Three smaller triangles are inscribed from the three vertices of a larger triangle in such a way that each side of each smaller triangle is \( \frac{2}{5} \) of the side of the adjacent larger triangle. The ratio of the total area of the three smaller triangles to the remaining part of the larger triangle is:
This problem involves the areas of similar triangles. When a smaller triangle is inscribed such that its sides are a fraction of the larger triangle's sides, the ratio of their areas is the square of the ratio of their corresponding sides. We will calculate the area of one small triangle, then the total area of the three small triangles, and finally the area of the remaining part of the larger triangle to find the required ratio.
Let the larger triangle be \(\triangle ABC\). Let its side length be \(s\). Its area can be denoted as \(\text{Area}(\triangle ABC)\).
Each smaller triangle has sides that are \( \frac{2}{5} \) of the side of the adjacent larger triangle. This means the smaller triangles are similar to the larger triangle.
The ratio of the sides of a small triangle to the large triangle is \( k = \frac{2}{5} \).
The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides. So, for one small triangle:
\[ \frac{\text{Area}(\text{small triangle})}{\text{Area}(\triangle ABC)} = k^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25} \]Let \(\text{Area}(\triangle ABC) = A\). Then the area of one small triangle is \( \frac{4}{25} A \).
There are three such smaller triangles. So, the total area of the three smaller triangles is:
\[ \text{Total Area of small triangles} = 3 \times \frac{4}{25} A = \frac{12}{25} A \]The remaining part of the larger triangle is the area of the larger triangle minus the total area of the three smaller triangles:
\[ \text{Area of remaining part} = A - \frac{12}{25} A = \left(1 - \frac{12}{25}\right) A = \left(\frac{25 - 12}{25}\right) A = \frac{13}{25} A \]Now, we need to find the ratio of the total area of the three smaller triangles to the remaining part of the larger triangle:
\[ \text{Ratio} = \frac{\text{Total Area of small triangles}}{\text{Area of remaining part}} = \frac{\frac{12}{25} A}{\frac{13}{25} A} = \frac{12}{13} \]Correct Option: A) 12 : 13
The question asks to identify the winner of the Ranji Trophy for a specific season, 2017-18. This requires knowledge of historical cricket tournament results, which is a factual recall question often found in general knowledge sections of competitive exams.
Correct Option: C) Vidarbha won the Ranji Trophy in the 2017-18 season. They defeated Delhi in the final to claim their first-ever title.
This problem involves understanding the concepts of free fall and the conservation of energy. We need to analyze how kinetic energy, potential energy, acceleration, and impulse behave for objects of different masses dropped from the same height under gravity, neglecting air resistance.
B) acceleration — In free fall, neglecting air resistance, all objects accelerate downwards at the same rate, which is the acceleration due to gravity (\(g\)), regardless of their mass. Therefore, at any height (including 10 m above the ground), their acceleration will be the same.
The problem presents a series of letter-pairs. To find the next letter-pair, we need to identify the pattern in how the letters in each position (first and second) change from one pair to the next.
The first letters are E, F, G, H.
We can observe a clear pattern: the first letter is incrementing by one position in the alphabet each time (E + 1 = F, F + 1 = G, G + 1 = H). Therefore, the next first letter should be H + 1 = I.
The second letters are N, O, P, Q.
Similarly, the second letter is also incrementing by one position in the alphabet each time (N + 1 = O, O + 1 = P, P + 1 = Q). Therefore, the next second letter should be Q + 1 = R.
The next first letter is I, and the next second letter is R. Combining them gives the letter-pair IR.
B) IR is the correct option because it follows the established pattern where both the first and second letters of each pair advance by one position in the alphabet.
To find the range of a set of data, we need to determine the difference between the largest and the smallest numbers in the set. We are given information about the mean of the three smallest numbers and the mean of the three largest numbers in a set of four numbers. We can set up equations based on this information to find the individual numbers or expressions for them, and then calculate the range.
Let the four numbers in ascending order be \(a, b, c, d\), where \(a \le b \le c \le d\).
The three smallest numbers are \(a, b, c\). Their mean is 19.
So, \(\frac{a + b + c}{3} = 19\)
This implies \(a + b + c = 19 \times 3\)
\[a + b + c = 57 \quad \text{(Equation 1)}\]The three largest numbers are \(b, c, d\). Their mean is 21.
So, \(\frac{b + c + d}{3} = 21\)
This implies \(b + c + d = 21 \times 3\)
\[b + c + d = 63 \quad \text{(Equation 2)}\]We need to find the range, which is the difference between the largest and the smallest number, i.e., \(d - a\).
Subtract Equation 1 from Equation 2:
\[(b + c + d) - (a + b + c) = 63 - 57\] \[b + c + d - a - b - c = 6\] \[d - a = 6\]The range of the set of data is \(d - a\).
B) 6 is the correct answer. The difference between the largest number (\(d\)) and the smallest number (\(a\)) is 6.
If x + y = 3, xy = 2 and x < y, then find the value of x3 - y3.
To find the value of \(x^3 - y^3\), we can use the algebraic identity for the difference of cubes: \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). We are given \(x + y = 3\) and \(xy = 2\). We first need to find the values of \(x - y\) and \(x^2 + y^2\) using the given information and then substitute them into the identity.
Find \(x - y\):
We know that \((x - y)^2 = (x + y)^2 - 4xy\).
Substitute the given values:
\[ (x - y)^2 = (3)^2 - 4(2) \]\[ (x - y)^2 = 9 - 8 \]\[ (x - y)^2 = 1 \]Take the square root of both sides:
\[ x - y = \pm\sqrt{1} \]\[ x - y = \pm 1 \]Given the condition \(x < y\), it implies that \(x - y\) must be negative. Therefore,
\[ x - y = -1 \]Find \(x^2 + y^2\):
We know that \(x^2 + y^2 = (x + y)^2 - 2xy\).
Substitute the given values:
\[ x^2 + y^2 = (3)^2 - 2(2) \]\[ x^2 + y^2 = 9 - 4 \]\[ x^2 + y^2 = 5 \]Calculate \(x^3 - y^3\):
Use the identity \(x^3 - y^3 = (x - y)(x^2 + xy + y^2)\).
Substitute the values we found and the given \(xy\):
\[ x^3 - y^3 = (-1)(5 + 2) \]\[ x^3 - y^3 = (-1)(7) \]\[ x^3 - y^3 = -7 \]A) -7 is the correct answer, as calculated above.
The question asks to classify 'Riccia' into its correct plant group. This requires knowledge of plant classification, specifically the major divisions of the plant kingdom.
Correct Option: A) Bryophyta
Riccia is a genus of liverworts, which are a type of bryophyte. Bryophytes are non-vascular plants that typically live in damp environments. They are characterized by a dominant gametophytic stage, the absence of true roots, stems, and leaves (instead having root-like rhizoids, stem-like axes, and leaf-like appendages), and reproduction through spores. Riccia fits all these descriptions.
To find out when the four bells will ring together again, we need to find the least common multiple (LCM) of their individual ringing intervals. The LCM will give us the minimum time after which all bells will ring simultaneously.
Correct Option: A) 8:24 AM is the time when all four bells will play together again.
This question tests understanding of basic energy transformations and the nature of thermal energy. We need to identify how a microwave oven heats coffee and the source of that energy.
B) It has a measurable amount of thermal energy obtained through the work done by the microwave oven, which in turn takes electrical energy from the electrical grid.
The question asks to identify the empire founded by the Sangama princes, Harihara and Bukka. This is a direct factual recall question from Indian history, specifically related to medieval South Indian kingdoms.
C) Vijaynagar: The Vijayanagara Empire was indeed founded by Harihara and Bukka in 1336 CE. They belonged to the Sangama dynasty, which was the first dynasty to rule the empire.