
36 m²
গ
42 m²
ঘ
48 m²

ব্যাখ্যা

Question: A cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is.

Solution:
The upper part of the calf is open. Here the area of the wetted part is asked.
Therefore, the upper opening has to be excluded from the total area.
Here,
Length = 6 m
Width = 4 m
Height = 1 m 25 cm = 1.25 m

So, the area of total wet surface = 2{(6 × 4) + (4 × 1.25) + (6 × 1.25)} - (6 × 4)
= (2 × 36.5) - 24
= 73 - 24
= 49 m²

১,৫০৬.

A line intersecting a circle in two points is called a _______.

ক
ক) Secant
খ
খ) Chord
গ
গ) Diameter
ঘ
ঘ) Tangent
ঙ
ঙ) None of these

ব্যাখ্যা

A line intersecting a circle in two points is called a Secant.

১,৫০৭.

If α and β are positive acute angles, sin(4α - β) = 1 and cos(2α + β) = 1/2, then the value of (α + 2β) is?

ক
40°
খ
55°
গ
45°
ঘ
50°

ব্যাখ্যা

Question: If α and β are positive acute angles, sin(4α - β) = 1 and cos(2α + β) = 1/2, then the value of (α + 2β) is?

Solution:
Given,
sin(4α - β) = 1
⇒ sin(4α - β) = sin 90°
⇒ 4α - β = 90° ..... (1)

again, cos(2α + β) = 1/2
⇒ cos(2α + β) = cos 60°
⇒ 2α + β = 60° .... (2)

(1) + (2) ⇒ 4α - β + 2α + β = 90° + 60°
⇒ 6α = 150°
⇒ α = 25°
From (2), β = 60° - (2 × 25°) = 10°
∴ (α + 2β) = 25° + (2 × 10°) = 45°

১,৫০৮.

If y = sin(sinx) then what is the value of dy/dx?

ক
cos(sinx)
খ
cosx.cos(cosx)
গ
cos2x
ঘ
cosx.cos(sinx)

ব্যাখ্যা

Question: If y = sin(sinx) then what is the value of dy/dx?

Solution:

১,৫০৯.

A median of a triangle divides it into two ___

ক
ক) congruent triangles
খ
খ) triangles of equal area
গ
গ) isosceles triangles
ঘ
ঘ) right triangles

ব্যাখ্যা

ত্রিভুজের যেকোন মধ্যমা ত্রিভুজটিকে সমান ক্ষেত্রফলবিশিষ্ট দুটি ত্রিভুজক্ষেত্রে বিভক্ত করে।

১,৫১০.

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

ক
9 ft
খ
16 ft
গ
24 ft
ঘ
26 ft

ব্যাখ্যা

Question: The area of a rectangle R with width 4 feet is equal to the area of a square S, which has a perimeter of 24 feet. The perimeter of the rectangle R is -

Solution:
ধরি,
চতুর্ভুজ, R এর দৈর্ঘ্য এবং প্রস্থ যথাক্রমে l, b.
বর্গের এক বাহু = a

প্রশ্নমতে,
4a = 24
a = 6

∴ চতুর্ভুজের ক্ষেত্রফল = বর্গের ক্ষেত্রফল
l × b = a²
l = a²/b
= 36/4
= 9

∴ চতুর্ভুজের পরিসীমা = 2(l + b)
= 2(9 + 4)
= 26 feet

১,৫১১.

The length of a garden is thrice its breadth. A playground measuring 180 sq. m occupies (1/15)th of the total area of the garden. The length of the garden is-

ক
81 m
খ
90 m
গ
120 m
ঘ
112 m

ব্যাখ্যা

Question: The length of a garden is thrice its breadth. A playground measuring 180 sq. m occupies (1/15)th of the total area of the garden. The length of the garden is

Solution:
Let L be the length and B is the breadth of the garden.
We have L = 3B.
Total area of the garden = 180 × 15 = 2700 sq m.
⇒ LB = 2700
⇒ 3B² = 2700
⇒ B² = 900
⇒ B = 30m

Hence the length of the garden = 30 × 3 = 90 m

১,৫১২.

sin²θ + sinθ = xcosθ - cos²θ, then what is the value of 2tanθ?

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

ব্যাখ্যা

Question: sin²θ + sinθ = xcosθ - cos²θ, then what is the value of 2tanθ?

Solution:
sin²θ + sinθ = xcosθ - cos²θ
⇒ sin²θ + cos²θ + sinθ = xcosθ
⇒ 1 + sinθ = xcosθ
⇒ (1 + sinθ)/cosθ = x
⇒ secθ + tanθ = x.....(1)

sec²θ - tan²θ = 1
⇒ (secθ + tanθ) (secθ - tanθ) = 1
⇒ x (secθ - tanθ) = 1
∴ (secθ - tanθ) = 1/x.......(2)

(1) - (2),
secθ + tanθ - secθ + tanθ = x - (1/x)
⇒ 2tanθ = (x² - 1)/x

১,৫১৩.

A person rides a bicycle around a circular path of radius 50m. The radius of the bicycle wheel of the bicycle is 50 cm. The cycle comes to the starting point for the first time in 1 hour. What is the number of revolutions of the wheel in 15 minutes?

ক
ক) 20
খ
খ) 25
গ
গ) 30
ঘ
ঘ) 35

ব্যাখ্যা

দেওয়া আছে,
বৃত্তকার মাঠের ব্যাসার্ধ r₁ = 50 m
বৃত্তকার মাঠের পরিধি, 2πr₁ = 2 × π × 50
= 100π m.
সাইকেলের চাকার ব্যাসার্ধ, r₂ = 50 cm = 0.5 m.
∴ সাইকেলের চাকার পরিধি, 2πr₂ = 2 × π × 0.5
= π m.
এখন চাকাটি π মিটার গেলে ঘুরবে = 1 বার।
∴ চাকাটি 100π গেলে ঘুরবে = 1 × 100×π
= 100 বার।
এখন, 60 মিনিটে চাকাটি ঘুরবে = 100 বার।
∴ 15 মিনিটে চাকাটি ঘুরবে = (100 × 15)/60
= 25 বার।

১,৫১৪.

The floor of a room is of size 5 m × 4m and its height is 3m. The walls and ceiling of the room require painting. What is the area to be painted?

ক
ক) 65 m²
খ
খ) 74 m²
গ
গ) 63 m²
ঘ
ঘ) 69 m²

ব্যাখ্যা

Area of Wall = ( 5 + 4 + 5 + 4 ) m. wall length × 3 m height
= 18 × 3 sq.m.
= 54 sq.m.

Area of Ceiling = 5 m × 4m
= 20 sq.m.

hence total painting area of walls and ceiling = 54 sq m + 20 sq m = 74 sq m
74 square meter area of walls and ceiling to be painted.

১,৫১৫.

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

ক
mn – 2p(m+n)
খ
mn – p²
গ
mn - p(m+n)
ঘ
(m - 2p)(n - 2p)
ঙ
(m - p)(n - p)

ব্যাখ্যা

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

Solution:

The area of the floor is mn
Length of rug = m - 2p (Distance of p on each side)
Width of rug = n - 2p (Distance of p on each side)

Thus area of rug = (m-2p)(n-2p)

১,৫১৬.

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

ক
ক) 30 cm
খ
খ) 35 cm
গ
গ) 60 cm
ঘ
ঘ) 70 cm

ব্যাখ্যা

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

Solution:
Let r be the radius of circle

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

The radius of the circle is 35 cm.
∴ Diameter = 2 × 35
= 70 cm

১,৫১৭.

If (sinθ + cosθ)/(sinθ - cosθ) = 7 then, secθ =?

ক
±1/3
খ
±2/3
গ
±5/3
ঘ
±5/2

ব্যাখ্যা

প্রশ্ন: If (sinθ + cosθ)/(sinθ - cosθ) = 7 then, secθ =?

সমাধান:
দেয়া আছে,
(Sinθ + Cosθ)/(Sinθ - Cosθ) = 7
⇒ (Sinθ + Cosθ) + (Sinθ - Cosθ)/(Sinθ + Cosθ) - (Sinθ - Cosθ) = (7 + 1)/(7 - 1)
⇒ (Sinθ + Cosθ + Sinθ - Cosθ)/(Sinθ + Cosθ - Sinθ + Cosθ) = 8/6
⇒ 2Sinθ/2Cosθ = 4/3
⇒ Sinθ/Cosθ = 4/3
⇒ tanθ = 4/3
⇒ tan²θ = 16/9
⇒ Sec²θ - 1 = 16/9
⇒ Sec²θ = (16/9) + 1
⇒ Sec²θ = (16 + 9)/9
⇒ Sec²θ = 25/9
⇒ Sec²θ = (16 + 9)/9
⇒ Sec²θ = 25/9
Secθ = ±5/3

১,৫১৮.

A tree casts a shadow 15√3 meters long when the sun's angle of elevation is 60°. Find the height of the tree.

ক
30 m
খ
45 m
গ
60 m
ঘ
40 m

ব্যাখ্যা

Question: A tree casts a shadow 15√3 meters long when the sun's angle of elevation is 60°. Find the height of the tree.
(সূর্যের উন্নতি কোণ 60° হলে একটি গাছের ছায়ার দৈর্ঘ্য 15√3 মিটার হয়। গাছটির উচ্চতা কত?)

Solution:

ধরি,
উচ্চতা = h
tan60° = h/15√3
⇒ √3 = h/15√3
⇒ h = 3 × 15
∴ h = 45

১,৫১৯.

Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9), and D(5, 4). What is the shape of the quadrilateral?

ক
Square
খ
Parallelogram but not a rhombus
গ
Rectangle but not a square
ঘ
Rhombus

ব্যাখ্যা

Question: Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9), and D(5, 4). What is the shape of the quadrilateral?

Solution:

∴ The shape of the quadrilateral is Rhombus.

১,৫২০.

One of the four angles of a rhombus is 90 degrees. If each side of the rhombus is 20 cm, what will be the length of the longer diagonal?

ক
20√2 cm
খ
25√2 cm
গ
30√2 cm
ঘ
35√2 cm
ঙ
None of these

ব্যাখ্যা

Question: One of the four angles of a rhombus is 90 degrees. If each side of the rhombus is 20 cm, what will be the length of the longer diagonal?

Solution:
A rhombus with one of its angle 90 degrees is a square. So, both the diagonals are of equal length.

Now, Diagonal of a square= √2 × side
Given Side = 20 cm
So, Diagonal = 20√2 cm

১,৫২১.

A hall, 20m long and 15m broad, is surrounded by a verandah of uniform width of 2.5m. The cost of flooring the verandah at Tk. 3.50 per square meter is

ক
ক) Tk. 500
খ
খ) Tk. 700
গ
গ) Tk. 600
ঘ
ঘ) Tk. 800

ব্যাখ্যা

Question: A hall, 20m long and 15m broad, is surrounded by a verandah of uniform width of 2.5m. The cost of flooring the verandah at Tk. 3.50 per square meter is-

Solution:
হলের দৈর্ঘ্য = 20 মিটার
হলের প্রস্থ = 15 মিটার
হলের ক্ষেত্রফল = 300 বর্গমিটার

বারান্দাসহ হলের দৈর্ঘ্য = 20 + (2 × 2.5) মিটার
= 25 মিটার
বারান্দাসহ হলের প্রস্থ = 15 + (2 × 2.5) মিটার
= 20 মিটার

বারান্দাসহ হলের ক্ষেত্রফল = 500 বর্গমিটার

বারান্দার ক্ষেত্রফল = (500 - 300) বর্গমিটার
= 200 বর্গমিটার

1 বর্গমিটারে খরচ = 3.5 টাকা
200 বর্গমিটারে খরচ = 3.5 × 200 = 700 টাকা

১,৫২২.

The difference between the radii of the bigger circle and the smaller circle is 14 cm and the difference between their areas is 1,056 cm2. The radius of the smaller circle is -

ক
ক) 7 cm
খ
খ) 5 cm
গ
গ) 9 cm
ঘ
ঘ) 3 cm

ব্যাখ্যা

ধরি,
ক্ষুদ্রতর বৃত্তের ব্যাসার্ধ r cm
∴ বৃহত্তর বৃত্তের ব্যাসার্ধ = (r + 14) cm
প্রশ্নমতে, Π(r + 14)²-Πr² = 1056
Π{r² + 2b × r × 14 + (14)² - r²} = 1056
r² + 28r + 196 - r² = 1056/Π
28r + 196 = 1056/(22/7)
= 1056 × (7/22)
= 336
28r = 336 - 196
= 140
∴ r = 140/28
= 5
অতএব, ক্ষুদ্রতর বৃত্তের ব্যাসার্ধ 5 cm.

১,৫২৩.

In the figure, if AB = 8, BC = 6, AC = 10 and CD = 9, then AD = ?

ক
ক) 12
খ
খ) 11
গ
গ) 13
ঘ
ঘ) 17

ব্যাখ্যা

From the measure of various sites of triangle ABC, we can see that it is a right angle triangle whose right angle is at Angle B.
Applying Pythagoras theorem to triangle ABD: we can find the measure of AD.

AD²=AB² + BD²
AD²= 8 × 8 + 15 × 15
AD = 17,
hence option D is the correct one.

১,৫২৪.

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

ক
ক) 30°
খ
খ) 45°
গ
গ) 60°
ঘ
ঘ) 90°

ব্যাখ্যা

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

Solution:

Let,
AB = height of tree
BC= Shadow of tree
angle of elevation = C
∴ BC = √3 AB

We know,
tanC = AB/BC
⇒ tanC = AB/√3AB
⇒ tanC = 1/√3
⇒ tanC = tan30°
∴ C = 30°

১,৫২৫.

If tanθ = 1, then the the value of

ক
3/2
খ
9/4
গ
1
ঘ
9/2
ঙ
2

ব্যাখ্যা

Question: If tanθ = 1, then the the value of

Solution:
Given,
tanθ = 1
⇒ tanθ = tan45°
∴ θ = 45°

Now,

১,৫২৬.

Which one is the complementary angle of 90°?

ক
ক) 0°
খ
খ) 180°
গ
গ) 270°
ঘ
ঘ) 90°

ব্যাখ্যা

Question: Which one is the complementary angle of 90°?

Solution:
দুটি কোণের সমষ্টি এক সমকোণ হলে, একটিকে অপরটির পূরক কোণ বলে।

৯০° এর পূরক কোণ = ৯০° - ৯০°
= ০°

১,৫২৭.

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

Question: 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 -

Solution:
ধরি,
চতুর্ভুজ, R এর দৈর্ঘ্য এবং প্রস্থ যথাক্রমে l, b.
বর্গের এক বাহু = a

প্রশ্নমতে,
4a = 24
a = 6

∴ চতুর্ভুজের ক্ষেত্রফল = বর্গের ক্ষেত্রফল
l × b = a²
l = a²/b
= 36/4
= 9

∴ চতুর্ভুজের পরিসীমা = 2(l + b)
= 2(9 + 4)
= 26 feet

১,৫২৮.

In the figure above, (a + b + c)/(s + t + u) = ?

ক
1/2
খ
1/3
গ
2/3
ঘ
1

ব্যাখ্যা

Question:

In the figure above, (a + b + c)/(s + t + u) = ?

Solution:
Here,
u = a + b
t = a + c
s = b + c

∴ s + t + u = b + c + a + c + a + b = 2(a + b + c)

Here,
(a + b + c)/(s + t + u)
= (a + b + c)/{2(a + b + c)}
= 1/2

১,৫২৯.

A square field is surrounded by a path of uniform width 5 meters. If the area of the path is 220 square meters, find the side length of the field.

ক
4 meters
খ
5 meters
গ
6 meters
ঘ
8 meters

ব্যাখ্যা

Question: A square field is surrounded by a path of uniform width 5 meters. If the area of the path is 220 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 × 5)
= x + 10 meters.

Area of path = Area of field with path - Area of field
⇒ 220 = (x + 10)² - x²
⇒ 220 = x² + 20x + 100 - x²
⇒ 220 = 20x + 100
⇒ 20x = 220 - 100
⇒ 20x = 120
⇒ x = 120/20
⇒ x = 6 meters

∴ Therefore, the side length of the field is 6 meters.

১,৫৩০.

The value of tan60° is:

ক
√3
খ
√3/2
গ
1
ঘ
1/2

ব্যাখ্যা

Question: The value of tan60° is:
(Officer General 2022 অনুযায়ী)

Solution:
cos60° = 1/2
sin60° = √3/2

Hence, tan60°
= sin60°/cos60°
= (√3/2)/(1/2)
= √3

১,৫৩১.

The external and internal diameters of a hemispherical bowl are 10 cm and 8 cm respectively. What is the total surface area of the bowl?

ক
284 cm²
খ
286 cm²
গ
274 cm²
ঘ
296 cm²

ব্যাখ্যা

Question: The external and internal diameters of a hemispherical bowl are 10 cm and 8 cm respectively. What is the total surface area of the bowl?

Solution:

here,
R = 5cm
r = 4cm

total surface area:
S = 2πR² + 2πr² + π(R² - r²)
= 2πR² + 2πr² + πR² - πr²
= 3πR² + πr²
= π {3 × (5)² + (4)²}
= (22/7) (75 + 16)
= (22/7) × 91
= 286 cm²

১,৫৩২.

asinθ = 1, acosθ = √3 then the value of √3tanθ - 1 = ?

ক
0
খ
1
গ
2
ঘ
√3 - 1

ব্যাখ্যা

Question: asinθ = 1, acosθ = √3 then the value of √3tanθ - 1 = ?

Solution:
asinθ = 1
acosθ = √3

Now,
asinθ/acosθ = 1/√3
⇒ tanθ = 1/√3
⇒ √3tanθ = 1
∴ √3tanθ - 1 = 0

১,৫৩৩.

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
ক) 8 centimetre
খ
খ) 7 centimetre
গ
গ) 12 centimetre
ঘ
ঘ) 6 centimetre

ব্যাখ্যা

Let, base = x
Then, height = 2x
Area = base × height
= x × 2x
= 2x²
Area is given as 72 cm²2x² = 72 cm²
⇒ x² = 36 cm²
⇒ x = 6 cm
Hence, the length of the base is 6 cm.

১,৫৩৪.

In triangle ABC, if AB = BC and ∠B = 90, ∠ A will be :

ক
ক) 40°
খ
খ) 45°
গ
গ) 110°
ঘ
ঘ) 130°

ব্যাখ্যা

Since, AB = BC
∠A = ∠C
let ∠A = x
so, ∠C is also equal to x
Now, ∠A + ∠B + ∠C = 180°
Or, 2x + 90 = 180
Or, x = (180-90)/2
Or, x = 45°

১,৫৩৫.

What is the angle between the hour hand and minute hand at 1 : 20 pm?

ক
ক) 80°
খ
খ) 90°
গ
গ) 120°
ঘ
ঘ) 45°

ব্যাখ্যা

Question: What is the angle between the hour hand and minute hand at 1 : 20 pm?

Solution:
কোণ = |১১ × মিনিট - ৬০ × ঘণ্টা|°/২
= |১১ × ২০ - ৬০ × ১|°/২
= |২২০ - ৬০|°/২
= ১৬০°/২
= ৮০°

১,৫৩৬.

If a ladder touches the roof of a wall and makes an angle of 30° with the 15 metre long wall, then the length of the ladder is-

ক
30 m
খ
7.5 m
গ
15 m
ঘ
17.3 m

ব্যাখ্যা

Question: If a ladder touches the roof of a wall and makes an angle of 30° with the 15 metre long wall, then the length of the ladder is-

Solution:

দেয়ালের উচ্চতা, h = 15 মিটার
দেয়ালের সাথে মইয়ের কোণ, θ = 30°
মইয়ের দৈর্ঘ্য, L = ?

আমরা জানি,
⇒ cos⁡θ = দেয়ালের উচ্চতা/মইয়ের দৈর্ঘ্য
⇒ cos30° = 15/L
⇒ √3/2 = 15/L
⇒ L = 30/√3
∴ L = 17.3 মিটার

১,৫৩৭.

A metallic cone of radius 18 cm and height 25 cm is melted and made into spheres of radius 3 cm each. How many spheres are there?

ক
ক) 36
খ
খ) 50
গ
গ) 75
ঘ
ঘ) 125

ব্যাখ্যা

Question: A metallic cone of radius 18 cm and height 25 cm is melted and made into spheres of radius 3 cm each. How many spheres are there?

Solution:
We know that,
Volume of the cone = (1/3)π × r² × h
and volume of the sphere = (4/3)π × r³

Here,
Radius of cone, r₁ = 18 cm
Height of cone, h = 25 cm
Radius of sphere, r₂ = 3 cm

Number of spheres = Volume of the cone/Volume of the sphere
= {(1/3)π × (r₁)² × h}/{(4/3)π × (r₂)³}
= {(1/3)π × 18 × 18 × 25}/{(4/3)π × 3 × 3 × 3}
= 75

১,৫৩৮.

∠X and ∠Y are complementary to each other. If ∠X = 25° + 2y and ∠Y = 3y, find the value of ∠Y.

ক
25°
খ
35°
গ
39°
ঘ
45°

ব্যাখ্যা

Question: ∠X and ∠Y are complementary to each other. If ∠X = 25° + 2y and ∠Y = 3y, find the value of ∠Y.

Solution:
Here,
∠X = 25° + 2y and ∠Y = 3y
For complementary angles,
∠X + ∠Y = 90°
⇒ (25° + 2y) + 3y = 90°
⇒ 25° + 5y = 90°
⇒ 5y = 65°
∴ y = 13°

So, ∠Y = 3 × 13° = 39°

১,৫৩৯.

If θ is a positive angle and 9sin²θ - 9 = 0, then the value of tan(θ - 30°) is equal to?

ক
1
খ
√3
গ
1/√3
ঘ
0

ব্যাখ্যা

Question: If θ is a positive angle and 9sin²θ - 9 = 0, then the value of tan(θ - 30°) is equal to?

Solution:
Given,
9sin²θ - 9 = 0
⇒ 9sin²θ = 9
⇒ sin²θ = 1
⇒ sinθ = 1
⇒ sinθ = sin90°
∴ θ = 90°

Now,
tan(θ - 30°) = tan(90° - 30°)
= tan60°
= √3

১,৫৪০.

The difference between the length and the perimeter of a rectangle is 100 cm. What is the breadth of the rectangle?

ক
80 cm
খ
60 cm
গ
100 cm
ঘ
Data Inadequate

ব্যাখ্যা

Question: The difference between the length and the perimeter of a rectangle is 100 cm. What is the breadth of the rectangle?

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

According to the question:
2(x + y) - x = 100
⇒ 2x + 2y - x = 100
⇒ x + 2y = 100

From this we cannot find 'y' (breadth), so the given data is inadequate.

১,৫৪১.

If sinA = cos2A, A =?

ক
30°
খ
40°
গ
50°
ঘ
60°

ব্যাখ্যা

প্রশ্ন: If sinA = cos2A, A =?

সমাধান:
sinA = cos2A
⇒ sinA = 1 - 2 sin²A
⇒ 2 sin²A + sinA - 1 = 0
⇒ 2 sin²A + 2 sinA - sinA -1 = 0
⇒ 2 sinA ( sinA + 1) -1 (sinA + 1) = 0
⇒ (sinA + 1) (2 sinA -1) = 0

হয়, sinA + 1 = 0
⇒ sinA = -1
⇒ A = 270°

অথবা 2 sinA - 1 = 0
⇒ sinA = 1/2
⇒ A = 30°

১,৫৪২.

A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is:

ক
50m
খ
100m
গ
125m
ঘ
200m

ব্যাখ্যা

Question: A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, then the altitude of the triangle is:

Solution:
In this question, we will denote ‘b’ as the base h1 and h2 as the altitudes of the triangle and parallelogram respectively.

Then, according to the data in the question:

½ × b × h1 = b × h2
⇒ h1 = 2 h2
⇒ h1 = 2 × 100 = 200m

∴ The altitude of the triangle is 200m.

১,৫৪৩.

In the figure ABCD is a square. If the length of the square is 10ft, then what will be the area of the triangle OCD?

ক
ক) 12.5 sft
খ
খ) 25 sft
গ
গ) 37.22 sft
ঘ
ঘ) Cannot be determined

ব্যাখ্যা

Question: In the figure ABCD is a square. If the length of the square is 10ft, then what will be the area of the triangle OCD?

Solution:

length of square 10 ft
area of square = 10² sq.ft
= 100 sq.ft

area of triangle OCD = (1/4) × 100 sq.ft
= 25 sq.ft

১,৫৪৪.

The volume of a rectangle with length, breadth, and height as 5x, 3x² and 7x⁴

ক
ক) 105x⁷
খ
খ) 105x²
গ
গ) 105x⁴
ঘ
ঘ) 105x

ব্যাখ্যা

The volume of the rectangle = 5x × 3x² × 7x⁴
= 105x^{(1 + 2 + 4)} = 105x⁷

১,৫৪৫.

The difference between the circumference and the radius of a circle is 111 cm. Find the radius of a circle is -

ক
ক) 14 cm
খ
খ) 21 cm
গ
গ) 28 cm
ঘ
ঘ) 35 cm

ব্যাখ্যা

Let r be the radius of circle

Given that,
2πr - r = 111
⇒ r(2π - 1) = 111
⇒ r{(44/7) - 1} = 111
⇒ r (44 - 7)/7 }=111
⇒ r(37/7) = 111
⇒ r = 111 (7/37)
r = 21

১,৫৪৬.

A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 8 cm. How many bottles will be needed to empty the bowl?

ক
ক) 20
খ
খ) 27
গ
গ) 38
ঘ
ঘ) 54

ব্যাখ্যা

Question: A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 8 cm. How many bottles will be needed to empty the bowl?

Solution:
অর্ধগোলকের আয়তন = (1/2)× 4πr³/3
= (2/3) π9³ ঘনসেমি

প্রতি সিলিন্ডার আকৃতির বোতলের আয়তন = π (3/2)² × 8
= 18π ঘনসেমি
ধরি, n সংখ্যক বোতল লাগবে।

n × 18π = (2/3) π9³
⇒ n = (2/3) π9³/18π
∴ n = 27

১,৫৪৭.

The radius of circle A is r, and the radius of circle B is 3r/4 . What is the ratio of the area of circle A to the area of circle B ?

ক
ক) 16 to 9
খ
খ) 9 to 16
গ
গ) 1 to 4
ঘ
ঘ) 4 to 3

ব্যাখ্যা

Question: The radius of circle A is r, and the radius of circle B is 3r/4 . What is the ratio of the area of circle A to the area of circle B ?

Solution:
A বৃত্তের ব্যাসার্ধ = r
A বৃত্তের ক্ষেত্রফল = πr²

B বৃত্তের ব্যাসার্ধ = 3r/4
B বৃত্তের ক্ষেত্রফল = π(3r/4)² = 9π²/16

A বৃত্তের ক্ষেত্রফল : B বৃত্তের ক্ষেত্রফল = πr² : 9π²/16
= 1 : 9/16
= 16 × 1 : (9/16)16
= 16 : 9

১,৫৪৮.

In the given figure, AB is the diameter of the circle with center O. If ∠BOD = 15° & ∠EOA = 85°, then find the value of ∠ECA.

ক
20°
খ
25°
গ
35°
ঘ
Can’t be determined

ব্যাখ্যা

Question: In the given figure, AB is the diameter of the circle with center O. If ∠BOD = 15° & ∠EOA = 85°, then find the value of ∠ECA.

Solution:
∠EOA = 85°, ∠BOD = 15°
∠EOD = 180° - (85° + 15°) = 80°
In ΔOED,
OE = OD (radii)
∠OED = ∠ODE = 50°

In ΔOEC,
∠EOC = 80°+15° = 95°, ∠OEC =50°
∴ ∠ECA = 180°- (95 + 50°) = 35°

১,৫৪৯.

sin²60° + 2tan45° - cos²30° - 2 =?

ক
ক) 1
খ
খ) - 2
গ
গ) 0
ঘ
ঘ) 2

ব্যাখ্যা

Question: sin²60° + 2tan45° - cos²30° - 2 =?

Solution:
sin²60° + 2tan45° - cos²30° - 2
= (sin60°)² + 2 × 1 - (cos30°)² - 2
= (√3/2)² + 2 - (√3/2)² - 2
= 0

১,৫৫০.

A triangular plot with sides of 25 feet, 40 feet and 55 feet is to be surrounded by a fence built on pillars set 5 feet apart. How many pillars will be required to surround the plot?

ক
ক) 21
খ
খ) 22
গ
গ) 23
ঘ
ঘ) 24

ব্যাখ্যা

Question: A triangular plot with sides of 25 feet, 40 feet and 55 feet is to be surrounded by a fence built on pillars set 5 feet apart. How many pillars will be required to surround the plot?

Solution:
ত্রিভুজ ক্ষেত্রটির পরিসীমা
= (25 + 40 + 55) ফুট
= 120 ফুট

একটি পিলার থেকে অন্য পিলারের দূরত্ব = 5 মিটার

মোট পিলার লাগবে
= (120/5) টি
= 24 টি

১,৫৫১.

If tan3A = √3, then A = ?

ক
20°
খ
30°
গ
45°
ঘ
60°

ব্যাখ্যা

Question: If tan3A = √3, then A = ?

Solution:
tan3A = √3
⇒ tan3A = tan60°
⇒ 3A = 60°
⇒ A = 60°/3
∴ A = 20°

১,৫৫২.

A copper sphere of radius 3 cm is beaten and drawn into a wire of diameter 0.2 cm. The length of the wire is-

ক
36 m
খ
18 m
গ
12 m
ঘ
9 m

ব্যাখ্যা

Question: A copper sphere of radius 3 cm is beaten and drawn into a wire of diameter 0.2 cm. The length of the wire is-

Solution:
ব্যাসার্ধ, r = 3 সেমি
গোলকটির আয়তন = (4/3) × π × r³
= (4/3) × π × 3³
= 36π

তারটি ব্যাসার্ধ = 0.2/2 = 0.1 সেমি
তারটির আয়তন = πr²l
= π × (0.1)² × l
= 0.01πl

শর্তমতে,
0.01πl = 36π
⇒ l = 36/0.01
⇒ l = 3600 cm
⇒ l = 36 m

১,৫৫৩.

If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one?

ক
1 : 8
খ
1 : 4
গ
1 : 2
ঘ
8 : 1

ব্যাখ্যা

Question: If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one?

Solution:
Let original radius = R
Then, new radius = R/2

Volume of reduced cylinder/Volume of original cylinder = π(R/2)²h/πR²h
= 1/4
= 1 : 4

১,৫৫৪.

Diagonal of a square is 16 inches long. What is the area of the square?

ক
256 square inch
খ
128 square inch
গ
64 square inch
ঘ
132 square inch

ব্যাখ্যা

If the length of the square is 'a', then diagonal = a√2
Therefore, a√2 = 16
⇒ a = 16/√2
⇒ a² = (16/√2)² = 128

১,৫৫৫.

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

ক
2520 m²
খ
2720 m²
গ
2250 m²
ঘ
2270 m²

ব্যাখ্যা

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

১,৫৫৬.

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

ক
5√3 m
খ
7√3 m
গ
9√3 m
ঘ
10√3 m

ব্যাখ্যা

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

Solution:

let, x is the height of the building.

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

১,৫৫৭.

The perimeter of the base of a cube is 48 cm. What is its volume?

ক
1728 cm³
খ
1232 cm³
গ
2744 cm³
ঘ
4096 cm³

ব্যাখ্যা

Question: The perimeter of the base of a cube is 48 cm. What is its volume?

Solution:
Let the side length of the cube be x.
So, 4x = 48
∴ x = 12 cm

Volume = (12)³
= 1728 cm³

১,৫৫৮.

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

ক
5
খ
5/2
গ
4√3
ঘ
3

ব্যাখ্যা

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

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

১,৫৫৯.

Salman converted a 8m length cube to 1m length cube. How many cubes can be formed?

ক
ক) 64
খ
খ) 256
গ
গ) 512
ঘ
ঘ) 1024

ব্যাখ্যা

Question: Salman converted a 8m length cube to 1m length cube. How many cubes can be formed?

Solution:
Here, the length of the big cube, L = 8m
so, the volume of the big cube is, V = 8³= 512m³

the length of the small cubes, l = 1m
so, the volume of one small cube, v = 1³ = 1m³

small cubes can be formed = 512/1 = 512

১,৫৬০.

A rectangular water reservoir contains 48000 liters of water. If the length of the reservoir is 6m and the breadth is 4m, the depth of the reservoir will be -

ক
ক) 1 m
খ
খ) 2 m
গ
গ) 3 m
ঘ
ঘ) 4 m

ব্যাখ্যা

Question: A rectangular water reservoir contains 48000 liters of water. If the length of the reservoir is 6m and the breadth is 4m, the depth of the reservoir will be -

Solution:
1 m³ = 1000 litre
⇒ 48000 litre = 48000/1000
= 48 m³

48 = 6 × 4 × depth
∴ depth = 48/24
= 2 m

১,৫৬১.

In a rectangle, the diagonal length is 13 and the width is 5. What is the perimeter of the rectangle?

ক
30
খ
34
গ
36
ঘ
48

ব্যাখ্যা

Question: In a rectangle, the diagonal length is 13 and the width is 5. What is the perimeter of the rectangle?

Solution:
Given that,
Diagonal of rectangle, d = 13
Width, w = 5

We know,
Pythagorean theorem, l² + w² = d²
⇒ 13² = l² + 5²
⇒ 169 = l² + 25
⇒ l² = 169 - 25
⇒ l² = 144 = 12²
∴ l = 12
∴ length In a rectangle is, l = 12

Perimeter of a rectangle, P = 2(l + w)
= 2(12 + 5)
= 2 × 17
= 34

So the perimeter of the rectangle is 34.

১,৫৬২.

If the volume of a sphere is 36π cm³, what is the surface area of the sphere?

ক
100π cm²
খ
144π cm²
গ
124π cm²
ঘ
36π cm²

ব্যাখ্যা

Question: If the volume of a sphere is 36π cm³, what is the surface area of the sphere?

Solution:
Given that the volume, V = 36π cm³
or, (4/3)πr³ = 36π
or, r³ = 27
∴ r = 3 cm

Surface area of a sphere, A = 4πr²
= 4π(3)²
= 36π cm²

১,৫৬৩.

A rectangular tank with a length of 8m and a width of 5m can store 60000 liters. What is the height of the tank?

ক
1.5 meters
খ
2 meters
গ
2.5 meters
ঘ
4 meters

ব্যাখ্যা

Question: A rectangular tank with a length of 8m and a width of 5m can store 60000 liters. What is the height of the tank?

Solution:
দেওয়া আছে, ট্যাংকের দৈর্ঘ্য (l) = 8m,
প্রস্থ (b) = 5 m, এবং
আয়তন (V) = 60000 লিটার।

ধরি, ট্যাংকটির উচ্চতা হল h মিটার।

আমরা জানি,
আয়তাকার ঘনবস্তুর আয়তন = দৈর্ঘ্য × প্রস্থ × উচ্চতা
= (8 × 5 × h) m³
= 40h m³

এখন, আমরা জানি, 1 m³ = 1000 লিটার।

প্রশ্নমতে,
40h × 1000 = 60000
⇒ 40h = 60000/1000
⇒ h = 60/40
∴ h = 1.5

সুতরাং, ট্যাংকটির উচ্চতা হলো 1.5 মিটার।

১,৫৬৪.

A frog is sitting on vertex A of a square ABCD. It starts jumping to the immediately adjacent vertex on either side in random fashion and stops when it reaches point C, in how many ways can it reach point C if it makes exactly 7 jumps?

ক
ক) 1
খ
খ) 5
গ
গ) 3
ঘ
ঘ) 0

ব্যাখ্যা

চিত্রে ABCD বর্গাকৃতির ক্ষেত্র যার A বিন্দুতে Frog রয়েছে।Frog টি A বিন্দু হতে লাভ শুরু করবে।
প্রথম লাফে Frog টি B বা D তে আসতে পারে।
দ্বিতীয় লাফে Frog টি A বা C তে আসতে পারে।
তৃতীয় লাফে Frog টি B বা D তে আসতে পারে।

এভাবে, প্রতিটি বিজোড় সংখ্যক বার লাফে অর্থাৎ ১ম, ৩য়, ৫ম বা ৭ম লাফে Frog টি হয় B বা D বিন্দুতে আসতে পারবে, C বিন্দুতে নয়।
কাজেই ৭ম লাফে C বিন্দুতে আসার কোনো Probability নেই।

তাই সঠিক উত্তর হবে অপশন d

১,৫৬৫.

The length of two parallel sides of a trapezium are 30 cm and 60 cm respectively, and the distance between the parallel sides is 8 cm. Find the area of the trapezium.

ক
320 cm²
খ
330 cm²
গ
360 cm²
ঘ
380 cm²

ব্যাখ্যা

Question: The length of two parallel sides of a trapezium are 30 cm and 60 cm respectively, and the distance between the parallel sides is 8 cm. Find the area of the trapezium.

Solution:
Area of the Trapezium = (1/2) × (Sum of the parallel sides) × (Distance between parallel sides)
= (1/2) × (30 + 60) × 8
= (1/2) × 90 × 8
= 360 cm²

∴ Area of the Trapezium = 360 cm²

১,৫৬৬.

The area of a rectangular field is 52000 sq. meters. This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field in the map is-

ক
1.6 m
খ
160 m
গ
160 cm
ঘ
1.6 cm

ব্যাখ্যা

Question: The area of a rectangular field is 52000 sq. meters. This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field in the map is -

Solution:
Length of the field = (3.25 × 100) m
= 325 m

We know,
Length × Breadth = 52000
Or, Breadth = 52000/325
∴ Breadth of the field = 160 m.

The breadth of the rectangular field in the map is = 160/100 cm
= 1.6 cm

১,৫৬৭.

The diameter of a circle is 14 cm. what is the circumference of the circle?

ক
ক) 33 cm
খ
খ) 44 cm
গ
গ) 55 cm
ঘ
ঘ) 66 cm

ব্যাখ্যা

Radius of the circle r = 14/2 = 7
The circumference of the circle = 2πr = 2 × (22/7) × 7 = 44 cm

১,৫৬৮.

What is the slope of a line perpendicular to the line whose equation is 3x + 4y = 8?

ক
2/3
খ
1/2
গ
3/5
ঘ
4/3

ব্যাখ্যা

প্রশ্ন: What is the slope of a line perpendicular to the line whose equation is 3x + 4y = 8?

সমাধান:
প্রদত্ত সরল রেখার সমীকরণ: 3x + 4y = 8

y = mx + c আকারে লিখি, যেখানে m হলো রেখার ঢাল।
4y = - 3x + 8
y = (- 3/4)x + 2

অতএব, মূল রেখার ঢাল (m) = - 3/4

আমরা জানি, কোনো রেখার উপর লম্ব রেখার ঢাল m = - 1/m
= -1/(- 3/4)
= 4/3

∴ লম্ব রেখার ঢাল = 4/3

১,৫৬৯.

sin(A + 45°) = √2/2, find the value of A.

ক
45°
খ
30°
গ
0°
ঘ
None

ব্যাখ্যা

Question: sin(A + 45°) = √2/2, find the value of A.

Solution:
sin(A + 45°) = √2/2
⇒ sin(A + 45°) = 1/√2
⇒ sin(A + 45°) = sin45°
⇒ A + 45° = 45°
⇒ A = 45° − 45°
∴ A = 0°

১,৫৭০.

What will be the value of 1 - 2sin²θ, if cos⁴θ - sin⁴θ = 2/3?

ক
1/4
খ
2/3
গ
1/√3
ঘ
1/√2

ব্যাখ্যা

Question: What will be the value of 1 - 2sin²θ, if cos⁴θ - sin⁴θ = 2/3?

Solution:
Given,
cos⁴θ - sin⁴θ = 2/3

Now, here we can apply the formula -
a⁴ - b⁴ = (a² - b²) (a² + b²)
⇒ (cos²θ - sin²θ) (cos²θ + sin²θ) = 2/3
⇒ 1 × (cos²θ - sin²θ) = 2/3 [because cos²θ + sin²θ = 1]
⇒ (1 - sin²θ) - sin²θ = 2/3
∴ 1 - 2sin²θ = 2/3

১,৫৭১.

What is the slope of a line perpendicular to the line whose equation is 15x - 3y = 9?

ক
- 5
খ
3
গ
- 1/12
ঘ
- 1/5

ব্যাখ্যা

Question: What is the slope of a line perpendicular to the line whose equation is 15x - 3y = 9?

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

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

এখন, প্রদত্ত সমীকরণটি হলো,
15x - 3y = 9
⇒ 3y = 15x - 9
⇒ y = (15x - 9) / 3
⇒ y = 5x - 3

(1) নং এর সাথে তুলনা করে পাই,
m = 5

∴ লম্ব (perpendicular) রেখার ঢাল হবে, m' = - (1/m)
= - 1/5

১,৫৭২.

If tanA = 3/4, then sinA = ?

ক
4/5
খ
2/5
গ
3/5
ঘ
1/5

ব্যাখ্যা

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

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

sinA = লম্ব/অতিভুজ
= 3/5

১,৫৭৩.

If the radius of cylinder is halved and height is doubled, then what will be the curved surface area?

ক
ক) increase by 1
খ
খ) the same
গ
গ) double
ঘ
ঘ) triple

ব্যাখ্যা

প্রশ্ন: If the radius of cylinder is halved and height is doubled, then what will be the curved surface area?

সমাধান:
আমরা জানি,
বেলনের ব্যাসার্ধ, r একক এবং উচ্চতা, h একক হলে
বেলনের বক্রতলের ক্ষেত্রফল ২πrh বর্গএকক

এখন,
ব্যাসার্ধ অর্ধেক হলে পাই, r/২ একক
উচ্চতা দ্বিগুণ হলে পাই, ২h একক
বেলনের বক্রতলের ক্ষেত্রফল হবে, ২π × (r/২) × ২h বর্গএকক
= ২πrh বর্গএকক

ক্ষেত্রফলের কোন পরিবর্তন হবে না।

১,৫৭৪.

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

ক
ক) 9 cm
খ
খ) 12 cm
গ
গ) 16 cm
ঘ
ঘ) 18 cm

ব্যাখ্যা

√(l² + b²) = √41.
l² + b² = 41
Also, lb = 20.

(l + b)² = (l² + b²) + 2lb
(l + b)² = 41 + 40 = 81
(l + b) = 9.

Perimeter = 2(l + b) = 18 cm.

১,৫৭৫.

The wheel of scooter has diameter 70 cm. How many revolutions per minute must the wheel make so that the speed of the scooter is kept at 26.4 km per hour?

ক
500
খ
250
গ
320
ঘ
200

ব্যাখ্যা

Question: The wheel of scooter has diameter 70 cm. How many revolutions per minute must the wheel make so that the speed of the scooter is kept at 26.4 km per hour?

Solution:
Distance travelled by wheel in one revolution = circumference of wheel = (22/7) × 70 = 220cm

Speed of scooter = 26.4km/hr = (26.4 × 1000 × 100)/60 = 44000cm/min

∴ The wheel has therefore got to travel 44000 cm in 1 min i.e. it has to perform 44000/220 revolution in 1min = 200 revolutions.

১,৫৭৬.

Find the midpoint of the line segment joining the points A₁(2, 5) and A₂(8, - 3).

ক
(2, - 5)
খ
(1, 1/3)
গ
(5, 1)
ঘ
(3, 6)

ব্যাখ্যা

Question: Find the midpoint of the line segment joining the points A₁(2, 5) and A₂(8, - 3).

Solution:

১,৫৭৭.

A rectangular courtyard has an area of 250 square meters. How many rectangular tiles of dimension (25 × 40) cm² will be needed to cover the courtyard completely?

ক
2000 tiles
খ
2300 tiles
গ
2500 tiles
ঘ
2800 tiles

ব্যাখ্যা

Question: A rectangular courtyard has an area of 250 square meters. How many rectangular tiles of dimension (25 × 40) cm² will be needed to cover the courtyard completely?

Solution:
Given that,
Area of the courtyard = 250 sq. meters
= (250 × 10000) cm² [1 m² = 10000 cm²]
= 2500000 cm²

Area of each tile = (25 × 40) cm²
= 1000 cm²

Therefore, total tiles required = Total area /Area of one tile
= 2500000/1000
= 2500 tiles

১,৫৭৮.

If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is-

ক
2.5m
খ
7.5m
গ
5m
ঘ
3.75m

ব্যাখ্যা

Question: If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is-

Solution:

let,
height of the girl DE = 1.5m
distance of the shadow CE = 4.5m
height of the lamp-post AB = h
BC = 3 + 4.5 = 7.5m

here,
tanθ = DE/CE
tanθ = AB/BC

∴ DE/CE = AB/BC
1.5/4.5 = h/7.5
h = 2.5m

১,৫৭৯.

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

ক
35°
খ
30°
গ
25°
ঘ
15°

ব্যাখ্যা

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

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

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

১,৫৮০.

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

ক
11 cm²
খ
44 cm²
গ
80 cm²
ঘ
121 cm²

ব্যাখ্যা

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

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

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

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

১,৫৮১.

If the total surface area of a cube is 726 square cm, find the volume of the cube is-

ক
ক) 1331 cm³
খ
খ) 1221 cm³
গ
গ) 1728 cm³
ঘ
ঘ) 1624 cm³

ব্যাখ্যা

Total surface area of cube 726 cm²
Total surface area of the cube = 6a²
Volume of cube = a³

Accordingly,
6a² = 726
a²= 121
a = 11

Volume of cube = a³ = 11³ ⇒ 1331 cm³

∴ The volume of the cube is 1331 cm³.

১,৫৮২.

A Triangle has a perimeter 13. The two shorter sides have integer lengths equal to x and x + 1. Which of the following could be length of the other side?

ক
ক) 3
খ
খ) 4
গ
গ) 10
ঘ
ঘ) 6

ব্যাখ্যা

Question: A Triangle has a perimeter 13. The two shorter sides have integer lengths equal to x and x+1. Which of the following could be length of the other side?

Solution:
The SHORTER sides have integral lengths equal to x and x + 1

Let the longest side be 'a'

So, a + x + (x +1) = 13
Or, a + 2x = 12 .......(1)

We know that the sum of the lengths of the shorter sides has to be more than the length of the longer one
Looking at the options, we can't have 8 or 10 as values for 'a'

Similarly, we can't have 2 or 4 as values for 'a' as it wouldn't be the longest side then.

So, the correct length of other side is 6

১,৫৮৩.

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

ক
256 cubic meters
খ
384 cubic meters
গ
512 cubic meters
ঘ
729 cubic meters

ব্যাখ্যা

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

Solution:
ধরি, ঘনকের বাহুর দৈর্ঘ্য = a মিটার।

আমরা জানি,
ঘনকের সম্পূর্ণ পৃষ্ঠের ক্ষেত্রফল = 6a²

প্রশ্নমতে,
6a² = 384
⇒ a² = 384/6
⇒ a² = 64
∴ a = 8 মিটার

এখন,
ঘনকের আয়তন = a³
= 8³
= 512 ঘন মিটার

অতএব, ঘনকটির আয়তন = 512 ঘন মিটার।

১,৫৮৪.

A pole 120 meters long breaks into two parts without complete separation and makes an angle of 30° with the ground. Find the length of the broken part of the pole.

ক
60 meters
খ
40 meters
গ
40√3 meters
ঘ
80 meters

ব্যাখ্যা

Question: A pole 120 meters long breaks into two parts without complete separation and makes an angle of 30° with the ground. Find the length of the broken part of the pole.

Solution:

খুঁটির মোট দৈর্ঘ্য = 120 মিটার
ধরি,ভাঙা অংশটির দৈর্ঘ্য = x মিটার
∴ অবশিষ্ট অংশটির দৈর্ঘ্য = (120 - x) মিটার
মই ভূমির সাথে কোণ তৈরি করে, θ = 30°

আমরা জানি,
sinθ = লম্ব/অতিভুজ
⇒ sin 30° =(120 - x)/x
⇒ 1/2 = (120 - x)/x
⇒ x = 2(120 - x)
⇒ x = 240 - 2x
⇒ 3x = 240
∴ x = 80 মিটার

অতএব, খুঁটির ভাঙা অংশটির দৈর্ঘ্য = 80 মিটার।

১,৫৮৫.

If we consider an anticlockwise direction, what is the time difference between 2 am and 10:30 pm?

ক
ক) 20 hours 30 minutes
খ
খ) 3 hours 30 minutes
গ
গ) 10 hours 30 minutes
ঘ
ঘ) 15 hours 30 minutes

ব্যাখ্যা

Question: If we consider an anticlockwise direction, what is the time difference between 2 am and 10:30 pm?

Solution:

anticlockwise means reverse direction of normal clock.
that means,
the clock will reversely show time like 2am then 1am then 12pm then 11pm then 10pm

so, the difference between 2am and 10.30pm will be = 3 hours and 30 minutes

১,৫৮৬.

In triangle ABC, AB = AC and ∠C = 45°. Find the measure of ∠A.

ক
100°
খ
95°
গ
90°
ঘ
80°

ব্যাখ্যা

Question: In triangle ABC, AB = AC and ∠C = 45°. Find the measure of ∠A.

Solution:

AB = AC
∴ ∠B= ∠C = 45°

∴ ∠A= 180° - ∠B - ∠C
= 180° - 45° - 45°
= 180° - 90°
= 90°

১,৫৮৭.

ক
30°
খ
45°
গ
60°
ঘ
90°

ব্যাখ্যা

Question:

Solution:

১,৫৮৮.

The ratio between the length and the 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 (in sq. m) is-

ক
15360
খ
153600
গ
30720
ঘ
307200

ব্যাখ্যা

Question: The ratio between the length and the 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 (in sq. m) is-

Solution:

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

Let
length = 3x metres and breadth = 2x metres.

Then,
2(3x + 2x) = 1600
⇒ 5x = 800
∴ x = 160.

∴ Length = 480 m and Breadth = 320 m.

∴ Area = (480 × 320) m² = 153600 m².

১,৫৮৯.

If tan(θ + 30°) = 1, find the value of cosθ.

ক
(√3 - 2)/(2√2)
খ
(√3 - 1)/(2√2)
গ
(√3 + 1)/(2√2)
ঘ
√3/(2√2)

ব্যাখ্যা

Question: If tan(θ + 30°) = 1, find the value of cosθ.

Solution:
Given,
tan(θ + 30°) = 1
⇒ tan(θ + 30°) = tan 45°
⇒ θ + 30° = 45°
⇒ θ = 45° - 30°
⇒ θ = 15°

Now,
cosθ = cos 15°
∴ cos 15° = cos(45° - 30°)
= cos45° cos30° + sin45° sin30°
= (1/√2 × √3/2) + (1/√2 × 1/2)
= (√3/2√2) + (1/2√2)
= (√3 + 1)/(2√2)

১,৫৯০.

How many ounces make 4.75 pounds?

ক
ক) 54
খ
খ) 57
গ
গ) 67
ঘ
ঘ) 69
ঙ
ঙ) 76

ব্যাখ্যা

1 pound = 16 ounces
∴ 4.75 pounds = 76 ounces

১,৫৯১.

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

ক
√5
খ
√45
গ
4√5
ঘ
3√5

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

১,৫৯২.

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

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

১,৫৯৩.

An equilateral triangle has a perimeter of 30 meters. What is its area?

ক
25√3 square meters
খ
30√3 square meters
গ
100 square meters
ঘ
60√3 square meters

ব্যাখ্যা

Question: An equilateral triangle has a perimeter of 30 meters. What is its area?

Solution:
দেওয়া আছে,
সমবাহু ত্রিভুজের পরিসীমা = 30 মিটার

সমবাহু ত্রিভুজের এক বাহুর দৈর্ঘ্য, a = পরিসীমা/3
= 30/3 মিটার
∴ a = 10 মিটার

সমবাহু ত্রিভুজের ক্ষেত্রফল = (√3/4) × (বাহুর দৈর্ঘ্য)²
= (√3/4) × 10² বর্গ মিটার
= (√3/4) × 100 বর্গ মিটার
= 100√3/4 বর্গ মিটার
= 25√3 বর্গ মিটার

অতএব, সমবাহু ত্রিভুজের ক্ষেত্রফল = 25√3 বর্গ মিটার।

১,৫৯৪.

If the area of a triangle with base x is equal to the area of a square with side x, then the altitude of the triangle is -

ক
x/2
খ
x
গ
2x
ঘ
3x

ব্যাখ্যা

Question: If the area of a triangle with base x is equal to the area of a square with side x, then the altitude of the triangle is -

Solution:
Let, altitude of triangle is h

Now
(1/2) × x × h = x²
⇒ h = 2x

১,৫৯৫.

The supplement of an angle exceeds twice the angle by 30°. Then the angle is equal to-

ক
30°
খ
45°
গ
50°
ঘ
60°

ব্যাখ্যা

Question: The supplement of an angle exceeds twice the angle by 30°. Then the angle is equal to-

Solution:
Let the angle be x
Then, its supplement = 180 - x

According to the question,
180 - x = 2x + 30
⇒ 180 - 30 = 3x
⇒ 150 = 3x
⇒ x = 50°

১,৫৯৬.

If B = 45° , then what is the value of (1 - cot²B)/(1 + cot²B)?

ক
1/2
খ
1
গ
2
ঘ
0

ব্যাখ্যা

Question: If B = 45° , then what is the value of (1 - cot²B)/(1 + cot²B)?

Solution:
Here, B = 45°

Now,
(1 - cot²B)/(1 + cot²B)
= {1 - (cot45°)²}/{1 + (cot45°)²}
= (1 - 1²)/(1 + 1²)
= 0/2
= 0

১,৫৯৭.

The sides of a triangle are in the ratio 10 : 24 : 26 and its perimeter is 300 m. What is its area?

ক
2500 m²
খ
3000 m²
গ
3500 m²
ঘ
4000 m²
ঙ
None of these

ব্যাখ্যা

Question: The sides of a triangle are in the ratio 10 : 24 : 26 and its perimeter is 300 m. What is its area?

Solution:
Let the sides are 10x, 24x, and 26x.
The perimeter is 300 m.
So, 10x + 24x + 26x = 300
⇒ 60x = 300
∴ x = 5

So, the sides are 10 × 5 = 50 meters
24 × 5 = 120 meters
26 × 5 = 130 meters

10² + 24² = 26²
⇒ 100 + 576 = 676
⇒ 676 = 676
so, it is a right triangle.

The area of a right triangle is = (1/2) × base × height
= (1/2) × 50 × 120
= 3000 m²

১,৫৯৮.

Which one is incorrect?

ক
ক) cosec²A = 1 + cot²A
খ
খ) sec²A = 1 + tan²A
গ
গ) tan²A = 1 - sec²A
ঘ
ঘ) cos²A = 1 - sin²A

ব্যাখ্যা

Question: Which one is incorrect?

Solution:
আমরা জানি,
sin²A + cos²A = 1
⇒ sin²A = 1 - cos²A
⇒ cos²A = 1 - sin²A

আবার,
sec²A - tan²A = 1
⇒ sec²A = 1 + tan²A
⇒ tan²A = sec²A - 1

এবং
cosec²A - cot²A = 1
⇒ cosec²A = 1 + cot²A
⇒ cosec²A - 1 = cot²A

১,৫৯৯.

A tree of height 4 meter casts a show of length 6.5 meter. What is the height of a house casting a shadow 26 meter long?

ক
ক) 14 meter
খ
খ) 17 meter
গ
গ) 15 meter
ঘ
ঘ) 16 meter

ব্যাখ্যা

If shadows length is 6.5 meters then tree's height is 4 meters
So, if shadows length is 26 meters, then house's height = 4×26 / 6.5 = 16 meters

১,৬০০.

If sinθ + cosθ = √3, then what is tanθ + cotθ equal to?

ক
1
খ
0
গ
2/√3
ঘ
1/√3

ব্যাখ্যা

Question: If sinθ + cosθ = √3, then what is tanθ + cotθ equal to?

Solution:
sinθ + cosθ = √3
⇒ (sinθ + cosθ)² = (√3)²
⇒ sin²θ+ 2sinθcosθ + cos²θ = 3

We know,
sin²θ + cos²θ = 1

∴ 1 + 2sinθcosθ = 3
⇒ 2sinθcosθ = 2
∴ sinθcosθ = 1

Now,
tanθ + cotθ
= sinθ/cosθ + cosθ/sinθ
= (sin²θ + cos²θ)/(sinθcosθ)
= 1/1
= 1