ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: The area of a square and a rhombus are equal. The diagonals of the rhombus are 8 meters and 9 meters, respectively. What is the length of one side of the square?

Solution:
The area of the rhombus = (1/2) × Product of the diagonals
= (1/2) × 8 × 9
= 36 square meters

The area of the square = 36 square meters.
∴ Length of one side of the square = √36 meters
= 6 meters

Question: From the top of a lighthouse which is 90 m above the sea, the angle of depression of a ship is 60°. How far is the ship from the lighthouse?

Solution:

Let the height of the lighthouse above sea be AC and it is given 90 m.
Ship is at point B so the distance between the base of lighthouse A and ship is AB.

Then, From ΔABC, 
AC/AB = tan 60° (√3)
⇒ 90/AB = √3
⇒ AB = 90/√3
⇒ AB = (30 · √3 · √3)/√3
∴ AB = 30√3 m

Question: A chord of length 16 cm is drawn in a circle of radius 10 cm. The distance of the chord from the center of the circle is-

Solution:
Given,
r = 10 cm
and c = 16 cm.
Half of the chord length = 16/2 = 8 cm

∴ Distance = √(10² - 8²)
= √(100 - 64) 
= √36
= 6 cm

The chord's distance from the circle's center is 6 cm.

Question: If θ is said to be an acute angle, and 7sin²θ + 3cos²θ = 4, then what is the value of tanθ?

Solution:
7sin²θ + 3cos²θ = 4
⇒ 7sin²θ + 3(1 - sin²θ) = 4
⇒ 7sin²θ + 3 - 3sin²θ = 4

Then, 4sin²θ = 1
⇒ sinθ = 1/2
⇒ θ = 30°

Now, put θ = 30° in tanθ, we will get,
tanθ = tan 30° = 1/√3

Question: A wheel of a car of radius 35 cm is rotating at 500 RPM. What is the speed of the car in km/hr?

Solution:
The radius of the wheel measures 35 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 × 35 = 2 × (22/7) × 35 = 220 cm.

In a minute, the distance covered by the wheel = circumference of the wheel × rpm
∴ this wheel will cover a distance of 220 × 500 = 110000 cm in a minute.

In an hour, the wheel will cover a distance of 110000 × 60 = 6600000 cm.
Therefore, the speed of the car = 6600000 cm/hr = 66 km/hr

Question: The radius of a circle is the same as the diagonal of a square whose area is 25 sq. cm. The area of the circle is -

Solution:
Area of square = 25
Side of square = √25 = 5

Diagonal of square = 5√2
So, the radius of the circle is 5√2 cm

Area of circle = πr²
= π(5√2)²
= 50π cm²
The area of the circle is 50π sq. cm.

Question: A tree 18 meters tall broke in such a way that the broken part makes a 30-degree angle with the ground. At what height did the tree break?

Solution:

sin30° = AC/BC
⇒ 1/2 = h/(18 - h)
⇒ 2h = 18 - h
⇒ 3h = 18
∴ h = 6
∴ গাছটি 6 মিটার উঁচুতে ভেঙেছিল।

Question: The area of a trapezium is 72 square cm. The lengths of its parallel sides are 12 cm and 6 cm. What is the distance between the parallel sides?

Solution:
We know,
The area of a trapezium = (1/2) × Sum of the lengths of the parallel sides × Distance between the parallel sides.

Let the distance between the parallel sides be d. Then,
d = (2 × Area of the trapezium)/Sum of the lengths of the parallel sides
= (2 × 72)/(12 + 6)
= 144/18
= 8 cm

Thus, the distance between the parallel sides 8 cm.

Question: The value of the following is:


Solution:

Question: What is the area of an isosceles triangle if two of its sides measure 10 and 8?

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 10 or be 8.

If the two equal sides of the triangle measure 10, the sides of the triangle become 8, 10, and 10.
The sum of the smaller two sides is greater than the third side, and hence, this is a valid configuration.

If the two equal sides of the triangle measure 8, the sides of the triangle become 8, 8, and 10.
However, the sum of the two smaller sides (8 + 8 = 16) is not greater than the third side (10).
∴ 8 is not a possible value of the third side.
Let, a = 10, b = 8
∴ Area = (b/4) × √(4a² - b²)
= (8/4) × √{4 × (10)² - (8)²}
= 2 × √{4 × 100 - 64}
= 2 × √{400 - 64}
= 2 × √336
= 2 × √(16 × 21)
= 2 × 4 × √21
= 8√21

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°

Question: If the area of a square is 676 square meters, what is the perimeter of the square? 

Solution:
Given,
The area of the square = 676 square meters.

Therefore,
The length of one side of the square = √676 meters = 26 meters.

We know,
The perimeter of a square = 4 × length of one side
= 26 × 4 meters
= 104 meters

Thus, the perimeter of the square is 104 meters.

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

Solution:
sin 17° = (x/y)
⇒ sin 73° = sin (90° – 17°)
= cos 17°

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

∴ sec 17° = y/{√(y² - x²)}

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

Solution:
We can find the longer diagonal by adding together the altitude of the top triangle and the altitude of the bottom triangle. To find these, use Pythagorean Theorem. We can use Pythagorean Theorem because one of the properties of a kite is that the two diagonals are perpendicular.

The top triangle has two sides of length 3 [labeled in the picture], and a base of 4 [provided in the written directions]. To figure out the altitude, split this triangle into 2 right triangles. The two legs are x [the altitude] and 2 [half of the base 4], and the hypotenuse is 3:
x² + 2² = 3²
⇒ x² + 4 = 9
⇒ x² = 5
∴ x = √5

We will do something similar for the bottom triangle. Consider one of the right triangles. It will have a hypotenuse of 7, one leg that we don't know, x [the altitude], and one leg 2 [half the shorter diagonal]. Set up the equation using the Pythagorean Theorem:
x² + 2² = 7²
⇒ x² + 4 = 49
⇒ x² = 45
∴ x = √45 = 3√5

∴ The length of the longer diagonal of this kite = 3√5 + √5 = 4√5

Question: The angle of elevation of an aeroplane from a point A on the ground is 60°. After a straight flight of the plane for 30 seconds, the angle of elevation becomes 30°. If the palne flies at a constant height of 3600√3 metre, what is the speed of plane?

Solution:

P and Q = Positions of plane
∠PAB = 60°, ∠QAB = 30°, PB = 3600√3 metre
In ∆ABP, tan 60° = BP/AB
⇒ √3 = 3600√3/AB
⇒ AB = 3600 metre

In ∆ACQ, tan 30° = CQ/AC
⇒ 1/√3 = 3600√3/AC
⇒ AC = 3600 × 3 = 10800 metre
∴ PQ = BC = AC – AB = 10800 – 3600 = 7200 metre
This distance is covered in 30 seconds.

∴ Speed of plane = 7200/30 = 240 m/sec

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

Question: If 0 ≤ θ ≤ 90° and 4 cos²θ - 4√3 cosθ + 3 = 0 then the value of θ is

Solution:
4 cos²θ – 4√3 cosθ + 3 = 0
⇒ (2cosθ)² – 2 · 2 cosθ · √3 + (√3)² = 0
⇒ (2cosθ – √3)² = 0
⇒ 2 cosθ – √3 = 0
⇒ 2 cosθ = √3
⇒ cosθ = √3/2
⇒ cosθ = cos 30°
∴ θ = 30°

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

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

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

Question: The area of a square is equal to the area of a parallelogram. If the base of the parallelogram is 49 meters and its height is 4 meters, what is the length of one side of the square?

Solution:
Area of the parallelogram = Base × Height
= 49 × 4
= 196 square meters

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

ATQ,
k² = 196
⇒ k = √196
∴ k = 14
Therefore, the length of one side of the square = 14 meters

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

Question: If 10 circles, all with different radii, are positioned in the same plane, what is the maximum possible number of distinct points where 2 or more of the circles intersect?

Solution:
From 10 circles, the number of pairs that be formed = ¹⁰C₂ = 95
ince each of these 45 pairs may intersect in at most two points, the maximum possible number of intersections = 45 × 2 = 90

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

Solution:
Given,
cos⁴θ - sin⁴θ = 2/3

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

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

Solution:

First of all we can notice that a triangle whose sides measure 9, 40 and 41 is a right triangle because 9² + 40² = 41²

A right triangle inscribed in a circle always has its hypotenuse as the diameter of the circle. Conversely, if the diameter of a circle forms one side of a triangle inscribed in the circle, that triangle is a right triangle.

Thus, the diameter of the circle is equal to the hypotenuse of the triangle, which is 41. Therefore, the radius of the circle is = Diameter/2
= 41/2
= 20.5