Question

The base and height of a triangle sum up to 15, and the area of the triangle is 24. What are the dimensions of the triangle?

Answer 1

you mean the base + height = 15 ?

Answer 2

Yes!

Answer 3

Okay,you just need to create a system!

Answer 4

\(\sf {The \ Base \Longrightarrow \ x} \)
\(\sf {The \ Height \Longrightarrow \ y} \)

Answer 5

and this is the system:

Answer 6

\(\sf \large x + y = 15\)

\(\sf \large\frac{ xy }{ 2 } = 24\)

Answer 7

understood?

Answer 8

Yes, thank you!

Answer 9

and now can you solve that system madam?

Answer 10

Yes, I just use the process of substitution correct?

Answer 11

yes,it will work :)

Answer 12

Okay, I will try as well! :)

Answer 13

nice;)

Answer 14

I got a crazy number :o

Answer 15

need help?

Answer 16

I got y=15-x and tried to substitute that into the other equation and just made a mess!

Answer 17

me too :P
are you sure that the question is not wrong?

Answer 18

it seems that it's a quadratic equation

Answer 19

i'm not sure about the question!
how did you find this question?

Answer 20

It's the right equation :/

Answer 21

An example I was given

Answer 22

@Anecia23

Answer 23

@ganeshie8 , what's your idea about it?

Answer 24

@hartnn

Answer 25

yes, it will be quadratic
product = 12
sum =15
x^2 + 15x+12 =0

Answer 26

but it will be \(\large \sqrt{177}\)

Answer 27

eh, in terms of sqrt 177....
the dimensions are not integers....but irrational numbers..

Answer 28

so don't you think that the question is wrong?

Answer 29

if the question just belongs to quadratic equations, then no harm in dimensions being irrational...

Answer 30

ohh..ok

Answer 31

let the height be (x)
so base is (15-x)
area = 1/2*base*height
24=1/2 (x)(15-x)
x^2-15x+48=0

Answer 32

and this time,it will be \(\large \sqrt{33}\)

Answer 33

yeah, still irrational

Answer 34

@hartnn how do u get 12

Answer 35

its 48 only
my bad

Answer 36

if u dont want a quadratic,

\( x+y = 15\) -------(1)
\(\frac{2}{2} xy = 24\)

\(x-y = \sqrt{15^2-4 \times 48}\)

\(x-y = \sqrt{33}\) ---(2)

Answer 37

ohh///.....okk

Answer 38

effect of doing thing in head.... :P

Answer 39

xy should be 48....

Answer 40

hehe......no problem same with me.....i missedd half

Answer 41

guys it's not important :D
the important thing is "the question is irrational"

Answer 42

wats problem wid irrational lengths ?

Answer 43

so there is no problem in it..

Answer 44

my height is irrational its 174.7834232398123442223... cm

Answer 45

lol

Answer 46

XD

Answer 47

how do you have so perfect instrument to measure it ? :P

Answer 48

but @ganeshie8 usually , teachers give "rational questions"

Answer 49

lets ask @Anecia23 what she got ?

Answer 50

or when they give "irrational questions" they will say "you can't find the exact value of ... "

Answer 51

how do you have so perfect instrument to measure it ? :P

Answer 52

ive put myself vertically before a scanning electron miscroscope... that gave my precision of 1nm... after that did a simple extrapolation to 23 sigfigs lol

Answer 53

seems legit

Answer 54

u got great height........=P

Answer 55

why is there so many people on one question

Answer 56

i don't know too :D sometimes this will happen :D

Answer 57

because beautiful profile pics attract more helpers.. :P

Answer 58

this an good idea too XD

Answer 59

haha....

Answer 60

i came here for that reason only.....not because i was called :P

Answer 61

so i won't call you again :D

Answer 62

call me only when the profic pic is of gorgeous girl ;)

Answer 63

:D

Answer 64

lol

here is my final attempt to make the gorgeous girl happy :)

the complete solution

\( x+y = 15\) -------(1)
\(\frac{1}{2} xy = 24\)

\(x-y = \sqrt{15^2-4 \times 48}\)
\(x-y = \sqrt{33}\) ---(2)

(1) + (2)

\(2x = 15 + \sqrt{33}\)

\(\color{red}{x = \frac{15 + \sqrt{33}}{2}}\)


(1) - (2)

\(2y = 15 - \sqrt{33}\)

\(\color{red}{y = \frac{15 - \sqrt{33}}{2}}\)


if the question asks for exact value, then @PFEH.1999 , as u said we must lev the answer in radical form as above.
otherwise we can plugit in calculator and approximate

Answer 65

nice job :)

Answer 66

where's the gorgeous girl now?
does she even know whats going on here :P

Answer 67

@Anecia23 , got it?

Answer 68

so you can't find the exact value of x

Answer 69

and y

Answer 70

i would take a moment to welcome her here ...
\[ \begin{array}l\color{red}{\text{W}}\color{orange}{\text{E}}\color{#e6e600}{\text{L}}\color{green}{\text{C}}\color{blue}{\text{O}}\color{purple}{\text{M}}\color{purple}{\text{E}}\color{red}{\text{ }}\color{orange}{\text{t}}\color{#e6e600}{\text{o}}\color{green}{\text{ }}\color{blue}{\text{O}}\color{purple}{\text{P}}\color{purple}{\text{E}}\color{red}{\text{N}}\color{orange}{\text{S}}\color{#e6e600}{\text{T}}\color{green}{\text{U}}\color{blue}{\text{D}}\color{purple}{\text{Y}}\color{purple}{\text{!}}\color{red}{\text{!}}\color{orange}{\text{}}\end{array} \ddot \smile \]

Answer 71

Sorry guys I fell asleep! :o

Answer 72

she's here! she's here :D

Answer 73

I appreciate the help! Thank you so much to all of you!

Answer 74

You're welcome :)

Answer 75

i forgot to welcome her to openstudy too :D besides that i forgot my the welcome code... XD

Answer 76

Hehe! First time using this website! I was having problems with this question but it makes sense now! Thanks again to everybody how helped! :*

Answer 77

:);)

Answer 78

hey, if you have any doubts browsing this site, you can ask me :)
*us

Answer 79

or post it in this group:
http://openstudy.com/study#/groups/openstudy%20feedback
:)

Answer 80

I'll definitely be back soon! :D

Answer 81

yaaay :)

Answer 82

:)

Answer 83

^___^

Answer 84

Any question ???

Answer 85

oh! @E.ali did you see last comments?
there's not any question :)

Answer 86

hehehe !