You would do pi x r squared. r=12, and 12 squared is 144. pi (3.14) times 144 equals 452.16!
Answer:
63° (same as Z)
Step-by-step explanation:
The question indicates that ΔXYZ is similar to ΔEDF.
By definition, when we indicate that triangle XYZ is similar to the triangle EDF, the order of the letters is important and match the congruent angles.
So, ∠X = ∠E, ∠Y = ∠D and ∠Z = ∠F
It's hard to see because the angles are quite similar in opening (56°, 63° and 61°), and the drawing doesn't seems to be to scale.
But according to the naming convention: ∠Z = ∠F
Since ∠Z = 63°, ∠F = 63°
<h2><em>Answer:</em></h2><h2><em>Answer:x = 7+4√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get1/√x = (2-√3)/(2-√3)(2+√3) = (2-√3)/(2²-√3²) =</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get1/√x = (2-√3)/(2-√3)(2+√3) = (2-√3)/(2²-√3²) =1/√x = 2-√3</em></h2><h2 /><h2><em>Answer:x = 7+4√3To find √x we proceed,√x = √(7+4√3)√x = √(7+2x2√3)√x = √(7+2√3x4)√x = √(3+4+2√3x4)….. {writing 7 = 3+4}If we observe RHS of √x we observe form of√(a² + b² +2ab) where a=√3 and b =√4Hence, √x =√(√3 +√4)² = √3 + √4 = 2+√3√x = 2+√31/√x = 1/(2+√3)Multiplying both numerator and denominator by 2 - √3, we get1/√x = (2-√3)/(2-√3)(2+√3) = (2-√3)/(2²-√3²) =1/√x = 2-√3Hence √x +1/√x = 2+√3 +2 -√3 = 4</em></h2><h2 />
<u>Collinear points</u>: Three points A, B and C are said to be collinear if they lie on the same straight line.
There points A, B and C will be collinear if AB + BC = AC as is clear from the adjoining figure.
In general, three points A, B and C are <u>collinear if the sum of the lengths of any two line segments among AB, BC and CA is equal to the length of the remaining line segment</u>, that is, either AB + BC = AC or AC +CB = AB or BA + AC = BC.
In other words,
<u>There points A, B and C are collinear if:</u>
(i) AB + BC = AC i.e.,
Or, (ii) AB + AC = BC i.e. ,
Or, AC + BC = AB i.e.,
