Hey guys! Today, we're diving into some exciting math problems that involve quadrilaterals and algebraic equations. We'll break down each problem step-by-step, making sure you grasp every concept along the way. So, let's put on our thinking caps and get started!
1. Unraveling the Mystery of Quadrilateral ABCD
The Challenge
We're given a quadrilateral ABCD with some interesting angle properties:
- ∠ACB = ∠BAD = 105°
- ∠ABC = ∠ADC = 45°
Our mission? To figure out the relationship between the sides AB, CD, and AD. Are they equal? Is one greater than the other? Let's find out!
Diving into the Solution
To tackle this, we'll use our knowledge of geometry, specifically the properties of triangles and quadrilaterals. The key to cracking this problem lies in understanding how the angles and sides of a shape relate to each other. Remember, the sum of angles in a triangle is always 180°, and in a quadrilateral, it's 360°.
First, let's focus on triangle ABC. We know ∠ACB = 105° and ∠ABC = 45°. We can find ∠BAC by subtracting these angles from 180°:
∠BAC = 180° - 105° - 45° = 30°
Now, let's shift our attention to quadrilateral ABCD. We know ∠ABC = 45° and ∠ADC = 45°. We also know ∠BAD = 105°. To find ∠BCD, we subtract the sum of the known angles from 360°:
∠BCD = 360° - 45° - 45° - 105° = 165°
Now, let's zoom in on triangle ACD. We know ∠ADC = 45° and we can find ∠ACD by subtracting ∠ACB from ∠BCD:
∠ACD = 165° - 105° = 60°
To find ∠CAD, we subtract ∠ADC and ∠ACD from 180°:
∠CAD = 180° - 45° - 60° = 75°
Now, let's compare triangles ABC and ACD. In triangle ABC, we have:
- ∠ABC = 45°
- ∠ACB = 105°
- ∠BAC = 30°
In triangle ACD, we have:
- ∠ADC = 45°
- ∠ACD = 60°
- ∠CAD = 75°
Notice that in triangle ABC, the side opposite the largest angle (∠ACB = 105°) is AB. In triangle ACD, the side opposite the largest angle (∠CAD = 75°) is CD. Since 105° is greater than 75°, we can infer that AB is likely greater than CD.
To further solidify this, let's think about the sine rule. The sine rule states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. Applying the sine rule to triangles ABC and ACD can give us a more precise comparison. However, based on our angle analysis alone, we have a strong indication that AB > CD.
Conclusion
After a thorough examination of the angles within quadrilateral ABCD, we've deduced that (C) AB > CD is the most likely true statement. Remember, guys, geometry problems often require us to piece together information from different parts of the figure and apply fundamental rules and theorems.
2. Cracking the Algebraic Code: Evaluating pq
The Challenge
We're presented with two equations:
- p + q = 25/4
- (1 + √p)(1 + √q) = 15/2
Our mission? To find the value of the product pq. This problem looks like a fun algebraic puzzle, so let's dive in!
Decoding the Equations
To solve this, we'll need to manipulate these equations strategically. The key here is to expand the second equation and see if we can relate it to the first equation. Expanding (1 + √p)(1 + √q) gives us:
1 + √p + √q + √(pq) = 15/2
Let's rearrange this to isolate the square roots:
√p + √q = 15/2 - 1 - √(pq) = 13/2 - √(pq)
Now, we need to find a way to use the first equation, p + q = 25/4. A clever trick is to square both sides of the equation √p + √q = 13/2 - √(pq). This will help us eliminate the square roots:
(√p + √q)² = (13/2 - √(pq))²
Expanding both sides, we get:
p + 2√(pq) + q = 169/4 - 13√(pq) + pq
We know that p + q = 25/4, so we can substitute that into the equation:
25/4 + 2√(pq) = 169/4 - 13√(pq) + pq
Now, let's rearrange the terms to group the square root terms together:
15√(pq) = 144/4 + pq
15√(pq) = 36 + pq
This equation looks a bit cleaner. To make it even simpler, let's make a substitution. Let x = √(pq), so pq = x². Our equation now becomes:
15x = 36 + x²
Rearranging this into a quadratic equation, we get:
x² - 15x + 36 = 0
Now, we can factor this quadratic equation. We're looking for two numbers that multiply to 36 and add up to -15. Those numbers are -12 and -3. So, we can factor the equation as:
(x - 12)(x - 3) = 0
This gives us two possible solutions for x:
x = 12 or x = 3
Since x = √(pq), we need to square both solutions to find pq:
If x = 12, pq = 12² = 144 If x = 3, pq = 3² = 9
Now, let's think critically about these solutions. We need to check if these values of pq are consistent with our original equations. Remember that √p + √q = 13/2 - √(pq). If pq = 144, then √(pq) = 12, and √p + √q = 13/2 - 12, which would be negative. However, since p and q are positive (due to the square roots), √p and √q must also be positive, so their sum cannot be negative. Therefore, pq = 144 is not a valid solution.
If pq = 9, then √(pq) = 3, and √p + √q = 13/2 - 3 = 7/2. This seems like a more plausible solution. Let's check if we can find p and q values that satisfy this. We have p + q = 25/4 and pq = 9. We can express q as 9/p and substitute this into the first equation:
p + 9/p = 25/4
Multiplying through by 4p, we get:
4p² + 36 = 25p
Rearranging, we get:
4p² - 25p + 36 = 0
Factoring this quadratic equation, we get:
(4p - 9)(p - 4) = 0
So, p = 9/4 or p = 4. If p = 9/4, then q = 9/(9/4) = 4. If p = 4, then q = 9/4. Either way, the values satisfy the original equations.
Conclusion
After a meticulous algebraic journey, we've determined that the value of (D) pq = 9. This problem highlights the importance of strategic equation manipulation and critical thinking to validate our solutions. Keep practicing, and you'll become a master of algebraic puzzles!
Final Thoughts
So, there you have it, guys! We've conquered a challenging quadrilateral problem and a tricky algebraic equation. Remember, math is all about practice and perseverance. Keep honing your skills, and you'll be amazed at what you can achieve. Until next time, keep those brains buzzing!