Tutor profile: Sumit K.
'a' and 'b' are the lengths of the base and height of a right angled triangle whose hypotenuse is 'h'. If the values of 'a' and 'b' are positive integers, which of the following cannot be a value of the square of the hypotenuse? (1) 13 (2) 23 (3) 37 (4) 41
The value of the square of the hypotenuse = h2 = a2 + b2 As the problem states that 'a' and 'b' are positive integers, the values of a2 and b2 will have to be perfect squares. Hence we need to find out that value amongst the four answer choices which cannot be expressed as the sum of two perfect squares. Choice 1 is 13. 13 = 9 + 4 = 32 + 22. Therefore, Choice 1 is not the answer as it is a possible value of h2 Choice 2 is 23. 23 cannot be expressed as the sum two numbers, each of which in turn happen to be perfect squares. Therefore, Choice 2 is the answer.
Area of a Rhombus of perimeter 56 cms is 100 sq cms. Find the sum of the lengths of its diagonals. A)33.40 B)34.40 C)31.20 D)32.30
Perimeter = 56. Let the side of the rhombus be “a”, then 4a = 56 => a =14. Area of Rhombus = Half the product of its diagonals. Let the diagonals be d1 and d2 respectively. 12*d1 * d2 = 100 => d1 * d2 = 200. By Pythagoras theorem, (d1)^2 + (d2)^2 = 4a^2 => (d1)^2 + (d2)^2 = 4*196 = 784. (d1)^2 + (d2)^2 + 2d1 * d2 = (d1+ d2)^2 = 784 +2*200 = 1184 => (d1+ d2) = √1184 = 34.40 Therefore, sum of the diagonals is equal to 34.40 cm . Correct Answer: 34.40
Once I had been to the post-office to buy stamps of five rupees, two rupees and one rupee. I paid the clerk Rs. 20, and since he did not have change, he gave me three more stamps of one rupee. If the number of stamps of each type that I had ordered initially was more than one, what was the total number of stamps that I bought?
The number of stamps that were initially bought were more than one of each type. Since the clerk gave me 3 one rupee stamps at the end, it means that I initially bought stamps for Rs17. Let I initially bought ‘x’ 5 Rs stamps, ‘y’ 2 Rs stamps and ‘z’ one rupee stamps. So, we have5x + 2y + z = 17. Now x cannot be three, so x = 2, also if x = 2, then y cannot be 3, otherwise z will be 1. Hence, y = 2 and z = 3. So, I bought x + y + z + 3 = 2 + 2 + 3 + 3 = 10 stamps.
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