Question

two cylindrical jars contain same amount of milk. If their diameters are in the ratio 3:4, find the ratio of their heights

Answer 1

A.T.Q.
V1 = V2
π(r1)^2h¹ = π(r2)^2h2
[(r1)^2]/[(r2)^2] = h2/h1
[r1/r2]^2 = h2/h1
( 3/4 )^2 = h2/h1
9/16 =h2/h1

Answer 2

Answer:

$(\boldsymbol{b}-\boldsymbol{c}) \times \boldsymbol{x}+(\boldsymbol{c}-\boldsymbol{a}) \times \boldsymbol{y}+(\boldsymbol{a}-\boldsymbol{b}) \times \boldsymbol{z}=0$

Solution:

Let us assume,

$\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=1$

We get,

$\frac{x}{b+c-a}=1$

$x=(b+c-a)$

$\frac{y}{c+a-b}=1$

$y=(c+a-b)$

$\frac{z}{a+b-c}=1$

$z=(a+b-c)$

To find: $(b-c)x+(c-a)y+(a-b)z$

$(b-c) \times x+(c-a) \times y+(a-b) \times z$

Substituting the values of x, y and z,

$(b-c)(b+c-a)+(c-a)(c+a-b)+(a-b)(a+b-c)$

On multiplying the terms we get,

$(b-c)(b+c-a)=b^{2}+b c-a b-b c-c^{2}+a c$

$(b-c)(b+c-a)=b^{2}+b c-a b-b c-c^{2}+a c$

$(a-b)(a+b-c)=a^{2}+a b-a c-a b-b^{2}+b c$

Equating all the terms we get,

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

$=b^{2}-c^{2}-a b+a c+c^{2}-a^{2}-b c+a b+a^{2}-b^{2}-a c+b c=0$