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Write ‘True’ or ‘False’ and justify your answer in the following:

An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surfacearea of cylinder

Answer True

Solution

            According to question, here is a metallic bucket in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet.

           

            The surface area of figure (A) is the surface area of (a), (b), (c)

            (a) = curved surface area of frustum

            (b) = area of circular base

            (c) = curved area of cylinder

            Total surface area = curved surface area of frustum of cone + area of circular base + curved surface are of cylinder.

            So the given statement is True.

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Write ‘True’ or ‘False’ and justify your answer in the following:

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Figure is \frac{\pi r^{2}}{3}[3h-2r]

 

AnswerTrue

Solution

            It is given that a cylindrical vessel of height h and radius r is raised upward with a hemispherical portion.

            From the figure radius of hemisphere = r cm

            The volume of the figure = volume of the cylinder – the volume of the hemisphere

                        =\pi r^{2}h - \frac{2}{3}\pi r^{3}

                        =\pi r^{2}\left [h-\frac{2}{3}r \right ]

                        =\pi r^{2}\left [\frac{3h -2r}{3} \right ]

                        =\frac{\pi r^{2}}{3}[3h-2r]

\text{ Hence the capacity of the vessel is }\frac{\pi r^{2}}{3}[3h-2r]

So the given statement is True.

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Write ‘True’ or ‘False’ and justify your answer in the following : 

A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is \frac{4}{3}\pi a ^{3}

It is given that the ball is exactly filled inside the cubical box of side a.

           

 Hence the diameter of the sphere = a

\text{ Radius of sphere} =\frac{a}{2}

 \text{ Volume of sphere} =\frac{4}{3} \pi r^{3}

=\frac{4}{3} \pi \left (\frac{a}{2} \right )^{3}

=\frac{4}{3} \pi \times \frac{a^3}{8}

\text{Volume of sphere}=\frac{\pi a ^{3}}{6}

\text{ Hence the volume of the sphere is not equal to }\frac{4}{3}\pi a ^{3}

Hence the given statement is False.

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Write ‘True’ or ‘False’ and justify your answer in the following :

A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone. The total  surface area of the combined solid is \pi r\left [ \sqrt{r^{2}+h^2}+3r+2h \right ]

It is given that there is a cone of radius r and height h and a cylinder of height h and radius r.

            

            Where A is placed on B

            

            Total surface area = Surface area of cone + Total surface area of cylinder – Surface area of part I – the surface area of part II

                        =\pi r (r +l)+2\pi r(r+h)-\pi r^{2}- \pi r^{2}

                        = \pi r (r +\sqrt{r^2 +h^2})+2 \pi r (r+h)-2\pi r ^2 (QI= \sqrt{r^2 +h^2})

                        = \pi r (r +\sqrt{r^2 +h^2}+2r+2h -2r)

                        = \pi r (r +\sqrt{r^2 +h^2}+r+2h )

            Total surface area is not equal to

\pi r\left [ \sqrt{r^{2}+h^2}+3r+2h \right ]

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Write ‘True’ or ‘False’ and justify your answer in the following :

A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4\pi rh + 4\pi r^2.

It is given that there is two cylinders of height h and radius r.

            

            Where one is placed on other then the shape formed is

            

The total surface area of the shape formed =2\pi r (r +2h)  

\text{ [Using the total surface area of cylinder }=2\pi r (r +h)]

                        =2\pi r (r +h)

                        =2\pi r^2 + 4 \pi r h

So the surface area is not equal to 4\pi rh + 4\pi r^2.

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Write ‘True’ or ‘False’ and justify your answer in the following: 

Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6 \pi r^2.

It is given that there is two hemispheres of radius r.

            Let A and B are two hemispheres of radius r.

            

            Join A and B along with their bases

            

            Now it is a full sphere of radius r

            The total surface area of a sphere =4 \pi r^{2}

            Here we found that the total surface area of the combination is 4 \pi r^{2} , but not 6 \pi r^{2}

           

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Write True or False and justify your answer:

In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having length 12 cm.

True

Let \; a\; =\; 11\; cm ;\; b\; =\; 12\; cm\; ;\; c = 13 cm

Semi perimeter (S) = \frac{a+b+c}{2}

                                         = \frac{11+12+13}{2}

                                        = \frac{36}{2}=18cm

                     

(Area\; of\; triangle \; ABC)\; by\; Heron's\; formula

=\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}

=\sqrt{18\left ( 18-11 \right )\left ( 18-12 \right )\left ( 18-13\right )}

= \sqrt{18\times 7\times 6\times 5}

= \sqrt{6\times 3\times 7\times 6\times 5}

= \sqrt{6^{2}\times 3\times 7\times 5}

= 6\sqrt{3\times 7\times 5}

= 6\sqrt{105}           \because \sqrt{105}=10.2469\simeq 10.25

= 6 \times 10.25

= 61.5 cm^{2}

Given\; altitude\; =\; 10.25\; cm\; and

Its \; corresponding\; base \; = \; 12\; cm

Area\; of \times triangle ABC = \frac{1}{2} \times Base \times corresponding\; Height

                                               = \frac{1}{2} \times 12 \times 10.25

                                                 =61.5cm^{2}

Hence the area obtained is the same.

Therefore the given statement is true.

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Write True or False and justify your answer:

The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs 3\; per\; m^{2} is Rs 918.

[True]

Let\; a = 51 \; m, b = 37\; m \; and \; c = 20\; m

                          

S=\frac{51+37+20}{2}=\frac{108}{2}=54m

We know that using Heron’s formula

Area\; of \; triangle AB=\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}

=\sqrt{54\left ( 54-51 \right )\left ( 54-37 \right )\left ( 54-20 \right )}

=\sqrt{54\times 3\times 17\times 34}

=\sqrt{9\times 3\times 2\times 3\times 17\times 17\times 2}

=\sqrt{3\times3\times 3\times 3\times 2\times 2\times 17\times 17}

=3\times 3\times 2\times 17

= 306m^{2}

To find cost :

 Cost of levelling 1 m2 = Rs 3

\therefore Cost\; of\; levelling\; 306\; m^{2} = \; 3 \times 306 = Rs.\; 918

Therefore the given statement is true.

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Write True or False and justify your answer:

The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.

[False]

According to question

Area\; of\; regular\; hexagon = sum\; of\; area \; of\; five\; equilateral \; triangles

                         

We know that a regular hexagon is divided into 6 equilateral triangles by its diagonals.

Area\; of\; 1\; equilateral\; triangle = \frac{\sqrt{3}}{4} \times a^{2}

Area\; of\; 6\; equilateral\; triangle =6\times \frac{\sqrt{3}}{4} \times a^{2}=\frac{3\sqrt{3}}{2}a^{2}

Area\; of\; 5\; equilateral\; triangle =5\times \frac{\sqrt{3}}{4} \times a^{2}=\frac{5\sqrt{3}}{2}a^{2}

Therefore the given statement is false.

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Write True or False and justify your answer:

The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 cm^{2}

[False]

We know that

Area\; of\; parallelogram = base \times height

                                             = 10 \times 3.5

                                              = 10 \times \frac{35}{10} =35cm^{2}

  

Hence, area of parallelogram is 35cm^{2}

Therefore the given statement is false.

 

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