Respuesta :

jiangjacky1234ozqz8r

Answer:

6) 6[tex]\sqrt{2}[/tex]

9) 40

10) [tex]\frac{5\sqrt{2} }{2}[/tex]

11) 13

Step-by-step explanation:

6)A right triangle rule: if 2 legs are equal, the hypotenuse is the length of that leg*[tex]\sqrt{2}[/tex]

9) Pythagorean Theorem

[tex]a^{2} +b^{2} =c^{2}[/tex]

We know the hypotenuse (41) so we substitute that for c and 9 for b now we need to find a

[tex]\sqrt{41^{2}-9^{2} }[/tex] which gives us 40

10) same with #6 but we do the opposite. SInce we have the hypotenuse, we can divide that by [tex]\sqrt{2}[/tex] because we know that if 2 legs are equal, the hypotenuse is multiplied by [tex]\sqrt{2}[/tex]. Multiply the numerator and denominator by [tex]\sqrt{2}[/tex] because we can't have a square root in the denominator.

11) like #9 we have the a and b but we need to find c

a=5 b=12 c=r

so [tex]\sqrt{5^{2}+12^{2} }[/tex] which gives us 13

DEVIL005

QUESTION:- FIND THE VARIABLES

[tex] \huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}[/tex]

6TH PART ->

GIVEN

  1. RIGHT ANGLE SO ITS A RIGHT ANGLED TRIANGLE SO WE CAN USE PYTHAGOUS THEOREM
  2. TWO SIDES ( BASE AND PERPENDICULAR) R EQUAL TO 6

SOLUTION->

[tex] {h}^{2} = {p}^{2} + {b}^{2} [/tex]

[tex] {c}^{2} = {6}^{2} + {6}^{2} \\ {c}^{2} = 36 + 36 \\ c = \sqrt{2( {6}^{2} )} \\ c = \sqrt{2}{\sqrt{ {6}^{{2}} } } \\ c = 6 \sqrt{2} \: \: \: ans[/tex]

9TH PART:-

GIVEN

  1. RIGHT ANGLE SO ITS A RIGHT ANGLED TRIANGLE SO WE CAN USE PYTHAGOUS THEOREM
  2. BASE = 9
  3. HYPOTENUSE= 41

SOLUTION->

[tex] {h}^{2} = {p}^{2} + {b}^{2} [/tex]

[tex] {41}^{2} = {x}^{2} + {9}^{2} \\ 1681 = {x}^{2} + 81 \\ 1681 - 81 = {x}^{2} \\ 1600 = {x}^{2} \\ x = \sqrt{40 \times 40} \\ x = 40 \: \: \: ans[/tex]

10 TH PART:-

GIVEN

  1. RIGHT ANGLE SO ITS A RIGHT ANGLED TRIANGLE SO WE CAN USE PYTHAGOUS THEOREM
  2. TWO SIDES ( BASE AND PERPENDICULAR) R EQUAL TO S
  3. HYPOTENUSE= 5

SOLUTION->

[tex]{h}^{2} = {p}^{2} + {b}^{2} [/tex]

[tex] {5}^{2} = {s}^{2} + {s}^{2} \\ 25 = 2 {s}^{2} \\ 12.5 = {s}^{2} \\ \sqrt{12.5} = s \\ 3.5 = s \: \: \: ans[/tex]

11TH PART:-

GIVEN

  1. RIGHT ANGLE SO ITS A RIGHT ANGLED TRIANGLE SO WE CAN USE PYTHAGOUS THEOREM
  2. BASE = 5
  3. PERPENDICULAR= 12

SOLUTION ->

[tex] {h}^{2} = {p}^{2} + {b}^{2} [/tex]

[tex] {r}^{2} = {12}^{2} + {5}^{2} \\ {r}^{2} \\ 144 + 25 \\ {r}^{2} = 169 \\ r = \sqrt{13 \times 13} \\ r = 13 \: \: \: \: ans[/tex]

HOPE IT HELPED

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \: DEVIL005 \: \star[/tex]

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