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VladimirAG [237]
3 years ago
14

Prove that a parallelogram is a square iff its diagonals are both congruent and perpendicular

Mathematics
1 answer:
Juliette [100K]3 years ago
6 0
Consider the parallelogram shown below.
The lengths of the sides are a and b.
The lengths of the diagonals are 2x and 2y.

Because the diagonals are both congruent and perpendicular, therefore there are two right triangles as shown.
Note that x = y.
Because x = y, each right triangle is isosceles and has the angles 90°, 45° and 45°.

From the Pythagorean theorem,
For one right triangle,
a² = x² + y² = x² + x² = 2x².
For the other right triangle,
b² = y² + x² = x² + x² = 2x²

Therefore
a² = b²
a = b

It follows that all sides of the parallelogram are equal and each angle is
45+45 = 90°

Therefore the parallelogram is a square.

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5. 3x + 5y + 5z =1<br> x - 2y = 5<br> 2x + 4y = 11
lilavasa [31]

Answer:

see explanation

Step-by-step explanation:

Given the 3 equations

3x + 5y + 5z = 1 → (1)

x - 2y = 5 → (2)

2x + 4y = 11 → (3)

Use (2) and (3) to solve for x and y

Multiply (2) by 2

2x - 4y = 10 → (4)

Add (3) and (4) term by term

4x = 21 ( divide both sides by 4 )

x = \frac{21}{4\\}

Substitute this value of x into (3)

2 × \frac{21}{4\\} + 4y = 11

\frac{21}{2\\} + 4y = 11 ( subtract \frac{21}{2\\} from both sides )

4y = \frac{1}{2} ( divide both sides by 4 )

y = \frac{1}{8\\}

Substitute the values of x and y into (1) and solve for z

3 × \frac{21}{4\\} + 5 × \frac{1}{8\\} + 5z = 1

\frac{63}{4} + \frac{5}{8} + 5z = 1

\frac{131}{8} + 5z = 1 ( subtract \frac{131}{8} from both sides )

5z = - \frac{123}{8} ( divide both sides by 5 )

z = - \frac{123}{40}

Solution is

x = \frac{21}{4\\}, y = \frac{1}{8\\}, z = - \frac{123}{40}

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3 years ago
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