1. In the given diagram, if area of triangle DOB is a, then find the total area of triangle ABC.
As, AD/DB = AE/EC
∴ DE || BC
Hence, ΔDOE ∼ ΔCOB
Now, in ΔBDC, ΔDOB and ΔBOC are formed such that they share the same base DC.
In such a formation, the ratio of the areas of the triangles is equal to the ratio of the respective bases.
Similarly, ΔDBC and ΔADC are formed from ΔABC and they share the same base AB.


As, AD/DB = AE/EC
∴ DE || BC
Hence, ΔDOE ∼ ΔCOB
Now, in ΔBDC, ΔDOB and ΔBOC are formed such that they share the same base DC.
In such a formation, the ratio of the areas of the triangles is equal to the ratio of the respective bases.
Similarly, ΔDBC and ΔADC are formed from ΔABC and they share the same base AB.
2. If p is a prime less than 500, how many number of the form 3p + 1 are perfect squares? 5may
Let 3p + 1 = m2 … (1)
∴ 3p = (m – 1) (m + 1)

Here a is the integer part of the number and f is the fractional part of the number.
Hence, f < 1


a’ = 0, f ’ < 1
Now, consider;

Now, if we expand both the exponents then we can observe that each term having odd

∴ 3p = (m – 1) (m + 1)
As p is a prime, we have three cases:
Case i:
m – 1 = 3 and m + 1 = p
∴ p = 5 and 3p + 1 = 16
Case ii:
m – 1 = p and m + 1 = 3
∴ p = 1, which is not true.
Case iii:
m + 1 = 3p and m – 1 = 1
∴ p = 1, which is not true.
∴ Equation (1) is true only in case i.
3. Find the 100th digit after the decimal point of
Here a is the integer part of the number and f is the fractional part of the number.
Hence, f < 1
a’ = 0, f ’ < 1
Now, consider;
Now, if we expand both the exponents then we can observe that each term having odd
Hence, k must be an integer.
Hence, a + f + f ’ is an integer.
∴ f + f ’ is an integer.
As both f and f ’ are greater than 0 but less than 1.
∴ 0 < f + f ’ < 2
∴ f + f ’ = 1
Now,
Hence, a + f + f ’ is an integer.
∴ f + f ’ is an integer.
As both f and f ’ are greater than 0 but less than 1.
∴ 0 < f + f ’ < 2
∴ f + f ’ = 1
Now,
Now, 2500 = 16125 > 10125
∴ 2–500 < 10–125
Hence, f ’ < 10–125
∴ 1 – f’ > 1 – 10–125
∴ f > 1 – 10–125
∴ f > 0.99… 125 times
∴ 100th digit after decimal of f is 9.
Hence, option 3.
∴ 2–500 < 10–125
Hence, f ’ < 10–125
∴ 1 – f’ > 1 – 10–125
∴ f > 1 – 10–125
∴ f > 0.99… 125 times
∴ 100th digit after decimal of f is 9.
Hence, option 3.
∴ Just one number of the form 3p +1 is a perfect square if p is a prime.
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