আমরা জানি, কার্পেটিংয়ের খরচ = দৈর্ঘ্য × প্রস্থ × প্রতি বর্গমিটারের খরচ।

ধরি, ঘরের প্রস্থ = x মিটার।

তাহলে, প্রথম অবস্থায়:

6 × x × প্রতি বর্গমিটারের খরচ = 270

দ্বিতীয় অবস্থায়:

6 × (x - 0.5) × প্রতি বর্গমিটারের খরচ = 240

এই দুইটি সমীকরণ থেকে আমরা প্রতি বর্গমিটারের খরচ এবং x নির্ণয় করতে পারি।

প্রথম সমীকরণ থেকে:

6x × প্রতি বর্গমিটারের খরচ = 270

⇒ প্রতি বর্গমিটারের খরচ = 270 / (6x)

দ্বিতীয় সমীকরণে এটি বসিয়ে:

6(x - 0.5) × (270 / (6x)) = 240

⇒ (x - 0.5) × (270 / x) = 240

⇒ 270(x - 0.5) = 240x

⇒ 270x - 135 = 240x

⇒ 270x - 240x = 135
⇒ 30x = 135

⇒ x = 135 / 30 = 4.5

অতএব, ঘরের প্রস্থ 4.5 মিটার।

*সঠিক উত্তর: d) 30
-2, 3, এবং 5 দ্বারা বিভাজ্য হলে,
-সংখ্যাটি 30 দ্বারা বিভাজ্য হবে (যেহেতু 30 = 2 × 3 × 5)।

Area of  rhombus = \({1 \over 2} \times {d_1} \times {d_2}\)

 

Where d­₁ and d₂ are the diagonals of the rhombus,

\(\eqalign{ & \Rightarrow 54\ {cm^2} = {1 \over 2}{d_1} \times {d_2} \cr & \Rightarrow {d_2} = 18\ cm \cr}\)

Hence, length of the second diagonal would be 18 cm.

১২৮ কিলোমিটার যায় = ৯ লিটার পেট্রোলে
∴ ১ কিলোমিটার যায় = (৯/১২৮) লিটার পেট্রোলে
                       = ০.০৭ লিটার পেট্রোলে

বড় ঘনকের আয়তন = ১x১x১ ঘন মি. = ১ ঘন মি.

ছোট ঘনকের ধার = ১০ সে.মি. = ০.১ মি.

সুতরাং, একটি ছোট ঘনকের আয়তন = ০.১ x ০.১ x ০.১ ঘন মি. = ০.০০১ ঘন মি.

ফলে, ছোট ঘনকের সংখ্যা = ১/০.০০১ = ১০০০ টি।

Let centre of circle be M.

Radius of circle = (1/2) × 20 cm = 10 cm

If PQR is isosceles, and QR is diameter, angle P will be right, as angle in a semicircle is right.

⇒ QP and RP will be equal sides of triangle PQR and hence will subtend equal angles at centre.

Together they subtend an angle of 180°.

⇒ PQ will subtend an angle of 90°.

∴ Area of minor segment = Area of sector – Area of triangle PMQ = ¼ π × 10 × 10 – ½ × 10 × 10 = 28.5 square cm

Let centre of ring be N.

A point P is taken on its inner boundary and a point Q is taken on its outer boundary such that P, Q and centre of ring are collinear.

⇒ PQ = QN – PN = 20 cm – 10 cm = 10 cm

⇒ PR = 2PQ = 20 cm

Volume of sphere = (4/3) × 3.14 × 20 × 20 × 20 = 33493.33 cubic cm

As we know, capacity of cylindrical tank is essentially equal to its volume.

Volume of a cylindrical tank = πr²h

Where, r = radius of the tank and h= height of the tank

∵ 1 litre = 1000 cm³ = 0.001 m³

∴ Capacity of cylindrical tank = 61.6 × 0.001 m³

⇒ πr²h = 61.6 × 0.001

\(\begin{array}{l} \Rightarrow \frac{{22}}{7} \times \;4\; \times \;{r^2}\; = \frac{{616}}{{10000}}\\ \Rightarrow \;{r^2}\; = \;\frac{{616 \times 7}}{{10000 \times 22 \times 4}}\; = \;0.0049 \end{array}\)

⇒ r = 0.07 m

Now, diameter = 2r

∴ Diameter = 2 × 0.07 = 0.14 m = 14 cm

Volume of a cuboid = length × breadth × height

After increment in the length of the sides, new sides of the room would be 11 m, 33 m and 22m respectively.

∴ New volume of the room = length × breadth × height

= 11 × 33 × 22 =7986 m³

বিস্তারিত পদ্ধতি:
দেওয়া আছে, বাইরের বর্গক্ষেত্রের বাহুর দৈর্ঘ্য = 15 cm.
এই বর্গক্ষেত্রের মধ্যবিন্দুগুলো যোগ করে একটি নতুন বর্গক্ষেত্র আঁকা হলো।
পিথাগোরাসের উপপাদ্য অনুযায়ী, ভেতরের বর্গক্ষেত্রের বাহুর দৈর্ঘ্য হবে = √((15/2)² + (15/2)²)
= √(225/4 + 225/4)
= √(450/4)
= √(225/2)
= 15 / √2 cm.

এখন, এই ভেতরের বর্গক্ষেত্রের ভেতরে একটি বৃত্ত আঁকা হলো যা বর্গক্ষেত্রের বাহুগুলোকে স্পর্শ করে।
তাহলে বৃত্তটির ব্যাস হবে ভেতরের বর্গক্ষেত্রের বাহুর দৈর্ঘ্যের সমান।
বৃত্তের ব্যাস = 15 / √2 cm.
সুতরাং, বৃত্তের ব্যাসার্ধ, r = (15 / √2) / 2 = 15 / (2√2) cm.

একটি নিরেট অর্ধগোলকের ব্যাসার্ধ বৃত্তের ব্যাসার্ধের সমান, অর্থাৎ r = 15 / (2√2) cm.
আমরা জানি, নিরেট অর্ধগোলকের সমগ্র তলের ক্ষেত্রফল (Total Surface Area) = 3πr²

মান বসিয়ে পাই,
TSA = 3 × 3.14 × (15 / (2√2))²
= 3 × 3.14 × (225 / 8)
= 9.42 × 28.125
= 264.9375 cm² (প্রায়)

বিকল্প হিসেবে, π = 3.14 ব্যবহার করলে, 3 × 3.14 × 225 / 8 = 2119.5 / 8 = 264.9375 ≈ 264.6 cm² (অপশন অনুযায়ী সঠিক মান কাছাকাছি হবে)।

শর্টকাট:
বাইরের বর্গের বাহু a হলে, অন্তর্বৃত্তের ব্যাসার্ধ r = a / (2√2).
অতএব, r = 15 / (2√2).
TSA = 3πr² = 3 × 3.14 × (225 / 8) = 264.9 cm². প্রদত্ত আপশনের মধ্যে 264.6 সঠিক উত্তর।

Given that, 4 times the sum of the areas of two circular faces is twice the area of its curved surface. The height of a right circular cylinder is 6 m. we know that,

Area of curved surface of the cylinder = 2 π r h

Area of right circular face = π r²

Then, 4 × Sum of the areas of its two circular faces = 2 × Area of curved surface of the cylinder

⇒ 4 × (π r² + π r²) = 2 ×2 π r h

⇒ 8 π r² = 4 π r h

⇒ 2 r = h

Or, r = h/2 = 6/2 = 3 m.

Let take a parallelogram.

Where ‘h’ is height and ‘b’ is base of parallelogram.

 As we know, area of parallelogram= b × h

In th given parallelogram,

Using above formula, b = 23 and h = 12

Area = 23 × 12 = 276 cm²

Given,

Radius of the base of the cone (R) = 10 m

Height of the Cone (H) = 24 m

∵ Slant height (l) = \(\sqrt {\left( {{R^2}{\rm{\;}} + {\rm{\;}}{H^2}} \right)}\)

∴ Slant Height = \(\sqrt {{{10}^2} + {{24}^2}} = 26{\rm{\;}}m\)

We know,

Curved Surface Area of the cone = πRl

∴ Curved Surface Area = π x 10 x 26 = 260π m²

Area of the cloth used = Curved Surface Area of the cone = 260π m²

Given, width of the cloth = 5.2 m

Length of the cloth used = Area of Cloth used/Width of Cloth

∴ Length of the cloth = 260π/5.2 = 50π 

Volume of the sphere is given by (4/3)π r³

Surface area of sphere is 4π r²

Given, volume of the sphere is \(179\frac{2}{3} = \frac{{539}}{3}\) m³

\(\begin{array}{l} \Rightarrow \left( {\frac{4}{3}} \right)\pi \;{r^3} = \left( {\frac{{539}}{3}} \right)\\ \Rightarrow r = {\left( {\frac{{539}}{{4\pi }}} \right)^{\frac{1}{3}}}m \end{array}\)

Now, Surface area of sphere is \(= \;4 \times \;\frac{{22}}{7} \times \;{\left( {\frac{{539}}{{4\pi }}} \right)^{\frac{2}{3}}} = 154{m^2}\)

he largest possible circle cut out of square has diameter equal to side of the square.

Area of circle = π × (d²/4) ;where d is the diameter of the circle

Area of circle = (3.14) × (10²/4) =78.5 sq.cm [Approx.]

Area of square = s × s ; where s is the side of the square

Area of square = 10 × 10 = 100 sq.cm

Ratio of circle to square = 78.5/100 = 0.785

= 0.8 [Approx.] = 4/5

Area = 3850m2

We know that area of a circle = πr², where r is the radius of the circle

Therefore, 3850 = πr²

22/7 r² = 3850   (taking π = 22/7)

r² = (3850 × 7)/22

r² = 1225

r = 35 m

We know that the cost of fencing = perimeter of the field × cost per meter

Perimeter of the field = 2 π r = 2 × 22/7 × 35           [perimeter of circle = 2 π r]

= 220 m

Therefore the cost of fencing = Rs. (220 × 4)

= Rs. 880

Hence, the cost of fencing the field is Rs. 880.

Given that, A plywood sheet is of rectangular shape with dimensions 38 m x 28 m

After cutting off the squares from corners:

(fig)

Clearly, l = 38 – 20 = 18 m,

b = 28 – 20 = 8 m,

h = 10 m.

we know that Volume of a box = l × b × h

∴ Volume of the box = (18 x 8 x 10) m³ = 1440 m³.

Let the edge of the cube and radius of the ball be x units and r units respectively

As we know, the ball fits exactly inside the cube

∴ Diameter of the ball = Edge of the cube

⇒ 2r = x

Now, Volume of cube = (edge)³ = x³ = 8r³

volume of ball = 4/3 × π r³

⇒ Required ratio \(= \frac{{8{r^3}}}{{\frac{4}{3}\pi {r^3}}} = \frac{6}{\pi }\)

⇒ Required ratio = 6 : π

Hence, the ratio of volumes of the cube and that of the ball is 6 : π

Let ‘s’ be the side of  the square.

Perimeter of square = 196 m

⇒ 4s = 196

⇒ s = 196/4

⇒ s = 49 m

Radius of circle = r = Side of a square with perimeter 196 sq.m = 49 m

∴ Area of circle = πr² = π × 49² = π × 49 × 49 = 7546 sq.m

Circumference of circle = 220 cm

∴ 2πr = 220

⇒ r = 35 cm

∴ Diameter = 2 × r = 2 × 35 = 70 cm

⇒ Side of square = diameter = 70 cm

⇒ Breadth of rectangle = (4/5) × side of square

⇒ Breadth of rectangle = (4 × 70)/5 = 56 cm

Length of rectangle = 60 cm

∴ Perimeter of rectangle = 2 × (length + breadth)

⇒ Perimeter of rectangle = 2 × (56 + 60) = 2 × 116

⇒ Perimeter of rectangle = 232 cm

Let the side of the square be a cm

∴ Perimeter of square = 4 × a

⇒ 4 × a = 28

⇒ a = 7 cm

∴ Area of square = a²

⇒ Area of square = 7² = 49 cm²

Now, diameter of circle = area of square

∴ Diameter of circle = 49 cm

Now, circumference of circle = π × (diameter)

∴ Circumference = π × 49

⇒ Circumference = 154 cm

Since the Cone, hemisphere and cylinder have same radius and same height

Volume of the cone is given by \(= {\rm{\;}}\frac{1}{3}{\rm{\pi }}{{\rm{r}}^2}{\rm{h}}\)

Volume of the hemisphere is given by \(= \frac{2}{3}{\rm{\pi }}{{\rm{r}}^3}\)

Volume of the cylinder is given by = πr²h

∴ Ratio of their volumes = Volume of the cone: Volume of the hemisphere: Volume of the cylinder

∴ Ratio of their volumes \(= {\rm{\;}}\frac{1}{3}{\rm{\pi }}{{\rm{r}}^2}{\rm{h\;}}:{\rm{\;}}\frac{2}{3}{\rm{\pi }}{{\rm{r}}^3}:{\rm{\pi }}{{\rm{r}}^2}{\rm{h\;}}\) [∵ r = h]

⟹ Ratio of their volumes \(= {\rm{\;}}\frac{1}{3}{\rm{\;}}:{\rm{\;}}\frac{2}{3}:1 = 1:2:3\)

Let the base and height of the triangle be B and H respectively.

And length and breadth of the rectangle be L and B1 respectively.

According to the given information:

The base of the triangle is 50% of the breadth of the rectangle

∴ B = ½ B1               ....... (1)

We know that, perimeter of rectangle = 2 × (length + breadth)

If the perimeter of the rectangle is 200 cm

Then, L + B1 = 100…………….. (2)

∵ Perimeter of the rectangle = 2(L + B)

Now, the area of a right-angled triangle is one-third of the area of a rectangle

∴ ½ × B × H = 1/3 × L × B1……………………. (3)

By analyzing above 3 equations and 4 unknowns

Hence, the value of H can’t be determined.

We know that, volume of cuboid = length × breadth × height

∴ Volume of earth dug out from the tank = 5m × 2.1 m × 4.5m = 47.25 m³

Also, Volume of the cuboidal tank = volume of earth dug out from the tank.

Now,

the exposed area of the field upon which the earth dug out is spread evenly = area of the total field – area of the top face of the tank

Area of the total field = 13.5 m × 2.5m = 33.75 sq. m.

Area of the top face of the tank = 5m × 2.1 m = 10.5 sq. m.

The exposed area of the field = (33.75 – 10.5) sq. ft. = 23.25 sq. m.

Now, the earth dug out is spread evenly over 23.25 sq. m. area.

If level of field raises by L m. then from volume constancy we can say,

23.25 × L = 47.25

⇒ L ≈ 2.03 m

We know that,

Total surface area of a closed cylinder = area of curved surface + area of circular top and bottom

∴ Total surface area of closed cylinder = 2πrh + 2πr² = 2πr(r+h) sq. units

Where, r = radius of base of cylinder, h = height of cylinder

According to the condition given in the question,

Curved surface area = (2/3) ×Total surface area

\(\therefore 2\pi rh\; = \frac{2}{3} \times 2\pi r\left( {r + h} \right)\)

⇒ 3h = 2 × (r + h)

⇒ h = 2r

Now, total surface area of a closed cylinder = 2πr(r+h) = 231 cm²

∴2πr(r+2r) = 231

6πr² = 231

\(\Rightarrow {r^2} = \frac{{231}}{{6\pi }} = \;\frac{{49}}{4}\;c{m^2}\)

⇒ r = 7/2 = 3.5 cm

∴ Volume of a right circular cylinder = πr²h cubic units

⇒ Volume of the given cylinder = π r² × 2r = 2π r³

⇒ Volume of the given cylinder = π r² × 2r = 2π r³ = 269.5 cm³

Let the length, breadth and height of old room be l. b and h respectively.

Painting is done on a surface area so cost of painting will be proportional to the surface area of the room, excluding the base area and top area.

Area to be painted = 2 × (bh + lh)

∵ Cost of Painting the above area = Rs. 600

Now, in the new room, the length, breadth and height are twice of those in the old room.

∴ New length = 2l; New breadth = 2b and new height= 2h

∴ New area to be painted = 2 × [(2b × 2h) + (2l × 2h)]

⇒ New area to be painted = 8 × (bh + lh)

∴ New cost of painting \(= \frac{{Old\;cost\;of\;painting}}{{Old\;Area}} \times New\;Area\)

⇒ New cost of painting \(= \frac{{600}}{{2\left( {bh + lh} \right)}} \times 8\left( {bh + lh} \right)\)

⇒ New cost of painting = 4 × 600 = Rs. 2400

The length of the rope tied to the horses should each be equal to half of the side of the square plot so that they just cannot reach one another.

∵ side of the square = 42 m,

Length of the rope = 42/2 = 21 m

The area grazed by each horse should be equal to the area of sector with radius 21 m (length of the rope)

Total area covered by the four horses = 4 × (area of sector of radius 21 m and central angle 90°) 

Total area left ungrazed = (Area of square field) – (Total area covered by four horses)

∴Total area left ungrazed  \(= \;{42^2}\;-\;4\; \times \left( {\frac{{90}}{{360}} \times \;\frac{{22}}{7}\; \times \;21\; \times \;21} \right)\)

⇒ Total area left ungrazed = 1764 – 1386 = 378 m²

We know that, circumference of circle = 2πr

Where r = radius of circle

As per information given in problem,

∴ 2πr – r = 111

⇒ r (2π – 1) = 111

\(\begin{array}{l} \Rightarrow \;r\; = \;\frac{{111}}{{2\pi - 1\;}}\\ \Rightarrow \;r\; = \;\frac{{111}}{{2\; \times \;\frac{{22}}{7}\; - \;1\;}} = \;\frac{{111}}{{\;\frac{{44}}{7}\; - \;1\;}}\\ \Rightarrow \;r\; = \;\frac{{111 \times \;7\;}}{{\;37\;}} \end{array}\)

⇒ r = 21 cm

Now, area of circle, A = πr²

∴A = (22/7) × 21 × 21

⇒ A = 1386 cm²