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kherson [118]
3 years ago
13

In a parallelogram PQRS, PQ = 8cm and QR = 10 cm. If the height corresponding to side PQ

Mathematics
1 answer:
hichkok12 [17]3 years ago
5 0

Answer:

4 cm

Step-by-step explanation:

  • PQ = 8 cm
  • QR = 10 cm
  • Height corresponding to PQ = 5 cm
  • Height corresponding QR = x

The area of parallelogram is equal to product of its base and height

It can be expressed as

  • A = 8*5 = 40

and

  • A= 10*x

Comparing them we get

  • 10x = 40 ⇒ x = 4 cm
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An equilateral triangle is inscribed in a circle of radius 6r. Express the area A within the circle but outside the triangle as
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Answer:

A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}

Step-by-step explanation:

We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

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\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}

We know that area of an equilateral triangle is equal to \frac{\sqrt{3}}{4}s^2, where s represents side length of triangle.

\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2

The area within circle and outside the triangle would be difference of area of circle and triangle as:

A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}

We can make a common denominator as:

A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}

A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}

Therefore, our required expression would be A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}.

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