\displaystyle \textbf{Question 1: }\text{A funnel is the combination of }\underline{\hspace{2cm}}\text{ of a cone and}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{A funnel is the combination of a frustum of a cone and a cylinder.}
\displaystyle \therefore \text{The blanks are: frustum and cylinder.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A cylindrical pencil sharpened at one edge is the combination of a}
\displaystyle \underline{\hspace{2cm}}\text{ and a }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{A cylindrical pencil sharpened at one edge is a combination of a cylinder and a cone.}
\displaystyle \therefore \text{The blanks are: cylinder and cone.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A surahi is the combination of a }\underline{\hspace{2cm}}\text{ and a}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{A surahi is the combination of a sphere and a cylinder.}
\displaystyle \therefore \text{The blanks are: sphere and cylinder.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A plumbline (sahul) is the combination of a }\underline{\hspace{2cm}}\text{ and a}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{A plumbline is the combination of a hemisphere and a cone.}
\displaystyle \therefore \text{The blanks are: hemisphere and cone.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The shape of a gilli, in the gilli-danda game, is a combination of}
\displaystyle \underline{\hspace{2cm}}\text{ and a }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{A gilli is the combination of a cylinder and two cones.}
\displaystyle \therefore \text{The blanks are: cylinder and two cones.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A shuttle cock used for playing badminton has the shape of the combination}
\displaystyle \text{of }\underline{\hspace{2cm}}\text{ and }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{A shuttle cock is the combination of a frustum of a cone and a hemisphere.}
\displaystyle \therefore \text{The blanks are: frustum of a cone and hemisphere.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Two identical solid cubes of side }a\text{ are joined end to end. The total}
\displaystyle \text{surface area of the resulting cuboid is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Dimensions of the resulting cuboid}=2a\times a\times a.
\displaystyle \text{Total surface area}=2(lb+bh+lh)
\displaystyle =2(2a^2+a^2+2a^2)=10a^2.
\displaystyle \therefore \text{The blank is }10a^2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If a solid cone of base radius }r\text{ and height }h\text{ is placed over a solid}
\displaystyle \text{cylinder having the same base radius and height as that of the cone, then the curved}
\displaystyle \text{surface area of the solid so formed is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Slant height of the cone}=\sqrt{r^2+h^2}.
\displaystyle \text{Curved surface area of cylinder}=2\pi rh.
\displaystyle \text{Curved surface area of cone}=\pi r\sqrt{r^2+h^2}.
\displaystyle \text{Curved surface area of the solid}
\displaystyle =2\pi rh+\pi r\sqrt{r^2+h^2}
\displaystyle =\pi r\left(2h+\sqrt{r^2+h^2}\right).
\displaystyle \therefore \text{The blank is }\pi r\left(2h+\sqrt{r^2+h^2}\right).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A solid ball is exactly fitted inside a cuboidal box of side }a.\text{ The volume}
\displaystyle \text{of the ball is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the ball}=a.
\displaystyle \therefore \text{Radius of the ball}=\frac{a}{2}.
\displaystyle \text{Volume of the ball}=\frac{4}{3}\pi\left(\frac{a}{2}\right)^3
\displaystyle =\frac{\pi a^3}{6}.
\displaystyle \therefore \text{The blank is }\frac{\pi a^3}{6}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A solid cylinder of radius }r\text{ and height }h\text{ is placed over another}
\displaystyle \text{cylinder of same height and radius. The total surface area of the shape so formed is}
\displaystyle \underline{\hspace{2cm}}\text{.} 
\displaystyle \text{Answer:}
\displaystyle \text{The two cylinders together form a cylinder of radius }r\text{ and height }2h.
\displaystyle \text{Total surface area}=2\pi r(2h)+2\pi r^2
\displaystyle =2\pi r(2h+r).
\displaystyle \therefore \text{The blank is }2\pi r(2h+r).
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Two identical solid hemispheres of equal base radius }r\text{ cm are stuck}
\displaystyle \text{together along their bases. The total surface area of the combination is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Two hemispheres joined along their bases form a sphere of radius }r.
\displaystyle \text{Total surface area of the sphere}=4\pi r^2.
\displaystyle \therefore \text{The blank is }4\pi r^2.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The capacity of a cylindrical vessel with a hemispherical portion raised}
\displaystyle \text{upward at the bottom as shown in Fig. 14.79 is }\underline{\hspace{2cm}}\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Capacity of the vessel}=\text{Volume of cylinder}-\text{Volume of hemisphere}.
\displaystyle =\pi r^2h-\frac{2}{3}\pi r^3
\displaystyle =\pi r^2\left(h-\frac{2r}{3}\right).
\displaystyle \therefore \text{The blank is }\pi r^2\left(h-\frac{2r}{3}\right).
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If a marble of radius }2.1\text{ cm is put into a cylindrical cup, full of}
\displaystyle \text{water, of radius }5\text{ cm and height }6\text{ cm, then the volume of water that flows out}
\displaystyle \text{of the cup is }\underline{\hspace{2cm}}\text{.} 
\displaystyle \text{Answer:}
\displaystyle \text{Volume of water that flows out}=\text{Volume of the marble}.
\displaystyle =\frac{4}{3}\pi r^3
\displaystyle =\frac{4}{3}\times\frac{22}{7}\times(2.1)^3
\displaystyle =38.808\text{ cm}^3.
\displaystyle \therefore \text{The blank is }38.808\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Three solid cubes have a face diagonal of }4\sqrt{2}\text{ cm each. Three other}
\displaystyle \text{solid cubes have a face diagonal of }8\sqrt{2}\text{ cm each. All the cubes are melted}
\displaystyle \text{together to form a cube. The side of the cube so formed is of length }\underline{\hspace{2cm}}\text{.}

\displaystyle \text{Answer:}
\displaystyle \text{Face diagonal of a cube}=a\sqrt{2}.
\displaystyle \therefore \text{Sides of the two types of cubes are }4\text{ cm and }8\text{ cm.}
\displaystyle \text{Let the side of the new cube be }x\text{ cm.}
\displaystyle x^3=3(4^3)+3(8^3)
\displaystyle =3(64)+3(512)=1728
\displaystyle \therefore x=\sqrt[3]{1728}=12\text{ cm.}
\displaystyle \therefore \text{The blank is }12\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In Fig. 14.80, a circle is inscribed in a square }ABCD\text{ and the square}
\displaystyle \text{is circumscribed by a circle. If the radius of the smaller circle is }r\text{ cm, then the area}
\displaystyle \text{of the shaded region in cm}^2\text{ is }\underline{\hspace{2cm}}\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Side of the square}=\text{Diameter of the smaller circle}=2r.
\displaystyle \text{Diagonal of the square}=2r\sqrt{2}.
\displaystyle \therefore \text{Radius of the outer circle}=r\sqrt{2}.
\displaystyle \text{Area inside outer circle but outside square}
\displaystyle =\pi(r\sqrt{2})^2-(2r)^2
\displaystyle =2\pi r^2-4r^2=2r^2(\pi-2).
\displaystyle \text{The four regions between the square and outer circle are equal.}
\displaystyle \therefore \text{Area of the shaded region}=\frac{1}{4}\times2r^2(\pi-2)
\displaystyle =\frac{r^2}{2}(\pi-2)\text{ cm}^2.
\displaystyle \therefore \text{The blank is }\frac{r^2}{2}(\pi-2).
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The volume of the greatest right circular cone, which can be cut from a}
\displaystyle \text{cube of side }4\text{ cm is }\underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{For the greatest cone, diameter of base}=\text{side of cube}=4\text{ cm.}
\displaystyle \therefore r=2\text{ cm},\qquad h=4\text{ cm.}
\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi r^2h
\displaystyle =\frac{1}{3}\pi(2)^2(4)=\frac{16\pi}{3}\text{ cm}^3.
\displaystyle \therefore \text{The blank is }\frac{16\pi}{3}\text{ cm}^3.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The volume of a cuboid is }24\sqrt{42}\text{ cm}^3.\text{ Its length is }5\sqrt{2}\text{ cm,}
\displaystyle \text{breadth and height are in the ratio }\sqrt{3}:\sqrt{7}.\text{ The height of the cuboid is}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the breadth}=k\sqrt{3}\text{ cm and height}=k\sqrt{7}\text{ cm.}
\displaystyle \text{Volume of cuboid}=\text{Length}\times\text{Breadth}\times\text{Height}.
\displaystyle 24\sqrt{42}=5\sqrt{2}\times k\sqrt{3}\times k\sqrt{7}
\displaystyle 24\sqrt{42}=5k^2\sqrt{42}
\displaystyle \therefore k^2=\frac{24}{5}.
\displaystyle \text{Height}=k\sqrt{7}=\sqrt{\frac{24}{5}}\times\sqrt{7}
\displaystyle =\sqrt{\frac{168}{5}}=\frac{2\sqrt{210}}{5}\text{ cm.}
\displaystyle \therefore \text{The blank is }\frac{2\sqrt{210}}{5}\text{ cm.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The sum of the length, breadth and height of a cuboid is }5\sqrt{3}\text{ cm and}
\displaystyle \text{the length of its diagonal is }3\sqrt{5}\text{ cm, then its total surface area is}
\displaystyle \underline{\hspace{2cm}}\text{.}
\displaystyle \text{Answer:}
\displaystyle l+b+h=5\sqrt{3}
\displaystyle l^2+b^2+h^2=(3\sqrt{5})^2=45.
\displaystyle (l+b+h)^2=l^2+b^2+h^2+2(lb+bh+hl)
\displaystyle (5\sqrt{3})^2=45+2(lb+bh+hl)
\displaystyle 75=45+2(lb+bh+hl)
\displaystyle 2(lb+bh+hl)=30.
\displaystyle \text{Total surface area of cuboid}=2(lb+bh+hl)=30\text{ cm}^2.
\displaystyle \therefore \text{The blank is }30\text{ cm}^2.
\displaystyle \\

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