IBA ফ্যাকাল্টি ভিত্তিক প্রস্তুতি

Question: The base of a right-angled triangle is 7 m and hypotenuse is 25 m. What is its area?

Solution:
Given,
Base = 7 m, Hypotenuse = 25 m

By Pythagoras' Theorem
Height² = Hypotenuse² - Base²
⇒ Height²= 25² - 7²
⇒ Height²= 625 - 49
⇒ Height² = 576
∴ Height = 24

We know,
Area = (1/2) × base × height
= (1/2) × 7 × 24
= 84 sq. meters

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°

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

Solution: 
In an isosceles triangle, two sides are equal. The possible third side can be either 6 cm or 12 cm.

If the equal sides are 6, the sides become 6, 6, and 12 — which violates the triangle inequality rule.
If the equal sides are 12, the sides become 6, 12, and 12 — which satisfies the triangle inequality.
∴ The valid third side is 12 cm.

Now,
a = 6 cm, b = 12 cm, c = 12 cm

∴ Semi-perimeter s =(6 + 12 + 12​)/2 =15 cm

We know
from "Heron’s formula"
Area of the triangle = √{s(s - a)(s - b)(s - c)}
= √{15(15 - 6)(15 - 12)(15 - 12)}
= √(15 × 9 × 3 × 3)
= 9√15 cm²

Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 7 cm and 15 cm. What is the area of the circle?

Solution:
The sides of the rectangle are 7 cm and 15 cm.
Perimeter of the rectangle =2(7 + 15) = 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

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

Solution:
The sides of the triangle are 9, 40, and 41.

Here,
9² + 40² = 81 + 1600 = 1681 = 41², this is a Pythagorean triplet,
∴  the triangle is right-angled, with 41 as the hypotenuse.

We know,
In a right-angled triangle, the radius of the circumscribed circle is half the hypotenuse.

∴ The radius of the circle that circumscribes the triangle =  41/2 = 20.5​ units

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

Solution:
Let,
the side of the square area used for growing cabbages this year = X ft.
∴ the area of the ground used for cultivation this year = X² sq. ft.

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

The cabbage field remained square-shaped in both years.
Given the increase of 211 cabbages (1 sq ft per cabbage),
∴ the area difference is: X² - Y² = 211
⇒ (X + Y)(X - Y) = 211

Since 211 is prime, the only factor pair is (211, 1), so:
X + Y = 211
X - Y = 1
→ Solving gives: X = 106 and Y = 105

This year's production = 106² = 11236 cabbages

Question: The ratio of the angles of a triangle is 5 : 15 : 16. What is the largest angle in degrees?

Solution: 
Given
The ratio of the angles of a triangle = 5 : 15 : 16

Let,
the angles = 5x , 15x  16x

ATQ,
5x + 15x + 16x = 180°
⇒ 36x  = 180°
⇒ x = 180°/36
∴ x = 5°

∴ the largest angle = 16 × 5° = 80°

Question: The ratio of length and breadth of a rectangular field is 5 : 3. A dog runs along the boundary of the field at a speed of 12 km/h, and takes 8 minutes to complete one full round. Find the area of the field in square meters.

Solution:
One round of the park is equal to the perimeter of the park.
So, by completing one round, the dog covers a distance equal to the perimeter of the park.
Now,
Distance or perimeter = speed × time
= 12 × (8/60)
= 8/5 km
= 1.6 km
= 1600 meters

Let
Length = 5x and breadth = 3x
So, Perimeter: 2(5x + 3x) = 1600
⇒ 2 × 8x = 1600
⇒ 16x = 1600
∴ x = 1600/16 = 100 meters

So, Length = 5 × 100 = 500 meters
And, Breadth = 3 × 100 = 300 meters

Area = Length × Breadth
= 500 × 300
= 150000 sq. m.

Question: A wire can be bent in the form of a circle of radius 49 cm. If it is bent in the form of a square, then its area will be -

Solution:
Given,
radius of the circle r = 49 cm

Circumference of the circle = 2πr 
 = 2 × (22/7) × 49 
= 2 × 22 × 7
= 308 cm 

The length of one side of the square = 308/4 = 77 cm

Area of the ​​square = (77)² cm²
= 5929 cm²

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 will be = (√3/2) × (28√3) = 42 cm

Question: The area of a circle is increased by 22 sq. cm if its radius is increased by 1 cm. The original radius of the circle is-

Solution:
Let,
the original radius of the circle = r cm

ATQ,
π(r + 1)² - πr² = 22
⇒ π[(r + 1)² - r²] = 22
⇒ r² + 2r + 1 - r² = 22/π
⇒ 2r + 1 = 22/(22/7)
⇒ 2r + 1 = 7
⇒ 2r = 7 - 1
⇒ 2r = 6
∴ r = 3

∴ the original radius of the circle = 3 cm

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

Question: The length of a rope, to which a cow is tied assume that the cow is able to move on all sides with equal ease, is increased from 21 m to 28 m. How much additional ground will it be able to graze? 

Solution:
We know,
Area of a circle = π × (radius)²

Given,
The cow can graze the area covered by the circle of radius 21 m initially, because the length of the rope is 21 m.
Therefore, the initial area that the cow can graze = (22/7) × 21² sq m.
= 1386 sq m.

When the length of the rope is increased to 28 m, grazing area becomes = (22/7) × 28² sq m.
= 2464 sq m.

The additional area it could graze when length is increased from 21 m to 28 m = (2464 - 1386) sq m.
= 1078 sq m.

Question: If the length of a rectangle is increased by 10% and its breadth is decreased by 10%, the change in its area will be-

Solution:
Let
the length and breadth be 100 unit and 100 unit respectively
∴ The area before change = (100 × 100) = 10000 square unit

The length after change = 100 + 10% of 100 = 100 + 10 = 110 unit
The breadth after change = 100 - 10% of 100 = 100 - 10 = 90 unit

The area after change = 110 × 90 = 9900 square unit

∴ Percentage change = [(10000 - 9900)/10000] × 100%
= (1/100) × 100%
= 1% 

∴ The area of the new rectangle is decreased by 1%.

Question: A wheel with 8 cogs is meshed with a larger wheel that has 16 cogs. If the smaller wheel makes 36 revolutions, how many revolutions will the larger wheel make?

Solution:
We know,
As the number of cogs increased, the revolutions decreased.
∴ More cogs (↑),Less revolutions (↓)

Hence, this is a problem related to indirect proportion.

Let
the number of revolutions of the larger wheel = x

ATQ,
16 : 8 : : 36 : x
⇒ 16 × x = 8 × 36
⇒ x = (8 × 36)/16
∴ x = 18

∴ The larger wheel will make 18 revolutions.

Question: A rectangular garden is 30 meters long and 18 meters wide. A walkway, 2.5 meters wide, is made all around the inside of the garden. What are the new length and width of the garden area left after building the walkway?

Solution: 

Given,
Total outer dimensions including the walkway:
Length = 30 meters
Width = 18 meters

Walkway is 2.5 meters wide on all sides, so:

Length of remaining garden = (30 - 2.5 - 2.5) meters
= (30 - 5) meters
=25 meters

Width of remaining garden = (18 - 2.5 - 2.5) meters
= (18 - 5) meters
=13 meters

Hence The dimensions of the remaining garden (excluding the walkway) are 25 meters by 13 meters

Question: If the length of each side of an equilateral triangle is increased by 2 meters, the area is found to be increased by 3 + √3 square meters. The length of each side of the triangle is:

Solution:
Let,
the length of each side of the equilateral triangle = a meters
∴ Its area = √3/4 × a² sq. meter

The area of the triangle when the length of each side increases by 2 meters = √3/4 × (a + 2)² sq. metre

ATQ,