Question

PLEASE HELP ME ON 6-11 AND SHOW WORK PLEASE!!

Answer 1

Answer:

6) 6$\sqrt{2}$

9) 40

10) $\frac{5\sqrt{2} }{2}$

11) 13

Step-by-step explanation:

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

9) Pythagorean Theorem

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

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

$\sqrt{41^{2}-9^{2} }$ which gives us 40

10) same with #6 but we do the opposite. SInce we have the hypotenuse, we can divide that by $\sqrt{2}$ because we know that if 2 legs are equal, the hypotenuse is multiplied by $\sqrt{2}$. Multiply the numerator and denominator by $\sqrt{2}$ 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 $\sqrt{5^{2}+12^{2} }$ which gives us 13

Answer 2

QUESTION:- FIND THE VARIABLES

$\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}$

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->

${h}^{2} = {p}^{2} + {b}^{2}$

${c}^{2} = {6}^{2} + {6}^{2} \\ {c}^{2} = 36 + 36 \\ c = \sqrt{2( {6}^{2} )} \\ c = \sqrt{2}{\sqrt{ {6}^{{2}} } } \\ c = 6 \sqrt{2} \: \: \: ans$

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->

${h}^{2} = {p}^{2} + {b}^{2}$

${41}^{2} = {x}^{2} + {9}^{2} \\ 1681 = {x}^{2} + 81 \\ 1681 - 81 = {x}^{2} \\ 1600 = {x}^{2} \\ x = \sqrt{40 \times 40} \\ x = 40 \: \: \: ans$

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->

${h}^{2} = {p}^{2} + {b}^{2}$

${5}^{2} = {s}^{2} + {s}^{2} \\ 25 = 2 {s}^{2} \\ 12.5 = {s}^{2} \\ \sqrt{12.5} = s \\ 3.5 = s \: \: \: ans$

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 ->

${h}^{2} = {p}^{2} + {b}^{2}$

${r}^{2} = {12}^{2} + {5}^{2} \\ {r}^{2} \\ 144 + 25 \\ {r}^{2} = 169 \\ r = \sqrt{13 \times 13} \\ r = 13 \: \: \: \: ans$

HOPE IT HELPED

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