In an equilateral triangle of side 10 cm, equilateral triangles of side 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) (a) Find the total number of triangles of side 1 cm each till 8^th row. (iii) (b) How many more number of triangles are there from 5^th row to 10^th row than in first 4 rows? Show working.

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2025: question 37

Set in the CBSE Class 10 board exam of 2025, paper 30/5/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions

In an equilateral triangle of side 10 cm, equilateral triangles of side 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?
1 mark
(ii)
How many triangles will be there in fourth row from the bottom?
1 mark
(iii)
(a) Find the total number of triangles of side 1 cm each till $8^{th}$ row.
2 marks
OR
(iii)
(b) How many more number of triangles are there from $5^{th}$ row to $1 0^{th}$ row than in first 4 rows? Show working.
2 marks

Figure in the paper: A large equilateral triangle subdivided into small unit equilateral triangles, alternately shaded grey and white. Rows are labelled from the top: 1st Row (1 triangle), 2nd Row (3), 3rd Row (5), continuing down to the 10th Row at the base, which has its bottom edge with 5 visible unit triangles across the base in the picture. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write the numbers of triangles per row as an A.P. and identify $a$ and $d$ . A side of 10 cm means 10 rows.

Hint 2

Use $a_{n} = a + (n - 1) d$ for one row. For the n-th row from the bottom, start from the last term and subtract $d$ each time.

Hint 3

For totals use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . The rows 5 to 10 together give $S_{10} - S_{4}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

The numbers of triangles in successive rows are $1, 3, 5, \dots$ , an A.P. with $a = 1$ , $d = 2$ and 10 rows (side 10 cm, each small side 1 cm).
(i) Bottom row: $a_{10} = a + 9 d = 1 + 9 \times 2 = 19$
1 mark
(ii) Counting upward from the bottom, the rows form an A.P. with first term 19 and $d = - 2$ . The fourth row from the bottom has $19 + 3 \times (- 2) = 13$ triangles (this is the 7th row from the top: $1 + 6 \times 2 = 13$ ).
1 mark
(iii)(a) $S_{8} = \frac{8}{2} [2 \times 1 + 7 \times 2]$
1 mark
(iii)(a) $= 4 \times 16 = 64$ triangles
1 mark
(iii)(b) Triangles from the 5th to the 10th row $= S_{10} - S_{4} = \frac{10}{2} [2 + 18] - \frac{4}{2} [2 + 6] = 100 - 16 = 84$
1 mark
(iii)(b) Triangles in the first 4 rows: $S_{4} = \frac{4}{2} [2 \times 1 + 3 \times 2] = 16$
½ mark
(iii)(b) Required difference $= 84 - 16 = 68$ more triangles
½ mark

Answer: (i) 19; (ii) 13; (iii)(a) 64; (iii)(b) 68 more triangles

Where the marks are

(i) $a_{10} = 1 + 9 \times 2 = 19$ (1 mark)
(ii) Fourth row from the bottom = 13 (1 mark)
(iii)(a) Correct use of $S_{n}$ formula for $n = 8$ (1 mark); $S_{8} = 64$ (1 mark)
(iii)(b) $S_{10} - S_{4} = 84$ (1 mark); $S_{4} = 16$ (½ mark); difference 68 (½ mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)
Topic: Finding nth Term from the Sum of n Terms

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 30/5/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2025: question 37

Hints

Step-by-step solution

Revise the chapter

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