Geometry: Mensuration, Trigonometry

External dimensions,
l = 50 cm,
b = 40 cm,
h = 23 cm

Internal dimension,
l' = 50 - (2 × 3) = 44 cm
b' = 40 - (2 × 3) = 34 cm
h' = 23 - 3 = 20 cm

The volume of the metal used in the box = External Volume - Internal Volume
= [( 50 × 40 × 23) - (44 × 34 × 20)] cm³
= 16080 cm³.

∴ Weight of the metal =
(16080 × 0.5)/1000 kg
= 8.04 kg

Question: A lamp post 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 lamp post break?

Solution:


Let,
height from ground (A) to broken part (C) = h
rest = 18 - h
The broken part makes a 30-degree angle with the ground at B.
It creates a triangle ABC where,
BC = 18 - h
AC = h

Now,
sin 30° = AC/BC
⇒ 1/2 = h/(18 - h)
⇒ 2h = 18 - h
⇒ 3h = 18
∴ h = 6

∴ the lamp post broke at 6 m height from the ground.

Question: Find the equation of the line with an x-intercept of - 6 and a y-intercept of 3.
​
​Solution:
​যখন একটি রেখা x-অক্ষকে 'a' বিন্দুতে এবং y-অক্ষকে 'b' বিন্দুতে ছেদ করে, 
​তখন এর সমীকরণ হয়: (x/a) + (y/b) = 1

এখানে, a = - 6 এবং b = 3
মানগুলো সমীকরণে বসিয়ে পাই,
x/(- 6) + y/3 = 1
⇒ (- x + 2y)/6 = 1
⇒ - x + 2y = 6
⇒ x - 2y + 6 = 0

Total Surface Area of a Cylinder = 2πr(r + h)
TSA = 2 × 22/7 × 4(4 + 10)
= (2 × 22 × 4 × 14)/7
= (2 × 22 × 4 × 2)
= 352 sq.cm

Question: Two ships are sailing in the sea on the two sides of the lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 45° and 45° respectively. If the lighthouse is 120 m high, the distance between the two ships is?

Solution:

Given that,
Height of the lighthouse = 120m 
Now,
In triangle ADC,
AD/DC = tan 45° 
⇒ AD/DC = 1      [tan 45° = 1]
⇒ AD = DC = 120m 
Again,
In triangle ABD,
AD/BD = tan 45° 
⇒ 120/BD = 1  [tan 45° = 1]
⇒ BD = 120 m

Now,
BC = BD + DC
= 120 + 120 = 240 m
∴ Total distance = 240 m

Question: Find the value of sin(7π/6).

Solution:
sin(7π/6)
= sin(π + π/6) [যেহেতু (π + θ) তৃতীয় চতুর্ভাগে পড়ে এবং তৃতীয় চতুর্ভাগে sin ঋণাত্মক, তাই sin(π + θ) = - sinθ]
= - sin(π/6)
= - sin(30°)
= - 1/2

Question: sin135° + sin45° = ?

Solution:

Given that,
sin135° + sin45°
= sin(180° - 45°) + sin45°
= sin45° + sin45°
= 2 × sin45°
= 2 × (1/√2)
= √2

∴ sin135° + sin45° = √2

Question: A square is drawn inside of a circle with a perimeter of 42π. What is the perimeter of that square?

Solution:
Circumference of the circle 2πr = 42π
r = 42π/2π
r = 21
So, Diameter of a circle d = 2r
= 2 × 21
= 42
If the square’s side is a, its diagonal distance
d = a√2
⇒ a = d/√2
⇒ a = 42/√2
⇒ a = (42 × √2)/(√2 × √2)
⇒ a = (42 × √2)/2
a = 21√2

So, perimeter 4a = 21√2 × 4
= 84√2

∴ Perimeter of a that square is  84√2

Question: A cuboid has its height , breadth, and length in the ratio 1 : 2 : 4, and its total surface area is 112 cm². Find the volume of the cuboid.

Solution:
দেয়া আছে,
 আয়তাকার ঘনবস্তুর মাত্রাগুলির অনুপাত = 1 : 2 : 4
এবং সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 112 cm²

ধরি , আয়তাকার ঘনবস্তুর মাত্রাগুলির অনুপাত যথাক্রমে x, 2x এবং 4x

আমরা জানি,
আয়তাকার ঘনবস্তুর সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 2(lb + bh + lh)
⇒ সমগ্র পৃষ্ঠতলের ক্ষেত্রফল = 2(x)(2x) + (2x)(4x) + (4x)(x)
⇒ 112 = 2(2x² + 8x² + 4x²)
⇒ 112 = 2(14x²)
⇒ 112 = 28x²
⇒ x² = 112/28
⇒ x² = 4
⇒ x = 2

সুতরাং, ঘনবস্তুটির মাত্রাগুলি হল,
দৈর্ঘ্য (l) = x = 2 cm
প্রস্থ (b) = 2x = 2 × 2 = 4 cm
উচ্চতা (h) = 4x = 4 × 2 = 8 cm

এখন, আয়তাকার ঘনবস্তুর আয়তন, V = l × b × h
⇒ V = 2 × 4 × 8
⇒ V = 64 ঘন সেমি।

সুতরাং, নির্ণেয় আয়তন হল 64 cm³

Question: 


Solution:

Question: A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:

Solution:

Question: There are two squares S₁ and S₂. The ratio of their areas is 9 : 16. If the side of S₁ is 12 cm. What is the side of S₂?

Solution:
Given that,
Two squares S_1​ and S₂​
Area of S₁ : Area of S₂ = 9 : 16
Side of S₁ = 12 cm

Now,
Let the side of S_2​ be x cm.
Then,
(Side of S₁)² : (Side of S₂)² = 9 : 16
⇒ 12² : x² = 9 : 16
⇒ 144/x² = 9/16
⇒ x² = (144 × 16)/9
⇒ x² = 256 
⇒ x² = 16²
∴ x = 16

So the side of S₂​ is 16 cm.

Question: A trapezium has parallel sides of length 15 m and 35 m. The distance between the sides is 12 m. Calculate the area of the trapezium.

Solution:
Given that,
Parallel sides of trapezium, a = 15 m and b = 35 m
Distance (height) between parallel sides, h = 12 m

We know,
Area of a trapezium =(1/2) × (sum of parallel sides) × height
= (1/2) × (a + b) × h
= (1/2) × (15 + 35) × 12 
= 50 × 6
= 300 m²

So the area of the trapezium is 300 m².

Given, Radius of circle, r=28 cm
then the circumference = 2πr
= 2 × 22/7 × 28
= 176 cm
 
Let 'a' be the side of the square,
 
circumference of circle = perimeter of the square
Or, 176=4a
Or, a = 176/4 = 44 cm
∴ area of square = a²
 = 44²
 = 1936 cm²

Question: Find the surface area of a cuboid 16 m long, 14 m broad and 7 m high.

Solution:
Where,
length, l = 16 m
breadth, b = 14 m
height, h = 7 m

We know, 
total surface area of a cuboid
= 2(lb + bh + hl)
= [2 (16 × 14 + 14 × 7 + 16 × 7)]
= 2 (224 + 98 + 112)
= (2 × 434)
= 868 m²

So the surface area of the cuboid is 868 square metres.

Question: If cotθ = 3/4, then secθ = ?

Solution:
দেওয়া আছে,
cotθ = 3/4 = ভূমি/লম্ব
∴ ভূমি = 3, লম্ব = 4

পিথাগোরাসের উপপাদ্য অনুযায়ী,
অতিভুজ = √(লম্ব² + ভূমি²)
= √(4² + 3²)
= √(16 + 9)
= √25
= 5

এখন,
secθ = অতিভুজ/ভূমি
∴ secθ = 5/3

মনে করি, আয়তক্ষেত্রের উচ্চতা x এবং ভূমি 3x
প্রশ্নমতে, 2(3x + x) = 64
⇒ 8x = 64
⇒ x = 8
∴ আয়তক্ষেত্রের উচ্চতা 8 এবং ভূমি 8×3  = 24 একক
∴ ক্ষেত্রফল = 8 × 24 = 192 একক 

Question: From a point Q on level ground, the angle of elevation of the top of a building is 45 degrees. If the building is 50 m high, find the distance of point Q from the foot of the building.

Solution:

Height of building AB = 50 m, angle of elevation ∠AQB = 45°

We know, tanθ = opposite/adjacent = AB/AQ
⇒ tan45° = 50/AQ
⇒ 1 = 50/AQ
⇒ AQ = 50 m

Thus, the distance from point Q to the foot of the building is 50 m.

Question: A ladder is leaning against a wall. It makes a 60° angle with the ground. If the length of the ladder is 10 meters, what is the distance between the foot of the ladder and the wall?

Solution:

ধরি, দেয়ালটি হলো AB এবং মইটি হলো AC।
মইটি ভূমির সাথে ∠ACB = 60° কোণ তৈরি করে।
মইয়ের দৈর্ঘ্য, AC = 10 মিটার।
মইয়ের গোড়া থেকে দেয়ালের দূরত্ব হলো BC।

এখন, ΔABC -এ
cos60° = BC/AC
⇒ 1/2 = BC/10
⇒ BC = 10/2
∴ BC = 5
∴ মইয়ের গোড়া থেকে দেয়ালের দূরত্ব 5 মিটার।

let the width of rectangle be x so the length & breath is increased by 2x.

so, new total area along with walkway is (60+2x)×(20+2x)

so, (60+2x)×(20+2x)-60×20 = 516
⇒ (60+2x)×(20+2x) = 1716
⇒ (30+x)×(10+x) = 429 = 33×13
⇒ x = 3

Question: If θ is a positive acute angle satisfying sin²θ + sin⁴θ = 1, then find the value of cot²θ + cot⁴θ.

Solution:
Given that,
sin²θ + sin⁴θ = 1 ..........(1)
⇒ sin⁴θ = 1 - sin²θ
⇒ sin⁴θ = cos²θ
⇒ sin²θ.sin²θ = cos²θ
⇒ sin²θ = cos²θ/sin²θ
⇒ sin²θ = cot²θ 

Now, putting sin²θ = cot²θ ..........(2)
∴  sin⁴θ = cot⁴θ ..........(3)
From equation (2) & (3) we get
sin²θ + sin⁴θ = cot²θ + cot⁴θ
∴ cot²θ + cot⁴θ = 1

Thus, the value of cot²θ + cot⁴θ = 1.

আমরা জানি,
বাক্সের উপরিতলের ক্ষেত্রফল = 2(ab + bc + ca)
= 2(2 × 3 + 3 × 4 + 4 × 2)
= 52 বর্গমিটার
∴ মোট খরচ = 52 × 3 = 156

Question: If C is the midpoint of the points A(1, 2) and B(7, 10), find the length of AC.

Solution:
দেওয়া আছে,
A(1, 2) এবং B(7, 10), এবং C হলো AB-এর মধ্যবিন্দু।

প্রথমে, দূরত্বের সূত্র ব্যবহার করে AB-এর দৈর্ঘ্য নির্ণয় করি।
AB = √{(x₂ - x₁)² + (y₂ - y₁)²}
AB = √{(7 - 1)² + (10 - 2)²}
AB = √(6² + 8²)
AB = √(36 + 64)
AB = √100
AB = 10

যেহেতু C হলো AB-এর মধ্যবিন্দু, তাই AC হবে AB-এর অর্ধেক।
∴ AC = AB/2
= 10/2
= 5

ধরি,
AB হচ্ছে দেয়াল এবং BC হচ্ছে মই।
এখানে ∠ACB = 60° এবং AC = 4.6 মিটার
তাহলে, AC/BC = cos 60° = 1/2 [ যেহেতু, cosΘ = ভূমি/অতিভুজ]
⇒ BC = 2 × AC
∴ BC = 2 × 4.6
= 9.2m

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

Question: A man looks into a mirror placed on the ground and sees the top of a tower. The mirror is 120 m away from the tower. If the man stands 0.6 m away from the mirror and his height is 1.8 m, find the height of the tower.

Solution:

Given that,
Distance from the mirror to the tower = 120 m
Distance from the man to the mirror = 0.6 m
Height of the man = 1.8 m
Height of the tower = H ?

Now,
Height of the man/Distance from man to mirror = Height of the tower/Distance from tower to mirror
⇒ 1.8/0.6 = H/120
⇒ 3 = H/120
⇒ H = 120 × 3 = 360 m

Question: The volume of a cylinder is 100π and its radius is 5. What is the lateral surface area?

Solution:
Given that,
Volume = 100π cubic units
Radius (r) = 5 units

We know,
Volume of a cylinder, V = πr²h
⇒ 100π = π × (5)² × h 
⇒ 100π = π × 25 × h
⇒ h = 100π/(25π)
⇒ h = 100/25
∴ h = 4 units

∴ Lateral surface area of a cylinder = 2πrh
= 2 × π × 5 × 4
= 40π

So the lateral surface area is 40π square units.

Question: If the difference between the circumference and diameter of a circle is 120 cm, then the diameter of the circle is -

Solution:
ধরি,
বৃত্তের ব্যাসার্ধ = r
বৃত্তের ব্যাস = 2r
বৃত্তের পরিধি = 2πr

প্রশ্নমতে,
2πr - 2r = 120
⇒ 2r(π - 1) = 120
⇒ r = (120/2){(22/7) - 1}
⇒ r = 60/(22 - 7)/7
⇒ r = (60 × 7)/15
∴ r = 28

∴ বৃত্তের ব্যাস = 2r = 2 × 28 = 56 সে.মি.

Question: The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

Solution:
Let the length and breadth of the rectangle be x cm and y cm.
We are given two conditions are,
Diagonal = √41 cm
By Pythagoras theorem:
√(x² + y²) = √41
⇒ x² + y² = 41 ........ (1)
And Area, xy = 20 cm² .......(2)

We know,
(x + y)² = x² + y² + 2xy
= 41 + 2 × 20
= 41 + 40
= 81
⇒ x + y = √81
∴ x + y = 9

∴ Perimeter = 2(x + y) = 2 × 9 = 18 cm

 So the perimeter of the rectangle is 18 cm.

Semi perimeter, s = (3 + 5 + 6)/2
= 7 cm
∴ Area = √{s(s - a)(s - b)(s - c)}
= √{7 (7 - 3) (7 - 5) (7 - 6)} Sq cm
= √(7 × 4 × 2 × 1)Sq cm
= 2√14 Sq cm

Say x is the height of the building.
a is a point 30 m away from the foot of the building.
Here, height is the perpendicular and distance between point a and foot of building is the base.
The angle of elevation formed is 30°.

Hence,
tan 30° = perpendicular/base = x/30
1/√3 = x/30
x = 30/√3

Question: If sec(3x - 40°) = cosec(50° - x), then the value of x is?

Solution:
sec(3x - 40°) = cosec(50° - x)
⇒ sec(3x - 40°) = cosec{90° - (40° + x)}
⇒ sec(3x - 40°) = sec(40° + x)
⇒ 3x - 40° = 40° + x
⇒ 2x = 80°

∴ x = 40°

Question: A square is inscribed in a circle of diameter 2a and another square is circumscribing circle. The difference between the areas of outer and inner squares is-

Solution:

দেওয়া আছে 
বৃত্তের ব্যাস = 2a
অভ্যন্তরীণ বর্গের কর্ণ = 2a 
অভ্যন্তরীণ বর্গের এক বাহু = x 

প্রশ্নমতে
√2x = 2a 
x = 2a/√2
x = √2a

অভ্যন্তরীণ বর্গের ক্ষেত্রফল = (√2a)²
= 2a²

অভ্যন্তরীণ বর্গের কর্ণ = বহিঃস্থ বর্গের এক বাহু = 2a
বহিঃস্থ বর্গের ক্ষেত্রফল = (2a)² = 4a² 

ক্ষেত্রফলের পার্থক্য = 4a² - 2a²
= 2a²

Let,
The edge of the third small cube be x cm
The volume of the cube = (edge)³
According to the question,
6³ + 8³ + x³ = 12³
⇒ 216 + 512 + x³ = 1728
⇒ x³ = 1728 - 728
= 1000
⇒ x = 1000^(1/3)
⇒ x = 10 cm.

r = √(13² - 12²)
= 5

So, the volume is V = 1/3∏r²h
= 1/3Π × 5² × 4
= 33.33Π cm³

Question: The volume of a cone is 300π cubic centimeters. If the radius of its base is 6 cm, what is the height of the cone?

solution:
দেওয়া আছে,
কোণকের আয়তন, V = 300π ঘন সে.মি.
ভূমির ব্যাসার্ধ, r = 6 সে.মি.
ধরি, কোণকের উচ্চতা = h সে.মি.

আমরা জানি,
কোণকের আয়তন, V = 1/3 × π × r² × h
∴ 300π = 1/3 × π × 6² × h
⇒ 300 = 1/3 × 36 × h (π উভয় পক্ষ থেকে বাদ দিয়ে)
⇒ 300 = 12h
⇒ h = 300 / 12
∴ h = 25 সে.মি.

অতএব, কোণকটির উচ্চতা = 25 সে.মি.

Question: When bent in the form of a circle, a wire has a radius of 28 cm. If the same wire is bent into the shape of a square, what will be its area in cm²?

Solution:
প্রদত্ত, বৃত্তের ব্যাসার্ধ, r = 28 cm

অতএব, পরিধি = 2πr
= 2 × (22/7) × 28 = 176 cm

ধরি, বর্গের বাহু = a cm
তাহলে, বর্গের পরিসীমা = 4a

এখন,
বৃত্তের পরিধি = বর্গের পরিসীমা
⇒ 176 = 4a
⇒ a = 176/4
= 44 cm

∴ বর্গের ক্ষেত্রফল = a²
= 44²
= 1936 cm² 

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

Number of bricks
=Courtyard area/1 brick area
=(2500×1600 / 20×10)=20000

Question: The greatest value of sin⁴2θ + cos⁴2θ is?

Solution:
sin²2θ + cos²2θ = 1

(sin²2θ + cos²2θ)² = 1²
⇒ sin⁴2θ + cos⁴2θ + 2 sin²2θ cos²2θ = 1
⇒ sin⁴2θ + cos⁴2θ = 1 − 2 sin²2θ cos²2θ [2 sin²2θ cos²2θ = 0 (when θ = 0° or 90°)]
∴ sin⁴2θ + cos⁴2θ = 1 

∴ greatest value = 1

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 = 2