A Question is given followed by two Statements I and II. Consider the Question and the Statements. Which one of the following is correct in respect of the above Question and the Statements?
Question:
ABC is an isosceles triangle with AB = AC = 10 units. If the area of the triangle is 48 square units, then what is the length of the base BC?
Statement-I : The length of BC is an even integer.
Statement-II: The height of the triangle is greater than the length of half of the base.
Analyzing the Isosceles Triangle Problem
The problem asks for the length of the base BC of an isosceles triangle ABC, given that the equal sides AB and AC are 10 units long and the area of the triangle is 48 square units. We also need to determine if two statements, Statement-I and Statement-II, are sufficient to answer the question.
Evaluating the Main Question Data
Let the isosceles triangle be ABC, where \(AB = AC = 10\). Let the base be \(BC = b\) and the height from vertex A to the base BC be \(h\). In an isosceles triangle, the altitude to the base bisects the base. Let D be the midpoint of BC. Then \(BD = DC = \frac{b}{2}\).
Triangle ADB is a right-angled triangle with hypotenuse \(AB = 10\), altitude \(AD = h\), and base \(BD = \frac{b}{2}\). According to the Pythagorean theorem:
\(h^2 + \left(\frac{b}{2}\right)^2 = AB^2\) \(h^2 + \frac{b^2}{4} = 10^2 = 100\)We are also given the area of the triangle:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\) \(48 = \frac{1}{2} \times b \times h\) \(b \times h = 96\)From this, we can express the height in terms of the base: \(h = \frac{96}{b}\).
Now, substitute this expression for \(h\) into the Pythagorean equation:
\(\left(\frac{96}{b}\right)^2 + \frac{b^2}{4} = 100\) \(\frac{9216}{b^2} + \frac{b^2}{4} = 100\)To solve for \(b\), let's clear the denominators by multiplying the entire equation by \(4b^2\):
\(4 \times 9216 + b^2 \times b^2 = 100 \times 4b^2\) \(36864 + b^4 = 400b^2\)Rearrange this into a standard form equation, treating it as a quadratic equation in terms of \(b^2\):
\(b^4 - 400b^2 + 36864 = 0\)Let \(x = b^2\). The equation becomes:
\(x^2 - 400x + 36864 = 0\)We can solve this quadratic equation for \(x\) using the quadratic formula \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\):
\(x = \frac{-(-400) \pm \sqrt{(-400)^2 - 4(1)(36864)}}{2(1)}\) \(x = \frac{400 \pm \sqrt{160000 - 147456}}{2}\) \(x = \frac{400 \pm \sqrt{12544}}{2}\)The square root of 12544 is 112:
\(x = \frac{400 \pm 112}{2}\)This gives two possible values for \(x\) (which is \(b^2\)):
- \(x_1 = \frac{400 + 112}{2} = \frac{512}{2} = 256\). So, \(b^2 = 256 \implies b = \sqrt{256} = 16\).
- \(x_2 = \frac{400 - 112}{2} = \frac{288}{2} = 144\). So, \(b^2 = 144 \implies b = \sqrt{144} = 12\).
Therefore, based on the information given in the question alone, the length of the base BC could be either 16 units or 12 units. The question cannot be answered uniquely from the given data alone.
Evaluating Statement-I
Statement-I: The length of BC is an even integer.
From our analysis of the main question, the possible lengths for BC are 16 and 12.
- Is 16 an even integer? Yes.
- Is 12 an even integer? Yes.
Since both possible values (12 and 16) satisfy the condition that BC is an even integer, Statement-I does not help us distinguish between the two possibilities. Therefore, Statement-I alone is not sufficient to answer the question.
Evaluating Statement-II
Statement-II: The height of the triangle is greater than the length of half of the base (\(h > \frac{b}{2}\)).
Let's test the two possible scenarios derived from the main question:
- Scenario 1: If \(BC = b = 16\). Then half the base is \(\frac{b}{2} = \frac{16}{2} = 8\). The corresponding height is \(h = \frac{96}{b} = \frac{96}{16} = 6\). Does this satisfy Statement II (\(h > \frac{b}{2}\))? Is \(6 > 8\)? No, this is false. So, \(BC=16\) is not a valid length under Statement II.
- Scenario 2: If \(BC = b = 12\). Then half the base is \(\frac{b}{2} = \frac{12}{2} = 6\). The corresponding height is \(h = \frac{96}{b} = \frac{96}{12} = 8\). Does this satisfy Statement II (\(h > \frac{b}{2}\))? Is \(8 > 6\)? Yes, this is true. So, \(BC=12\) is the only valid length under Statement II.
Statement-II allows us to uniquely determine that the length of the base BC must be 12 units. Therefore, Statement-II alone is sufficient to answer the question.
Final Conclusion
Based on the analysis:
- The main question data provides two possible answers for BC (12 and 16), so it's insufficient alone.
- Statement-I is consistent with both possibilities, hence it is insufficient alone.
- Statement-II allows us to eliminate one possibility and arrive at a unique answer (BC=12), hence it is sufficient alone.
This situation matches the description: "The Question can be answered by using one of the Statements alone, but cannot be answered using the other Statement alone."
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