Geometry: Mensuration, Trigonometry

Area of the circle =  πr² 
= π(√5)² 
= 5π
= 15.708

Question: A cube has a total surface area of 72 m². Determine the length of its diagonal.

Solution:
মনে করি,
ঘনকটির ধার = a 
ঘনকের সম্পূর্ণ পৃষ্ঠের ক্ষেত্রফল = 6a²

প্রশ্নমতে,
6a²  = 72
⇒ a² = 72/6
⇒ a² = 12
⇒ a² = (4 × 3)
⇒ a = 2√3 

∴ ঘনকটির কর্ণের দৈর্ঘ্য = a√3
= 2√3 × √3 
= 2 × (√3)2
= 2 × 3
= 6

∴ ঘনকটির কর্ণের দৈর্ঘ্য = 6 মিটার।

Question: The surface area of a cube is 96 cm². Its volume is-

Solution: 
Here, 
Surface area, 6a² = 96
⇒ a² = 96/6
⇒ a² = 16
⇒ a = √16
⇒ a = 4

∴ Volume = a³
= 4³
= 64 cm³

Question:

Solution:

Question: Find the equation of the vertical line passing through the point (- 3, 5).

Solution:

একটি উল্লম্ব রেখা (vertical line) হলো এমন একটি সরলরেখা যা Y-অক্ষের সমান্তরাল। এই ধরনের রেখার একটি বিশেষ বৈশিষ্ট্য হলো, রেখার উপর অবস্থিত প্রতিটি বিন্দুর x-স্থানাঙ্ক (x-coordinate) একই থাকে, কিন্তু y-স্থানাঙ্ক (y-coordinate) পরিবর্তিত হতে পারে।

উল্লম্ব রেখার সাধারণ সমীকরণ হলো: x = a, যেখানে a একটি ধ্রুবক সংখ্যা এবং রেখার প্রতিটি বিন্দুর x এর মান একই থাকে।

প্রশ্নে বলা হয়েছে রেখাটি (- 3, 5) বিন্দুর মধ্য দিয়ে যায়।
 যেহেতু এই বিন্দুর x-স্থানাঙ্ক হলো - 3,
সুতরাং রেখাটির সমীকরণ হবে: x = - 3

Question: The area of the floor of a museum is 500 square meters. How many unbroken tiles of dimension 10 x 20 cm² will be required to cover the floor completely?

Solution: 
Area of the floor = 500 square meters

Area of one tile = 10 × 20 cm²
=(10/100) × (20/100) m²
= .10 × .20 m²
= 0.02 m²

Therefore,
Tiles required = 500/.02
= 500/(2/100)
= 250 × 100
= 25,000 tiles

Question: What is the weight of 1 cubic meter of water?

Solution: 
We know, 
The weight of 1 cubic meter of water is 1000 kg.

Question: A 60-meter cable is attached from the top of a vertical pole down to the ground. If the cable makes an angle of 30 degrees with the ground, find the height of the pole.

Solution:
 
Let,
Height, AB = h

Given that, 
AC = 60m                    
∠ACB = 30°

∴ sin30°= AB/AC
⇒ 1/2 = h/60
⇒ h = 60 × (1/2)
∴ h = 30 m

So the height of the pole is 30 meters

Let, Breadth = x 
So, length = x + 15% of x = x + 3x/20 = 23x/20
ATQ,
23x/20 × x = 460
Or, x² = (460×20)/23 = 400
Or, x = 20
So, length = (23×20)/20 = 23 m 

Question: The ratio between the perimeter and the length of a rectangle is 5 : 2. If the area of the rectangle is 484 sq. cm, what is the length of the rectangle?

Solution: 
Let Length = l
& Breadth = b
Perimeter of a rectangle = 2(l + b)

Now,
2(l + b)/l = 5/2
⇒ 4(l + b) = 5l
⇒ l = 4b

Area, l × b = 484
⇒ 4b × b = 484
⇒ b² = 121
⇒ b = 11

∴ l = 4 × 11 = 44
The length of the rectangle is 44 cm. 

Question: The slope of a line perpendicular to one with slope 2 is:

Solution:
আমরা জানি,
যেকোনো সরলরেখার ঢাল যদি m হয়, তাহলে তার উপর লম্ব রেখার ঢাল হবে: - 1/m

এখানে,
ঢাল m = 2
∴  লম্ব রেখার ঢাল = - 1/2 

Question: The three angles of a triangle are x/3, x/3, and 4x/3 respectively. What is the sum of the two smallest angles?

Solution:
দেওয়া আছে,
ত্রিভুজের তিনটি কোণ যথাক্রমে x/3, x/3, এবং 4x/3

প্রশ্নমতে,
(x/3) + (x/3) + (4x/3) = 180°
⇒ (x + x + 4x)/3 = 180°
⇒ 6x/3 = 180°
⇒ 2x = 180°
⇒ x = 180°/2
⇒ x = 90°

এখন,
১ম কোণ = 90°/3 = 30° 
২য় কোণ = 90°/3 = 30°
৩য় কোণ = (4 × 90°)/3 = 120° 

∴ ক্ষুদ্রতম কোণ দুইটির সমষ্টি = 30° + 30° = 60°

একটি সরলরেখা অপর একটি সরলরেখার উপর তখনই লম্ব হবে যখন ঢালদ্বয়ের গুণফল -1 হবে। 
y = x - 2 সমীকরণের ঢাল হলো 1 [ঢাল = x এর সহগ]
এখন দেখতে হবে কোন সমীকরণের ঢাল -1
অপশন b তে, 2y = - 2x - 5
⇒ y = - x - 5/2
অতএব, y = - x - 2 সরলরেখার উপর লম্ব রেখার সমীকরণ  2y = - 2x - 5

Question: If xsinA = 1, xcosA = √3, then the value of (√3tanA + 2) = ?

Solution:
Given that
xsinA =1, xcosA = √3

xsinA/xcosA = 1/√3
⇒ tanA = 1/√3
⇒ √3tanA = 1
⇒ √3tanA + 2 = 1 + 2
∴ √3tanA + 2 = 3

1 hectare = 10,000 m²
So, Area = (1.5 x 10000) m² = 15000 m².
Depth =5/100 m=1/20m.
Volume = (Area x Depth) =15000 x1/20m³= 750 m^3.

Question: A rectangular sheet of paper, 10cm long and 8cm wide has squares of side 2cm cut from each of its corner. The sheet is then folded to form a tray of depth 2cm. What is the volume of this tray?

Solution: 
Length of tray = 10 - (2 × 2) = 10 - 4 = 6 cm.
Breadth of tray = 8 - (2 × 2) = 4 cm.
Depth of tray = 2 cm.
∴ Volume of tray = 6 × 4 × 2 = 48 cm³

Question: The radius of circle A is r, and the radius of circle B is 2r/3. What is the ratio of the area of circle A to the area of circle B?

Solution:
The radius of circle A = r
The area of circle A = πr²

The radius of circle B = 2r/3
The area of circle B = π(2r/3)² = 4πr²/9

∴ The ratio of the area of circle A to the area of circle B = πr² : 4πr²/9
= 1 : 4/9
= 9 : 4

Question: In a trapezoid, the lengths of the two parallel bases are 8 cm and 16 cm. If the height of the trapezoid is 6 cm, find the area of the trapezoid.

Solution:
Given that,
Trapezoid with bases a = 8 cm and b = 16 cm
Height, h = 6 cm

We know,
Area of trapezoid = (1/2) × (sum of bases) × height
= (1/2) × (a + b) × h
= (1/2) × (8 + 16) × 6
= (1/2) × 24 × 6
= 12 × 6
= 72

∴ The area of the trapezoid is 72 sq. cm

Question: The supplement of an angle exceeds twice the angle by 30°. Then the angle is equal to-

Solution:
Let the angle be x
Then, its supplement = 180 - x

According to the question,
180 - x = 2x + 30
⇒ 180 - 30 = 3x
⇒ 150 = 3x
⇒ x = 50°

tanθ = লম্ব/ভূমি
[এখানে, লম্ব = খুঁটির দৈর্ঘ্য এবং ভূমি = ছায়ার দৈর্ঘ্য] 
⇒ tanθ = 6/2√3
⇒ tanθ = tan 60°
∴ θ = 60°

Question: The area of the largest triangle that can be inscribed in a semi-circle of radius r, is-

Solution:
 Largest triangle inscribed in a semicircle has,
Base = diameter of semicircle = 2r
Height = radius = r (vertex at the top of the semicircle)

We know,
Area of triangle = (1/2) × base × height
= (1/2) × 2r × r
= r²

Question: If tan(θ + 30°) = √3, what is the value of sinθ?

Solution:
Given that,
tan(θ + 30°) = √3
⇒ tan(θ + 30°) = tan 60°
⇒ θ + 30° = 60°
⇒ θ = 60° - 30°
⇒ θ = 30°

Now,
sinθ 
= sin 30°
= 1/2

Question: A hall is 30m long and 10m broad. If the sum of the areas of the floor and the ceiling is equal to of the areas of four walls, the volume of the hall is-

Solution:
Let,
The height of the hall = h

Given that, 
Length of the room = 30 m
Breadth of the room = 10 m

According to the question,
2 × (30 × 10) = 2 × (30 + 10) × h
⇒ 2 × 40 × h = 2 × (30 × 10)
⇒ 40h = 300
⇒ h = 300/40
⇒ h = 30/4
∴ h = 15/2

We know,
 Volume = 30 × 10 × (15/2)
= 2250 m³

∴ Total distance of shadow of boy and distance from base of lamp post = 2.4 + 4.5 = 6.9 m
Let the height of lamp post be 'h' m

According to question,
⇒ 1.3/2.4 = h/6.9
⇒ h = (6.9 × 1.3)/2.4
⇒ h = 3.7375 = 3.74m

So, The height of the lamp post is 3.74 meters.

Question: If tan 53° = 4/3, then, what is the value of tan8°?

Solution:
Given that,
tan 53° = 4/3

We know,
tan(A - B) = (tanA - tanB)/(1 + tanA tanB)

Now,
8° = 53° - 45°
tan8° = tan(53° - 45°)
⇒ tan8° = (tan53° - tan45°)/(1 + tan53° tan45°)
⇒ tan8° = {(4/3) - 1}/{1 + (4/3) × 1}
⇒ tan8° = (1/3)/(7/3)
⇒ tan8° = 1/7

Question: In the figure below, AB is perpendicular to BC and DB = DC. If AD = √7 cm and AC = 5 cm, what is the value of BC?

Solution:

ΔABD-এ, পিথাগোরাসের উপপাদ্য অনুযায়ী,
BD² + AB² = AD²
⇒ BD² + AB² = (√7)²
⇒ BD² + AB² = 7 ... (1)

আবার, ΔABC-এ,
BC² + AB² = AC²
⇒ (BD + DC)² + AB² = 5²
⇒ (2DC)² + AB² = 25 [যেহেতু BD = DC, তাই BC = BD + DC = 2DC]
⇒ 4DC² + AB² = 25
⇒ 3DC² + (DC² + AB²) = 25 [যেহেতু BD = DC, তাই BD² = DC²]
⇒ 3DC² + 7 = 25 [সমীকরণ (1) থেকে মান বসিয়ে]
⇒ 3DC² = 25 - 7
⇒ 3DC² = 18
⇒ DC² = 6
⇒ DC = √6

অতএব, BC = 2DC (যেহেতু BD = DC)
= 2√6 cm

∴ BC এর মান 2√6 cm

sec (x − 30°) = 2/√3
Or, sec (x - 30°) = sec 30°
Or, x - 30° = 30°
Or, x = 60°
∴ tan 60° = √3

দেওয়া আছে,
sin2(A + B + C) = 1
⇒ sin 2(A + B + C) = sin 90°
⇒ 2(A + B + C) = 90°
∴ (A + B + C) = 45°............(i)
আবার, tan (B + C - A) = √3
⇒ tan (B + C - A) = tan60°(tan 60° = √3)
B + C - A = 60°............(ii)
এখন, (i) ও (ii) নং সমীকরণ যোগ করে পাই,
(A + B - C) + (B + C - A) = 45° + 60°
⇒2B = 105°
⇒ B = 52(1/2)°

Question: A rectangular floor of dimensions 18 m × 12 m is to be covered with a carpet 60 cm wide. Calculate how many metres of carpet are required.

Solution:
Given that,
Floor dimensions = 18 m × 12 m
Carpet width = 60 cm = 0.6 m [1m = 100cm]

Now, Area of floor = length × breadth = 18 × 12 = 216m²

And,
Width of carpet = 0.6m Length of carpet required = L m 
Area covered by L m of carpet = 0.6 × L  m²

This must equal the area of the floor, 0.6 × L = 216
L = 216/0.6 = 360 m

So 360 metres of carpet will be required.

Question: A cube has a total surface area of 294 square meters. What is the volume of the cube?

Solution:
ধরি, ঘনকের বাহুর দৈর্ঘ্য = a মিটার।

আমরা জানি,
ঘনকের সম্পূর্ণ পৃষ্ঠের ক্ষেত্রফল = 6a²

প্রশ্নমতে,
6a² = 294
⇒ a² = 294/6
⇒ a² = 49
∴ a = 7 মিটার

এখন,
ঘনকের আয়তন = a³
= 7³
= 343 ঘন মিটার

অতএব, ঘনকটির আয়তন = 343 ঘন মিটার।

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.

Let the longer side of the right angled triangle be x
Then, hypotenuse = (x + 2) and shorter side = (x - 7)
Here,
• h = x + 2
• p = x
• b = x - 7
Using Pythagoras Theorem :
 (Hypotenuse)² = (Perpendicular)² + (Base)²
⇒ (x + 2)² = (x)² + (x - 7)²
⇒ x² + 2(x)(2) + 2² = x² + x² - 2(x)(7) + (7)²
⇒ x² + 4x + 4 = x² + x² - 14x  + 49
⇒ x² - 18x  +  45 = 0
⇒ x² - 15x - 3x + 45 = 0
⇒ x(x - 15) - 3(x - 15) = 0
⇒ (x - 3) (x - 15)
The value of x can't be 3 as (x - 7) would be negative.
Hence, the value of the longer side is 15 cm.
So, the hypotenuse is 15 + 2 = 17 cm

r = √(5² - 4²)
= 3

So, the volume is V = 1/3∏r²h
= 1/3Π × 3² × 4
= 12Π cm³

Question: A cube has a total surface area of 72 m². Determine the length of its diagonal.

Solution:
মনে করি,
ঘনকটির ধার = a 
ঘনকের সম্পূর্ণ পৃষ্ঠের ক্ষেত্রফল = 6a²

প্রশ্নমতে,
6a²  = 72
⇒ a² = 72/6
⇒ a² = 12
⇒ a² = (4 × 3)
⇒ a = 2√3 

∴ ঘনকটির কর্ণের দৈর্ঘ্য = a√3
= 2√3 × √3 
= 2 × (√3)²
= 2 × 3
= 6

∴ ঘনকটির কর্ণের দৈর্ঘ্য = 6 মিটার।

Question: The height of a cylinder is five times the radius of the cylinder. If the volume of the cylinder is 625π cm^3, what is the height of the cylinder?

Solution: 
Given, radius = r and
Height = 5r

ATQ, πr² × 5r = 625π
⇒ r³ = 125
⇒ r = 5

∴ Height = 5 × 5
= 25 cm 
= 0.25 m

Question: An observer 1.6 m tall stands 20 meters away from a tree. The angle of elevation from his eye to the top of the tree is 45°. What is the height of the tree?

Solution:
 

মনে করি, 
গাছটির উচ্চতা AB। পর্যবেক্ষকের চোখ C বিন্দুতে আছে এবং তার উচ্চতা CD = 1.6 m
পর্যবেক্ষক থেকে গাছটির দূরত্ব BD = 20 m
এখানে, A, C এবং E বিন্দু দ্বারা গঠিত ACE হলো একটি সমকোণী ত্রিভুজ, যার ∠C = 45°।

আমরা জানি,
tan θ = লম্ব/ভূমি
এখানে, লম্ব = AE এবং ভূমি = CE
∴ tan 45° = AE/20
∴ 1 = AE/20
∴ AE = 20 মিটার

গাছটির মোট উচ্চতা, AB = AE + EB
= 20 + 1.6
= 21.6 মিটার

সুতরাং, গাছটির উচ্চতা হলো 21.6 মিটার।

Question: In an isosceles triangle, each of the equal sides is 16 cm long and the included angle between the equal sides is 30°. What is the area of the triangle?

Solution:
Given that,
The included angle between the equal sides, θ = 30°
Each of the equal sides of the isosceles triangle, a = b = 16 cm

We know that,
Area of the triangle = (1/2) × a × b × sin θ
= (1/2) × 16 × 16 × sin 30°
= (1/2) × 16 × 16 × (1/2)
= 8 × 8
= 64 square cm

∴ The area of the triangle is 64 cm².

Area or outer space = 2a x 2a = 4a²
From ABCD rectangle, In ΔBAD,
BD² = (AD)² + (AB)²
⇒ (2a)² = x² + x²
⇒ 4a² = 2x² 
⇒ 2a² = x² 
∴ x = √2a

Area of inner square = √2a × √2a = 2a² 

∴ The difference between the areas of outer and inner squares is = 4a² - 2a² = 2a² 

Question: The height of a cylinder is four times its radius. If the cylinder’s volume is 32π cm³, what is the radius of the cylinder?

Solution:
ধরি,
সিলিন্ডারের ব্যাসার্ধ = r সেমি
∴ সিলিন্ডারের উচ্চতা = 4r সেমি

আমরা জানি,
সিলিন্ডারের আয়তন = πr²h ঘন সেমি
প্রশ্নমতে,
πr²h = 32π
⇒ πr² × 4r = 32π
⇒ 4πr³ = 32π
⇒ 4r³ = 32
⇒ r³ = 32/4
⇒ r³ = 8
⇒ r = 2