CBSE 2025 · Region 5 · Set 1 · Q37 · 4 marks
In an equilateral triangle of side $\displaystyle 10$ cm, equilateral triangles of side $\displaystyle 1$ cm are formed as shown in the figure below, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row and so on.
Based on given information, answer the following questions using Arithmetic Progression.(i)How many triangles will be there in bottom most row ?(ii)How many triangles will be there in fourth row from the bottom?(iii)Find the total number of triangles of side $\displaystyle 1$ cm each till $\displaystyle 8^{\text {th }}$ row.How many more number of triangles are there from $\displaystyle 5^{\text {th }}$ row to $\displaystyle 10^{\text {th }}$ row than in first $\displaystyle 4$ rows? Show working.
In an equilateral triangle of side $\displaystyle 10$ cm, equilateral triangles of side $\displaystyle 1$ cm are formed as shown in the figure below, such that there is one triangle in the first row, three triangles in the second row, five triangles in the third row and so on.
Based on given information, answer the following questions using Arithmetic Progression.
(i)
How many triangles will be there in bottom most row ?
(ii)
How many triangles will be there in fourth row from the bottom?
(iii)
Find the total number of triangles of side $\displaystyle 1$ cm each till $\displaystyle 8^{\text {th }}$ row.
How many more number of triangles are there from $\displaystyle 5^{\text {th }}$ row to $\displaystyle 10^{\text {th }}$ row than in first $\displaystyle 4$ rows? Show working.
Marking-scheme solution
Given A.P. is $\displaystyle 1$, $\displaystyle 3$, 5. . .
(i) \(\displaystyle \mathrm{a}_{10}=1+9 \times 2=19\)
(ii) \(\displaystyle \mathrm{a}_{4}(\) from bottom \(\displaystyle )=19+3 \times(-2)=13\)
(iii) (a) \(\displaystyle \mathrm{S}_{8}=\frac{8}{2} \times[2 \times 1+7 \times 2]\)
\[\text { = } 64
\]
OR
(iii) (b) Number of triangles from \(\displaystyle 5^{\text {th }}\) row to \(\displaystyle 10^{\text {th }}\) row \(\displaystyle =\mathrm{S}_{10}-\mathrm{S}_{4}\)
\[\begin{aligned}
& =\frac{10}{2} \times[2 \times 1+9 \times 2]-\frac{4}{2} \times[2 \times 1+3 \times 2] \\
& =84
\end{aligned}
\]
\[\text { Number of triangles in first } \left.4 \text { rows, } \mathrm{S}_{4}=\frac{4}{2} \times[2 \times 1+3 \times 2] .\right] .
\]
\[\text { Required number of triangles }=84-16=68
\]
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.