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hram777 [196]
3 years ago
15

Find a area of right angle triangle with base 10cm and hypotenuse 26cm

Mathematics
1 answer:
Ivahew [28]3 years ago
7 0

Answer:

The area of triangle is 120 cm².

Step-by-step explanation:

First, you have to calculate the height of the triangle using Pythogorem Theorem :

{a}^{2}  +  {b}^{2}  =  {c}^{2}

Let a = base = 10 cm,

Let b = height = h cm,

Let c = hypotenuse = 26 cm,

{10}^{2}  +  {h}^{2}  =  {26}^{2}

{h}^{2}  =  {24}^{2}  -  {10}^{2}

{h}^{2}  = 576

h =  \sqrt{576}

h = 24 \: cm

We have found out that the height is 24 cm. Next, we have to use the area of triangle formula :

area =  \frac{1}{2}  \times base \times height

Let base = 10cm,

Let height = 24cm,

area =  \frac{1}{2}  \times 10 \times 24

area =  \frac{1}{2}  \times 240

area = 120 \:  {cm}^{2}

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Find the absolute maximum and minimum values of f on the set D. f(x,y)=2x^3+y^4, D={(x,y) | x^2+y^2&lt;=1}.
castortr0y [4]

Answer:

absolute maximum is f(1, 0) = 2 and the absolute minimum is f(−1, 0) = −2.

Step-by-step explanation:

We compute,

$ f_x = 6x^2, f_y=4y^3 $

Hence, $ f_x = f_y = 0 $  if and only if (x,y) = (0,0)

This is unique critical point of D. The boundary equation is given by

$ x^2+y^2=1$

Hence, the top half of the boundary is,

$ T = \{ x, \sqrt{1-x^2} : -1 \leq x \leq 1\}

On T we have, $ f(x, \sqrt{1-x^2} = 2x^3 +(1-x^2)^2 = x^4 +2x^3-2x^2+1  \text{ for}\ -1 \leq x \leq 1$

We compute

$ \frac{d}{dx}(f(x, \sqrt{1-x^2}))= 4x^3+6x^2-4x = 2x(2x^2+3x-2)=2x(2x-1)(x+2)=0$

0 if and  only if x=0, x= 1/2 or x = -2.

We disregard  $ x = -2 \notin [-1,1]$

Hence, the critical points on T are (0,1) and $(\frac{1}{2}, \sqrt{1-(\frac{1}{2})^2}=\frac{\sqrt3}{2})$

On the bottom half, B, we have

$ f(x, \sqrt{1-x^2})= f(x,-\sqrt{1-x^2})$

Therefore, the critical points on B are (0,-1) and $( 1/2, -\sqrt3/2)  

It remains to  evaluate f(x, y) at the points $ (0,0), (0 \pm1), (1/2, \pm \sqrt3/2) \text{ and}\  (\pm1, 0)$ .

We should consider  latter two points, $(\pm1,0)$, since they are the boundary points for the T and also  B. We compute $ f(0,0)=0, \ \f(0 \pm1)=1, \ \ f(0, \pm \sqrt3/2)=9/16, \ \ f(1,0 )= 2 \text{ and}\ \ f(-1,0)= -2 $

We conclude that the  absolute maximum = f(1, 0) = 2

And the absolute minimum = f(−1, 0) = −2.

6 0
3 years ago
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