Geometry: Mensuration, Trigonometry

PrepBank · পাতা ২০ / ২১ · ১,৯০১–২,০০০ / ২,০৮৫

১,৯০১.

ক
secA + tanA
খ
secA - tanA
গ
secA + cotA
ঘ
cosA + tanA

ব্যাখ্যা

Question:

Solution:

১,৯০২.

ABCD is a square and one of its sides AB is also a chord of the circle as shown in the figure. What is the area of the square?

ক
12
খ
9
গ
12√2
ঘ
18

ব্যাখ্যা

Question: ABCD is a square and one of its sides AB is also a chord of the circle as shown in the figure. What is the area of the square?

Solution:
চিত্রানুসারে, O হলো বৃত্তের কেন্দ্র এবং OA ও OB হলো বৃত্তের ব্যাসার্ধ, যার দৈর্ঘ্য 3।
AOB একটি সমকোণী ত্রিভুজ, যেখানে ∠AOB = 90° এবং অতিভুজ = AB

পিথাগোরাসের উপপাদ্য অনুসারে,
AB² = OA² + OB²
AB² = 3² + 3²
AB² = 9 + 9
AB² = 18

আমরা জানি, বর্গক্ষেত্রের ক্ষেত্রফল = বাহুর দৈর্ঘ্য^২
যেহেতু ABCD একটি বর্গ, তাই এর ক্ষেত্রফল হলো AB²
সুতরাং, বর্গটির ক্ষেত্রফল হলো 18

১,৯০৩.

The cost of cultivating a square field at the rate of Tk. 685 per hectare is Tk. 6165. The cost of putting a fence around it at the rate of Tk. 48.75 per meter would be -

ক
Tk. 23400
খ
Tk. 52650
গ
Tk. 58500
ঘ
Tk. 117000

ব্যাখ্যা

Question: The cost of cultivating a square field at the rate of Tk. 685 per hectare is Tk. 6165. The cost of putting a fence around it at the rate of Tk. 48.75 per meter would be -

Solution:
Area of the field = 6165/685 hectare
= 9 hectare
= (9 × 10000) m²
= 90000 m²

∴ Side of the field = √90000 m
= 300 m

Perimeter of the field = (300 × 4) m
= 1200 m.

Cost of fencing = Tk. (1200 × 48.75)
= Tk. 58500

১,৯০৪.

Nirob cuts a 4 cm cube into 1 cm cubes. What is the percentage increase in the surface area after cutting?

ক
200%
খ
250%
গ
300%
ঘ
340%

ব্যাখ্যা

Question: Nirob cuts a 4 cm cube into 1 cm cubes. What is the percentage increase in the surface area after cutting?

Solution:
শুরুতে পৃষ্ঠতলের ক্ষেত্রফল = 6 × 4²
= 6 × 16
= 96

ধরি, n সংখ্যক ঘনক তৈরি করা হলো।
n 1³ = 4³
n = 64

64 টি ঘনকের প্রত্যেকটির পৃষ্ঠতলের ক্ষেত্রফল = 6 × 1²
= 6 cm²

মোট পৃষ্ঠতলের ক্ষেত্রফল = 64 × 6
= 384 cm²

বৃদ্ধি = 384 - 96
= 288 cm²

∴ শতকরা বৃদ্ধি = (288/96) × 100%
= 300%

১,৯০৫.

From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is-

ক
120 m
খ
120√2 m
গ
200 m
ঘ
100√3 m

ব্যাখ্যা

Question: From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is-

Solution:

Given that,
Hight of tower, AB = 100 m
Angle of elevation from point P, ∠APB = 30°

We know,
tanθ = opposite/adjacent = AB/PA
⇒ tan30° = 100/PA
⇒ 1/√3 = 100/PA
∴ PA = 100√3 m

Thus, the distance from point P to the foot of the tower is 100√3 m.

১,৯০৬.

The number of parallellograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is -

ক
ক) 6
খ
খ) 9
গ
গ) 12
ঘ
ঘ) 18

ব্যাখ্যা

আমরা দেখতে পাচ্ছি যে চারটি পরস্পর সমান্তরাল রেখাকে তিনটি পরস্পর সমান্তরাল রেখা ছেদ করলে মোট আয়তক্ষেত্রের সংখ্যা হবেঃ
6 + 5 + 4 + 3 + 2 + 2 + 1 = 18

১,৯০৭.

In a right triangle, the length of one of the legs is 6 and the length of the hypotenuse is 10. What is the length of the other leg?

ক
8 units
খ
12 units
গ
5 units
ঘ
15 units

ব্যাখ্যা

Question: In a right triangle, the length of one of the legs is 6 and the length of the hypotenuse is 10. What is the length of the other leg?

Solution:
Hypotenuse, c = 10 units
One leg, a = 6 units
Other leg, b = ?

Using the Pythagorean theorem,
⇒ a² + b² = c²
⇒ b² + 6² = 10²
⇒ b² + 36 = 100
⇒ b² = 100 - 36
⇒ b² = 64
⇒ b² = 8²
∴ b = 8

∴ The length of the other leg is 8 units.

১,৯০৮.

If the measures of the angles in a triangle are in the ratio of 3 : 7 : 8, then the degrees in the largest angle:

ক
120°
খ
80°
গ
90°
ঘ
100°

ব্যাখ্যা

Question: If the measures of the angles in a triangle are in the ratio of 3 : 7 : 8, then the degrees in the largest angle:

Solution:
মনে করি, ত্রিভুজের কোণ তিনটি যথাক্রমে 3x, 7x এবং 8x

আমরা জানি, ত্রিভুজের তিন কোণের সমষ্টি 180°

শর্তমতে,
3x + 7x + 8x = 180°
⇒ 18x = 180°
⇒ x = 180°/18
⇒ x = 10°

বৃহত্তম কোণটি হলো 8x
∴ বৃহত্তম কোণের মান = 8 × 10° = 80°

১,৯০৯.

Two buildings are 40 m apart. The angle of depression of the top of one building of height 100 m with the top of second building of unknown height is 60°. Find the height of second building?

ক
30.8 m
খ
60 m
গ
76.8 m
ঘ
40.5 m

ব্যাখ্যা

Question: Two buildings are 40 m apart. The angle of depression of the top of one building of height 100 m with the top of second building of unknown height is 60°. Find the height of second building?

Solution:

Le t the height of the second building AD be h.
EC = 100 - h
DC = AB = 40

From ΔECD we get,
EC/DC = tan60°
⇒ (100 - h)/40 = √3
⇒ 100 - h = 40 × √3
⇒ - h = - 100 + 40 × 1.73
⇒ - h = - 100 + 69.2
∴ h = 30.8 m

১,৯১০.

60π cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in meters will be -

ক
24 meters
খ
88 meters
গ
196 meters
ঘ
240 meters
ঙ
300 meters

ব্যাখ্যা

Question: 60π cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in meters will be -

Solution:
Let the length of the wire be h
Radius, r =(1/2) mm =1/20 cm.
We know,
Volume of a cylinder(wire) = πr²h

∴ π × (1/20) × (1/20) × h = 60π
⇒ h = 60 × 20 × 20
⇒ h = 24000

∴ The length of the wire = 24000 cm
= (24000/100) meters
= 240 meters

১,৯১১.

The ratio of the volumes of two spheres is 27 : 8. What is the ratio of their surface areas?

ক
3√3 : 2√2
খ
3 : 2
গ
27 : 8
ঘ
9 : 4

ব্যাখ্যা

Question: The ratio of the volumes of two spheres is 27:8. What is the ratio of their surface areas?

Solution:
Volume of a Sphere: V = (4/3)π(r)³
Surface Area of a Sphere: S = 4πr²

Given,

১,৯১২.

Find the equation of the line with x-intercept = 4 and y-intercept = 3.

ক
4x - 3y - 12 = 0
খ
4x + 3y - 12 = 0
গ
3x - 4y + 12 = 0
ঘ
3x + 4y - 12 = 0

ব্যাখ্যা

Question: Find the equation of the line with x-intercept = 4 and y-intercept = 3.

Solution:
Given, x-intercept = 4,
So, the line passes through (4, 0).
y-intercept = 3,
So, the line passes through (0, 3).

We know,
The intercept form of a line is:
(x/a) + (y/b) = 1, where a = x-intercept, b = y-intercept.
⇒ (x/4) + (y/3) = 1
⇒ (3x + 4y)/12 = 1
⇒ 3x + 4y = 12
⇒ 3x + 4y - 12 = 0

∴ The equation of the line is 3x + 4y - 12 = 0

১,৯১৩.

If the length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, then the length of the median to its greatest side is -

ক
4.8 cm
খ
8 cm
গ
6 cm
ঘ
5 cm

ব্যাখ্যা

Question: If the length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, then the length of the median to its greatest side is -

Solution:
According to question,
Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle.

To find the length of the median to the greatest side of a triangle, we can use the formula:
Median = 1/2 * √(2b^2 + 2c^2 - a^2)
Where a, b, and c are the lengths of the sides of the triangle, and a is the greatest side.
In this case, the lengths of the sides are 6 cm, 8 cm, and 10 cm. The greatest side is 10 cm.
Using the formula:
Median = 1/2 * √(2 * 8² + 2 * 6² - 10²)
= 1/2 * √(128 + 72 - 100)
= 1/2 * √(100)
= 1/2 * 10
= 5 cm

So, the length of the median to the greatest side of the triangle is 5 cm.

১,৯১৪.

The height of a cylinder is four times the radius of the cylinder. If the volume of the cylinder is 256π cm³, what is the radius of the cylinder?

ক
ক) 4 cm
খ
খ) 8 cm
গ
গ) 16 cm
ঘ
ঘ) 24 cm

ব্যাখ্যা

ধরি
সিলিন্ডারের ব্যাসার্ধ r cm
সিলিন্ডারের উচ্চতা 4r cm

আমরা জানি,
সিলিন্ডারের আয়তন= πr²h ঘন একক

প্রশ্নমতে,
πr²× 4r = 256π
4r³ = 256
r³= 64
r³ = 4³
r = 4

১,৯১৫.

The volume of a cuboid with length, breadth and height as 11x³, 12x⁵ and 13x⁷ respectively is:

ক
1716x¹⁵
খ
1015x⁹
গ
1050x
ঘ
1172x¹⁵

ব্যাখ্যা

Question: The volume of a cuboid with length, breadth and height as 11x³, 12x⁵ and 13x⁷ respectively is:

Solution:
দেওয়া আছে
আয়তাকার ঘনবস্তুর (cuboid) দৈর্ঘ্য = 11x³
আয়তাকার ঘনবস্তুর (cuboid) প্রস্থ = 12x⁵
আয়তাকার ঘনবস্তুর (cuboid) উচ্চতা = 13x⁷

আয়তাকার ঘনবস্তুর আয়তন = (11x³) × (12x⁵) × (13x⁷)
= 1716x¹⁵

১,৯১৬.

The volume of a cube is numerically equal to the sum of its edges. What is its total surface area in square units?

ক
ক) 84
খ
খ) 66
গ
গ) 72
ঘ
ঘ) 36

ব্যাখ্যা

Question: The volume of a cube is numerically equal to the sum of its edges. What is its total surface area in square units?

Solution:
Let,
Length of one side = a
Volume = a³
Total sides = 12

ATQ
a³ = 12a
∴ a² = 12

∴ Total surface = 6a²
= 6 × 12
= 72

১,৯১৭.

The base of a right prism is a trapezium whose lengths of two parallels sides are 10 cm and 6 cm and distance between them is 5 cm. If the height of the prism is 8 cm, its volume is -

ক
ক) 320 cm³
খ
খ) 300 cm³
গ
গ) 310 cm³
ঘ
ঘ) 300.5 cm³

ব্যাখ্যা

Length of the parallel sides of prism = 10 cm and 6 cm
Height of prism = 8 cm
∴ Volume of prism = (1/2){(10 + 6) × 5 × 8}
= (1/2) × 16 × 5 × 8
= 320 cm³

১,৯১৮.

A towel, when bleached, was found to have lost 10% of its length and 10% of its breadth. The percentage of decrease in area is:

ক
28%
খ
25%
গ
19%
ঘ
15%
ঙ
None

ব্যাখ্যা

Question: A towel, when bleached, was found to have lost 10% of its length and 10% of its breadth. The percentage of decrease in area is:

Solution:
Let,
The length be 10x cm
The breadth be 10y cm
The area of the rectangle be 10x × 10y = 100xy

After reduction the length is = 10x(90/100) = 9x
After reduction the breadth is = 10y(90/100) = 9y
New area after reduction = 9x × 9y = 81xy

Change in area = 100xy - 81xy = 19xy
Change in Percentage = (19xy/100 × xy) × 100 = 19%
∴ Decreases in the area is 19%

১,৯১৯.

The dimensions of a hall are 40 m, 25 m and 20 m. If each person requires 200 cubic meters, find the number of persons who can be accommodated in the hall.

ক
150
খ
140
গ
120
ঘ
100

ব্যাখ্যা

Question: The dimensions of a hall are 40 m, 25 m and 20 m. If each person requires 200 cubic meters, find the number of persons who can be accommodated in the hall.

Solution:
Length of the hall = 40 m
Breadth of hall= 25 m
Height of hall = 20 m
Volume of the hall = 40 × 25 × 20 = 20000 m³
Space occupied by each person = 200 m³
Number of person that can accommodate in the hall = 20000/200 = 100

১,৯২০.

The dimensions of a box are 2,3 and 4 meters. the cost of painting the outer sides of the box at the rate of tk. 3 per square meter is ___

ক
ক) Tk. 156
খ
খ) Tk. 120
গ
গ) Tk. 136
ঘ
ঘ) Tk. 160

ব্যাখ্যা

আমরা জানি,
বাক্সের উপরিতলের ক্ষেত্রফল = 2(ab + bc + ca)
= 2(2 × 3 + 3 × 4 + 4 × 2)
= 52 বর্গমিটার
∴ মোট খরচ = 52 × 3 = 156

১,৯২১.

Calculate the area of a rhombus if the length of its side is 2 cm and one of its angles A is 30 degrees.

ক
2 cm²
খ
4 cm²
গ
3 cm²
ঘ
5 cm²

ব্যাখ্যা

Question: Calculate the area of a rhombus if the length of its side is 2 cm and one of its angles A is 30 degrees.
(যদি একটি রম্বসের একটি বাহুর দৈর্ঘ্য ২ সেন্টিমিটার এবং একটি কোণ 'A' ৩০ ডিগ্রী হয়, তবে তার ক্ষেত্রফল কেমন হবে?)

Solution:
দেওয়া আছে,
বাহু = s = 2 cm
কোণ A = 30 degrees
বাহুর বর্গ = 2 × 2 = 4

ক্ষেত্রফল, A = s² × sin (30°)
⇒ A = 4 × (1/2)
∴ A = 2 cm²

১,৯২২.

Write an equation of the line with slope 2 and x-intercept (- 4, 0).

ক
y = 2x + 8
খ
y = - 2x + 4
গ
y = (1/2)x + 8
ঘ
(- 1/2)x + 8

ব্যাখ্যা

Question: Write an equation of the line with slope 2 and x-intercept (- 4, 0).

Solution:
Given that,
Slope m = 2
x-intercept (- 4, 0)

We know,
y - y₁ = m(x - x1)
⇒ y - 0 = 2{x - (- 4)}. ; [Here, (x₁, y₁) = (- 4, 0) and m = 2]
⇒ y = 2(x + 4)
∴ y = 2x + 8

So the equation of the line is y = 2x + 8.

১,৯২৩.

A rectangular plot of land has a fence along three of its four sides, the unfenced side and the side opposite the unfenced side have a length that is three times the length of the other two sides. If the area of the plot is 432 square feet, what is the total length of the fence is feet?

ক
60
খ
72
গ
80
ঘ
90
ঙ
None of these

ব্যাখ্যা

প্রশ্ন: A rectangular plot of land has a fence along three of its four sides, the unfenced side and the side opposite the unfenced side have a length that is three times the length of the other two sides. If the area of the plot is 432 square feet, what is the total length of the fence is feet?

সমাধান:
ধরি
আয়তাকার জমির প্রস্থ = x
আয়তাকার জমির দৈর্ঘ্য = 3x

প্রশ্নমতে,
3x × x = 432
⇒ 3x² = 432
⇒ x² = 144
⇒ x² = 12²
∴ x = 12

মোট বেড়ার দৈর্ঘ্য = (x + 3x + x) ফুট
= 5x ফুট
= 5 × 12 ফুট
= 60 ফুট

১,৯২৪.

The top and bottom of a flag on a building subtend angles of 60° and 30° respectively at a point B which is 48 meter away from the building. Find the height of the flag?

ক
29√2
খ
25√3
গ
32√3
ঘ
41√2

ব্যাখ্যা

Question: The top and bottom of a flag on a building subtend angles of 60° and 30° respectively at a point B which is 48 meter away from the building. Find the height of the flag?

Solution:

Let height of building be AC = X and height of flag be CD = h.

In ΔDAB
tan60° = (X + h)/48
⇒ √3 = (X + h)/48
⇒ X + h = 48√3
∴ h = 48√3 - X ..................(1)

In ΔCAB
tan30° = X/48
⇒ 1/√3 = X/48
∴ X = 48/√3

From (1) we get,
h = 48√3 - 48/√3
= (48 × 3 - 48)/√3
= (144 - 48)/√3
= 96/√3
= (32 × 3)/√3
= 32√3

১,৯২৫.

ক
7/3
খ
3/7
গ
0
ঘ
Unidentified

ব্যাখ্যা

Question:

Solution:

১,৯২৬.

Find an equation of the vertical line containing the point (9, - 3).

ক
x = 9
খ
y = - 3
গ
y = 9
ঘ
x = - 3

ব্যাখ্যা

Question: Find an equation of the vertical line containing the point (9, - 3).

Solution:
দেওয়া আছে,
প্রদত্ত বিন্দুটি হলো (9, -3)।

উল্লম্ব রেখার একটি প্রধান বৈশিষ্ট্য হলো, এই রেখার উপর অবস্থিত প্রতিটি বিন্দুর x-স্থানাঙ্ক সর্বদা একই থাকে। যেহেতু রেখাটি (9, -3) বিন্দু দিয়ে যায়, তাই রেখাটির উপর অবস্থিত প্রতিটি বিন্দুর x-এর মান হবে 9।

সুতরাং, নির্ণেয় সমীকরণটি হবে x = 9.

১,৯২৭.

How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?

ক
ক) 5600
খ
খ) 6000
গ
গ) 6400
ঘ
ঘ) 7200

ব্যাখ্যা

Number of bricks =Volume of the wall/ Volume of 1 brick
=800 x 600 x 22.5/25 x 11.25 x 6
= 6400.

১,৯২৮.

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

ক
ক) 3/5
খ
খ) 2/7
গ
গ) 1/2
ঘ
ঘ) 1/3

ব্যাখ্যা

The ratio of the length of an arc of a circle to its circumference is always the same as the ratio of a sector to its area.
(r × θ) / (2π × r) = θ / 2π

ATQ,
sector area : total area = (θ/2π)×πr² : πr² = 3 : 5
(θ/2π)×πr² / πr² = 3 / 5
θ/2π = 3 / 5
(θ/2π) × 2π = 3/5 × 2π
∴ θ = 6π / 5
Arc length = 6π/5 × r
Circumference = 2πr

∴ Arc length/Circumference = (6π/5 × r) / 2πr = 3/5

১,৯২৯.

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

ক
ক) 2230 m²
খ
খ) 2450 m²
গ
গ) 2350 m²
ঘ
ঘ) 2520 m²

ব্যাখ্যা

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

Solution:
According to the question,
The difference between the length and breadth of a rectangle is 23 m
⇒ l - b = 23..............(1)
Its perimeter is 206 m
⇒ 2(l + b) = 206 m
⇒ l + b = 103 m.................(2)

Adding both equation (1) and (2)
⇒ 2l = 126
⇒ l = 63 m

Putting l = 63 in Equation (2)
⇒ b = 103 - 63
⇒ b = 40 m
Area of rectangle = Length(l) × Breadth(B)
⇒ 63 × 40
⇒ 2520 m²
∴ Its area is 2520 m².

১,৯৩০.

Two cans have the same height equal to 21 cm. One can is cylindrical, the distance of whose base is 10 cm. The other can has square base of side 10 cm. What is the difference in their capacities?

ক
ক) 300 cm³
খ
খ) 350 cm³
গ
গ) 400 cm³
ঘ
ঘ) 450 cm³

ব্যাখ্যা

Question: Two cans have the same height equal to 21 cm. One can is cylindrical, the distance of whose base is 10 cm. The other can has square base of side 10 cm. What is the difference in their capacities?

Solution:
Difference in capacities = Volume of cubodial can - Volume of cylindrical can
= [ (10 × 10 × 21) - ((22/7) × 5 × 5 × 21))] cm³
= (2100 - 1650) cm³
= 450 cm³

১,৯৩১.

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

ক
30°
খ
95°
গ
90°
ঘ
120°

ব্যাখ্যা

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

Solution:

AB = AC
∴ ∠B = ∠C = 30°

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

১,৯৩২.

Find the area of the rhombus having each side equal to 17 cm and one of its diagonals equal to 16 cm.

ক
180
খ
200
গ
220
ঘ
240

ব্যাখ্যা

ABCD is a rhombus in which AB = BC = CD = DA = 17 cm

Diagonal AC = 16 cm (with O being the diagonal intersection point)

Therefore, AO = 8 cm

In ∆ AOD,

AD² = AO² + OD²⇒ 17² = 8² + OD²⇒ 289 = 64 + OD²⇒ 225 = OD²⇒ OD = 15

Therefore, BD = 2 × OD
= 2 × 15
= 30 cm

Now, area of rhombus

= 1/2 × d₁ × d₂= 1/2 × 16 × 30
= 240 cm²

১,৯৩৩.

Find the volume of a cone with radius 9 cm and height 10 cm.

ক
392π cubic cm
খ
270π cubic cm
গ
553.33 cubic cm
ঘ
None of these

ব্যাখ্যা

Question: Find the volume of a cone with radius 9 cm and height 10 cm.

Solution:
Volume = (1/3) × π × r² × h
Volume = (1/3) × π × 9² × 10
= 270π cubic cm

১,৯৩৪.

The circumcentre of a triangle ABC is O. If ∠BAC = 85° and ∠BCA = 75°, then the value of ∠AOC is-

ক
20°
খ
30°
গ
40°
ঘ
50°

ব্যাখ্যা

Question: The circumcentre of a triangle ABC is O. If ∠BAC = 85° and ∠BCA = 75°, then the value of ∠AOC is-

Solution:

∠BAC = 85°
∠BCA = 75°
∠ABC = 180° - 85° - 75°
= 20°

∠AOC = 2 ∠ABC
= 2 × 20°
= 40°

১,৯৩৫.

The scale of a local map is 1cm to 5km. An area is represented on the map by a rectangle of dimensions 5cm × 9cm. The actual area in km² is -

ক
1024
খ
1038
গ
1048
ঘ
1125

ব্যাখ্যা

Question: The scale of a local map is 1cm to 5km. An area is represented on the map by a rectangle of dimensions 5cm × 9cm. The actual area in km² is -

Solution:
The actual area = (5 × 5) × (9 × 5) km² [1 cm = 5 km]
= 25 × 45 km²
= 1125 km²

১,৯৩৬.

What would be the measure of the perimeter of a square whose area is equal to 324 square cm?

ক
36 cm
খ
72 cm
গ
64 cm
ঘ
48 cm

ব্যাখ্যা

Question: What would be the measure of the perimeter of a square whose area is equal to 324 square cm?

Solution:
দেওয়া আছে
বর্গক্ষেত্রের ক্ষেত্রফল = 324 square cm
বর্গের এক বাহুর দৈর্ঘ্য = a cm

প্রশ্নমতে
a² = 324
⇒ a² = (18)²
∴ a = 18

বর্গক্ষেত্রের পরিসীমা = 4a
= 4 × 18
= 72 cm

১,৯৩৭.

If the ratio of the areas of 2 squares is 2 : 1, then the ratio of the perimeters of the squares is

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

ব্যাখ্যা

Question: If the ratio of the areas of 2 squares is 2 : 1, then the ratio of the perimeters of the squares is-

Solution:
Let,
Area of first Square is 2x²
Area of Second Square is x²

So, The length of arm of first Square is √2x
The length of arm of Second Square is x

The perimeter of first Square is 4√2x
The perimeter of Second Square is 4x

So, The ratio of the perimeters of the squares is 4√2x : 4x
= √2 : 1

১,৯৩৮.

The measure of an angle is such that its complementary angle is 25° less than one-third of its supplementary angle. What is the measure of the angle?

ক
95°
খ
102.5°
গ
75.5°
ঘ
82.5°
ঙ
None

ব্যাখ্যা

Question: The measure of an angle is such that its complementary angle is 25° less than one-third of its supplementary angle. What is the measure of the angle?

Solution:
Let
the required angle be x°

∴ Supplementary angle = 180 − x
∴ Complementary angle = 90 − x

ATQ,
Complementary angle = (1/3 of supplementary angle) − 25
⇒ 90 − x = {(1/3) × (180 − x)} - 25
⇒ 90 − x = {(180 − x) - 75}/3
⇒ 3(90 − x) = 180 − x - 75
⇒ 270 − 3x = 105 − x
⇒ 3x − x = 270 − 105
⇒ 2x = 165
∴ x = 82.5°

∴ The required angle is 82.5°

১,৯৩৯.

If tan(θ - 30°) = √3, then find sinθ.

ক
√3/2
খ
1/√2
গ
1/2
ঘ
1

ব্যাখ্যা

Question: If tan(θ - 30°) = √3, then find sinθ.

Solution:
Given,
tan(θ - 30°) = √3
⇒ tan(θ - 30°) = tan60°
⇒ θ - 30° = 60°
∴ θ = 90°

Now,
sinθ = sin90° = 1

১,৯৪০.

From point P on level ground, the angle of elevation of the top tower is 30º. If the tower is 300 m high, the distance of point P from the foot of the tower is:

ক
498 m
খ
584 m
গ
451 m
ঘ
519 m

ব্যাখ্যা

Question: From point P on level ground, the angle of elevation of the top tower is 30º. If the tower is 300 m high, the distance of point P from the foot of the tower is:

Solution:

tan30° = RQ/PQ
⇒ 1/√3 = 300/PQ
⇒ PQ = 300√3
⇒ PQ ≈ 300 × 1.73
⇒ PQ ≈ 519 m

১,৯৪১.

The cube shown below has a volume of 64 cubic inches. What is the length of the line segment AB?

ক
4√3
খ
4√2
গ
3√3
ঘ
√3

ব্যাখ্যা

Question: The cube shown below has a volume of 64 cubic inches. What is the length of the line segment AB?

Solution:
Let side length = a

Here,
Volume of the cube, a³ = 64
⇒ a³ = 4³
⇒ a = 4

∴ Length of the line segment (space diagonal) AB = √(a² + a² + a²)
= √(3a²)
= a√3
= 4√3

১,৯৪২.

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
16 : 9
খ
15 : 8
গ
10 : 3
ঘ
9 : 13
ঙ
None

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:
The radius of circle A = r
The area of circle A = πr²

The radius of circle B = 3r/4
The area of circle B = π(3r/4)² = 9πr²/16

∴ The ratio of the area of circle A to the area of circle B = πr² : 9πr²/16
= 1 : 9/16
= 16 : 9

১,৯৪৩.

A cylindrical rod of iron, whose height is equal to its radius, is melted and cast into spherical balls whose radius is half the radius of the rod. Find the number of balls.

ক
ক) 3
খ
খ) 4
গ
গ) 5
ঘ
ঘ) 6

ব্যাখ্যা

Question: A cylindrical rod of iron, whose height is equal to its radius, is melted and cast into spherical balls whose radius is half the radius of the rod. Find the number of balls.

Solution:
ধরি,
সিলিন্ডারের ব্যাসার্ধ = r
তাহলে, উচ্চতা = r

∴ আয়তন = π × r²× r = πr³

গোলাকার বলের ব্যাসার্ধ = r/2
আয়তন = (4/3)π(r/2)³
= πr³/6

∴ গোলকের সংখ্যা = πr³/(πr³/6)
= 6

১,৯৪৪.

Three solid spheres of radii 3 cm, 4 cm, and 5 cm respectively are melted and converted into a single solid sphere. Find the radius of this sphere.

ক
ক) 3 cm
খ
খ) 4 cm
গ
গ) 6 cm
ঘ
ঘ) 12 cm

ব্যাখ্যা

Question: Three solid spheres of radii 3 cm, 4 cm, and 5 cm respectively are melted and converted into a single solid sphere. Find the radius of this sphere.

Solution:
মনে করি,
নতুন গোলকটির ব্যাসার্ধ = r cm

প্রশ্নমতে,
(4/3) × π × r³ = {(4/3) × π × 3³} + {(4/3) × π × 4³} + {(4/3) × π × 5³}
⇒ (4/3) × π × r³ = (4/3) × π × {27 + 64 +125)
⇒ r³ = 6³
⇒ r³= 216
⇒ r = 6

১,৯৪৫.

The radius of a cylinder is three times its height. If the volume of the cylinder is 243π cm³, what is the height of the cylinder?

ক
9 cm
খ
6 cm
গ
3 cm
ঘ
3.5 cm

ব্যাখ্যা

Question: The radius of a cylinder is three times its height. If the volume of the cylinder is 243π cm³, what is the height of the cylinder?

Solution:
Given that,
Volume, V = 243π cm³
Radius, r = 3h

We know,
V = πr²h
⇒ πr²h = 243π
⇒ (3h)²h = 243
⇒ 9h³ = 243
⇒ h³ = 243/9
⇒ h³ = 27
⇒ h³ = 3³
∴ h = 3

So the height is 3 cm

১,৯৪৬.

If each side of the square is increased by 50%, what will be the ratio between the new area and the original area of the square?

ক
5 : 4
খ
9 : 4
গ
4 : 5
ঘ
4 : 9

ব্যাখ্যা

Question: If each side of the square is increased by 50%, what will be the ratio between the new area and the original area of the square?

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

The side of new square is x + 50% of x = x + x/2 = 3x/2
∴ The area of new square is (9x²)/4

∴ The ratio between the new area and the original area of the square = (9x²)/4 : x²
= (9/4) : 1
= 9 : 4

১,৯৪৭.

If θ = 45°, then find the value of (3 + cot2θ)/(3 - cot2θ).

ক
0
খ
1
গ
1/√3
ঘ
√3
ঙ
undefined

ব্যাখ্যা

Question: If θ = 45°, then find the value of (3 + cot2θ)/(3 - cot2θ).

Solution:
(3 + cot2θ)/(3 - cot2θ)
= [3 + cot(2 × 45°)]/[3 - cot(2 × 45°)]
= (3 + cot90)°/(3 - cot90)°
= (3 + 0)/(3 - 0)
= 3/3
= 1

১,৯৪৮.

If 1 + tan²θ = 4 and θ < 90°, than what is the value of θ = ?

ক
75°
খ
60°
গ
55°
ঘ
30°

ব্যাখ্যা

Question: If 1 + tan²θ = 4 and θ < 90°, than what is the value of θ = ?

Solution:
Given that,
1 + tan²θ = 4 and θ < 90°
⇒ sec²θ = 4 ; [sec²θ = 1 + tan²θ]
⇒ (secθ)² = (2)²
⇒ secθ = 2
⇒ secθ = sec60°
⇒ θ = 60°

১,৯৪৯.

The greatest value of sin⁴θ + cos⁴θ is?

ক
1/2
খ
1
গ
2
ঘ
1/3

ব্যাখ্যা

Question: The greatest value of sin⁴θ + cos⁴θ is?

Solution:
We know,
sin²θ + cos²θ = 1
⇒ (sin²θ + cos²θ)² = 1² ; Squaring both sides
⇒ sin⁴θ + cos⁴θ + 2sin²θ.cos²θ = 1
⇒ sin⁴θ + cos⁴θ = 1 - 2sin²θ.cos²θ
⇒ sin⁴θ + cos⁴θ = 1 - 0 ; [Put θ = 90° , cos90° = 0]
∴ sin⁴θ + cos⁴θ = 1

So the greatest value of sin⁴θ + cos⁴θ is 1.

১,৯৫০.

The area of a hemisphere is 48Π m². What is the radius of the hemisphere?

ক
2m
খ
4m
গ
6m
ঘ
12m

ব্যাখ্যা

The area of a hemisphere is 48Π m². What is the radius of the hemisphere?
Let the radius of the hemisphere be r.
Therefore, 3Πr² = 48Π
or, r² = 16
or, r = 4m

১,৯৫১.

Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 long.

ক
75°
খ
30°
গ
45°
ঘ
60°

ব্যাখ্যা

Question: Find the angle of elevation of the sun when the shadow of a pole of 18 m height is 6√3 long.

Solution:

shadow = base = 6√3 m
height =18 m

let, angle of elevation θ

tanθ = 18/6√3
⇒ tanθ = 3/√3
⇒ tanθ = √3
⇒ tanθ = tan60°
∴ θ = 60°

১,৯৫২.

If sin 17° = (x/y), then sec 17° is equal to

ক
{√(y² - x²)}/y
খ
{√(y² - x²)}/x
গ
y/{√(y² - x²)}
ঘ
x/{√(y² - x²)}

ব্যাখ্যা

Question: If sin 17° = (x/y), then sec 17° is equal to

Solution:
Given,
sin 17° = (x/y)

We know,
sin² θ + cos²θ = 1
∴ cos θ = √(1 - sin θ)

∴ cos 17° = √(1 - sin²17°)
= √{1 - (x²/y²)}
= √{(y² - x²)/y²}
= {√(y² - x²)}/y

∴ sec 17° = 1/cos 17°
= 1/[{√(y² - x²)}/y]
= y/{√(y² - x²)}

১,৯৫৩.

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

ক
ক)
খ
খ) 70√3 m
গ
গ) 70 m
ঘ
ঘ) 35√3 m

ব্যাখ্যা

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

Solution:

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

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

Height of the tower 70√3 meter.

১,৯৫৪.

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

ক
34 m
খ
1156 m
গ
136 m
ঘ
Cannot be determined

ব্যাখ্যা

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

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

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

One side of the square = 34 m.
So, perimeter of the square = 4 × 34
= 136 m

১,৯৫৫.

Find the maximum distance between two points on the perimeter of a rectangular garden whose length and breadth are 24 m and 7 m.

ক
25 m
খ
17 m
গ
31 m
ঘ
62 m

ব্যাখ্যা

Question: Find the maximum distance between two points on the perimeter of a rectangular garden whose length and breadth are 24 m and 7 m.

Solution:
একটি আয়তক্ষেত্রের পরিসীমার উপর অবস্থিত দুটি বিন্দুর মধ্যে সর্বাধিক দূরত্ব হলো এর কর্ণের দৈর্ঘ্য। কর্ণের দৈর্ঘ্য পিথাগোরাসের সূত্র ব্যবহার করে নির্ণয় করা যায়।

কর্ণের দৈর্ঘ্য = √(দৈর্ঘ্য^২ + প্রস্থ^২)
= √(24² + 7²)
= √(576 + 49)
= √625
= 25 মিটার

সুতরাং, দুটি বিন্দুর মধ্যে সর্বাধিক দূরত্ব হলো 25 মিটার।

১,৯৫৬.

A circle and a rectangle have the same perimeter. The sides of the rectangle are 9 cm and 13 cm. What is the area of the circle?

ক
ক) 154 cm²
খ
খ) 144 cm²
গ
গ) 124 cm²
ঘ
ঘ) 114 cm²

ব্যাখ্যা

The sides of the rectangle are 9 cm and 13 cm.
Perimeter of the rectangle =2(9 + 13)=44 cm
Circumference of circle =44 cm.

Here
2πr = 44
(22/7)r = 22
r/7 = 1
r = 7
Area of circle = πr² = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm²

১,৯৫৭.

A circle and a rectangle have the same perimeter. The sides of the rectangle are 16 cm and 50 cm. What is the area of the circle?

ক
1256 sq. cm.
খ
1386 sq. cm.
গ
1540 sq. cm.
ঘ
1078 sq. cm.

ব্যাখ্যা

Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 16 cm and 50 cm. What is the area of the circle?

Solution:
আয়তক্ষেত্রটির পরিসীমা = 2 × (16 + 50) সেমি
= 2 × 66 সেমি
= 132 সেমি।

যেহেতু বৃত্তের পরিধি ও আয়তক্ষেত্রের পরিসীমা সমান, তাই বৃত্তের পরিধিও 132 সেমি।

∴ বৃত্তের পরিধি, C = 2πr
⇒ 2πr = 132
⇒ r = 132/(2 × 22/7)
∴ r = 21 সেমি।

বৃত্তের ক্ষেত্রফল, A = πr²= (22/7) × (21)²= (22/7) × 441
= 1386 বর্গ সেমি।

সুতরাং, বৃত্তটির ক্ষেত্রফল হলো 1386 বর্গ সেমি।

১,৯৫৮.

A rectangular sheet of paper, 10 cm long and 8 cm wide has squares of side 2 cm cut from each of its corner. The sheet is then folded to form a tray of depth 2 cm. What is the volume of this tray?

ক
32cm³
খ
49 cm³
গ
54 cm³
ঘ
48 cm³

ব্যাখ্যা

Question: A rectangular sheet of paper, 10 cm long and 8 cm wide has squares of side 2 cm cut from each of its corner. The sheet is then folded to form a tray of depth 2 cm. What is the volume of this tray?

Solution:
Length of tray = 10 - (2 × 2) = 10 - 4 = 6 cm.
Breadth of tray = 8 - (2 × 2) = 4 cm.
Depth of tray = 2 cm.
∴ Volume of tray = 6 × 4 × 2 = 48 cm³

১,৯৫৯.

A cistern 8m long and 5m wide contains water up to a depth of 1m 50cm. The total area of the wet surface is:

ক
64 m²
খ
49 m²
গ
59 m²
ঘ
79 m²

ব্যাখ্যা

Question: A cistern 8m long and 5m wide contains water up to a depth of 1m 50cm. The total area of the wet surface is:

Solution:
দেওয়া আছে,
চৌবাচ্চার দৈর্ঘ্য (l) = 8 m, প্রস্থ (b) = 5 m, এবং গভীরতা (h) = 1 m 50 cm = 1.5 m

ভেজা পৃষ্ঠের (wet surface) ক্ষেত্রফল নির্ণয়ের জন্য আমরা সমগ্র পৃষ্ঠের ক্ষেত্রফল থেকে উপরের তলের ক্ষেত্রফল বাদ দেবো।

আমরা জানি,
সমগ্র পৃষ্ঠের ক্ষেত্রফল = 2(lb + bh + lh) বর্গ একক

∴ ভেজা পৃষ্ঠের ক্ষেত্রফল = 2(lb + bh + lh) - lb বর্গ একক
= 2{(8 × 5) + (5 × 1.5) + (8 × 1.5)} - (8 × 5)
= 2(40 + 7.5 + 12) - 40
= 2(59.5) - 40
= 119 - 40
= 79 m²

সুতরাং, ভেজা পৃষ্ঠের (wet surface) মোট ক্ষেত্রফল 79 বর্গ মিটার।

১,৯৬০.

The area of a rectangle R with a width of 4 feet is equal to the area of a square S, which has a perimeter of 32 feet. The perimeter of the rectangle R is:

ক
21 ft
খ
24 ft
গ
36 ft
ঘ
40 ft

ব্যাখ্যা

Question: The area of a rectangle R with a width of 4 feet is equal to the area of a square S, which has a perimeter of 32 feet. The perimeter of the rectangle R is:

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

প্রশ্নমতে,
4a = 32
∴ a = 8

এখানে,
আয়তক্ষেত্রের ক্ষেত্রফল = বর্গের ক্ষেত্রফল
∴ l × b = a²
⇒ l = a²/b
= 64/4
= 16

∴ আয়তক্ষেত্রের পরিসীমা = 2(l + b)
= 2(16 + 4)
= 40 feet

১,৯৬১.

Find the value of cosec(- π/6)

ক
- 1
খ
- 2
গ
- 1/2
ঘ
- 3/2

ব্যাখ্যা

Question: Find the value of cosec(- π/6)

Solution:
cosec(- π/6)
= - cosec(π/6)
= - 1/sin(π/6)
= - 1/sin30°
= - 1/(1/2)
= - 2

১,৯৬২.

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
ক) Tk. 579.92
খ
খ) Tk. 682.50
গ
গ) Tk. 702
ঘ
ঘ) Tk. 702.60

ব্যাখ্যা

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

১,৯৬৩.

A cylinder has a radius of 5 cm and a height of 7 cm. What is its volume?

ক
1040 cm³
খ
1240 cm³
গ
540 cm³
ঘ
550 cm³

ব্যাখ্যা

Question: A cylinder has a radius of 5 cm and a height of 7 cm. What is its volume?

Solution:
Radius, r = 5 cm
Height, h = 7 cm

We know,
Volume = πr²h
= (22/7) × (5)² × 7
= 550 cm³

১,৯৬৪.

The perimeter of one face of a cube is 36 cm. Its volume must be-

ক
216 cm³
খ
592 cm³
গ
343 cm³
ঘ
729 cm³

ব্যাখ্যা

Question: The perimeter of one face of a cube is 36 cm. Its volume must be-

Solution:
perimeter of one face is = 36 cm

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

ATQ,
⇒ 4a = 36
⇒ a = 36/4
= 9 cm

∴ volume = a³
= 9³
= 729 cm³

১,৯৬৫.

100 cubic centimeters of silver is drawn into a wire 2 mm in diameter. The length of the wire in meters will be:

ক
31.83 meters
খ
30 meters
গ
32.33 meters
ঘ
84 meters

ব্যাখ্যা

Let the length of the wire be 'h'

Radius of the wire, r = 2/2 mm
= 1 mm
= 1/10 cm

Therefore, πr²h = 100
or, π (1/10)²h =100
or, h = 100 × 100/3.1416
= 3,183.09 cm
= 31.83 meters

১,৯৬৬.

What is the measure of each interior angle in a regular hexagon?

ক
80°
খ
100°
গ
105°
ঘ
120°

ব্যাখ্যা

Question: What is the measure of each interior angle in a regular hexagon?

Solution:
সুষম বহুভুজের বাহুর সংখ্যা n হলে তার কোণগুলোর সমষ্টি (2n - 4) সমকোণ।
সুতরাং সুষম ষড়ভুজের ছয় কোণের সমষ্টি = (2 × 6 - 4) সমকোণ
= (12 - 4) × 90°
= 8 × 90°
= 720°
সুতরাং সুষম ষড়ভুজের একটি শীর্ষ কোণ = 720°/6
= 120°

১,৯৬৭.

What is the value of tan40° tan50° tan60°?

ক
ক) 1
খ
খ) 1/√2
গ
গ) √3
ঘ
ঘ) - √3

ব্যাখ্যা

tan40° tan50° tan60°
= tan(90°− 50°) tan50° tan60°
=cot50° tan50° tan60°
= (1/ tan50°)tan50° tan60°
= tan60°
= √3

১,৯৬৮.

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

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

ব্যাখ্যা

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

Solution:
The area of a square inscribed in a circle is 140 cm²
side of square = √140 cm = 2√35 cm
diagonal of the square = √2 × 2√35
= 2√70 cm

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

area of semi-circle = 220/2
= 110 cm²

১,৯৬৯.

The base of a right-angled triangle is 24 and hypotenuse is 26. What is its area?

ক
220 sq. meters
খ
80 sq. meters
গ
140 sq. meters
ঘ
120 sq. meters
ঙ
90 sq. meters

ব্যাখ্যা

Question: The base of a right-angled triangle is 24 and hypotenuse is 26. What is its area?

Solution:
The area of a right angled triangle = (1/2) × base × height

Given that,
Base = 24 and Hypotenuse = 26
Height² = Hypotenuse² - Base²
= 26² - 24²
= 676 - 576
Height² = 100
∴ Height = 10

Area = (1/2) × base × height
= (1/2) × 24 × 10
= 120 sq. meters

১,৯৭০.

If tan45° = 2x, what is the value of x²?

ক
1/2
খ
1/4
গ
1/6
ঘ
1/8

ব্যাখ্যা

Question: If tan45° = 2x, what is the value of x²?

Solution:
tan45 = 2x
⇒ 1 = 2x
⇒ x = 1/2
⇒ x² = 1/4

১,৯৭১.

A box having height 2m, length 10m and width 7m have the top lid open. What is the surface area of the box?

ক
138
খ
118
গ
104
ঘ
208

ব্যাখ্যা

Question: A box having height 2m, length 10m and width 7m have the top lid open. What is the surface area of the box?

Solution:
বাক্সের দৈর্ঘ্য a = 10 m
বাক্সের প্রস্থ b = 7 m
বাক্সের উচ্চতা c = 2 m

উপরের অংশ বাদে পৃষ্ঠের ক্ষেত্রফল = 2(ab + bc + ca) - ab
= 2(10 × 7 + 7 × 2 + 2 × 10) - 10 × 7
= 2(70 + 14 + 20) - 70
= 2 × 104 - 70
= 208 - 70
= 138

১,৯৭২.

What is the length of the diagonal of a square whose side is √18 cm?

ক
√6 cm
খ
72 cm
গ
6 cm
ঘ
6√2 cm
ঙ
None of these

ব্যাখ্যা

Question: What is the length of the diagonal of a square whose side is √18 cm?

Solution:
We know that,
Diagonal of a square = √2 × each side
= √2 × √18
= √36
= 6

So, the length of the diagonal is 6 cm.

১,৯৭৩.

If the diagonal and the area of a rectangle are 25 m² and 168 m², what is the length of the rectangle?

ক
ক) 12 m
খ
খ) 17 m
গ
গ) 24 m
ঘ
ঘ) 31 m

ব্যাখ্যা

Let the length of the rectangle be x metre.
Then, a breath of the rectangle = (168/x)

∴ √{(x)² + (168/x)²} = 25
⇒ √{(x² + (28224/x²)} = 25
⇒ {(x² + (28224/x²)} = 625
⇒ x⁴ - 625x² + 28224 = 0
⇒ x²(x² - 576) - 49(x² - 576) = 0
⇒ (x² - 576)(x² - 49) = 0
⇒ x² = 576 or x² = 49
⇒ x = 24 or x = 7

Hence, length = 24 cm. and breadth = 7 m.

১,৯৭৪.

If the volume of a sphere is divided by its surface area, the result is 25 cm. the radius of the sphere is -

ক
50 cm
খ
70 cm
গ
81 cm
ঘ
75 cm

ব্যাখ্যা

Question: If the volume of a sphere is divided by its surface area, the result is 25 cm. the radius of the sphere is -

Solution:
Let,
the radius of the sphere is r cm
∴ the volume of a sphere = (4/3)πr³
∴ the surface area of a sphere = 4πr²

ATQ,
{(4/3)πr³}/(4πr²) = 25
⇒ r/3 = 25
∴ r = 75

∴ the radius of the sphere is 75 cm

১,৯৭৫.

A rectangular carpet has an area of 120 sq. meters and a perimeter of 46 meters. The length of its diagonal is-

ক
17 m
খ
20 m
গ
15 m
ঘ
16 m

ব্যাখ্যা

Question: A rectangular carpet has an area of 120 sq. meters and a perimeter of 46 meters. The length of its diagonal is-

Solution:
Let, the length of carpet be x m and breadth the y m.

ATQ,
2(x + y) = 46
x + y = 23
and xy = 120

Diagonal = √(x² + y²)
= √{(x + y)² - 2xy}
= √(23² - 2 . 120)
= √(529 - 240)
= √289
= 17 m

১,৯৭৬.

If the difference between the circumference and diameter of a circle is 60 cm, then the radius of the circle is-

ক
ক) 6 cm
খ
খ) 7 cm
গ
গ) 14 cm
ঘ
ঘ) 20 cm

ব্যাখ্যা

Question: If the difference between the circumference and diameter of a circle is 60 cm, then the radius of the circle is-

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

প্রশ্নমতে,
2πr - 2r = 60
2r(π - 1) = 60
2r{(22/7) - 1} = 60
2r{(22 - 7)/7} = 60
2r(15/7) = 60
r = (60 × 7)/30
r = 14

১,৯৭৭.

In the figure above, how many of the points on line segment PQ have coordinates that are both integers?

ক
5
খ
8
গ
10
ঘ
11

ব্যাখ্যা

Question:

In the figure above, how many of the points on line segment PQ have coordinates that are both integers?

Solution:
The equation of a straight line passing through points P(x₁, y₁) and Q(x₂, y₂) is: (y − y₁)/(x - x₁) = (y₁ - y₂)/(x₁ - x₂)
or P(0, 30) and Q(50, 0):
(y - 30)/(x - 0) = (30 - 0)/(0- 50)
⇒ 50(y - 30) = 30x
⇒ 50y - 1500 = 30x
⇒ 3x + 5y = 150

If x is a multiple of 5, then y will be an integer. x ranges from 0 to 50, inclusive. There are total of 11 multiples of 5 in this range: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50.

১,৯৭৮.

When the sun's altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?

ক
ক) 60.55m
খ
খ) 140 m
গ
গ) 35 m
ঘ
ঘ) 20.2 m

ব্যাখ্যা

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m
∠ABD = 30°
∠ACD = 60°
Let CD = x, AD = h
From the right ⊿CDA
tan 60° = AD/CD
√3 = h/x ......eq(1)
From the right ⊿BDA
tan30° = AD/BD
1/√3 = h/(70 + x) ........ eq(2)
eq(1)/eq(2)
√3/(1/√3) = (h/x)/{h/(70+x)}
3 = 70 + x/x
2x = 70
x = 35
Substituting this value of x in eq:1, we have
√3 = h/35
h = 35√3
= 60.55 m

১,৯৭৯.

If each side of a rectangle is increased by 20% the increase in the area of a rectangle will be ____ .

ক
ক) 50%
খ
খ) 35%
গ
গ) 40%
ঘ
ঘ) 44%

ব্যাখ্যা

ধরি,
আয়তকারক্ষেত্রের দৈর্ঘ্য = x
আয়তকার ক্ষেত্রের প্রস্থ = y
∴ ক্ষেত্রফল = xy বর্গএকক

20% বৃদ্ধিতে নতুন দৈর্ঘ্য = x + x এর 20%
= x + এর 20/100
= x + x/5
= 6x/5

20% বৃদ্ধিতে নতুন প্রস্থ = y+y এর 20%
= y + এর 20/100
= y + y/5
= 6y/5
∴ নতুন ক্ষেত্রফল = (6x/5) × (6y/5) বর্গএকক
= 36xy/25 বর্গএকক

ক্ষেত্রফল বৃদ্ধি = {(36xy/25) - xy} বর্গএকক
= 11xy/25বর্গএকক

∴ ক্ষেত্রফল বৃদ্ধির হার = (11xy × 100/25 × xy)%
= 44%

১,৯৮০.

If 3cosec²θ - 2 = 2, find the value of θ.

ক
0°
খ
30°
গ
60°
ঘ
90°

ব্যাখ্যা

Question: If 3cosec²θ - 2 = 2, find the value of θ.

Solution:
Here,
3cosec²θ - 2 = 2
⇒ 3cosec²θ = 4
⇒ cosec²θ = 4/3
⇒ cosecθ = 2/√3
⇒ 1/sinθ = 2/√3
⇒ sinθ = √3/2
⇒ sinθ = sin60°
∴ θ = 60°

১,৯৮১.

If rSinθ = √3 and rCosθ = 1, then the value of (√3Cotθ + 1) = ?

ক
1
খ
2
গ
3
ঘ
0

ব্যাখ্যা

Question: If rSinθ = √3 and rCosθ = 1, then the value of (√3Cotθ + 1) = ?

Solution:
Given that,
rSinθ = √3 and rCosθ = 1
⇒ rCosθ/rSinθ = 1/√3
⇒ Cotθ = 1/√3
⇒ √3Cotθ = 1

(Add 1 both sides)
√3Cotθ + 1 = 1 + 1
∴ √3Cotθ + 1 = 2

১,৯৮২.

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

ক
0
খ
1/√2
গ
1/2
ঘ
√3/2

ব্যাখ্যা

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

Solution:
Given that,
tan(θ + 15°) = √3
⇒ tan(θ + 15°) = tan 60°
⇒ θ + 15° = 60°
⇒ θ = 60° - 15°
⇒ θ = 45°

Now,
sinθ
= sin45°
= 1/√2

১,৯৮৩.

If sinθ + cosθ = √2 sin(90° - θ), then what is the value of tanθ?

ক
√2 - 1
খ
√2 + 1
গ
√2
ঘ
1

ব্যাখ্যা

Question: If sinθ + cosθ = √2 sin(90° - θ), then what is the value of tanθ?

Solution:
Given,
sinθ + cosθ = √2 sin(90° - θ)
⇒ sinθ + cosθ = √2 cosθ
⇒ (sinθ + cosθ)/cosθ = √2
⇒ (sinθ/cosθ) + (cosθ/cosθ) = √2
⇒ tanθ + 1 = √2
∴ tanθ = √2 - 1

১,৯৮৪.

If the volume of a cube is 2744 cm³, then the surface area of the cube will be -

ক
784 cm²
খ
1176 cm²
গ
1136 cm²
ঘ
1276 cm²

ব্যাখ্যা

Question: If the volume of a cube is 2744 cm³, then the surface area of the cube will be -

Solution:
দেওয়া আছে,
আয়তন, a³ = 2744
⇒ a³ = 14³
⇒ a = 14

পৃষ্ঠের ক্ষেত্রফল = 6a²
= 6 × 14²
= 6 × 196
= 1176 cm²

১,৯৮৫.

If the area of a triangle is 1125 cm² and its base : corresponding altitude is 2 : 5, what is the base of the triangle?

ক
ক) 30 cm
খ
খ) 45 cm
গ
গ) 60 cm
ঘ
ঘ) 75 cm

ব্যাখ্যা

Question: If the area of a triangle is 1125 cm² and its base : corresponding altitude is 2 : 5, what is the base of the triangle?

Solution:
Let, the base = 2x cm
and altitude = 5x cm

ATQ,
(1/2) × 2x × 5x = 1125
⇒ 5x² = 1125
⇒ x² = 225
∴ x = 15

∴ Base of the triangle = (2 × 15) cm = 30 cm

১,৯৮৬.

If measures of the angles in a triangle are in the ratio of 1 : 2 : 6, then the degrees in the largest angle:

ক
20°
খ
40°
গ
90°
ঘ
120°

ব্যাখ্যা

Question: If measures of the angles in a triangle are in the ratio of 1 : 2 : 6, then the degrees in the largest angle:

Solution:
Given that,
The angles of a triangle are in the ratio 1 : 2 : 6
Let,
x, 2x, 6x

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

Now
x + 2x +6x = 180°
9x = 180°
x = 180°/9 = 20°
∴ x = 20°

∴ Largest angle = 6x = 6 × 20 = 120°

১,৯৮৭.

What is the volume of a cube whose surface area is 96?

ক
ক) 64
খ
খ) 52
গ
গ) 48
ঘ
ঘ) 60

ব্যাখ্যা

Question: What is the volume of a cube whose surface area is 96?

Solution:
ধরি,
ঘনকের বাহুর দৈর্ঘ্য = ক
তাহলে,
৬ক^২ = ৯৬
ক^২ = ১৬
ক = ৪

∴ আয়তন = ক^৩ = (৪)^৩ = ৬৪

১,৯৮৮.

The area of the base of a rectangular tank is 6500 sq. cm and the volume of water contained in it is 2.6 cubic meters. The depth of water is?

ক
2 m
খ
40 m
গ
42 m
ঘ
4 m
ঙ
3 m

ব্যাখ্যা

Question: The area of the base of a rectangular tank is 6500 sq. cm and the volume of water contained in it is 2.6 cubic meters. The depth of water is?

Solution:
Let depth = D cm.
We know,
Volume Base Area x Depth
and, 1 cubic meter = 1,000,000 cubic centimeters

Then,
D × 6500 = 2.6 × 100 × 100 × 100
∴ D = (2.6 × 100 × 100 × 100)/6500 cm
= 400 cm
= 4 m

১,৯৮৯.

The circumcentre of a triangle ABC is 'O'. If ∠BAC = 85° and ∠BCA = 75°, then the value of ∠OAC is -

ক
70°
খ
40°
গ
90°
ঘ
60°

ব্যাখ্যা

Question: The circumcentre of a triangle ABC is 'O'. If ∠BAC = 85° and ∠BCA = 75°, then the value of ∠OAC is -

Solution:
According to the question,
Given:
∠BAC = 85°
∠BCA = 75°
∠OAC = ?

∠ABC + ∠BCA + ∠CAB = 180°
⇒ ∠ABC + 75° + 85° = 180°
∴ ∠ABC = 20°

∠COA = 2 × ∠ABC
⇒ ∠COA = 2 × 20 = 40°

In ΔAOC,
We know OC = OA
∴ ∠OAC = ∠OCA
∴ ∠OAC + ∠OCA + ∠COA = 180°
⇒ 2∠OAC = 180° - 40°
⇒ 2∠OAC = 140°
∴ ∠OAC = 70°

১,৯৯০.

A swimming pool is 25 m long and 18 m broad. When a number of men dive into the pool, the height of the water rises by 1 cm. If the average amount of water displaced by one of the men be 0.1 cubic meter, how many men are there in the pool?

ক
ক) 34
খ
খ) 42
গ
গ) 45
ঘ
ঘ) 47

ব্যাখ্যা

Question: A swimming pool is 25 m long and 18 m broad. When a number of men dive into the pool, the height of the water rises by 1 cm. If the average amount of water displaced by one of the men be 0.1 cubic meter, how many men are there in the pool?

Solution:
Volume of water displaced= 25 × 18 × (1/100) m³= 9/2 m³
Volume of water displaced by 1 man = 0.1m³
∴ Number of men are in the pool = (9/2)/0.1 = (9 × 10)/2 = 45 [0.1 m = 10 cm]

১,৯৯১.

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

ক
10 cm
খ
16 cm
গ
12 cm
ঘ
18 cm

ব্যাখ্যা

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

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

Let
the other parallel side = b cm

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

∴ The other parallel side is 12 cm.

১,৯৯২.

The ratio of the areas of a square of side 6 cm and an equilateral triangle of side 6 cm is:

ক
ক) 2 : √3
খ
খ) 4 : √3
গ
গ) 2 : √2
ঘ
ঘ) 3 : √3

ব্যাখ্যা

Required ratio : =(6 × 6) : (√3 × 6 × 6/4)
=1 : √3/4
= 4 : √3

১,৯৯৩.

If the size of a tile is 9" by 9", how many tiles are required to cover a 12 ft. wide and 18 ft. long floor?

ক
384
খ
216
গ
32
ঘ
24

ব্যাখ্যা

Question: If the size of a tile is 9" by 9", how many tiles are required to cover a 12 ft. wide and 18 ft. long floor?

Solution:
Side of the tile is 9" = 9/12 ft.
∴ Area of the tile is (9/12)² = 81/144 sq. ft.

Area of the floor = 12 × 18 = 216 sq. ft.

Number of tiles = 216/(81/144) = (216 × 144)/81 = 384

১,৯৯৪.

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

ক
ক) 5 m
খ
খ) 7.5 m
গ
গ) 9 m
ঘ
ঘ) 10 m

ব্যাখ্যা

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

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

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

∴ The length of the ladder is 9 m.

১,৯৯৫.

If A = π/3, B = A/2 then sin (A + B) =?

ক
0
খ
1
গ
2
ঘ
3

ব্যাখ্যা

Question: If A = π/3, B = A/2 then sin (A + B) =?

Solution:
B = A/2
= π/6

A + B = ( π/3) + ( π/6)
= π/2

sin(π/2 )
= 1

১,৯৯৬.

A square park is surrounded by a path of uniform width 3 meters all around it. The area of the path is 240 sq. meters. Find the perimeter of the park.

ক
52 m
খ
56 m
গ
60 m
ঘ
68 m

ব্যাখ্যা

Question: A square park is surrounded by a path of uniform width 3 meters all around it. The area of the path is 240 sq. meters. Find the perimeter of the park.

Solution:
Let, one side of the park is = x meter.
So, one side of the park with path = x + (3 + 3)
= x + 6

We know,
Area of the park = x²
Area of the path, (x + 6)² - x² = 240
⇒ x² + 12x + 36 - x² = 240
⇒ 12x + 36 = 240
⇒ 12x = 240 - 36
⇒ 12x = 204
⇒ x = 204/12
∴ x = 17

One side of the square = 17 m.
So, perimeter of the square = 4 × 17
= 68 m

১,৯৯৭.

The perimeter of an equilateral triangle is 84√3 cm. Find its height.

ক
44 cm
খ
52 cm
গ
42 cm
ঘ
41 cm

ব্যাখ্যা

Question: The perimeter of an equilateral triangle is 84√3 cm. Find its height.

Solution:
Given,
The perimeter of the equilateral triangle = 84√3 cm.

∴ Each side of the equilateral triangle
= (84√3)/3
= 28√3 cm.

We know,
The height of the equilateral triangle = (a√3)/2

∴ The height of the equilateral triangle will be
= (√3/2) × (28√3)
= 42 cm

১,৯৯৮.

The area of a square field is 24200 sq m. What is the perimeter of the square field?

ক
440√2 m.
খ
440 m
গ
110 m
ঘ
110√2 m

ব্যাখ্যা

Question: The area of a square field is 24200 sq m. What is the perimeter of the square field?

Solution:
The area of a square field is 24200 sq m.
∴ Length of the field √24200 m = √(2 × 121 × 100) = √(2 × 11² × 10²) = 110√2 m.

∴ The perimeter of the square = 4 × 110√2 = 440√2 m.

১,৯৯৯.

If secθ - tanθ = 1/√3, the value of secθ.tanθ = ?

ক
2/√3
খ
2/3
গ
1/2
ঘ
2

ব্যাখ্যা

Question: If secθ - tanθ = 1/√3, the value of secθ.tanθ = ?

Solution:
sec²θ - tan²θ = 1
⇒ (secθ - tanθ) (secθ + tanθ) = 1
⇒ 1/√3 (secθ + tanθ) = 1
⇒ secθ + tanθ = √3

2secθ = (1/√3) + √3 = 4/√3
⇒ secθ = 2/√3
⇒ secθ = sec30
⇒ θ = 30°

secθ.tanθ = (2/√3) × (1/√3)
= 2/3

২,০০০.

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
35π sq. meter
খ
50π sq. meter
গ
60π sq. meter
ঘ
90π sq. meter

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 = 10 and h = 8
r = √(l² - h²)
= √(10² - 8²)
= √(100 - 64)
= √36
= 6

∴ Curved surface area = πrl
= π × 6 × 10
= 60π sq. meter