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Lana71 [14]
2 years ago
7

Prove the diagonals of a square are perpendicular

Mathematics
1 answer:
zysi [14]2 years ago
6 0

Answer:

Given :- ABCD is a square.

To proof :- AC = BD and AC ⊥ BD

Proof :- In △ ADB and △ BCA

AD = BC [ Sides of a square are equal ]

∠BAD = ∠ABC [ 90° each ]

AB = BA [ Common side ]

△ADB ≅ △BCA [ SAS congruency rule ]

⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]

In △AOB and △AOD

OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]

AB = AD [ Sides of a square are equal ]

AO = AO [ Common side ]

△AOB ≅ △ AOD [ SSS congruency rule ]

⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]

∠AOB + ∠AOD = 180° [ Linear pair ]

∠ AOB = ∠AOD = 90°

⇒ AO ⊥ BD

⇒ AC ⊥ BD

Hence proved, AC = BD and AC ⊥BD

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The area of each circle
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Answer:

8. 25\pi in^2

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Step-by-step explanation:

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r = d/2

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r = 10/2

r = 5 in

Area = \pi 5^2

Area = 25\pi in^2

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r = 6/2

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4 0
3 years ago
Eliminate the parameter.<br><br> x = 3 cos t, y = 3 sin t
ElenaW [278]

Answer:

x^2+y^2 = 3^2

Step-by-step explanation:

We need to eliminate the parameter t

Given:

x = 3 cos t

y = 3 sin t

Squaring the above both equations

(x)^2=(3 cos t)^2

(y)^2 =(3 sin t)^2

x^2 = 3^2 cos^2t

y^2=3^2 sin^2t

Now adding both equations

x^2+y^2=3^2 cos^2t+3^2 sin^2t

Taking 3^2 common

x^2+y^2=3^2 (cos^2t+sin^2t)

We know that cos^2t+sin^2t = 1

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x^2+y^2 = 3^2

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4 0
2 years ago
The figure shows the layout of a symmetrical pool in a water park. What is the area of this pool rounded to the tens place? Use
k0ka [10]

Answer:

2489ft^{2}

Step-by-step explanation:

The pool are is divided into 4 separated shapes: 2 circular sections and 2 isosceles triangles. Basically, to calculate the whole area, we need to find the area of each section. Due to its symmetry, both triangles are equal, and both circular sections are also the same, so it would be enough to calculate 1 circular section and 1 triangle, then multiply it by 2.

<h3>Area of each triangle:</h3>

From the figure, we know that <em>b = 20ft </em>and <em>h = 25ft. </em>So, the area would be:

A_{t}=\frac{b.h}{2}=\frac{(20ft)(25ft)}{2}=250ft^{2}

<h3>Area of each circular section:</h3>

From the figure, we know that \alpha =2.21 radians and the radius is R=30ft. So, the are would be calculated with this formula:

A_{cs}=\frac{\pi R^{2}\alpha}{360\°}

Replacing all values:

A_{cs}=\frac{(3.14)(30ft)^{2}(2.21radians)}{6.28radians}

Remember that 360\°=6.28radians

Therefore, A_{cs}=994.5ft^{2}

Now, the total are of the figure is:

A_{total}=2A_{t}+2A{cs}=2(250ft^{2} )+2(994.5ft^{2})\\A_{total}=500ft^{2} + 1989ft^{2}=2489ft^{2}

Therefore the area of the symmetrical pool is 2489ft^{2}

3 0
3 years ago
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Which statement about decision making analysis is false
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Step-by-step explanation:

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