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) Find the total number of triangles of side 1 cm each till 8th row.
(iv) How many more number of triangles are there from 5th to 10th row than in first 4 rows? Show working.
PYQ: 36 – CBSE 2025 – Code 30 – Series 5 – Set 1
Ans: Here, 1st row has 1 triangle
2nd row has 3 triangles
3rd row has 5 triangles
Since here, number of triangle are increasing at same interval of 2 in each next row, It will form an AP where
1st term, a = 1
and common difference, d = 2
(i) Number of triangles in bottom most row:
Since the side length of the triangle is 10 cm,
∴ bottom most row ill be 10th row,
∴ n = 10
We need to find number of triangles in 10th row, it means we need to calculate value of 10th term
Since in an AP, value of nth term is given by Tn = a + (n – 1) d
∴ T10 = 1 + (10 – 1) 2 = 1 + 9 x 2 = 19
Therefore, there will be 19 triangles in the bottom most row.
(ii) Number of triangles in the fourth row from the bottom:
Note: This is tricky one, read carefully.
Since 4th row from bottom, is 7th row from top (10 – 4 + 1)
Hence we will calculate number of triangles in 7th row from the top.
Like in part (i), here n = 7 and we will calculate value of T7.
Since in an AP, value of nth term is given by Tn = a + (n – 1) d
∴ T7 = 1 + (7 – 1) 2 = 1 + 6 x 2 = 13
Therefore, there will be 13 triangles in the 4th row from the bottom.
(iii) Total number of triangles of side 1 cm each till 8th row:
Since, we need to calculate total number of triangles till 8th row, we will calculate sum of first 8 terms
Since in an AP, sum of nth term is given by Sn = (n/2) [2a + (n – 1) d]
∴ S7 = (8/2) [2 x 1 + (8 – 1) 2] = 4 (2 + 7 x 2) = 64
Therefore, till 8th row, total 64 triangles of side 1 cm each.
(iv) More number of triangles from 5th to 10th row than in first 4 rows:
Difference in number of Triangles = number of Triangles from 5th to 10th row – number of Triangles from 1st to 4th row
Since number of Triangles from 5th to 10th row = sum of Triangles till 10th row – sum of Triangles till 5th row
∴ Difference = (sum of Triangles till 10th row – sum of Triangles till 5th row) – sum of Triangles till 4th row …….. (i)
Let’s calculate all 3 values one by one and then solve for calculating the difference:
sum of Triangles till 10th row, S10 = (10/2) [2 x 1 + (10 – 1) 2] = 4 (2 + 9 x 2) = 80
sum of Triangles till 5th row, S5 = (5/2) [2 x 1 + (5 – 1) 2] = 4 (2 + 4 x 2) = 40
sum of Triangles till 4th row, S4 = (4/2) [2 x 1 + (4 – 1) 2] = 4 (2 + 3 x 2) = 32
Next, we substitute these values in equation (i), we get:
∴ Difference = (sum of Triangles till 10th row – sum of Triangles till 5th row) – sum of Triangles till 4th row
= (80 – 40) – 32
= 8
Therefore, 8 triangles are more in 5th to 10th row than in first 4 rows.
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