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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 4 · Set 1 · Q29
(a)To protect plants from heat, a shed of iron rods covered with green cloth is made. The lower part of the shed is a cuboid mounted by semi-cylinder as shown in the figure. Find the area of the cloth required to make this shed, if dimensions of the cuboid are $\displaystyle 14$ m × $\displaystyle 25$ m × $\displaystyle 16$ m
(b)The internal and external radii of a hollow hemisphere are $\displaystyle 5 \sqrt{2} \mathrm{~cm}$ and $\displaystyle 10$ cm respectively. A cone of height $\displaystyle 5 \sqrt{7} \mathrm{~cm}$ and radius $\displaystyle 5 \sqrt{2} \mathrm{~cm}$ is surmounted on the hemisphere as shown in the figure. Find the total surface area of the object in terms of $\displaystyle \pi$. (Use $\displaystyle \sqrt{2}=1.4$ )
(a)
To protect plants from heat, a shed of iron rods covered with green cloth is made. The lower part of the shed is a cuboid mounted by semi-cylinder as shown in the figure. Find the area of the cloth required to make this shed, if dimensions of the cuboid are $\displaystyle 14$ m × $\displaystyle 25$ m × $\displaystyle 16$ m
(b)
The internal and external radii of a hollow hemisphere are $\displaystyle 5 \sqrt{2} \mathrm{~cm}$ and $\displaystyle 10$ cm respectively. A cone of height $\displaystyle 5 \sqrt{7} \mathrm{~cm}$ and radius $\displaystyle 5 \sqrt{2} \mathrm{~cm}$ is surmounted on the hemisphere as shown in the figure. Find the total surface area of the object in terms of $\displaystyle \pi$. (Use $\displaystyle \sqrt{2}=1.4$ )
Marking-scheme solution
Area of cloth required to cover four walls \(\displaystyle =2(14 \times 16+25 \times 16)\)
\[=1248 \mathrm{~m}^{2}
\]
Radius \(\displaystyle =\frac{14}{2}=7 \mathrm{~m}\)
Area of cloth required to cover cylindrical part \(\displaystyle =\frac{22}{7} \times 7 \times 25+\frac{22}{7} \times 7^{2}\)
\[=704 \mathrm{~m}^{2}
\]
∴ Area of total cloth required \(\displaystyle =1248+704=1952 \mathrm{~m}^{2}\)
Let internal and external radii be \(\displaystyle \mathrm{r}_{1}=5 \sqrt{2} \mathrm{~cm}\) and \(\displaystyle \mathrm{r}_{2}=10 \mathrm{~cm}\) respectively.
∴ Slant height (l) of the cone \(\displaystyle =\sqrt{(5 \sqrt{2})^{2}+(5 \sqrt{7})^{2}}=15 \mathrm{~cm}\)
Now, the total surface area of the object \(\displaystyle =2 \pi \mathrm{r}_{2}{ }^{2}+\pi \mathrm{r}_{1} \mathrm{l}+\pi\left(\mathrm{r}_{2}{ }^{2}-\mathrm{r}_{1}{ }^{2}\right)\)
\[\begin{aligned}
& =\pi\left(2 \times 10^{2}+5 \sqrt{2} \times 15+10^{2}-(5 \sqrt{2})^{2}\right) \\
& =355 \pi \mathrm{~cm}^{2}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.