Ratio and Proportion - If x/(y+z)=y/(z+x)=z/(x+y) then prove the value of each ratio is 1/2 or -1 calculator

SolutionHelp

1. If $a:b:c=2:3:5$ then find value of $(a^2+b^2+c^2)/(ab+bc+ca)$

1. If

$a/b=2/3$ , then find

$(2a-3b)/(a-2b)$

2. If

$a:2=b:3=c:5$ , then find

$(a^2+b^2+c^2)/(ab+bc+ca)$

3. If

$a/4=b/15$ , then find

$(a+b)/(a-b)$

4. If

$(2a+3b)/(a+b)=13/5$ , then find

$a/b$

5. If

$(x^2+1)/(x^2-1)=2$ , then find

$x$

6. If

$x:30=12:20$ , then find

$x$

2. If $a:b=2:3,b:c=4:5$ then find $a:b:c$

1. If

$a:b=2:3,b:c=4:5$ , then find

$a:b:c$

2. If

$b:c=4:5,c:d=5:6$ , then find

$b:c:d$

3. If

$a:b=2:3,b:c=4:5,c:d=5:6$ , then find

$a:b:c:d$

4. If

$2a=3b$ , then find

$a:b$

5. If

$2a=3b=7c$ , then find

$a:b:c$

3. If $a/b=c/d=e/f$ then prove that $(2a+3c-4e)/(2b+3d-4f)=(5a-4c+3e)/(5b-4d+3f)$

1. If

$a/b=c/d=e/f$ , then prove that

$(2a+3c-4e)/(2b+3d-4f)=(5a-4c+3e)/(5b-4d+3f)$

2. If

$x/(b^2-c^2)=y/(c^2-a^2)=z/(a^2-b^2)$ , then prove that

$x+y+z=0$

4. If $x/(y+z)=y/(z+x)=z/(x+y)$ then prove the value of each ratio is $1/2$ or $-1$

1. If

$x/(y+z)=y/(z+x)=z/(x+y)$ , then prove each ratio =

$1/2,-1$

2. If

$(5a+6b)/(7c)=(6b+7c)/(5a)=(7c+5a)/(6b)$ , then prove each ratio =

$2,-1$

3. If

$(ax+by)/(x+y)=(ay+bz)/(y+z)=(az+bx)/(z+x)$ , then prove each ratio =

$(a+b)/2$

5. Geometric Mean

1. Geometric Mean of

$49/4$ and

$16/9$

2. Geometric Mean of

$8a^2$ and

$32b^2$

3. Geometric Mean of

$(a-b)^2$ and

$(a+b)^2$

4. If geometric mean

$6x^2$ and x is

$9x$ , find value of x

6. Ratios(duplicate, triplicate) and proportional(mean, third, fourth)

1. Duplicate ratio of 2:3

2. Triplicate ratio of 2:3

3. Sub-Duplicate ratio of 16:25

4. Sub-Triplicate ratio of 8:27

5. Compounded ratio of 2:5,3:4

6. Compounded ratio of 2:3,3:4,4:5

7. Mean proportional of 8,32

8. Third proportional of 16,36

9. Fourth proportional of 3,7,15

10. Compare ratios of 2:3,3:4,4:5

Example

1. If $x/(y+z)=y/(z+x)=z/(x+y)$ then prove the value of each ratio is $1/2,-1$

Solution:
Here

$x/(y+z)=y/(z+x)=z/(x+y)$

Case-1 : If

$x+y+z!=0$ , then

Each ratio

$=(x+y+z)/(y+z+z+x+x+y)$

$=(x+y+z)/(2y+2z+2x)$

$=(x+y+z)/(2(y+z+x))$

Cancel the common factor

$(x+y+z)$

$=(1)/(2)$

Case-2 : If

$x+y+z=0$ , then

$y+z=-x$

Then, the first ratio

$=(x)/(y+z)$

$=(x)/(-x)$

Cancel the common factor

$-x$

$=-1$

Hence, each ratio

$=-1$ .

Thus, the value of each ratio is

$(1)/(2)$ or

$-1$ .