nilkhanth ki mratyu kese hoi hogi? anuman lagakar likhe
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Answers
Answer:
Relation - Cube
Let me write the complete question for a better understanding. There is something missing in the question.
Complete Question:
Area of total surface of a cube is s square units and length of diagonal is d units, then relation between s and d will be:
a. \: 2d {}^{2} = sa.2d
2
=s
b. \: d {}^{2} = sb.d
2
=s
c. \: 2s {}^{2} = dc.2s
2
=d
d. \: s { }^{2} = dd.s
2
=d
We are given that, total surface area of a cube is s square units and length of diagonal is s units. With this information, we are asked to find out the relation between s and d.
Let's consider aa units be the edge of side of cube.
We know that, the total surface are of cube is, s = 6a². Therefore,
\begin{gathered}\implies s = 6 {a}^{2} \\ \\ \implies {a}^{2} = \frac{s}{6} \\ \\ \implies a = \sqrt{ \frac{s}{6} }\end{gathered}
⟹s=6a
2
⟹a
2
=
6
s
⟹a=
6
s
We know that, the diagonal of cube is, d = √3a. Therefore,
\begin{gathered}\implies d = \sqrt{3}a \\ \\ \implies a = \dfrac{d}{ \sqrt{3}}\end{gathered}
⟹d=
3
a
⟹a=
3
d
The relation between ss and dd are;
\implies \sqrt{\dfrac{s}{6}} = \dfrac{d}{ \sqrt{3} }⟹
6
s
=
3
d
Now, On squaring both sides, we get:
\begin{gathered}\implies \bigg( \sqrt{\dfrac{s}{6}}\bigg)^2 = \bigg(\dfrac{d}{ \sqrt{3} }\bigg)^2 \\ \\ \implies \dfrac{s}{ \cancel{ \: 6 \: }} = \dfrac{ {d}^{2} }{ \cancel{ \: 3 \: }} \\ \\ \implies \dfrac{s}{2} = {d}^{2}\\ \\ \implies \boxed{\bf{s = 2 {d}^{2}}}\end{gathered}
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