পেট্রোবাংলা প্রিলি ও লিখিত সমন্বিত প্রস্তুতি [Archived]

Question: The base of a right-angled triangle is 16 and hypotenuse is 20. What is its area?

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

Base = 16, Hypotenuse = 20
Height² = Hypotenuse² - Base²
= 20² - 16²
= 400 - 256
Height² = 144
∴ Height = 12

Area = (1/2) × base × height
= (1/2) × 16 × 12
= 96 sq. meters

Question: If area of circle is equal to volume of sphere with equal radii, find the radius.

Solution:
Let r be the radius, We have
πr² = (4/3)πr³
⇒ r = 3/4

Question: The dimensions of a rectangular floor are 16 feet by 20 feet. When a rectangular rug is placed on the floor, a strip of floor 3 feet wide is exposed on all sides. What are the dimensions of the rug, in feet?

Solution:

Given 3 feet wide is exposed on all sides. Hence all 4 sides will have 3 feet gap.

Length of floor = 20
Length of rug = 20 - 3 -3 =14

Width of floor = 16
Width of rug =16 - 3 - 3 =10

Hence dimensions of rug = 10 by 14

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

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

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

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

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

Question: If the ratio of radius of two spheres is 4 : 7, the ratio of their volume is-

Solution:
Ratio of radii of 2 spheres is 4 : 7.
∴ratio of their volume = 4³ : 7³ = 64 : 343

Question: If the perimeter of square region S and the perimeter of circular region C are equal, then the ratio of the area of S to the area of C is closest to-

Solution:
If Perimeter of square = Perimeter of Circle,
then:
4a = 2πr, where a = side of square and r is radius of the circle
a/r = π/2

Area of S/Area of C = a²/(πr²)
= (π/2)² × (1/π)
= π/4
= 3.14/4
≈ 3/4

Question: The ratio of length and breadth of a rectangular park is 4 : 2. If a cat running along the boundary of the park at the speed of 18 km/hr completes one round in 10 minutes, find the area of the park in square meters.

Solution:
One round of the park is equal to the perimeter of the park.
So, by completing one round, the cat covers a distance equal to the perimeter of the park.
Now,
Distance or perimeter = speed × time
= 18 × (10/60)
= 3 km
= 3000 meters

Let Length = 4x and breadth = 2x
So, Perimeter:
2(4x + 2x) = 3000
⇒ 8x + 4x = 3000
⇒ 12x = 3000
∴ x = 3000/12 = 250 meters

So, Length = 4 × 250 = 1000 meters
And, Breadth = 2 × 250 = 500 meters

Area = Length × Breadth
= 1000 × 500
= 500000 sq. m.

Question: The slant height of a right circular cone is 13 m and its height is 5 m. Find area of the curved surface.

Solution:
Area of curved surface = πrl
Now
r = √(13² - 5²)
= √(169 - 25)
= √144
= 12m

∴ Required Area= (22/7) × 13 × 12
= 490.28 m²

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

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

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

Question: If the diagonal of a square field is 16 m, what is its area?

Solution:
Area of a square:
(Side)² = (1/2) × (diagonal)²
= (1/2) × (16)²
= (1/2) × 256
= 128 m²

Question: The perimeter of an equilateral triangle is 96√3 cm. Find its height.

Solution:
Perimeter of the equilateral triangle is 96√3 cm.
Each of the side of the equilateral triangle is (96√3/3) = 32√3 cm.
The height of the equilateral triangle will be = (√3/2) × (32√3) = 48 cm

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

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

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

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

Question: The area of a rectangle and square are equal. The side of the square is 5 cm and the smaller side of the rectangle is half that of the square. The length of the other side of the rectangle would be-

Solution:
Side of Square = 5 cm, and length of one side of rectangle = 5/2 = 2.5 cm
Let the length of the other side of the rectangle = B

As per the question:
Area of rectangle = Area of square
Length × Breadth = Side × Side
⇒ 2.5 × B = 5 × 5
⇒ B = 25/2.5
∴ B = 10 cm

Question: Ratio of Volumes of cube and Sphere is 6/π. Find the ratio of side of cube and radius of sphere.

Solution:
Let the side of cube is 'a' and radii of sphere is 'r'.
Now Volume of cube= a³
Volume of sphere= (4/3)πr³

a³/{(4/3)πr³} = 6/π
⇒ a³/r³ = (6 × 4)/3
⇒ a³/r³ = 8/1
⇒ a/r = 2/1
Hence the answer is 2 : 1

Question: The length of one side of a square inscribed in a circle is 2. What is the area of the circle?

Solution:
বৃত্তের অন্তর্লিখিত বর্গের বাহুর দৈর্ঘ্য ২ একক
∴ বর্গের কর্ণের দৈর্ঘ্য = ২√২ একক

এখানে বর্গের কর্ণ বৃত্তটির ব্যাসের সমান।
∴ বৃত্তের ব্যাসার্ধ = (২√২)/২ একক = √২ একক

বৃত্তের ক্ষেত্রফল = π(√২)^২ বর্গএকক
= ২π বর্গএকক

Question: The complementary angle of supplementary angle of 130°-

Solution:
For supplementary angle: The sum of two angles is 180°.
For complementary angle: The sum of two angles is 90°.

The supplement angle of 130° = 180° - 130° = 50°
The complement angle of 50° = 90° - 50° = 40°

∴ The complement angle of the supplement angle of 130° is 40°

Question: If the area of the trapezium, whose parallel sides are 6 cm and 10 cm is 32 sq. cm, what will be the distance between the parallel sides?

Solution:
Parallel sides of a trapezium = 6 cm, and 10 cm
Area of trapezium = (1/2)(sum of the parallel sides) × distance between the parallel sides
32 = (1/2)(6 + 10 ) × distance 
⇒ 32 = 8 × distance
⇒ distance = 32/8 = 4 cm

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

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°

Question: Find the area of an isosceles triangle whose sides are 10 cm, 6 cm and 6 cm.

Solution:
Semi perimeter of triangle s = (10 + 6 + 6)/2 = 11.
Area of triangle = √{11 × (11 - 10) × (11 - 6) × (11 - 6)}
= √(11 × 5 × 5)
= 5√11 cm²

Question: If length and width of a rectangular plot were each increased by 20%, what would be the percentage increase in the area of the plot?

Solution:
মনে করি,
দৈর্ঘ্য = x একক এবং প্রস্থ = y একক
∴ ক্ষেত্রফল = xy বর্গ একক

20% বৃদ্ধিতে
নতুন দৈর্ঘ্য = x + x এর 20%
= 12x/10 একক

20% বৃদ্ধিতে
প্রস্থ = y + y এর 20%
= 12y/10 একক
∴ নতুন ক্ষেত্রফল = (12x/10) ×( 12y/10) = 144xy/100 বর্গ একক

ক্ষেত্রফল বৃদ্ধি =(144xy/100) - xy
=(144xy - 100xy)/100
= 44xy/100

শতকরা ক্ষেত্রফল বৃদ্ধি = {(44xy/100) × (1/xy) × 100}% = 44%

Question: In a ∆ABC, AB = BC, ∠B= x° and ∠A = (2x - 20)°. Then, ∠B= ?

Solution:
AB = BC ⇒ ∠C = ∠A = (2x - 20)°.

∠A + ∠B + ∠C =180⁰
⇒ (2x - 20) + x + (2x - 20 ) = 180
⇒ 5x - 40 =180
⇒ 5x = 220
⇒ x = 44.
∴ ∠B = 44°.

Question: In a cyclic quad. ABCD, ∠A = 80°. Then ∠C = ?

Solution:
Opposite angles of a cyclic quadrilateral are supplementary.
∴ ∠A + ∠C = 180°
⇒ 80° + ∠C =180°
⇒ ∠C = 100°.

Question: A ladder is placed in such a way that its foot is 15 m away from a wall and its top reaches a window 20 m above the ground. The length of the ladder is-

Solution:

Let BC be the wall and AC be the ladder.
Then , BC = 20 m and AB =15m
∴ AC² = BC² + AB² = (20)² + (15)² = (400 + 225) = 625
⇒ AC = √625 = 25m. 

Question: If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is-

Solution:
Since, ΔABC and ΔPQR are similar triangles.
then, ∠B = ∠Q = 83°
Thus, in ΔABC,
∠C = 180° - (∠A + ∠ B)
⇒ ∠C = 180° - (47° + 83°)
∴ ∠C = 50°

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


Solution:

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

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

∴ ML = OM + OL = (5 +12 ) cm = 17cm.

Question: The length of the longest rod that can be placed in a room 30 m long, 24 m broad and 18 m high, is-

Solution:
Length of room = 30 m
Breadth of room = 24 m
Height of room = 18 m

Length of the longest rod = Diagonal of the room = √(30² + 24² + 18²)
= √(900 + 576 + 324)
= √(1800)
= √(900 × 2)
= 30√2

Question: A water tank is 30 m long, 20 m wide and 12 m deep. It is made of iron sheet which is 3 m wide. The tank is open at the top. If the cost of iron sheet is TK. 10 per meter, what is the total cost of iron sheet required to build the tank?

Solution:
Length of water tank = 30 m
Width of water tank = 20 m
Depth of water tank = 12 m

Area of water tank = 2(lb + bh +hl) - lb
= 2 (30 × 20 + 20 × 12 + 12 × 30) - 30 × 20
= 2 (600 + 240 + 360) - 600
= 2 (1200) - 600
= 2400 - 600
= 1800 m²

Area of iron sheet (L × B) = area of tank = 1800 m²
⇒ L × B = 1800
⇒ L × 3 = 1800
∴ L = 1800/3 = 600 m

Given that, the cost of iron sheet is Tk. 10 per meter.
So, the total cost of iron sheet to build the tank = 10 × 600 = Tk. 6000

Question: What is the volume of a hemisphere having radius 3 cm?

Solution:
The radius of a hemisphere is 3cm
A hemisphere's volume is equal to (2/3)πr³

Volume = (2/3)π.3³)
⇒ 2π × 9
⇒ 18π

∴ The volume of a hemisphere having a radius 3 cm is 18π cm³

Question: The curved surface area of a cylindrical pillar is 264 m² and its volume is 924 m³. Find the ratio of its diameter to its height.

Solution:
(πr²h)/(2πrh) = 924/264
⇒ r = (924/264) × 2
∴ r = 7

And,
2πrh = 264 
⇒ h = (264/2) × (7/22) × (1/7)
∴ h = 6

∴ Required ratio = (2r)/h
= 14/6
= 7 : 3.

Question: In the adjoining figure ABCD is a rhombus. If ∠A = 70° then ∠CDB =?

Solution:
Let ∠CDB= x°.
then , CD = CB ⇒ ∠CBD = ∠CDB = x°.
∠BCD = ∠BAD = 70° (opp. s of a rhombus)
∴ x + x + 70 = 180° (sum of the angles of a ∆ is 180°)
⇒ 2x = 110°
⇒ x = 55°
∴ ∠CDB = 55°.