Solution

Given:
\[ a + b + c = 0 \]

We need to find the value of:
\[ \frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab} \]

Since \(a + b + c = 0\), we can write \(a = -(b + c)\).

Step 1: Substitute \(a = -(b + c)\)

\[ \frac{(-b – c)^2}{(-b – c)^2 – bc} + \frac{b^2}{b^2 – c(-b – c)} + \frac{c^2}{c^2 – b(-b – c)} \]

Step 2: Simplify

\[ \frac{(b + c)^2}{(b + c)^2 – bc} + \frac{b^2}{b^2 + bc – c^2} + \frac{c^2}{c^2 + bc – b^2} \]

Step 3: Simplify further

\[ \frac{(b + c)^2}{b^2 + c^2 + 2bc – bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]

\[ \frac{(b + c)^2}{b^2 + c^2 + bc} + \frac{b^2}{b^2 + c^2 + bc} + \frac{c^2}{b^2 + c^2 + bc} \]

Step 4: Combine the fractions

\[ \frac{(b + c)^2 + b^2 + c^2}{b^2 + c^2 + bc} = \frac{b^2 + c^2 + 2bc + b^2 + c^2}{b^2 + c^2 + bc} \]
\[ = \frac{2b^2 + 2c^2 + 2bc}{b^2 + c^2 + bc} \]
\[ = 2 \frac{b^2 + c^2 + bc}{b^2 + c^2 + bc} \]
\[ = 2 \]

Conclusion

The value of \(\frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab}\) is 2.

Solution

Given:
\[ \sin \alpha + (\sin \alpha)^2 = 1 \]

Step 1: Simplify the given equation

\[ \sin \alpha = 1 – (\sin \alpha)^2 \]
\[ \sin \alpha = (\cos \alpha)^2 \] (Since \((\sin \alpha)^2 + (\cos \alpha)^2 = 1\))

Step 2: Substitute \(\sin \alpha = (\cos \alpha)^2\) into the expression

\[ (\cos \alpha)^{12} + 3(\cos \alpha)^{10} + 3(\cos \alpha)^8 + (\cos \alpha)^6 – 1 \]
\[ = \left((\cos \alpha)^4 + (\cos \alpha)^2\right)^3 – 1 \]
\[ = \left((\sin \alpha)^2 + (\cos \alpha)^2\right)^3 – 1 \]
\[ = 1^3 – 1 \]
\[ = 1 – 1 \]
\[ = 0 \]

Conclusion

The value of \((\cos \alpha)^{12} + 3(\cos \alpha)^{10} + 3(\cos \alpha)^8 + (\cos \alpha)^6 – 1\) is 0.

Solution

Given:
– In \(\triangle ABC\), \(\angle A = \angle B = 60^\circ\), \(AC = \sqrt{34} \text{ cm}\).
– Lines \(AD\) and \(BD\) intersect at \(D\) with \(\angle D = 90^\circ\).
– \(DB = 3 \text{ cm}\).

Step 1: Determine the Type of Triangle

Since \(\angle A = \angle B = 60^\circ\), \(\angle C\) must also be \(60^\circ\) (as the sum of angles in a triangle is \(180^\circ\)). Therefore, \(\triangle ABC\) is an equilateral triangle.

Step 2: Find the Length of AB

Since \(\triangle ABC\) is equilateral, \(AB = AC = \sqrt{34} \text{ cm}\).

Step 3: Use Pythagoras’ Theorem

In \(\triangle ADB\), which is a right-angled triangle, we can use Pythagoras’ theorem:
\[ AB^2 = AD^2 + DB^2 \]
\[ (\sqrt{34})^2 = AD^2 + 3^2 \]
\[ 34 = AD^2 + 9 \]
\[ AD^2 = 34 – 9 \]
\[ AD^2 = 25 \]
\[ AD = \sqrt{25} \]
\[ AD = 5 \text{ cm} \]

Conclusion

The length of \(AD\) is 5 cm.

Solution

In a regular hexagon, all sides are equal, and all internal angles are 120°. The diagonals of a regular hexagon divide it into six equilateral triangles.

Let’s denote the side length of the hexagon as \(a\).

Area of the Hexagon PQRSTU:

The area of an equilateral triangle with side length \(a\) is given by:
\[ \text{Area of equilateral triangle} = \frac{\sqrt{3}}{4}a^2 \]

Since the hexagon is made up of six equilateral triangles, the area of the hexagon is:
\[ \text{Area of hexagon} = 6 \times \frac{\sqrt{3}}{4}a^2 = \frac{3\sqrt{3}}{2}a^2 \]

Area of Quadrilateral PQOU:

Quadrilateral PQOU is made up of two equilateral triangles, POQ and UOQ. Therefore, the area of quadrilateral PQOU is:
\[ \text{Area of quadrilateral PQOU} = 2 \times \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{2}a^2 \]

Ratio of Areas:

The ratio of the area of quadrilateral PQOU to the area of hexagon PQRSTU is:
\[ \text{Ratio} = \frac{\text{Area of quadrilateral PQOU}}{\text{Area of hexagon}} = \frac{\frac{\sqrt{3}}{2}a^2}{\frac{3\sqrt{3}}{2}a^2} = \frac{1}{3} \]

Conclusion

The ratio of the area of quadrilateral PQOU to the area of hexagon PQRSTU is 1:3.

Solution

Given:
– The mixture of two varieties of sugar is sold at ₹80 per kg for a profit of 25%.
– The cost of the variety of sugar in larger quantity is ₹40 per kg.
– The ratio of the two varieties of sugar is 3:2.

Step 1: Find the cost price of the mixture

The selling price of the mixture is ₹80 per kg, and the profit is 25%. Therefore, the cost price (CP) of the mixture is:
\[ CP = \frac{\text{Selling Price}}{1 + \text{Profit Percentage}} = \frac{80}{1 + 0.25} = \frac{80}{1.25} = ₹64 \text{ per kg} \]

Step 2: Calculate the cost of the other variety of sugar

Let the cost of the other variety of sugar be ₹\(x\) per kg. Using the weighted average formula for the cost price of the mixture:
\[ CP_{\text{mixture}} = \frac{(3 \times 40) + (2 \times x)}{3 + 2} \]
\[ 64 = \frac{120 + 2x}{5} \]
\[ 320 = 120 + 2x \]
\[ 2x = 200 \]
\[ x = ₹100 \text{ per kg} \]

Conclusion

The cost of the other variety of sugar is ₹100 per kg.