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Yogita Ingle 6 years, 1 month ago
Let ABC be equilateral triangle.
Let AD be perpendicular bisector from A on to BC. So BD = CD = 1/2 BC
ADC is a right angle triangle. So AC² = AD² + DC²
AC² = AD² + (1/2 AC)²
AD² = 3/4 AC²
4 AD² = 3 AC²
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Yogita Ingle 6 years, 1 month ago
The first three digit number which is divisible by 7 is 105 and last three digit number which is divisible by 7 is 994.
This is an A.P. in which <i>a</i> = 105, d = 7 and l = 994.
Let the number of terms be n . Then<i> </i>tn = 994.
nth term of A.P = tn = a + (n - 1)d.
⇒ 994 = 105 + (n -1)7.
⇒ 889 = 7(n-1)
⇒ n -1 = 127
∴ n = 128.
∴ There are128 three digit numbers which are divisible by 7.
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