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solve all questions from this pdf with explanation

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### **Problem 1: Similar Triangles**
Two triangles \( \Delta ABC \) and \( \Delta DEF \) are similar. The smallest side of \( \Delta ABC \) is 15 units. The sides of \( \Delta ABC \) are in the ratio 3:4:5. The area of \( \Delta DEF \) is half of the area of \( \Delta ABC \). What is the largest side of \( \Delta DEF \) in units?

#### **Solution:**
1. **Determine the sides of \( \Delta ABC \):**
   - The smallest side \( AB = 15 \) units.
   - Ratio of sides = \( 3:4:5 \).
   - If \( 3x = 15 \), then \( x = 5 \).
   - Therefore, sides of \( \Delta ABC \) are: \( 15 \) units, \( 20 \) units, and \( 25 \) units.

2. **Determine the ratio of areas of \( \Delta ABC \) and \( \Delta DEF \):**
   - \( \text{Area of } \Delta DEF = \frac{1}{2} \times \text{Area of } \Delta ABC \).
   - Since the triangles are similar, the ratio of the areas is the square of the ratio of corresponding sides.
   - Let the scale factor be \( k \). Thus, \( k^2 = \frac{1}{2} \), so \( k = \frac{1}{\sqrt{2}} \).

3. **Determine the sides of \( \Delta DEF \):**
   - The sides of \( \Delta DEF \) are \( 15 \times \frac{1}{\sqrt{2}} \), \( 20 \times \frac{1}{\sqrt{2}} \), and \( 25 \times \frac{1}{\sqrt{2}} \).
   - Largest side = \( 25 \times \frac{1}{\sqrt{2}} = \frac{25\sqrt{2}}{2} \) units.

   **Answer: The largest side of \( \Delta DEF \) is \( \frac{25\sqrt{2}}{2} \) units.**

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### **Problem 2: Trapezium Diagonals**
AB is parallel to DC in a trapezium \( ABCD \). \( AB > DC \) and the diagonals \( AC \) and \( BD \) intersect at \( O \). Given \( AO = 3x-15 \), \( OB = x+9 \), \( OC = x-5 \), and \( OD = 5 \). If \( x \) has two values \( x_1 \) and \( x_2 \), then find the value of \( x_1^2 + x_2^2 \).

#### **Solution:**
1. **Use the section formula:**
   - Since \( O \) is the intersection point of the diagonals in the trapezium, by the property of trapeziums, the product of the segments of the diagonals are equal.
   - \( AO \times OC = OB \times OD \).

2. **Substitute the values:**
   - \( (3x-15) \times (x-5) = (x+9) \times 5 \).
   - Expand both sides: \( 3x^2 - 15x - 15x + 75 = 5x + 45 \).

3. **Solve the quadratic equation:**
   - Simplify to \( 3x^2 - 30x + 75 = 5x + 45 \).
   - Move all terms to one side: \( 3x^2 - 35x + 30 = 0 \).

4. **Find \( x_1 \) and \( x_2 \):**
   - Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
   - Here \( a = 3 \), \( b = -35 \), and \( c = 30 \).

5. **Compute \( x_1^2 + x_2^2 \):**
   - \( x_1 + x_2 = \frac{-(-35)}{3} = \frac{35}{3} \), and \( x_1x_2 = \frac{30}{3} = 10 \).
   - \( x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1x_2 \).
   - Compute \( \frac{35^2}{9} - 20 = 1355/9 - 20 = 56 \).

   **Answer: The value of \( x_1^2 + x_2^2 \) is 56.**

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Would you like to continue with the next set of questions, or do you need further clarifications on the above solutions? Here are some follow-up questions you might find useful:

1. How does the area ratio between similar triangles affect their corresponding side lengths?
2. What are the properties of diagonals in trapeziums, especially regarding intersection points?
3. How do you derive the quadratic formula for solving equations?
4. What other methods can be used to solve quadratic equations?
5. Can you explain the section formula and its application in more complex geometric shapes?

**Tip:** When dealing with similar triangles, always remember that the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Explore detailed solutions to mathematical problems involving similar triangles and trapezium diagonals. This page covers topics such as calculating areas of similar triangles, properties of trapezium diagonals, and solving quadratic equations. Suitable for high school students.

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