Geometry: Mensuration, Trigonometry

Question: 

Solution: 

Question: ABCD is a square and one of its sides AB is also a chord of the circle as shown in the figure. What is the area of the square?


Solution:
চিত্রানুসারে, O হলো বৃত্তের কেন্দ্র এবং OA ও OB হলো বৃত্তের ব্যাসার্ধ, যার দৈর্ঘ্য 3।
 AOB একটি সমকোণী ত্রিভুজ, যেখানে ∠AOB = 90° এবং অতিভুজ = AB

পিথাগোরাসের উপপাদ্য অনুসারে,
AB² = OA² + OB²
AB² = 3² + 3²
AB² = 9 + 9
AB² = 18

আমরা জানি, বর্গক্ষেত্রের ক্ষেত্রফল = বাহুর দৈর্ঘ্য^২ 
যেহেতু ABCD একটি বর্গ, তাই এর ক্ষেত্রফল হলো AB²
সুতরাং, বর্গটির ক্ষেত্রফল হলো 18

Question: From a point P on a level ground, the angle of elevation of the top of a tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is-

Solution:

Given that,
Hight of tower, AB = 100 m
Angle of elevation from point P, ∠APB = 30°

We know,
tanθ = opposite/adjacent ​= AB/PA
⇒ tan30° = 100/PA
⇒ 1/√3 = 100/PA
∴ PA = 100√3 m

Thus, the distance from point P to the foot of the tower is 100√3 m.

আমরা দেখতে পাচ্ছি যে চারটি পরস্পর সমান্তরাল রেখাকে তিনটি পরস্পর সমান্তরাল রেখা ছেদ করলে মোট আয়তক্ষেত্রের সংখ্যা হবেঃ
6 + 5 + 4 + 3 + 2 + 2 + 1 = 18

Question: If the measures of the angles in a triangle are in the ratio of 3 : 7 : 8, then the degrees in the largest angle:

Solution:
মনে করি, ত্রিভুজের কোণ তিনটি যথাক্রমে 3x, 7x এবং 8x

আমরা জানি, ত্রিভুজের তিন কোণের সমষ্টি 180°

শর্তমতে,
3x + 7x + 8x = 180°
⇒ 18x = 180°
⇒ x = 180°/18
⇒ x = 10°

বৃহত্তম কোণটি হলো 8x
∴ বৃহত্তম কোণের মান = 8 × 10° = 80°

Question: 60π cubic centimeters of silver is drawn into a wire 1 mm in diameter. The length of the wire in meters will be -

Solution:
Let the length of the wire be h
Radius, r =(1/2) mm =1/20 cm.
We know, 
Volume of a cylinder(wire) = πr²h

∴ π × (1/20) × (1/20) × h = 60π
⇒ h = 60 × 20 × 20
⇒ h = 24000  
 
∴ The length of the wire = 24000 cm 
= (24000/100) meters
= 240 meters

Question: The ratio of the volumes of two spheres is 27:8. What is the ratio of their surface areas?

Solution: 
Volume of a Sphere: V = (4/3)π(r)³
Surface Area of a Sphere: S = 4πr²

Given, 

Question: Find the equation of the line with x-intercept = 4 and y-intercept = 3.

Solution:
Given, x-intercept = 4,
So, the line passes through (4, 0).
y-intercept = 3,
So, the line passes through (0, 3).

We know,
The intercept form of a line is:
(x/a) + (y/b) = 1, where a = x-intercept, b = y-intercept.
⇒ (x/4) + (y/3) = 1
⇒ (3x + 4y)/12 = 1
⇒ 3x + 4y = 12
⇒ 3x + 4y - 12 = 0

∴ The equation of the line is 3x + 4y - 12 = 0

Question: If the length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, then the length of the median to its greatest side is -

Solution:
According to question,
Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle.

To find the length of the median to the greatest side of a triangle, we can use the formula:
Median = 1/2 * √(2b^2 + 2c^2 - a^2)
Where a, b, and c are the lengths of the sides of the triangle, and a is the greatest side.
In this case, the lengths of the sides are 6 cm, 8 cm, and 10 cm. The greatest side is 10 cm.
Using the formula:
Median = 1/2 * √(2 * 8² + 2 * 6² - 10²)
= 1/2 * √(128 + 72 - 100)
= 1/2 * √(100)
= 1/2 * 10
= 5 cm

So, the length of the median to the greatest side of the triangle is 5 cm.

Question: The volume of a cuboid with length, breadth and height as 11x³, 12x⁵ and 13x⁷ respectively is:

Solution:
দেওয়া আছে
আয়তাকার ঘনবস্তুর (cuboid) দৈর্ঘ্য = 11x³
আয়তাকার ঘনবস্তুর (cuboid) প্রস্থ = 12x⁵
আয়তাকার ঘনবস্তুর (cuboid) উচ্চতা = 13x⁷ 

আয়তাকার ঘনবস্তুর আয়তন = (11x³) × (12x⁵) × (13x⁷)
= 1716x¹⁵

Length of the parallel sides of prism = 10 cm and 6 cm
Height of prism = 8 cm
∴ Volume of prism = (1/2){(10 + 6) × 5 × 8}
= (1/2) × 16 × 5 × 8
= 320 cm³

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

Question: Write an equation of the line with slope 2 and x-intercept (- 4, 0).

Solution: 
Given that, 
Slope m = 2
x-intercept (- 4, 0)

We know, 
y - y₁ ​= m(x - x1​)
⇒ y - 0 = 2{x - (- 4)}. ; [Here, (x₁, y₁) = (- 4, 0) and m = 2]
⇒ y = 2(x + 4)
∴ y = 2x + 8

So the equation of the line is  y = 2x + 8.

Question: Find an equation of the vertical line containing the point (9, - 3).

Solution:
দেওয়া আছে, 
প্রদত্ত বিন্দুটি হলো (9, -3)।

উল্লম্ব রেখার একটি প্রধান বৈশিষ্ট্য হলো, এই রেখার উপর অবস্থিত প্রতিটি বিন্দুর x-স্থানাঙ্ক সর্বদা একই থাকে। যেহেতু রেখাটি (9, -3) বিন্দু দিয়ে যায়, তাই রেখাটির উপর অবস্থিত প্রতিটি বিন্দুর x-এর মান হবে 9।

সুতরাং, নির্ণেয় সমীকরণটি হবে x = 9.

Number of bricks =Volume of the wall/ Volume of 1 brick
=800 x 600 x 22.5/25 x 11.25 x 6
= 6400.

The ratio of the length of an arc of a circle to its circumference is always the same as the ratio of a sector to its area.
(r × θ) / (2π × r) = θ / 2π

ATQ,
sector area : total area = (θ/2π)×πr² : πr² = 3 : 5
(θ/2π)×πr² / πr² = 3 / 5
θ/2π = 3 / 5
(θ/2π) × 2π = 3/5 × 2π
∴ θ = 6π / 5
Arc length = 6π/5 × r
Circumference = 2πr

∴ Arc length/Circumference = (6π/5 × r) / 2πr = 3/5

ABCD is a rhombus in which AB = BC = CD = DA = 17 cm

Diagonal AC = 16 cm (with O being the diagonal intersection point)

Therefore, AO = 8 cm

In ∆ AOD,

AD² = AO² + OD²
⇒ 17² = 8² + OD²
⇒ 289 = 64 + OD²
⇒ 225 = OD²
⇒ OD = 15

Therefore, BD = 2 × OD
= 2 × 15
= 30 cm

Now, area of rhombus

= 1/2 × d₁ × d₂
= 1/2 × 16 × 30
= 240 cm²

Question: If tan(θ - 30°) = √3, then find sinθ.

Solution:
Given,
tan(θ - 30°) = √3
⇒ tan(θ - 30°) = tan60°
⇒ θ - 30° = 60°
∴ θ = 90°

Now,
sinθ = sin90° = 1

Question: The cube shown below has a volume of 64 cubic inches. What is the length of the line segment AB? 



Solution:
Let side length = a

Here,
Volume of the cube, a³ = 64
⇒ a³ = 4³
⇒ a = 4

∴ Length of the line segment (space diagonal) AB = √(a² + a² + a²)
= √(3a²) 
= a√3
= 4√3

Question: If θ = 45°, then find the value of (3 + cot2θ)/(3 - cot2θ). 

Solution:
(3 + cot2θ)/(3 - cot2θ)
= [3 + cot(2 × 45°)]/[3 - cot(2 × 45°)]
= (3 + cot90)°/(3 - cot90)°
= (3 + 0)/(3 - 0)
= 3/3
= 1

Question: If 1 + tan²θ = 4 and θ < 90°, than what is the value of θ = ?

Solution:
Given that,
1 + tan²θ = 4 and θ < 90°
⇒ sec²θ = 4    ; [sec²θ = 1 + tan²θ]
⇒ (secθ)² = (2)²
⇒ secθ = 2
⇒ secθ = sec60°
⇒ θ = 60°

Question: The greatest value of sin⁴θ + cos⁴θ is?

Solution:
We know,
sin²θ + cos²θ = 1
⇒ (sin²θ + cos²θ)² = 1²    ; Squaring both sides
⇒ sin⁴θ + cos⁴θ + 2sin²θ.cos²θ = 1 
⇒ sin⁴θ + cos⁴θ = 1 - 2sin²θ.cos²θ
⇒ sin⁴θ + cos⁴θ = 1 - 0    ; [Put θ = 90° , cos90° = 0]
∴ sin⁴θ + cos⁴θ = 1 

So the greatest value of sin⁴θ + cos⁴θ is 1.

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

Solution:
Given,
sin 17° = (x/y)

We know,
sin² θ + cos² θ = 1
∴ cos θ = √(1 - sin θ)

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

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

Question: Find the maximum distance between two points on the perimeter of a rectangular garden whose length and breadth are 24 m and 7 m.

Solution:
একটি আয়তক্ষেত্রের পরিসীমার উপর অবস্থিত দুটি বিন্দুর মধ্যে সর্বাধিক দূরত্ব হলো এর কর্ণের দৈর্ঘ্য। কর্ণের দৈর্ঘ্য পিথাগোরাসের সূত্র ব্যবহার করে নির্ণয় করা যায়।

কর্ণের দৈর্ঘ্য = √(দৈর্ঘ্য^২ + প্রস্থ^২)
= √(24² + 7²)
= √(576 + 49)
= √625
= 25 মিটার

সুতরাং, দুটি বিন্দুর মধ্যে সর্বাধিক দূরত্ব হলো 25 মিটার।

Question: A circle and a rectangle have the same perimeter. The sides of the rectangle are 16 cm and 50 cm. What is the area of the circle?

Solution:
আয়তক্ষেত্রটির পরিসীমা = 2 × (16 + 50) সেমি
= 2 × 66 সেমি
= 132 সেমি।

যেহেতু বৃত্তের পরিধি ও আয়তক্ষেত্রের পরিসীমা সমান, তাই বৃত্তের পরিধিও 132 সেমি।

∴ বৃত্তের পরিধি, C = 2πr
⇒ 2πr = 132
⇒ r = 132/(2 × 22/7)
∴ r = 21 সেমি।

বৃত্তের ক্ষেত্রফল, A = πr²
= (22/7) × (21)²
= (22/7) × 441
= 1386 বর্গ সেমি।

সুতরাং, বৃত্তটির ক্ষেত্রফল হলো 1386 বর্গ সেমি।

Question: A cistern 8m long and 5m wide contains water up to a depth of 1m 50cm. The total area of the wet surface is:

Solution:
দেওয়া আছে,
চৌবাচ্চার দৈর্ঘ্য (l) = 8 m, প্রস্থ (b) = 5 m, এবং গভীরতা (h) = 1 m 50 cm = 1.5 m

ভেজা পৃষ্ঠের (wet surface) ক্ষেত্রফল নির্ণয়ের জন্য আমরা সমগ্র পৃষ্ঠের ক্ষেত্রফল থেকে উপরের তলের ক্ষেত্রফল বাদ দেবো।

আমরা জানি,
সমগ্র পৃষ্ঠের ক্ষেত্রফল = 2(lb + bh + lh) বর্গ একক

∴ ভেজা পৃষ্ঠের ক্ষেত্রফল = 2(lb + bh + lh) - lb বর্গ একক
= 2{(8 × 5) + (5 × 1.5) + (8 × 1.5)} - (8 × 5)
= 2(40 + 7.5 + 12) - 40
= 2(59.5) - 40
= 119 - 40
= 79 m²

সুতরাং, ভেজা পৃষ্ঠের (wet surface) মোট ক্ষেত্রফল 79 বর্গ মিটার।

Question: The area of a rectangle R with a width of 4 feet is equal to the area of a square S, which has a perimeter of 32 feet. The perimeter of the rectangle R is:

Solution:
ধরি, আয়তক্ষেত্র R এর দৈর্ঘ্য এবং প্রস্থ যথাক্রমে l, b.
বর্গের এক বাহু = a

প্রশ্নমতে,
4a = 32
∴ a = 8

এখানে,
আয়তক্ষেত্রের ক্ষেত্রফল = বর্গের ক্ষেত্রফল 
∴ l × b = a²
 ⇒ l = a²/b
= 64/4
= 16

∴ আয়তক্ষেত্রের পরিসীমা = 2(l + b)
= 2(16 + 4)
= 40 feet

Question: Find the value of cosec(- π/6)

Solution:
cosec(- π/6)
= - cosec(π/6)
= - 1/sin(π/6)
= - 1/sin30°
= - 1/(1/2)
= - 2

Area of the carpet :
= [(6.20 - 0.20) × (8 - 0.20)] m² 
= (6 × 7.8) m² 
= 46.8 m² 
∴ Cost of carpeting :
= Tk. (46.8 × 15)
= Tk. 702

Question: A cylinder has a radius of 5 cm and a height of 7 cm. What is its volume?

Solution: 
Radius, r = 5 cm 
Height, h = 7 cm

We know, 
Volume = πr²h
= (22/7) × (5)² × 7
= 550 cm³

Question: The perimeter of one face of a cube is 36 cm. Its volume must be-

Solution:
perimeter of one face is = 36 cm

let, length of one side is = a cm
∴ perimeter = 4a cm

ATQ,
⇒ 4a = 36
⇒ a = 36/4
= 9 cm

∴ volume = a³
= 9³
= 729 cm³

Let the length of the rectangle be x metre.
Then, a breath of the rectangle = (168/x)

∴ √{(x)² + (168/x)²} = 25
⇒ √{(x² + (28224/x²)} = 25
⇒ {(x² + (28224/x²)} = 625
⇒ x⁴ - 625x² + 28224 = 0
⇒ x²(x² - 576) - 49(x² - 576) = 0
⇒ (x² - 576)(x² - 49) = 0
⇒ x² = 576 or x² = 49
⇒ x = 24 or x = 7

Hence, length = 24 cm. and breadth = 7 m.

Question: If the volume of a sphere is divided by its surface area, the result is 25 cm. the radius of the sphere is -

Solution:
Let,
the radius of the sphere is r cm
∴ the volume of a sphere = (4/3)πr³
∴ the surface area of a sphere = 4πr²

ATQ,
{(4/3)πr³}/(4πr²) = 25
⇒ r/3 = 25
∴ r = 75

∴ the radius of the sphere is 75 cm

Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that BC = 70 m
∠ABD = 30°
∠ACD = 60°
Let CD = x, AD = h
From the right ⊿CDA
tan 60° = AD/CD
√3 = h/x ......eq(1)
From the right ⊿BDA
tan30° = AD/BD
1/√3 = h/(70 + x) ........ eq(2)
eq(1)/eq(2)
√3/(1/√3) = (h/x)/{h/(70+x)}
3 = 70 + x/x
2x = 70
x = 35
Substituting this value of x in eq:1, we have
√3 = h/35
h = 35√3
= 60.55 m

Question: If rSinθ = √3 and rCosθ = 1, then the value of (√3Cotθ + 1) = ?

Solution:
Given that,
rSinθ = √3 and rCosθ = 1
⇒ rCosθ/rSinθ = 1/√3
⇒ Cotθ = 1/√3
⇒ √3Cotθ = 1

(Add 1 both sides)
√3Cotθ + 1 = 1 + 1
∴ √3Cotθ + 1 = 2

Question: If tan(θ + 15°) = √3, what is the value of sinθ?

Solution:
Given that,
tan(θ + 15°) = √3
⇒ tan(θ + 15°) = tan 60°
⇒ θ + 15° = 60°
⇒ θ = 60° - 15°
⇒ θ = 45°

Now,
sinθ
= sin45°
= 1/√2

Question: The area of the base of a rectangular tank is 6500 sq. cm and the volume of water contained in it is 2.6 cubic meters. The depth of water is?

Solution:
Let depth = D cm.
We know,
Volume Base Area x Depth
and, 1 cubic meter = 1,000,000 cubic centimeters

Then,
D × 6500 = 2.6 × 100 × 100 × 100 
∴ D = (2.6 × 100 × 100 × 100)/6500 cm 
= 400 cm
= 4 m

Question: The area of a trapezium is 96 square cm. The length of one of the parallel sides is 12 cm, and the distance between the parallel sides is 8 cm. Find the length of the other parallel side.

Solution:
Given,
Area of the trapezium = 96 cm²
One parallel side a = 12 cm
Distance between the parallel sides h = 8 cm

Let
the other parallel side = b cm

We know,
The area of a trapezium = (1/2) × (a + b) × h
⇒ 96 = (1/2) × (12 + b) × 8
⇒ 96 = (12 + b) × 4
⇒ (12 + b) = 96/4
⇒ 12 + b = 24
⇒ b = 24 - 12
∴ b = 12

∴ The other parallel side is 12 cm.

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 = (a√3)/2

∴ The height of the equilateral triangle will be 
= (√3/2) × (28√3)
= 42 cm

Question: The slant height of a right circular cone is 10 m, and its height is 8 m. Find the area of its curved surface.

Solution: 
Here, l = 10 and h = 8
r = √(l² - h²) 
= √(10² - 8²) 
= √(100 - 64) 
= √36
= 6

∴ Curved surface area = πrl 
= π × 6 × 10
= 60π sq. meter