ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

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

Solution:

∴ The shape of the quadrilateral is Rhombus.

Question: What is the area of an isosceles triangle if two of its sides measure 6 and 12?

Solution:
The given triangle is an Isosceles triangle and hence, two of the three sides of the triangle are equal.
Hence, the third side of the triangle can either be 6 or be 12.

If the two equal sides of the triangle measure 6, the sides of the triangle become 6, 6, and 12.
However, the sum of the two smaller sides is not greater than the third side.
∴ 6 is not a possible value of the third side.

If the two equal sides of the triangle measure 12, the sides of the triangle become 6, 12, and 12.
The sum of the smaller two sides is greater than the third side, and hence, the value of the third side is 12.

Let, a = 12, b = 6
∴ Area = (b/4) × √(4a² - b²)
= (6/4) × √{4 × (12)² - (6)²}
= (3/2) × √(576 - 36)
= (3/2) ×√540
= (3/2) × √(36 × 15)
= (3/2) × 6 × √15
= 9√15

Question: What is the measure of the radius of the circle that circumscribes a triangle whose sides measure 9, 40, and 41?

Solution:
Here,
9² + 40² = 81 + 1600 = 1681 = 41²
∴ 9, 40, and 41 is a Pythagorean triplet. So, the triangle is right angled.

Key property about right triangles:
In a right angled triangle, the radius of the circle that circumscribes the triangle is half the hypotenuse.
In the given triangle, the hypotenuse = 41.


Therefore, the radius of the circle that circumscribes the triangle = 41/2 = 20.5 units.

Question: A wheel of a car of radius 21 cm is rotating at 600 RPM. What is the speed of the car in km/hr?

Solution:
The radius of the wheel measures 21 cm.

In one rotation, the wheel will cover a distance which is equal to the circumference of the wheel.
∴ in one rotation this wheel will cover 2 × π × 21 = 2 × (22/7) × 21 = 132 cm.

In a minute, the distance covered by the wheel = circumference of the wheel × rpm
∴ this wheel will cover a distance of 132 × 600 = 79200 cm in a minute.

In an hour, the wheel will cover a distance of 79200 × 60 = 4752000 cm.

Therefore, the speed of the car = 4752000 cm/hr = 47.52 km/hr

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.

Question: The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? [Assume that the cow is able to move on all sides with equal ease.]

Solution:
The cow can graze the area covered by the circle of radius 19 m initially, because the length of the rope is 19 m.
Area of a circle = π × (radius)²
Therefore, the initial area that the cow can graze = (22/7) × 19² sq. m.
When the length of the rope is increased to 30 m, grazing area becomes = (22/7) × 30² sq. m.
The additional area it could graze when length is increased from 19 m to 30 m = (22/7) × (30² - 19²) sq. m.
= (22/7) × (30 + 19)(30 - 19) = (22/7) × 49 × 11 = 1694 sq. m.

Question: A lady grows cabbage in her garden that is in the shape of a square. Each cabbage takes 1 square foot of area in her garden. This year, she has increased her output by 211 cabbages when compared to last year. The shape of the area used for growing the cabbage has remained a square in both these years. How many cabbages did she produce this year?

Solution:
The shape of the area used for growing cabbages has remained a square in both the years.
Let the side of the square area used for growing cabbages this year be X ft.
Therefore, the area of the ground used for cultivation this year = X² sq.ft.

Let the side of the square area used for growing cabbages last year be Y ft.
Therefore, the area of the ground used for cultivation last year = Y² sq.ft.

As the number of cabbages grown has increased by 211, the area would have increased by 211 sq ft because each cabbage takes 1 sq ft space.
Hence,
X² - Y² = 211
⇒(X + Y)(X - Y) = 211.
211 is a prime number and hence it will have only two factors. i.e., 211 and 1.
Therefore, 211 can be expressed as product of 2 numbers in only way = 211 × 1
i.e., (X + Y)(X - Y) = 211 × 1

So, (X + Y) should be 211 and (X - Y) should be 1.
Solving the two equations we get X = 106 and Y = 105.

Therefore, number of cabbages produced this year = X² = 106² = 11236.

Question: A cube of side 5 cm is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted?

Solution:
When a 5 cubic centimeter cube is sliced into 1 cubic centimeter cubes, we will get 5 × 5 × 5 = 125 cubes of 1 cubic centimeter.

In each side of the larger cube, the smaller cubes on the edges will have more than one of their sides painted.
Therefore, the cubes which are not on the edge of the larger cube and that lie on the facing sides of the larger cube will have exactly one side painted.

In each face of the larger cube, there will be 5 × 5 = 25 cubes.
Of these, the cubes on the outer rows will be on the edge. 16 such cubes exist on each face.
If we count out the two outer rows on either side of a face of the cube, we will be left with 3 × 3 = 9 cubes which are not on the edge in each face of the cube.

Therefore, there will be 9 cubes of 1-cc volume per face that will have exactly one of their sides painted.
In total, there will be 9 × 6 = 54 such cubes.

Question: Calculate the surface area of a cylinder with radius 6 cm and height 15 cm.

Solution:
Here,
radius r = 6 cm
height h = 15 cm
Surface Area = (2πrh + 2πr²)
Surface Area = 2 × π × 6 × 15 + 2 × π × 6² = 252π square cm

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

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

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

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

Question: Find the area of the trapezium if height is 5 cm and AB and CD are given as 10 and 6 cm respectively.

Solution:
Given,
AB = 10cm, CD = 6cm, height = 5cm

According to the formulae,
Area of Trapezium = (1/2) h (AB+CD)
⇒ Area of Trapezium = 1/2 × 5 × (10 + 6)
⇒ Area of Trapezium = 40 cm²

Question: Given the area of the rhombus is 120-meter square then find the length of one of the diagonals if the other diagonal is of length 12 m.

Solution: 
Since we know that,
Area of Rhombus = (1/2) × Diagonal₁ × Diagonal₂ 

Putting all the known values, we get
120 = (1/2) × Diagonal₁ × Diagonal₂
⇒ 240 = 12 × Diagonal₂
⇒  Diagonal₂ = 240/12
∴ Diagonal₂ = 20 m

Question: A spherical water tank has a radius of 15 m. Find the volume of the tank.

Solution:
Let,
radius, r = 15 m
Volume spherical = (4/3) × π × r³
Volume = (4/3) × π × 15³ = 4500π cubic m

Question: The length of a rectangle is thrice its breath, and its perimeter is 112 meters. What is its area?

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

Perimeter of a rectangle = 2 (Length + Breadth)
So, 2(3x + x) = 112
⇒ 6x + 2x = 112
⇒ 8x = 112
∴ x = 112/8 = 14

Now, Breadth = 14, so, length = 14 × 3 = 42

So, its area = Length × Breadth
= 42 × 14 = 588 sq. m.