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3 years ago

Find the area of the isosceles triangle.

Mathematics

1 answer:

Allushta [10]3 years ago

7 0

Answer:

60in^2

Step-by-step explanation:

Area of a triangle = Height x Base X 1/2

Base = 10

Other 2 sides = 13

Now, let's split the Base into 2 parts.

=> 1 part = 5 and 2 part = 5

We can find the height using Pythagorean theorem

=> Isosceles triangle

=> (1 part of the base)^2 + Height^2 = The Hypotenuse^2 = the longer side of a triangle = 13^2

=> 5^2 + Height ^2 = 13^2

=> 25 + Height^2 = 169

=> 25 - 25 + Height ^2 = 169 - 25

=> Height^2 = 144

=> Height = Square root of 144

=> Height = 12

Area = 12 x 10 x 1/2

=> 12 x 5

=> 60 in^2

So, the Area is 60 in ^2

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Step-by-step explanation:

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3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for y with respect to x, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

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$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$y=b\sin\theta\implies\dfrac{\mathrm dy}{\mathrm d\theta}=b\cos\theta$

$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

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3 years ago

Find the area of the isosceles triangle.​

Find the area of the isosceles triangle.