The triangle inside the three squares shows the proof of the Pythagorean Theorem
Area of one triangle is 25 sq units, so one side = 5
Area of second triangle is 144 sq units, so the side = 12
Area of third triangle is 169sq units, so one side = 13
Therefore three side of the triangle are, 5, 12 and 13
proof of the Pythagorean Theorem
Base = 12
Height = 5
Hypotenuse = 13
5^2 + 12^2 = 25 +169 =13^2
The Pythagorean Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In the triangle inside the three squares, the hypotenuse is the longest side, which is 13 units long.
The other two sides are 5 units and 12 units long.
We can see that the square with side length 13 has an area of 169 square units.
The square with side length 5 has an area of 25 square units, and the square with side length 12 has an area of 144 square units.
The sum of the areas of the two smaller squares is 25 + 144 = 169 square units.
This is equal to the area of the square with side length 13, which is the hypotenuse of the triangle.
Therefore, the Pythagorean Theorem is proved.
Here is a diagram of the proof:
right triangle inside three squares. The hypotenuse of the triangle is the side of the largest square. The other two sides of the triangle are the sides of the two smaller squares.
The hypotenuse of the triangle is the side of the largest square. The other two sides of the triangle are the sides of the two smaller squares.
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B) 48 square feet
C) 72 square feet
D) 108 square feet
Answer:
Step-by-step explanation: 50 por ciento serio la mitad del 6 y del 8 oseas que los numeros quedarian en 9 y 12 multiplicas 9 por 12 y es 108
There are 30 weeks in 210 days. 210/7=30
Answer:
We can conclude that Δ GHI ≅ Δ JKL by SAS postulate.
Step-by-step explanation:
Δ GHI and Δ JKL are congruents because:
1. Their sides GH and JK are equal (9 units = 9 units)
2. Their included angles ∠G and ∠J are equal (62° = 62°)
3. Their sides GI and JL are equal (17 units = 17 units)
Now, we can conclude that Δ GHI ≅ Δ JKL by SAS postulate.
The coordinates of the midpoint of two points A and B are the averages of the corresponding coordinates of A and B.
So if one point is (a, b) and another point is (c, d), the point exactly between them has coordinates ((a+b)/2, (c+d)/2).
In this example, the points
A(-3a, b)
B(3a, b)
have midpoint
F((-3a+3a)/2, (b+b)/2) = F(0, 2b/2) = F(0, b)
always 2a units. Find the equation to the locus
of the moving point.
Answer:
Step-by-step explanation:
Let P be the moving point and the given points be A (c,0) and B (-c, 0). If (h, k) be the co-ordinates of any position of P on its locus then:
PA + PB = 2a
PA = 2a - PB
PA² = 4a² + PB² – 4a ∙ PB
PA² – PB² = 4a² – 4a ∙ PB
[(h - c)² +(k - 0)²] - [(h + c)² +(k - 0)²] = 4a² – 4a. PB
-4hc = 4a² – 4a∙PB
a ∙ PB = a² + hc
Taking square
a² ∙ PB² = (a² + hc)²
a² [(h + c)² + (k - 0)²] = (a² + hc)²
a² [h² + c² + 2hc + k²] = a⁴ + 2a²hc + h²c²
a²h² – h²c² + a²k² = a⁴ – a²c²
(a² – c²)h² + a²k² = a² (a² – c²)
h²/a² + k²/(a² – c²) = 1
Therefore, the required equation to the locus of P is