Question

I need help on theses questions

Answer 1

whats that../

Answer 2

okay so for the first question in the picture I think its a scalene triangle and to find the perimeter for the triangle do i have to find the mid point to all the sides ?

Answer 3

no..just find the lengths of all sides and sum up all..its for perimetre..

Answer 4

can you help me find the perimeter ?

Answer 5

and answer the rest of the questions ?

Answer 6

to find AB BC CA use distance formula..as this..

\[d=\sqrt{(x2-x1)^2+(y2-y1)^2}\]

for all use this..

Answer 7

okay I'm going to find the distance for them give me a second

Answer 8

for the distance ab i got 0 i think i did something wrong lol

Answer 9

ummm show your work..

Answer 10

AB=2sqrt10
how you get zero..??

Answer 11

idk lol

Answer 12

\[D = \sqrt{(10 - 0) + ( 0-10)}\]

Answer 13

im guessing thats wrong

Answer 14

sorry I'm really bad at math

Answer 15

see..

AB=d=sqrt{(0-10)^2+(0-10)^2}
=>AB=d=sqrt{100+100}
=>AB=d=sqrt200
=>AB=sqrt2*100
=>AB=10sqrt2

Answer 16

it also has square at (10-0)^2 and (0-10)^2

Answer 17

oh lol

Answer 18

now do it..!!!

Answer 19

\[D=\sqrt{(-5 - 10)^2 + (5 - 10)^2}\]

Answer 20

did i write that one correctly for BC

Answer 21

yeeppp..xactly..

Answer 22

\[D= \sqrt{(-15)^2 +(-5)^2}\]

Answer 23

right...

Answer 24

\[D=\sqrt{(225) + (25)}\]

Answer 25

right..

Answer 26

\[D=\sqrt{255}\]

Answer 27

no...add again...

Answer 28

\[D=\sqrt{250}\]

Answer 29

yes..now.???

Answer 30

what do i do know?

Answer 31

now*

Answer 32

now sqrt250=5sqrt10

okay..???you have to write it as this..

\[5\sqrt{10}\]

Answer 33

now find CA similar way and then proceed..

Answer 34

\[D=\sqrt{(0-(-5)^2 + (0-5)^2}\]

Answer 35

right...

Answer 36

\[D=\sqrt{(5)^2 +(-5)^2}\]

Answer 37

right..

Answer 38

\[D=\sqrt{(25) + (25)}\]

Answer 39

\[D=\sqrt{50}\]

Answer 40

is that right

Answer 41

yes now write it as..

\[\sqrt{50}=5\sqrt{2}\]

Answer 42

now find perimeter by just adding them..!!!all the three distances..

Answer 43

how do i write the one for BC it was 250

Answer 44

sqrt{250}=5sqrt10

Answer 45

ok so it would be \[Perimeter = 10\sqrt{2} + 5 \sqrt{10} + 5 \sqrt{2}\]

Answer 46

=\[=15\sqrt{2}+5\sqrt{10}\]

Answer 47

how do i add it with the sqrt ? sorry I'm really bad at this

Answer 48

okay..??

Answer 49

\[10\sqrt{2}+5\sqrt{2}=(10+5)\sqrt{2}=15\sqrt{2}\]

Answer 50

Aren't I suppose to add all sides why just 2

Answer 51

coz rest term has sqrt{10} which can not be taken as common as we take sqrt{2} from both...

Answer 52

oh lol

Answer 53

i dont understand though what would be the perimeter than

Answer 54

it the perimeter that you get ..the sum of all sides is called perimeter..

Answer 55

so i would add \[10 \sqrt{2}\]

Answer 56

to 15sqrt 2

Answer 57

yes you can..

Answer 58

come at skype..i will tell you there how to do..!!all.

Answer 59

\[15\sqrt{2} +10\sqrt{2}\]

Answer 60

\[25\sqrt{4}\]

Answer 61

??

Answer 62

no its 25sqrt2

Answer 63

oh lol so is that the perimeter

Answer 64

yes..iti is..

Answer 65

now to prove it right triangle..

use this..

(BC)^2=(AC)^2+(AB)^2

if it satisfies then it is right triangle..if not then it is not a right triangle..

Answer 66

it is phythagoras's thorem...used for right triangles..

Answer 67

my brother said its \[(16)^2 + (15)^2 + (7)^2\]

Answer 68

is that correct

Answer 69

??

Answer 70

@zaibali.qasmi

Answer 71

no...use that values you get in step I for AB BC and CA..in above which i tell you..

Answer 72

\[5\sqrt{10}^2 + 5\sqrt{2}^2 +10\sqrt{2}^2\]

Answer 73

is that correct

Answer 74

sorry I'm so bad at this

Answer 75

yes..but write it inn clear...as..


i write it as.

bc^2=ac^2+ab^2
see it carefully and plug in values..

Answer 76

im confused

Answer 77

to what..??

Answer 78

see
AB=10sqrt2
BC=5sqrt10
CA=5sqrt2

now
{5sqrt10}^2={10sqrt2}^2+{5sqrt2}^2
=>25*10=100*2+25*2
=>250=200+50
=>250=250

this shows that it is a right triangle...

Answer 79

now to find the simply multiply

AC and AB...

Answer 80

to find area///just multiply AC and AB

Answer 81

okay thank you so is ac still 5 sqrt 2 and ab 10 sqrt 2 and and i would just multiply them

Answer 82

@zaibali.qasmi

Answer 83

yes...

Answer 84

???

Answer 85

AC and AB have still same values and you just have to multiply them..

Answer 86

so it would be 25 * 200 ?

Answer 87

??

Answer 88

@zaibali.qasmi

Answer 89

no..

AB=10sqrt2
AC=5sqrt2

so

AB*AC=10sqrt2*5sqrt2=50*2=100

Answer 90

so the area is 100

Answer 91

?

Answer 92

yes,,,

Answer 93

now how do i find the slope of bc

Answer 94

?

Answer 95

to find the slope of BC use this..

m=\[m=\frac{ y2-y1 }{ x2-x1 }\]

here m is slope...

Answer 96

okay so it would be \[M = 5 - 10 \over -5 - 10\]