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Geometry: Mensuration, Trigonometry

মোট প্রশ্ন২,০৮৫এই পাতা১০০প্রতি পাতা১০০
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উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

Geometry: Mensuration, Trigonometry

PrepBank · পাতা ৩ / ২১ · ২০১–৩০০ / ২,০৮৫

২০১.
The measurement of a rectangle is 16 feet by 12 feet. What is the area of the smallest circle that can cover this rectangle entirely (so that no part of the rectangle is outside the circle)?
  1. ক
    ক) 100π sq.ft
  2. খ
    খ) 100 sq.ft
  3. গ
    গ) 192 sq.ft
  4. ঘ
    ঘ) 128π sq.ft
ব্যাখ্যা
Question: The measurement of a rectangle is 16 feet by 12 feet. What is the area of the smallest circle that can cover this rectangle entirely (so that no part of the rectangle is outside the circle)?

Solution:

ABCD একটি বৃত্তস্থ আয়তক্ষেত্র 
যেখানে AB = CD = 16 ফুট 
AD = BC = 12 ফুট 
 
ΔABC সমকোণী 
AC2 = AB2 + BC2
AC2 = 162 + 122
AC2 = 256 + 144 
AC2 = 400
AC2 = 202
AC = 20 

ABCD আয়তক্ষেত্রের কর্ণ ACই হলো বৃত্তের ব্যাস 
বৃত্তের ব্যাসার্ধ r = 20/2 = 10 ফুট 

বৃত্তের ক্ষেত্রফল = π × 102
= 100π
২০২.
A pole of 48m long breaks such that parts are not completely separated and the upper part makes on angle 30° with the ground. At what height did the pole break?
  1. ক
    ক) 16 m
  2. খ
    খ) 14 m
  3. গ
    গ) 18 m
  4. ঘ
    ঘ) 20 m
ব্যাখ্যা
Question: A pole of 48m long breaks such that parts are not completely separated and the upper part makes on angle 30° with the ground. At what height did the pole break?

Solution:

খুঁটিটি মাটি হতে x মিটার উঁচুতে ভেঙ্গেছিল।

আমরা জানি,
sin30° = AB/AC
⇒ sin30° = x/(48 - x)
⇒ 1/2 = x/(48 - x)
⇒ 48 - x = 2x
⇒ 3x = 48
∴ x = 16
২০৩.
find the value of sin221° + cos221
  1. ক
    1
  2. খ
    2
  3. গ
    1/2
  4. ঘ
    4
ব্যাখ্যা

Question: find the value of sin221° + cos221°

Solution: 
sin221° + cos221°
= 1 [sin2θ + cos2θ = 1]

২০৪.
Three cubes with sides in the ratio 3 : 4 : 5 are melted fo form a single cube whose diagonal is 12√3 cm. The sides of the cubes are-
  1. ক
    3 cm, 4 cm, 8 cm
  2. খ
    4 cm, 8 cm, 10 cm
  3. গ
    6 cm, 8 cm, 10 cm
  4. ঘ
    None of these
ব্যাখ্যা
Question: Three cubes with sides in the ratio 3 : 4 : 5 are melted fo form a single cube whose diagonal is 12√3 cm. The sides of the cubes are-

Solution:
Let, the sides of the three cubes be 3x, 4x and 5x.
Then, Volume of the new cube = (3x)3 + (4x)3 + (5x)3
= 216x3

Edge of the new cube, ‍a3 = 216x3
⇒ a = (216x3)1/3
∴ a = 6x

The diagonal of the new cube = 6√3x

ATQ,
6√3x = 12√3
∴ x = 2

So, the side of cubes are (3 × 2) cm, (4 × 2) cm, (5 × 2) cm Or, 6 cm, 8 cm and 10 cm
২০৫.
Find the greatest value of sin4A + cos4A.
  1. ক
    1
  2. খ
    2
  3. গ
    0
  4. ঘ
    4
ব্যাখ্যা

Question: Find the greatest value of sin4A + cos4A.

Solution:
We know,
sin2A + cos2A = 1
⇒(sin2A + cos2A)2 = 12
⇒ (sin2A)2 + (cos2A)2 + 2sin2Acos2A = 1
⇒ sin4A + cos4A = 1 - 2sin2Acos2A
⇒ sin4A + cos4A = 1 - 2sin290°cos290° [since we need maximum value]
⇒ sin4A + cos4A = 1 - 2(1 × 0)
⇒ sin4A + cos4A = 1 - 0
∴ sin4A + cos4A = 1 

২০৬.
The diameter of a circle is equal to the perimeter of a square whose area is 576cm2. What is the circumference of the circle?
  1. ক
    ক) 24π
  2. খ
    খ) 36π
  3. গ
    গ) 72π
  4. ঘ
    ঘ) 96π
ব্যাখ্যা
Area of square = 576 cm2
Side of squared = √576
                          = 24 cm
Perimeter of square :
= 4a
= (4 × 24 cm
= 96 cm
= diameter of circle

∴ Circumference of circle :
           =πd =96×π=96π cm
২০৭.
In square ABCD below, what is the value of (AC)(AD)/(AB)(DC) =?
  1. ক
    ক) 1
  2. খ
    খ) 2
  3. গ
    গ) √2
  4. ঘ
    ঘ) 1/2
ব্যাখ্যা
প্রশ্ন: In square ABCD below, what is the value of (AC) (AD) / (AB) (DC) =?


সমাধান: 
ধরি, AB = BD = CD = AC = x

AD = √(AC2 + CD2)
= √(x2 + x2)
= √(2x2)
= x√2

(AC) (AD) / (AB) (DC)
= x.x√2/x.x
= √2
২০৮.
The area of a regular hexagon of side 3√2 cm is:
  1. ক
    24√3 cm2
  2. খ
    27√3 cm2
  3. গ
    21√3 cm2
  4. ঘ
    18√3 cm2
ব্যাখ্যা

Question: The area of a regular hexagon of side 3√2 cm is:

Solution:
A regular hexagon consists of 6 equilateral triangle
Area of regular hexagon
= 6 × (√3/4) × (side)2
= 6 × (√3/4) × (3√2)2
= 27√3 cm2

২০৯.
You are looking at a billboard 40m away with an angle of elevation of 30°. At what height is the billboard?
  1. ক
    ক) 20
  2. খ
    খ) 30
  3. গ
    গ) 40
  4. ঘ
    ঘ) 50
ব্যাখ্যা

Question: You are looking at a billboard 40m away with an angle of elevation of 30°. At what height is the billboard?

Solution:


ধরি,
উচ্চতা = XZ = h
অতিভুজ = XY = 40 মিটার

প্রশ্নমতে,
Sin30° = লম্ব/অতিভুজ
⇒ 1/2 = h/40
∴ 2h = 40 মিটার
h = 20

২১০.
If sinθ/cosθ = 1, then find θ.
  1. ক
    60°
  2. খ
    30°
  3. গ
    45°
  4. ঘ
    90°
ব্যাখ্যা

Question: If sinθ/cosθ = 1, then find θ.

Solution:
sinθ/cosθ = 1
⇒ tanθ = 1
⇒ tanθ = tan45°
∴ θ = 45°

২১১.
If the side of a square is increased by 10%, by what percent will the area be increased?
  1. ক
    16%
  2. খ
    21%
  3. গ
    32%
  4. ঘ
    100%
ব্যাখ্যা

Question: If the side of a square is increased by 10%, by what percent will the area be increased?

Solution:
Let the original side length = 10 units.

∴ Area = 10 × 10 = 100 square units

Again, 
After a 10% increase, the new side length = 10 + 10% of 10
= 10 + 1 = 11 units

∴ New area = 11 × 11 = 121 square units

∴ Increase in area = 121 - 100 square units
= 21 square units

∴ Percentage increase in area = (21/100) × 100%
= 21%

So the area will increase by 21%

২১২.
The side of an equilateral triangle is 4m. What is the height of the triangle?
  1. ক
    √3 m
  2. খ
    4√3 m
  3. গ
    3√3 m
  4. ঘ
    2√3 m
ব্যাখ্যা
Question: The side of an equilateral triangle is 4m. What is the height of the triangle?

Solution:
Given,
The side of an equilateral triangle = 4m

We know,
Area of an equilateral triangle = (√3/4) × 42
= (√3/4) × 16
= 4√3

Let,
the height of the triangle = h

We also know,
(1/2) × base × height = area
⇒ (1/2) × 4 × h = 4√3
⇒ h = (4√3 × 2)/4
∴ h = 2√3

So, the height of the triangle = 2√3 m
২১৩.
What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
  1. ক
    ক) 814
  2. খ
    খ) 820
  3. গ
    গ) 840
  4. ঘ
    ঘ) 844
ব্যাখ্যা

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 x 41) cm2.
Required number of tiles 
= (1517 x 902) / (41 x 41)= 814.

২১৪.
If tan3A = √3, then what is the value of A?
  1. ক
    ক) 60°
  2. খ
    খ) 40°
  3. গ
    গ) 30°
  4. ঘ
    ঘ) 20°
ব্যাখ্যা
Question: If tan3A = √3, then what is the value of A?

Solution:
Given,
tan3A = √3
⇒ tan3A = tan60°
⇒ 3A = 60°
∴ A = 20°
২১৫.
If the radius of a circle is increased by 100%, by what % is the area of the circle increased?
  1. ক
    100%
  2. খ
    200%
  3. গ
    300%
  4. ঘ
    400%
ব্যাখ্যা
Question: If the radius of a circle is increased by 100%, by what % is the area of the circle increased?

Solution:
ধরি, 
বৃত্তের ব্যাসার্ধ, r = 10 
∴ বৃত্তের ক্ষেত্রফল = πr2
= π (10)2 
= 100π 

আবার, 
বৃত্তের ব্যাসার্ধ 100% বৃদ্ধিতে, 
বৃত্তের নতুন ব্যাসার্ধ = 10 + 10 এর 100%
= 10 + 10 এর 100/100
= 20
∴ বৃত্তের নতুন ক্ষেত্রফল = πr2
= π (20)2 
= 400π

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

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

∴ বৃত্তের ক্ষেত্রফল শতকরা 100% বৃদ্ধি পায়
২১৬.
ΔABC is a triangle where AB is perpendicular to BC. Which equation is true?
  1. ক
    ক) AB2 = AC2 + BC2
  2. খ
    খ) BC2 = AB2 + AC2
  3. গ
    গ) AC2 = AB2 - BC2
  4. ঘ
    ঘ) AC2 = AB2 + BC2
ব্যাখ্যা
Question: ΔABC is a triangle where AB is perpendicular to BC. Which equation is true?

Solution: 


so, 
AC2 = AB2 + BC2
২১৭.
If both the length and the breadth of a rectangle are increased by 20%, what is the percentage increase in its area?
  1. ক
    40%
  2. খ
    36%
  3. গ
    42%
  4. ঘ
    None of these
ব্যাখ্যা

Question: If both the length and the breadth of a rectangle are increased by 20%, what is the percentage increase in its area?

Solution:
Let the original length = x
and breadth = y
∴ Original area = x × y = xy

After 20% increase,
 New length = x + 20% of x = x × (1 + 20/100)
= x × 1.2 = 1.2x
and new breadth = y × 1.2 = 1.2y

∴ New area = (1.2x) × (1.2y) = 1.44xy

∴ Increase in area = New area - Original area = 1.44xy - xy = 0.44xy

∴ Percentage increase in area = (Increase in area / Original area) × 100%
= (0.44xy/xy) × 100%
= 0.44 × 100%
= 44%

২১৮.
The area of a triangle is 216 cm2 and sides are in the ratio 3 : 4 : 5. The perimeter of the triangle is-
  1. ক
    12 cm
  2. খ
    36 cm
  3. গ
    72 cm
  4. ঘ
    144 cm
ব্যাখ্যা

Question: The area of a triangle is 216 cm2 and sides are in the ratio 3 : 4 : 5. The perimeter of the triangle is- 

Solution: 
32 + 42 = 52 

It is a right-angled triangle.
let, the sides 3x, 4x, 5x

(1/2) × 3x × 4x = 216 
⇒ 12x2 = 432 
⇒ x2 = 36 
⇒ x = 6

perimeter = (3 × 6) + (4 × 6) + (4 × 6)
= 72 cm

২১৯.
An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:
  1. ক
    ক) 2%
  2. খ
    খ) 2.02%
  3. গ
    গ) 4%
  4. ঘ
    ঘ) 4.04%
ব্যাখ্যা

100 cm is read as 102 cm.
A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
Percentage error =404/(100 x 100) x 100%= 4.04%

২২০.
If the area of the trapezium whose parallel sides are 6 cm and 10 cm is 32 sq. cm, then the distance between the parallel sides is:
  1. ক
    2 cm
  2. খ
    3 cm
  3. গ
    4 cm
  4. ঘ
    5 cm
ব্যাখ্যা
Question: If the area of the trapezium whose parallel sides are 6 cm and 10 cm is 32 sq. cm, then the distance between the parallel sides is:

Solution: 
Let the required distance be P cm
Then,
½ × (6 + 10) × P = 32
⇒ P = 4 cm
২২১.
A circle of radius 3 cm is drawn inscribed in a right angle triangle ABC, right angled at C. If AC is 10 Find the value of CB.
  1. ক
    10.5 cm
  2. খ
    (20/7)√58 cm
  3. গ
    23/2 cm
  4. ঘ
    None of these
ব্যাখ্যা
Solution: A circle of radius 3 cm is drawn inscribed in a right angle triangle ABC, right angled at C. If AC is 10 Find the value of CB.

Solution:

(Tangents from the external point are equal)
(x + 7)2 = 102 + (x + 3)2
⇒ x2 + 14x + 49 = 100 + x2 + 6x + 9
⇒ 8x = 60
∴ x = 7.5cm

∴ CB = x + 3 = 7.5 + 3 = 10.5 cm 
২২২.
একটি সিলিন্ডারের ভূমির ক্ষেত্রফল 100π ব‍র্গমিটার এবং আয়তন 900π ঘনমিটার হলে সিলিন্ডারের উচ্চতা কত?
  1. ক
    3 মিটার
  2. খ
    6 মিটার
  3. গ
    9 মিটার
  4. ঘ
    12 মিটার
  5. ঙ
    কোনটিই নয়
ব্যাখ্যা
প্রশ্ন: একটি সিলিন্ডারের ভূমির ক্ষেত্রফল 100π ব‍র্গমিটার এবং আয়তন 900π ঘনমিটার হলে সিলিন্ডারের উচ্চতা কত?

সমাধান:
ধরি, 
সিলিন্ডারের ব্যাসা‍র্ধ = r
আমরা জানি,
সিলিন্ডারের ভূমির ক্ষেত্রফল = πr2

সিলিন্ডারের ভূমির আয়তন = πr2h

প্রশ্নমতে,
πr2h/πr2 = 900π/100π
∴ h = 9
২২৩.
A pole 6m high casts a shadow 2√3m long on the ground, then the Sun's elevation is-
  1. ক
    ক) 60°
  2. খ
    খ) 45°
  3. গ
    গ) 30°
  4. ঘ
    ঘ) 90°
ব্যাখ্যা
প্রশ্ন: A pole 6m high casts a shadow 2√3m long on the ground, then the Sun's elevation is-

সমাধান:

tanθ = লম্ব/ভূমি
বা, tanθ = 6/(2√3)
বা, tanθ = √3
বা, tanθ = tan60°
∴ θ = 60°
২২৪.
A solid cube of side 6 is first painted pink and then cut into smaller cubes of side 2. How many of the smaller cubes have Paint on exactly two sides.
  1. ক
    ক) 13
  2. খ
    খ) 27
  3. গ
    গ) 30
  4. ঘ
    ঘ) 12
ব্যাখ্যা
Question: A solid cube  of side  6 is first painted pink and then cut into smaller cubes  of side 2.  How many of the smaller cubes  have  Paint on exactly two sides.

Solution: 
A solid cube of side 6 is first painted pink and then cut into smaller cubes of side 2.

This means, we get a 3 × 3 × 3 cube
To get the the sides with just 2 faces painted, picture the edges of the each of the sides of the cube.
Apart from the corners (which have 3 faces painted), all other small cubes have 2 faces painted.

In total, we have 12 cubes of side 2 with exactly 2 sides painted
২২৫.
The area of a square-shaped garden is 3600 square meters. What is the total length of the fence around it?
  1. ক
    140 m
  2. খ
    240 m
  3. গ
    280 m
  4. ঘ
    360 m
ব্যাখ্যা
Question: The area of a square-shaped garden is 3600 square meters. What is the total length of the fence around it?
(একটি বর্গাকার বাগানের ক্ষেত্রফল ৩৬০০ বর্গমিটার। বাগানের চারদিকে বেড়া দেওয়া আছে। বেড়ার মোট দৈর্ঘ্য কত?)

Solution: 
আমরা জানি,
বর্গক্ষেত্রের বাহু = √ক্ষেত্রফল
∴ বাহু = (√৩৬০০)
= ৬০ মিটার

আবার,
বর্গক্ষেত্রের পরিসীমা = ৪ × বাহু
= (৪ × ৬০) মিটার
= ২৪০ মিটার

∴ বেড়ার মোট দৈর্ঘ্য = ২৪০ মিটার।
২২৬.
If a 44 meter ladder is placed against a 22 meter wall such that it just reaches the top of the wall, the angle of elevation of the wall is-
  1. ক
    60º
  2. খ
    90º
  3. গ
    30º
  4. ঘ
    45º
ব্যাখ্যা

Question: If a 44 meter ladder is placed against a 22 meter wall such that it just reaches the top of the wall, the angle of elevation of the wall is-

Solution:
Given that,
Ladder's length = 44 m
Wall's height = 22 m

Perpendicular = Wall's height = 22 m
Hypotenuse = Ladder's length = 44 m

We know,
Sinθ = Perpendicular/Hypotenuse
⇒ Sinθ = 22/44
⇒ Sinθ = 1/2
⇒ Sinθ = sin30º
∴ θ = 30º

২২৭.
Find the area of a triangle whose sides are 15 cm, 8 cm, and 17 cm. 
  1. ক
    40 cm2
  2. খ
    30 cm2
  3. গ
    20 cm2
  4. ঘ
    60 cm2
ব্যাখ্যা

Question: Find the area of a triangle whose sides are 15 cm, 8 cm, and 17 cm.

Solution:
ত্রিভুজের তিন বাহু যথাক্রমে,
a = 15 cm, b = 8 cm, c = 17 cm

অর্ধপরিধি, S = (a + b + c)/2
= (15 + 8 + 17)/2
= 40/2 = 20 cm

আমরা জানি,
ত্রিভুজের ক্ষেত্রফল = √{S(S - a)(S - b)(S - c)}
= √{20(20 - 15)(20 - 8)(20 - 17)}
= √{20 × 5 × 12 × 3}
= √3600
= 60 cm2

অতএব, ত্রিভুজের ক্ষেত্রফল = 60 cm2

২২৮.
In a right triangle, the length of one of the legs is 4 and the length of the hypotenuse is 5. What is the length of the other leg?
  1. ক
    3
  2. খ
    6
  3. গ
    4
  4. ঘ
    3.5
ব্যাখ্যা
Question: In a right triangle, the length of one of the legs is 4 and the length of the hypotenuse is 5. What is the length of the other leg?

Solution: 
সমকোণী ত্রিভুজের অতিভুজ = 5
সমকোণ সংলগ্ন এক বাহু = 4
সমকোণ সংলগ্ন অপর বাহু = a 

পিথাগোরাসের উপপাদ্য অনুসারে,
a2 + 42 = 52
⇒ a2 + 16 = 25
⇒ a2 = 25 - 16
⇒ a2 = 9
⇒ a2 = 32
∴ a = 3
২২৯.
From a point C on a level ground, the angle of elevation of the top of a tower is 30 degree. If the tower is 100 meter high, find the distance from point C to the foot of the tower.
  1. ক
    ক) 170 meter
  2. খ
    খ) 172 meter
  3. গ
    গ) 173 meter
  4. ঘ
    ঘ) 167 meter
ব্যাখ্যা
.


Let AB be the tower
then∠ACB=30º
AB=100 meter
AB/AC= tan30º
=>100/AC= 1/√3
=>AC= √3/100
=>AC= 1.73×100
=>AC= 173meter

২৩০.
The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.
  1. ক
    66π m2
  2. খ
    60π m2
  3. গ
    50π m2
  4. ঘ
    48π m2
ব্যাখ্যা
Question: The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

Solution: 
Here, l = 10m
h = 8m

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

∴Curved surface area = πrl
= (π × 6 × 10) m2 = 60π m2
২৩১.
The area of a square is equal to the area of a parallelogram. If the base of the parallelogram is 45 meters and its height is 5 meters, what is the length of one side of the square?
  1. ক
    18 meters
  2. খ
    15 meters
  3. গ
    7 meters 
  4. ঘ
    12 meters 
ব্যাখ্যা

Question: The area of a square is equal to the area of a parallelogram. If the base of the parallelogram is 45 meters and its height is 5 meters, what is the length of one side of the square?

Solution:
Area of the parallelogram = Base × Height
= 45 × 5
= 225 square meters

Let,
Side of the square = k meters
∴ Area of the square = k2 square meters

ATQ,
k2 = 225
⇒ k = √225
∴ k = 15
Therefore, the length of one side of the square = 15 meters

২৩২.
The height of a cone is 12 cm and the radius of its base is 5 cm. What is the approximate volume of the cone in cubic centimeters?
  1. ক
    70π cm3
  2. খ
    50π cm3
  3. গ
    100π cm3
  4. ঘ
    None of the above
ব্যাখ্যা
Question: The height of a cone is 12 cm and the radius of its base is 5 cm. What is the approximate volume of the cone in cubic centimeters?
(একটি সমবৃত্তভূমিক কোণকের উচ্চতা ১২ সে.মি. এবং ভূমির ব্যাসার্ধ ৫ সে.মি. হলে, কোণকটির আয়তন প্রায় কত ঘন সে.মি.?)

Solution:
ধরি,
কোণকের ব্যাসার্ধ, r = ৫ সে.মি.
কোণকের উচ্চতা, h = ১২ সে.মি. 

আমরা জানি,
কোণকের আয়তন = (1/3)πr2h.
= (১/৩) × π × ৫২ × ১২
= ১০০π ঘন সে.মি.
২৩৩.
A square lawn with side 50 m long has a circular flower bed in the center. If the area of the lawn, excluding the flower bed, is 1884 m2  what is the radius of the circular flower bed?
  1. ক
    ক) 12 meters
  2. খ
    খ) 14 meters
  3. গ
    গ) 15 meters
  4. ঘ
    ঘ) 16 meters
ব্যাখ্যা
Question: A square lawn with side 50 m long has a circular flower bed in the center. If the area of the lawn, excluding the flower bed, is 1884 m2  what is the radius of the circular flower bed?

Solution: 
Area of the flower bed = Area of square lawn - Area of lawn excluding circular bed
= [(50)2 - 1884] m2 = 616 m2

Let the radius of the circular flower bed is r meters. 
Area of the circular flower bed is πr2

ATQ,
πr2 = 616
⇒ r2 = (7/22) × 616
⇒ r2 = 196
∴ r = 14
২৩৪.
If a regular square pyramid has a base of size 8 cm and height 30 cm. What is its volume?  
  1. ক
    ক) 256 cc
  2. খ
    খ) 350 cc
  3. গ
    গ) 425 cc
  4. ঘ
    ঘ) 640 cc
ব্যাখ্যা
Question: If a regular square pyramid has a base of size 8 cm and height 30 cm. What is its volume?  

Solution: 
Volume of the pyramid = (1/3) × 82 × 30 cm3
= 640 cm3 or 640 cc
২৩৫.
From the figure, which of the following must be true?
(I) x + y = 90
(II) x is 35 units greater than y
(III) x is 35 units less than y

  1. ক
    I only
  2. খ
    II only
  3. গ
    III only
  4. ঘ
    I and III only
ব্যাখ্যা

Question: From the figure, which of the following must be true?
(I) x + y = 90
(II) x is 35 units greater than y
(III) x is 35 units less than y


Solution:
চিত্রে কোণ x হলো ত্রিভুজটির একটি বহিঃস্থ কোণ। সুতরাং, এর মান বিপরীত অন্তঃস্থ কোণ দুটি, 35 এবং y-এর সমষ্টির সমান।

অর্থাৎ, x = y + 35

 এই সমীকরণ থেকে বোঝা যায় যে x এর মান y এর চেয়ে 35 একক বেশি। তাই, (II) বিবৃতিটি সত্য এবং (III) মিথ্যা।

এখন, যদি x একটি স্থূলকোণ (x > 90) হয়, তাহলে x + y এর মান 90 এর চেয়ে বেশি হবে। সুতরাং, x + y যে অবশ্যই 90 এর সমান হবে, এমন কোনো কথা নেই। তাই, (I) বিবৃতিটি অনিবার্যভাবে সত্য নয়।

অতএব, শুধুমাত্র (II) অবশ্যই সঠিক।

২৩৬.
A square and a circle have the same perimeter. The length of the side of the square is 22 cm. What is the area of the circle?
  1. ক
    576 square cm
  2. খ
    616 square cm
  3. গ
    720 square cm
  4. ঘ
    784 square cm
ব্যাখ্যা

Question: A square and a circle have the same perimeter. The length of the side of the square is 22 cm. What is the area of the circle?

Solution:
বর্গের পরিসীমা = 4 × বাহুর দৈর্ঘ্য
= 4 × 22 সেমি
= 88 সেমি

প্রশ্নমতে, বর্গ এবং বৃত্তের পরিসীমা সমান।
সুতরাং, বৃত্তের পরিধি = 88 সেমি

আমরা জানি,
বৃত্তের পরিধি = 2πr
⇒ 2πr = 88
⇒ 2 × (22/7) × r = 88
⇒ (44/7) × r = 88
⇒ r = 88 × (7/44)
∴ r = 14 সেমি

এখন, বৃত্তের ক্ষেত্রফল = πr2
= (22/7) × (14)2
= (22/7) × 196
= 22 × 28
= 616 বর্গ সেমি

২৩৭.
An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
  1. ক
    356 cm2
  2. খ
    576 cm2
  3. গ
    27342/23 cm2
  4. ঘ
    22275/28 cm2
  5. ঙ
    None of the above
ব্যাখ্যা
Question: An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
 
The area between two consecutive ribs is subtending = 360/8 = 45° at the center of the assumed flat circle.

Area between two consecutive ribs of circle
২৩৮.
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.
  1. ক
    60°
  2. খ
    45°
  3. গ
    40°
  4. ঘ
    35°
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.
(প্রদত্ত চিত্র অনুসারে, AB বৃত্তের ব্যাস এবং O কেন্দ্র। ∠BOD = ১৫° এবং ∠EOA = ৮৫° হলে, ∠ECA এর মান কত?)

Solution:
∠EOA = 85°, ∠BOD = 15°
∠EOD = 180° - (85° + 15°) = 80°

In ΔOED,
OE = OD (ব্যাসার্ধ)
∠OED = ∠ODE = 50°

In ΔOEC,
∠EOC = 80°+15° = 95°, ∠OEC =50°
∴ ∠ECA = 180°- (95 + 50°) = 35°
২৩৯.
A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq.ft per day, they approximately what time will be taken by the cow to graze the whole field ?
  1. ক
    4.86 days
  2. খ
    5.2 days
  3. গ
    6.16 days
  4. ঘ
    8 days
ব্যাখ্যা

Question: A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq.ft per day, they approximately what time will be taken by the cow to graze the whole field ?

Solution: 
Area of the field = π 142 sq.ft 
= 616  sq.ft 

Time required =616/100 
= 6.16 days 

২৪০.
There are two squares S1 and S2. The ratio of their areas is 4 : 25. If the side of S1 is 6 cm. What is the side of S2?
  1. ক
    20 cm
  2. খ
    5 cm
  3. গ
    15 cm
  4. ঘ
    12 cm
ব্যাখ্যা
Question: There are two squares S1 and S2. The ratio of their areas is 4 : 25. If the side of S1 is 6 cm. What is the side of S2?

Solution:
Let,
The Area of S1 = 4x2
The Area of S2 = 25x2

∴ Side of S1 = 2x
∴ Side of S2 = 5x

ATQ,
2x = 6
∴ x = 3

∴ Side of S2 = 5 × 3 = 15 cm
২৪১.
The floor of a company's office has an area of 20,000 square feet. If the floor is in the shape of a square, approximately how many feet long is each side?
  1. ক
    140
  2. খ
    450
  3. গ
    500
  4. ঘ
    1000
  5. ঙ
    None of these
ব্যাখ্যা
Question: The floor of a company's office has an area of 20,000 square feet. If the floor is in the shape of a square, approximately how many feet long is each side?

Solution:
The floor is in the shape of a square
Area of the floor 20000 square feet

∴ Length of each side = √20000 feet
= 141.42 feet

∴ We can say Length of each side is approximately 140 feet
২৪২.

If ∠XYZ in the figure above is a right angle, what is the value of x?
  1. ক
    155°
  2. খ
    145°
  3. গ
    125°
  4. ঘ
    110°
ব্যাখ্যা
Question:

If ∠XYZ in the figure above is a right angle, what is the value of x?

Solution:

∠XYZ = 90°
∠XYA = 90° - 55° = 35°
 
∠x + ∠XYA = 180°
⇒ ∠x = 180° - 35°
∴ ∠x = 145°
 
২৪৩.
The area of the triangle whose vertices are given by the coordinates (1, 2), (- 4, -3) and (4, 1) is:
  1. ক
    10 sq. units
  2. খ
    14 sq. units
  3. গ
    8 sq. units
  4. ঘ
    12 sq. units
ব্যাখ্যা

Question: The area of the triangle whose vertices are given by the coordinates (1, 2), (- 4, -3) and (4, 1) is:

Solution:
Given that, 
Vertices of triangle = (1, 2), (- 4, -3), (4, 1)

We know,
Area of triangle = (1/2) × |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| [whose vertices are (x1, y1), (x2, y2) and (x3, y3)]
= (1/2) × |1(- 3 - 1) + (- 4) (1 - 2) + 4{2 - (- 3)}|
= (1/2) × |(- 4) + 4 + 20|
= 20/2
= 10 sq. units

So the area of the triangle is 10 sq. units.

২৪৪.
A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. What is the volume of the cone so formed?
  1. ক
    7π cm3
  2. খ
    8π cm3
  3. গ
    9π cm3
  4. ঘ
    10π cm3
  5. ঙ
    12π cm3
ব্যাখ্যা

Given that, radius, r = 3 cm and height, h = 4 cm
Therefore, volume, V = (1/3) × πr2h
                                  = (1/3) × π × 32 × 4
                                  = 12π cm3
২৪৫.
The distance from the centre of a circle of radius 5 cm to a chord of lengh 8 cm is-
  1. ক
    ক) 5 cm
  2. খ
    খ) 4 cm
  3. গ
    গ) 3 cm
  4. ঘ
    ঘ) 2 cm
ব্যাখ্যা

Let AB = 8 cm. And, be the chord of the circle with radius AO = 5 cm

Draw OP⊥AB, join OA
according to the theorem
AP = 1/2 AB = 1/2 × 8 = 4 cm
In △APO, ∠A = 90°
∴ (AO)2 = (AP)2 + (OP)2
OP = (AO)2 − (AP)2
= √(52 − 42)
= √9
∴ OP = 3cm

২৪৬.
The area of a square and rectangle are equal. The length of the rectangle is greater than the length of any side of the square by 6 cm and the breadth is less than 4 cm. Find the perimeter of the rectangle.
  1. ক
    66 cm
  2. খ
    52 cm
  3. গ
    48 cm
  4. ঘ
    42 cm
  5. ঙ
    None
ব্যাখ্যা
Question: The area of a square and rectangle are equal. The length of the rectangle is greater than the length of any side of the square by 6 cm and the breadth is less than 4 cm. Find the perimeter of the rectangle.

Solution:
Let,
the length of each side of the square be x cm.
Then, the length of rectangle = (x + 6) cm
and its breadth = (x - 4) cm

ATQ,
(x + 6)(x - 4) = x2
⇒ x2 + 6x - 4x - 24 = x2
⇒ 2x = 24
∴ x = 12

Length = 12 + 6 = 18 cm
Breadth = 12 - 4 = 8 cm

∴ Perimeter = 2(length + breadth) = 2 (18 + 8) = 2 × 26 = 52 cm
২৪৭.
একটি সামন্তরিকের দুইটি সন্নিহিত বাহুর দৈর্ঘ্য যথাক্রমে 7 সেন্টিমিটার এবং 5 সেন্টিমিটার হলে, এর পরিসীমার অর্ধেক কত?
  1. ক
    12
  2. খ
    20
  3. গ
    24
  4. ঘ
    28
ব্যাখ্যা
প্রশ্ন: একটি সামন্তরিকের দুইটি সন্নিহিত বাহুর দৈর্ঘ্য যথাক্রমে 7 সেন্টিমিটার এবং 5 সেন্টিমিটার হলে, এর পরিসীমার অর্ধেক কত?

সমাধান:
আমরা জানি,
সামান্তরিকের পরিসীমা = 2 × সন্নিহিত বাহুদ্বয়ের সমষ্টি
= 2 × (7 + 5)
= 24 সে.মি.

∴ পরিসীমার অর্ধেক = 24/2 সে.মি.
= 12 সে.মি.
২৪৮.
A closed box made of wood of uniform thickness has length, breadth and height as 14cm, 12cm and 10cm, respectively. If the thickness of the wood is 1cm, the inner surface area is:
  1. ক
    ক) 296 cm2
  2. খ
    খ) 316 cm2
  3. গ
    গ) 376 cm2
  4. ঘ
    ঘ) 592 cm2
ব্যাখ্যা
Internal dimensions of the box are : 
Length =(14 - 2)cm =12cm
Breadth =(12 - 2)cm =10cm
Height =(10 - 2)= 8cm
∴ Inner surface area =2×(12 × 10 + 8 × 10 + 8 × 12)cm2
                                  = 2 × (120 + 80 + 96) cm2 
                                  = 592 cm2 
২৪৯.
If , what is the value of A?
  1. ক
    30°
  2. খ
    45°
  3. গ
    60°
  4. ঘ
    90°
ব্যাখ্যা

Question: If , what is the value of A?

Solution:

২৫০.
If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?
  1. ক
    60°
  2. খ
    45° 
  3. গ
    30° 
  4. ঘ
    70°
ব্যাখ্যা

Question: If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?

Solution:
 
খুঁটির উচ্চতা AB = 20 m
খুঁটির ছায়ার দৈর্ঘ্য BC =20√3 m

ΔABC হতে পাই,
tanθ = লম্ব/ভূমি
বা, tanθ = AB/BC
বা, tanθ = 20/(20√3)
বা, tanθ = 1/√3
বা, tanθ = tan30°
∴ θ = 30°

২৫১.
The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle:
  1. ক
    ক) 81.9 cm
  2. খ
    খ) 25.8 cm
  3. গ
    গ) 33.5 cm
  4. ঘ
    ঘ) 72.7 cm
  5. ঙ
    ঙ) 85.3 cm
ব্যাখ্যা

Radius of incircle = a/2√3.
Area of incircle = (π × a2)/12 cm2
∴ πa2/12 = 154
⇒ a2 = (154 × 12 × 7)/22
⇒ a = 14√3
∴ Perimeter of the triangle = (3 × 14√3) cm
= (14 × 1.732) cm
= 72.7 cm (approx.)

২৫২.
How many bricks are required to build a wall that is 8 m long, 6 m high, and 11 cm thick, if each brick measures 25 cm × 60 cm × 11 cm?
  1. ক
    220
  2. খ
    320
  3. গ
    120
  4. ঘ
    240
ব্যাখ্যা

Question: How many bricks are required to build a wall that is 8 m long, 6 m high, and 11 cm thick, if each brick measures 25 cm × 60 cm × 11 cm?

Solution:
Given,
Wall long = 8 m = 800 cm
Wall thick = 11 cm
Wall high = 6 m = 600 cm

∴ Volume of the wall = (800 × 600 × 11) cm3

∴ Volume of the brick = 25 cm × 11 cm × 60 cm

∴ bricks need to build the wall = Volume of the wall ÷ Volume of the brick
= (800 × 600 × 11) ÷ (25 × 11 × 60)
= 320

২৫৩.
What is the value of cos 120°? 
  1. ক
    2 
  2. খ
    1/2
  3. গ
    - 1/2
  4. ঘ
    1
ব্যাখ্যা

Question: What is the value of cos 120°?

Solution: 
cos 120°
= cos(90° + 30°)
= - sin 30°
= - 1/2

২৫৪.
A 314 cm long copper strip is bent into a round wheel. What is the wheel’s diameter? 
  1. ক
    50 cm
  2. খ
    10 cm
  3. গ
    100 cm
  4. ঘ
    80 cm
ব্যাখ্যা

Question: A 314 cm long copper strip is bent into a round wheel. What is the wheel’s diameter?

Solution:
Length of the strip = Circumference of the wheel = 314 cm

We know,
Circumference C = π × d

wheel’s diameter, d = C ÷ π = 314 ÷ 3.14 = 100 cm

২৫৫.
If a 40√3 meter ladder is placed against a 60 meter wall such that it just reaches the top of the wall, the angle of elevation of the wall is -
  1. ক
    30°
  2. খ
    45°
  3. গ
    60°
  4. ঘ
    90°
ব্যাখ্যা

Question: If a 40√3 meter ladder is placed against a 60 meter wall such that it just reaches the top of the wall, the angle of elevation of the wall is -

Solution:

এখানে,
দেয়ালের উচ্চতা (লম্ব) = 60 মিটার
মইয়ের দৈর্ঘ্য (অতিভুজ) = 40√3 মিটার
উন্নতি কোণ (angle of elevation) = θ

আমরা জানি,
sinθ = লম্ব/অতিভুজ
⇒ sinθ = 60/40√3
⇒ sinθ = 3/2√3
⇒ sinθ = √3/2
⇒ sin θ = sin60°
∴ θ = 60°

সুতরাং, দেয়ালের উন্নতি কোণ ((angle of elevation) = 60°

২৫৬.
The perimeter of a rectangle is 30cm. If the breadth of the rectangle is 6cm, then the ratio of the length and breadth will be-
  1. ক
    3 : 2
  2. খ
    2 : 3
  3. গ
    3 : 4
  4. ঘ
    4 : 3
ব্যাখ্যা

Question: The perimeter of a rectangle is 30cm. If the breadth of the rectangle is 6cm, then the ratio of the length and breadth will be-

Solution: 
Let the length be x cm.

Now
2(6 + x) = 30 
⇒ 6 + x = 15
⇒ x = 15 - 6
⇒ x = 9
∴ Length = 9 cm

∴ Required ratio = Length : Breadth
= 9 : 6
= 3 : 2

২৫৭.
A square and a circle have the same perimeter. The side of the length of square is 22 cm, what is the area of the circle?
  1. ক
    375 sq. cm
  2. খ
    225 sq. cm
  3. গ
    154 sq. cm
  4. ঘ
    616 sq. cm
ব্যাখ্যা
Question: A square and a circle have the same perimeter. The side of the length of square is 22 cm, what is the area of the circle?

Solution:
Perimeter of the square = 4 × 22
= 88 cm

∴ Circumference of circle = 88 cm
⇒ 2πr = 88
⇒ r = 88/2π
⇒ r = (88 × 7)/(2 × 22)
∴ r = 14 cm

∴ Area of circle = πr2
= (22/7) × (14)2
= 616 sq. cm
২৫৮.
What is the solution of 2cos2θ + 3sinθ - 3 = 0; where θ is an acute angle.
  1. ক
    ক) 90°
  2. খ
    খ) 30°
  3. গ
    গ) 60°
  4. ঘ
    ঘ) 45°
ব্যাখ্যা
Question: What is the solution of 2cos2θ + 3sinθ - 3 = 0; where θ is an acute angle.

Solution:
2cos2θ + 3sinθ - 3 = 0
⇒ 2(1 - sin2θ) + 3sinθ - 3 = 0
⇒ 2{(1 + sinθ) (1 - sinθ)} - 3(1 - sinθ) = 0
⇒ (1 - sinθ) {2(1 + sinθ) - 3} = 0
⇒ (1 - sinθ) (2sinθ - 1) = 0

হয়,
1 - sinθ = 0
⇒ sinθ = 1
⇒ sinθ = sin90°
θ = 90°

অথবা,
2sinθ - 1 = 0
⇒ sinθ = 1/2
⇒ sinθ = ‍sin30°
∴ θ = 30°
যেহেতু, θ সূক্ষ্মকোণ,  θ = 30°
২৫৯.
If the diameter of a circle is decreased by 100%, by what percentage is the area of the circle decreased?
  1. ক
    500%
  2. খ
    400%
  3. গ
    300%
  4. ঘ
    200%
  5. ঙ
    None
ব্যাখ্যা
প্রশ্ন: If the diameter of a circle is decreased by 100%, by what percentage is the area of the circle decreased?

সমাধান:
ধরি,
বৃত্তের ব্যাস = 2 সে.মি.
তাহলে, ব্যাসার্ধ = 2/2 = 1 সে.মি.
তাহলে, বৃত্তের ক্ষেত্রফল = π(1)2 = π বর্গ সে.মি.

100% হ্রাসে,
নতুন ব্যাস = 2 - {2 × (100/100)} সেমি = 0 সে.মি.
নতুন ব্যাসার্ধ = 0
নতুন ক্ষেত্রফল = π(0)2 = 0 বর্গ সে.মি.

ক্ষেত্রফল হ্রাস = π - 0 = π বর্গ সে.মি.

∴ ক্ষেত্রফল শতকরা হ্রাস পায় = π  × 100%
২৬০.

If O is the center of the circle above and the length of arc RSP is twice the length of arc PQR, then x equals-
  1. ক
    60°
  2. খ
    90°
  3. গ
    100°
  4. ঘ
    150°
  5. ঙ
    120°
ব্যাখ্যা
Question:

If O is the center of the circle above and the length of arc RSP is twice the length of arc PQR, then x equals-

Solution:
The ratio of smaller arc (PQR) to the larger arc (PSR) = 1 : 2.
∴ The ratio of angle will be same.
Total angle is 360°
∴ x = (1/3) × 360°
∴ x = 120°
২৬১.
A circle has a circumference of 24π. A square is inscribed inside the circle. What is the perimeter of the square?
  1. ক
    96π
  2. খ
    72√2
  3. গ
    108
  4. ঘ
    48√2
ব্যাখ্যা

Question: A circle has a circumference of 24π. A square is inscribed inside the circle. What is the perimeter of the square?

Solution: 
Given that, 
The circumference of the circle is 24π

We know, 
Circumference, 2πr = 24π
⇒ 2r = 24
∴ diameter = 24
and radius = 24/2 = 12
Since the square is inscribed in the circle, the diagonal of the square equals the diameter of the circle.
∴ Diagonal of the square = 24

Let the side length of the square be s.
For a square, diagonal = s√2
So,
⇒ s√2 = 24
⇒ s = 24/√2
⇒ s = 24√2/2
∴ s = 12√2
∴ Perimeter of the square = 4 × side = 4 × 12√2 = 48√2

So the perimeter of the square is 48√2.

২৬২.
In the triangle ABC, AB = 12 cm and AC = 8 cm, and ∠BAC = 60°. What is the value of the length of the side BC? 
  1. ক
    6√3 cm
  2. খ
    10 cm
  3. গ
    10√2 cm
  4. ঘ
    4√7 cm
ব্যাখ্যা

Question: In the triangle ABC, AB = 12 cm and AC = 8 cm, and ∠BAC = 60°. What is the value of the length of the side BC? 

Solution:
Given that,
In the triangle, ABC, AB = 12 cm and AC = 8 cm, and ∠BAC = 60°.


We know,
According to the law of cosine, if a, b, and c are three sides of a triangle ΔABC and ∠A is the angle between AC and AB then, a2 = b2 + c2 - 2bc × cos∠A

According to the concept,
BC2 = AB2 + AC2 - 2 × AB × AC × cos60°
⇒ BC2 = 122 + 82 - 2 × 12 × 8 × (1/2)
⇒ BC2 = 144 + 64 - 96 = 112
⇒ BC = √112 = √(16 × 7)
⇒ BC = 4√7

∴ The measure of BC is 4√7 cm.

২৬৩.
A right-angled triangle has hypotenuse 50 units and one side 30 units. Find the area.
  1. ক
    600 sq. units
  2. খ
    400 sq. units
  3. গ
    550 sq. units
  4. ঘ
    480 sq. units
ব্যাখ্যা
Question: A right-angled triangle has hypotenuse 50 units and one side 30 units. Find the area.

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

Given that,
Base = 30 and Hypotenuse = 50

Form Pythagoras Theorem,
Height2 = Hypotenuse2 - Base2
= 502 - 302
= 2500 - 900
⇒ Height2 = 1600
∴ Height = 40

Area = (1/2) × base × height
= (1/2) × 30 × 40
= 600 sq. units
২৬৪.
Question:
  1. ক
    1/2
  2. খ
    - 3
  3. গ
    5
  4. ঘ
    3/5
ব্যাখ্যা
Question:


Solution:
২৬৫.
If 3 cm is the length of a median of an equilateral triangle, then the area is -
  1. ক
    ক) 2/√3 sq cm
  2. খ
    খ) 3/√3 sq cm
  3. গ
    গ) 3√3 sq cm
  4. ঘ
    ঘ) 2√3 sq cm
ব্যাখ্যা
Consider an equilateral triangle ABC having sides a and a median AD of length x unit'.
In an equilateral triangle, the median is always the perpendicular bisector of the triangle. 
So, BD=a/2

In triangle ABD, by Pythagoras theorem, we have
AB2=AD2+BD2
⟹a2=x2+(a​/2)2
⟹a2=x2+ a2/4​
⟹3a2​/4=x2
or,a2=4x2/3​

Now, area of equilateral triangle =√3​a2​/4
                                                     =​​(√3​​/4) × (4x2/3​)
                                                   =√3x2/3

 length of a median of an equilateral triangle = 3 cm

so,  area of equilateral triangle = √3x2/3 = √3×32/3 = 3√3 sq cm
২৬৬.
If cos(θ - 15°) = √3/2 then, sin2θ = ?
  1. ক
    0
  2. খ
    1
  3. গ
    √3/2
  4. ঘ
    1/2
ব্যাখ্যা
Question: If cos(θ - 15°) = √3/2 then, sin2θ = ?

Solution:
Given,
cos(θ - 15°) = √3/2
⇒ cos(θ - 15°) = cos30°
⇒ θ - 15° = 30°
∴ θ = 45°

Now,
sin2θ = (sin 45°)2
= (1/√2)2
= 1/2
২৬৭.
The angle of elevation of a ladder learning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is -
  1. ক
    ক) 8.2 m
  2. খ
    খ) 6.5 m
  3. গ
    গ) 4.6 m
  4. ঘ
    ঘ) 9.2 m
ব্যাখ্যা
Question: The angle of elevation of a ladder learning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is -

Solution:

Let AB be the wall and BC be the ladder.

 Then, ∠ACB = 60°
and AC = 4.6 m

We know,
cos∠ACB = AC/BC
⇒ cos60° = AC/BC
⇒ AC/BC= 1/2
⇒ BC = 2 × AC
⇒ BC = 2 × 4.6
∴ BC = 9.2 m
২৬৮.
If the length of diagonal AC of a square ABCD is 5.2 cm, then the area of the square is:
  1. ক
    ক) 10.52 sq.cm
  2. খ
    খ) 11.52 sq.cm
  3. গ
    গ) 12.52 sq.cm
  4. ঘ
    ঘ) 13.52 sq.cm
ব্যাখ্যা
p>Area of the square :
=[1/2×(5.2)2] sq.cm
= (1/2×27.04) sq.cm
=13.52 sq.cm

২৬৯.
The ratio of length and breadth of a rectangular park is 7 : 5. A man runs along its boundary at 8 km/hr and takes 9 minutes for one round. Find its area in sq. meters.
  1. ক
    8750 sq. m.
  2. খ
    65500 sq. m.
  3. গ
    87500 sq. m.
  4. ঘ
    7500 sq. m.
ব্যাখ্যা
Question: The ratio of length and breadth of a rectangular park is 7 : 5. A man runs along its boundary at 8 km/hr and takes 9 minutes for one round. Find its area in sq. meters.

Solution:
One round of the park is equal to the perimeter of the park.
So, by completing one round, the man covers a distance equal to the perimeter of the park.
Now,
Distance or perimeter = speed × time
= 8 × (9/60)
= 1.2 km
= 1200 meters

Let,
Length = 7x and breadth = 5x
So, Perimeter,
2(7x + 5x) = 1200
⇒ 24x = 1200
∴ x =1200/24 = 50 meters

So, Length = 7 × 50 = 350 meters
And, Breadth = 5 × 50 = 250 meters

Area = Length × Breadth
= 350 × 250
= 87500 sq. m.
২৭০.
If secθ + tanθ = x, then tanθ is -
  1. ক
    ক) (x2 + 1)/x
  2. খ
    খ) (x2 - 1)/x
  3. গ
    গ) (x2 + 1)/2x
  4. ঘ
    ঘ) (x2 - 1)/2x
ব্যাখ্যা

We know that,
sec2θ - tan2θ = 1
⇒ (secθ + tanθ)(secθ - tanθ) =1
⇒ x (secθ - tanθ) = 1
⇒ secθ - tanθ = 1/x ...... (i)
again, secθ + tanθ = x ...... (ii)
From (ii) - (i)
2tanθ = x - 1/x = (x2 - 1)/x
⇒ tanθ = (x2 - 1)/2x

২৭১.
If the diagonal and the area of a rectangle are 25 m and 168 m2, what is the width of the rectangle?
  1. ক
    ক) 12 m
  2. খ
    খ) 10 m
  3. গ
    গ) 7 m
  4. ঘ
    ঘ) 5 m
ব্যাখ্যা
Question: If the diagonal and the area of a rectangle are 25 m and 168 m2, what is the width of the rectangle?

Solution: 
let, length x and width y 
area = xy = 168 

diagonal = √(x2 + y2) = 25
⇒ x2 + y2 = 252

(x + y)2 = x2 + 2xy + y2
= 625 + 2 × 168
= 625 + 336
= 961

x + y = √961 = 31 

(x - y)2 = x2 - 2xy - y2
= 625 - 2 × 168
= 625 - 336
= 289

x - y = √289 
= 17 

x + y - x + y = 31 - 17 
⇒ 2y = 14
∴ y = 7 m
২৭২.

In the figure above, what is the value of d?
  1. ক
    ক) 2
  2. খ
    খ) 2.5
  3. গ
    গ) 3.5
  4. ঘ
    ঘ) 2.75
ব্যাখ্যা
Question: 

In the figure above, what is the value of d?

Question: 
AC2 = AB2 + BC2 
⇒ (d + 3d)2 = 62 + 82
⇒ (4d)2 = 36 + 64
⇒ 16d2 = 100
⇒ d2 = 100/16
⇒ d = 10/4
∴ d = 2.5 
২৭৩.
A square room has a square carpet symmetrically placed in it. This leaves an uncovered area of 9 meter2. The area of the whole room is 25 meter2. What is the length of the one side of the carpet?
  1. ক
    2 meter
  2. খ
    4 meter
  3. গ
    6 meter
  4. ঘ
    8 meter
ব্যাখ্যা
Question: A square room has a square carpet symmetrically placed in it. This leaves an uncovered area of 9 meter2. The area of the whole room is 25 meter2. What is the length of the one side of the carpet?

Solution:
মনেকরি 
কার্পেটের এক বাহুর দৈর্ঘ্য x মিটার 
 
প্রশ্নমতে,
25 - 9 = x2
⇒ 16 = x2
⇒ 42 = x2
∴ x = 4
২৭৪.
A rectangular water reservoir contains 75000 liters of water. If the length of the reservoir is 5 m and the breadth is 3 m, the depth of the reservoir will be - 
  1. ক
    3 m
  2. খ
    4 m
  3. গ
    5 m
  4. ঘ
    7 m
ব্যাখ্যা
Question: A rectangular water reservoir contains 75000 liters of water. If the length of the reservoir is 5 m and the breadth is 3 m, the depth of the reservoir will be - 

Solution: 
1 m3 = 1000 litre
⇒ 75000 litre = 75000/1000
= 75 m3 

75 =3 × 5 × depth 
∴ depth = 75/15
= 5 m
২৭৫.
PQRS has an area equal to 28 m2. QR is parallel to PS. QP is perpendicular to PS. If QR is 6 m and PS is 8 m, then what is RS?
  1. ক
    ক) √5 m
  2. খ
    খ) 2√5 m
  3. গ
    গ) 4 m
  4. ঘ
    ঘ) 5√2 m
ব্যাখ্যা
Question: PQRS has an area equal to 28 m2. QR is parallel to PS. QP is perpendicular to PS. If QR is 6 m and PS is 8 m, then what is RS?


Solution: 
area of trapezoid = (1/2) (QR + PS) QP = 28
⇒ (6 + 8) QP = 56 
⇒ 14 × QP = 56 
∴ QP = 56/14 = 4 


ST = PS - PT
= PS - QR
= 8 - 6
= 2m

RS2 = RT2 + ST2 = 42 + 22 
= 16 + 4
= 20 m

∴ RS = √20 m
= √(4×5)
= 2√5 m
২৭৬.
If each side of the square is increased by 20%, what will be the ratio between the new area and the original area of the square?
  1. ক
    9 : 5 
  2. খ
    15 : 7 
  3. গ
    27 : 13 
  4. ঘ
    36 : 25
ব্যাখ্যা

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

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

The side of new square is x + 20% of x
= x + (x/5)
= 6x/5

∴ The area of new square is
= (6x/5)2
= (36x2)/25

∴ The ratio between the new area and the original area of the square = {(36x2)/25} : x2
= 36/25 : 1
= 36 : 25

২৭৭.
A circular floor with a radius of 7 cm is 60% carpeted. What is the area of the uncovered section?
  1. ক
    60.60  cm2
  2. খ
    61.60  cm2
  3. গ
    63.60  cm2
  4. ঘ
    62.60  cm2
ব্যাখ্যা
Question: A circular floor with a radius of 7 cm is 60% carpeted. What is the area of the uncovered section?

Solution : 
Area of the floor = πr2 
= (22/7) × 72 cm2
= (22/7) × 49 cm2
= 154 cm2

If 60% of its area is carpeted, then uncovered area = (100 - 60)%
= 40% 

∴ the area of the uncovered section = 154 × 40%  cm2
= 154 × (40/100) cm2
= 6160/100 cm2
= 61.60  cm2
২৭৮.
If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?
  1. ক
    60°
  2. খ
    45°
  3. গ
    30°
  4. ঘ
    70°
ব্যাখ্যা
Question: If a 20-meter tall pole creates a shadow of length 20√3 meters, what is the angle of elevation of the sun?

Solution:

খুঁটির উচ্চতা AB = 20 m
খুঁটির ছায়ার দৈর্ঘ্য BC =20√3 m

ΔABC হতে পাই,
tanθ = লম্ব/ভূমি
বা, tanθ = AB/BC
বা, tanθ = 20/(20√3)
বা, tanθ = 1/√3
বা, tanθ = tan30°
∴ θ = 30°
২৭৯.
A wheel rotates 10 times per minutes and moves 25m during each rotation. How many meters does the wheel moves in 45 minutes?
  1. ক
    8756m
  2. খ
    11050m
  3. গ
    11250m
  4. ঘ
    None of these
ব্যাখ্যা
Question: A wheel rotates 10 times per minutes and moves 25m during each rotation. How many meters does the wheel moves in 45 minutes?

Solution:
1 মিনিটে চাকাটি অতিক্রম করে = 10 × 25 = 250 মিটার
45 মিনিটে চাকাটি অতিক্রম করে = (250 × 45) = 11250 মিটার

∴ চাকাটি 45 মিনিটে 11250 মিটার পথ অতিক্রম করবে ।


২৮০.
What is the length of the diagonal of a square whose area is 4 times of another square with diagonal as 5√2 cm?
  1. ক
    10 cm
  2. খ
    10√2 cm
  3. গ
    12√2 cm
  4. ঘ
    10√5 cm
ব্যাখ্যা
Question: What is the length of the diagonal of a square whose area is 4 times of another square with diagonal as 5√2 cm?

Solution:
Area of square = (1/2) × (length of diagonal)2

Area of square2 =(1/2) × (5√2)2  
= 25 cm2

Area of square1 = 4 × 25 = 100 cm2

∴ Length of diagonal of square1 = √(2 × area)
= √(2 × 100)
= 10√2 cm
২৮১.
The total surface area of a rectangular solid cube of length 5 cm, width 3 cm and height 2 cm is - 
  1. ক
    31 cm
  2. খ
    62 cm
  3. গ
    62 cm3
  4. ঘ
    62 cm2
ব্যাখ্যা
Question: The total surface area of a rectangular solid cube of length 5 cm, width 3 cm and height 2 cm is - 

Solution : 
the formula for the total surface area A is:
A = 2(lw + lh + wh)

The total surface area of a rectangular solid cube of length 5 cm, width 3 cm and height 2 cm is = 2{(5 × 3) + (3 × 2) + (2 × 5)} cm2
= 2(15 + 6 + 10) cm2
= 62 cm2 
২৮২.
If Asinθ = 1 and Acosθ = √2, find the value of (3√2/tanθ) + 1 =?
  1. ক
    6
  2. খ
    8
  3. গ
    7
  4. ঘ
    9
ব্যাখ্যা
Question: If Asinθ = 1 and Acosθ = √2, find the value of (3√2/tanθ) + 1 =?

Solution: 
Asinθ = 1
Acosθ = √2

∴ Asinθ/Acosθ = 1/√2
tanθ = 1/√2

(3√2/tanθ) + 1
= 3√2/(1/√2) + 1
= 6 + 1
= 7
২৮৩.
If sec(x − 30°) = 2, then tan x = ?
  1. ক
    √3
  2. খ
    1/√2
  3. গ
    1/√3
  4. ঘ
    None of the above
ব্যাখ্যা
sec (x − 30°) = 2
Or, sec (x - 30°) = sec 60°
Or, x - 30° = 60°
Or, x = 90°
∴ tan 90° = not defined
২৮৪.
What is the slope of a line perpendicular to the line whose equation is 3x + 4y = 12?
  1. ক
    4/3
  2. খ
    - 3/4
  3. গ
    5/2
  4. ঘ
    3/5
ব্যাখ্যা

Question: What is the slope of a line perpendicular to the line whose equation is 3x + 4y = 12?

Solution:
প্রদত্ত সরল রেখার সমীকরণ: 3x + 4y = 12
y = mx + c আকারে লিখি, যেখানে m হলো রেখার ঢাল।

4y = - 3x + 12
⇒ y = (- 3/4)x + 3

অতএব, মূল রেখার ঢাল (m) = - 3/4

আমরা জানি, কোনো রেখার উপর লম্ব রেখার ঢাল m1 = - 1/m
= - 1/(- 3/4)
= 4/3

∴ লম্ব রেখার ঢাল = 4/3

২৮৫.
A cylindrical tank has a radius of 6 meters and a height of 7 meters. If the metal used to make the cylinder costs Tk. 40 per cubic meter, find the total cost of the metal required.
  1. ক
    Tk. 11,680
  2. খ
    Tk. 21,680
  3. গ
    Tk. 31,680
  4. ঘ
    Tk. 41,680
  5. ঙ
    None
ব্যাখ্যা

Question: A cylindrical tank has a radius of 6 meters and a height of 7 meters. If the metal used to make the cylinder costs Tk. 40 per cubic meter, find the total cost of the metal required.

Solution:
Given,
Radius of the cylinder, r = 6 m
Height of the cylinder, h = 7 m
Cost per cubic metre = Tk. 40

The volume of the cylinder:
V = πr2h = (22/7) × (6)2 × 7 = (22/7) × 36 × 7 = 22 × 36 = 792 cubic metres

Total cost = Volume × Cost per cubic metre = 792 × 40 = 31,680

∴ The cost of the cylinder is Tk. 31,680

২৮৬.
A rectangular block 8 cm by 12 cm by 16 cm is cut up in to an exact number of equal cubes. Find the least possible number of cubes.
  1. ক
    ক) 24
  2. খ
    খ) 26
  3. গ
    গ) 27
  4. ঘ
    ঘ) 28
ব্যাখ্যা
প্রশ্ন :A rectangular block 8 cm by 12 cm by 16 cm is cut up in to an exact number of equal cubes. Find the least possible number of cubes.
সমাধান : 
Sides of the rectangular block = 8 cm, 12 cm, 16 cm

The HCF of the sides of the cuboid is the side of the cube

Volume of the cuboid = Length × breadth × height
Volume of the cube = side3

Let n be the number of cubes
HCF( 8, 12, 16) = 4 cm

side of the square = 4 cm
Volume of the cuboid = n × Volume of the cube
Length × breadth × height = n × side3
8 × 12 × 16 = n ×  4 ×  4 ×  4
n = 2 ×  3 ×  4
n = 24

number of cubes is 24
২৮৭.
rsinθ = 1, rcosθ = √3 then the value of (√3tanθ - 1) = ?
  1. ক
    0
  2. খ
    - 1
  3. গ
    2
  4. ঘ
    - 2
ব্যাখ্যা
Question: rsinθ = 1, rcosθ = √3 then the value of (√3tanθ - 1) = ?

Solution:
rsinθ = 1
rcosθ = √3

Now,
rsinθ/rcosθ = 1/√3
⇒ tanθ = 1/√3
⇒ √3tanθ = 1
⇒ √3tanθ - 1 = 1 - 1
∴ √3tanθ - 1 = 0
২৮৮.
  1. ক
    1
  2. খ
    4
  3. গ
    3
  4. ঘ
    1/2
ব্যাখ্যা
Question:

Solution:
২৮৯.
If xtan60° + cos45° = sec45°, then the value of x2 + 1 is?
  1. ক
    ক) 1/6
  2. খ
    খ) 5/6
  3. গ
    গ) 6/7
  4. ঘ
    ঘ) 7/6
ব্যাখ্যা
Question: If xtan60° + cos45° = sec45°, then the value of x2 + 1 is?

Solution:
xtan60° + cos45° = sec45°
⇒ x . √3 + (1/√2) = √2
⇒ √6x + 1 = 2
⇒ √6x = 1
⇒ x = 1/√6
⇒ x2 = (1/√6)2
⇒ x2 = 1/6
⇒ x2 + 1 = (1/6) + 1
∴ x2 + 1 = 7/6
২৯০.
The 2nd angle of a right triangle is 30 degrees. Then how many degrees is the 3rd angle?
  1. ক
    ক) 50
  2. খ
    খ) 60
  3. গ
    গ) 70
  4. ঘ
    ঘ) 80
ব্যাখ্যা
আমরা জানি 
সমকোণী ত্রিভুজের একটি কোণ 90° 
২য় কোণ 30°  হলে 
৩য় কোণ =180° - (90° + 30°) = 180° - 120° = 60°
২৯১.
The ratio of the angles of a triangle is 2 : 3 : 4. What is the largest angle in degrees?
  1. ক
    80°
  2. খ
    75°
  3. গ
    100°
  4. ঘ
    90°
ব্যাখ্যা
Question: The ratio of the angles of a triangle is 2 : 3 : 4. What is the largest angle in degrees?

Solution: 
ত্রিভুজের কোণগুলোর অনুপাত =  2 : 3 : 4
ধরি 
কোণগুলো = 2x , 3x  4x

প্রশ্নমতে,
2x + 3x + 4x = 180°
বা, 9x  = 180°
বা, x = 180°/9
x = 20°

বৃহত্তম কোণ = 4 × 20° = 80°
২৯২.
The side length of a square inscribed in a circle is 2. What is the area of circle?
  1. ক
    π
  2. খ
    π√2
  3. গ
    2π
  4. ঘ
    π√3
ব্যাখ্যা

Let the radius of the circle be 'r'
Therefore, the side length of the square is r√2
According to the question,
√2r = 2
or, r = √2
area of circle = πr2 = π(√2)2 = 2π
২৯৩.
The cube root of 1331 is? 
  1. ক
    ক) 13
  2. খ
    খ) 11
  3. গ
    গ) 19
  4. ঘ
    ঘ) 17
ব্যাখ্যা
The cube root of 1331 is : ∛1331 
                                         = (1331)1/3
                                         = (113)1/3
                                          = 11
২৯৪.
The ratio of the angles of a triangle is 2 : 3 :4. What is the largest angle in degrees?
  1. ক
    ক) 30
  2. খ
    খ) 60
  3. গ
    গ) 80
  4. ঘ
    ঘ) 90
ব্যাখ্যা
Question: The ratio of the angles of a triangle is 2 : 3 : 4. What is the largest angle in degrees?

Solution: 
একটি ত্রিভুজে কোণগুলির অনুপাত ২ : ৩ : ৪
কোণগুলি ২x, ৩x, ৪x

ত্রিভুজের তিনকোণের সমষ্টি ১৮০° 

২x + ৩x + ৪x = ১৮০° 
⇒ ৯x = ১৮০° 
∴ x = ২০°

অতএব, বৃহত্তম কোণটি ৪ × ২০°
= ৮০° 
২৯৫.
If sec2θ + tan2θ = 7, then the value of θ when 0° ≤ θ ≤ 90° is?
  1. ক
    60°
  2. খ
    30°
  3. গ
    0°
  4. ঘ
    90°
ব্যাখ্যা

Question: If sec2θ + tan2θ = 7, then the value of θ when 0° ≤ θ ≤ 90° is?
 
solution: 
Given that, 
sec2θ + tan2θ = 7
⇒ 1 + tan2θ + tan2θ = 7.  ; [sec2θ = 1 + tan2θ]
⇒ 2tan2θ = 6
⇒ tan2θ = 3
⇒ tanθ = √3
⇒ θ = 60∘

২৯৬.
The base of a triangle is twice its height, which is 5cm. What is the area, in square centimeters, of the triangle?
  1. ক
    ক) 6.25
  2. খ
    খ) 10
  3. গ
    গ) 12.5
  4. ঘ
    ঘ) 25
ব্যাখ্যা
Question: The base of a triangle is twice its height, which is 5cm. What is the area, in square centimeters, of the triangle?
Solution:
ত্রিভুজের ভূমি = 5 cm
ত্রিভুজের উচ্চতা =5/2 = 2.5 cm

ত্রিভুজের ক্ষেত্রফল = (1/2) × 5 × 2.5 =  (1/2) × 12.5  = 6.25 cm2
২৯৭.
A storm breaks a tree. The broken part of tree bends so that the top of the tree touches the ground and makes an angle of 60° with the horizontal plane. If the distance between the base of the tree and the point where top of tree touches the ground is 10 m, find the height of the tree?
  1. ক
    17.32 m
  2. খ
    24.7 m
  3. গ
    37.3 m
  4. ঘ
    27.3 m
ব্যাখ্যা
Question: A storm breaks a tree. The broken part of tree bends so that the top of the tree touches the ground and makes an angle of 60° with the horizontal plane. If the distance between the base of the tree and the point where top of tree touches the ground is 10 m, find the height of the tree?

Solution:

PQ = 10 and let RQ be X.

RQ/PQ = tan60°
⇒ X/10 = √3
∴ X = 10√3

Now,
PR2 = X2 + (10)2
PR2 = (10√3)2 + (10)2 = 300 + 100
PR2 = 400
∴ PR = 20

Height of tree = RQ + PR
= X + 20
= 10√3 + 20
= 10 × 1.73 + 20
= 17.3 + 20
= 37.3 meters
২৯৮.
Find the maximum distance between two points on the perimeter of a rectangular garden whose length and breadth are 12 m and 5 m. 
  1. ক
    7 m
  2. খ
    9 m
  3. গ
    13 m
  4. ঘ
    17 m
ব্যাখ্যা
Question: Find the maximum distance between two points on the perimeter of a rectangular garden whose length and breadth are 12 m and 5 m. 

Solution: 
 maximum distance = diagonal length 
= √(122 + 52 )
= √(144 + 25) 
= √169 
= 13 m 
২৯৯.
একটি দড়িকে বৃত্তাকারে বাঁকালে তার ব্যাসার্ধ হয় ৭ সেমি। যদি দড়িটিকে একটি বর্গের আকারে বাঁকানো হয়, তবে এর ক্ষেত্রফল কত হবে?
  1. ক
    ১২১ বর্গ সেমি
  2. খ
    ১৪৪ বর্গ সেমি
  3. গ
    ১৩২ বর্গ সেমি
  4. ঘ
    ১০০ বর্গ সেমি
  5. ঙ
    ১৬৯ বর্গ সেমি
ব্যাখ্যা

প্রশ্ন: একটি দড়িকে বৃত্তাকারে বাঁকালে তার ব্যাসার্ধ হয় ৭ সেমি। যদি দড়িটিকে একটি বর্গের আকারে বাঁকানো হয়, তবে এর ক্ষেত্রফল কত হবে?

সমাধান:
এখানে,
বৃত্তের ব্যাসার্ধ, r = ৭ সেমি
∴ বৃত্তের পরিধি = ২πr
= ২ × (২২/৭) × ৭
= ৪৪ সেমি

ধরি,
বর্গের প্রতিটি বাহুর দৈর্ঘ্য = ক সেমি
∴ বর্গের পরিসীমা = ৪ক

এখন,
বৃত্তের পরিধি = বর্গের পরিসীমা
⇒ ৪৪ = ৪ক
⇒ ক = ৪৪/৪ = ১১ সেমি

∴ বর্গের ক্ষেত্রফল = ক2 = ১১2 = ১২১ বর্গ সেমি

৩০০.
If sinθ = cosθ then what is the value of θ?
  1. ক
    30°
  2. খ
    0°
  3. গ
    45°
  4. ঘ
    90°
ব্যাখ্যা

Question: If sinθ = cosθ then what is the value of θ?

Solution:
sinθ = cosθ
∴ sinθ/cosθ = 1
⇒ tanθ = 1 
⇒ tanθ = tan45°
∴ θ = 45°