find the total area of 12 squares whose sides are 12cm,13cm,14cm,...23cm respectively
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Answered by
2
= (1² + 2² + ......... + 23²) - (1² + 2² + ......... + 11²)
= [23(23+1)(2x23+1)]/6 - [11(11+1)(2x11+1)]/6
= (23 x 24 x 47)/6 - (11 x 12 x 23)/6
= 4324 - 506
= 3818
= [23(23+1)(2x23+1)]/6 - [11(11+1)(2x11+1)]/6
= (23 x 24 x 47)/6 - (11 x 12 x 23)/6
= 4324 - 506
= 3818
Answered by
2
Area of square is side²
All sides are in cm
a=12
b=13
c=14
d=15
e=16
f=17
g=18
h=19
i=20
j=21
k=23
Let the another l = 22
Total area of square = a²+b²+c²+....l²
=>we know the sum of square of natural number is n(n+1)(2n+1)/6
We have to find sum of squares from 1 up to 23 and subtract the sum of squares of series from 1 up to 11 from that.
= (1² + 2² + ......... + 23²) - (1² + 2² + ......... + 11²)
= [23(23+1)(2x23+1)]/6 - [11(11+1)(2x11+1)]/6
= (23 x 24 x 47)/6 - (11 x 12 x 23)/6
= 4324 - 506
= 3818 cm²
Total area of 12 squares are 3818 cm²
☺️☺️☺️ I hope you get correct solution
All sides are in cm
a=12
b=13
c=14
d=15
e=16
f=17
g=18
h=19
i=20
j=21
k=23
Let the another l = 22
Total area of square = a²+b²+c²+....l²
=>we know the sum of square of natural number is n(n+1)(2n+1)/6
We have to find sum of squares from 1 up to 23 and subtract the sum of squares of series from 1 up to 11 from that.
= (1² + 2² + ......... + 23²) - (1² + 2² + ......... + 11²)
= [23(23+1)(2x23+1)]/6 - [11(11+1)(2x11+1)]/6
= (23 x 24 x 47)/6 - (11 x 12 x 23)/6
= 4324 - 506
= 3818 cm²
Total area of 12 squares are 3818 cm²
☺️☺️☺️ I hope you get correct solution
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