Prove that "The area of the equilateral triangle described on a side of a square is half of the
area of the equilateral triangle described on its diagonal."
Answers
Answer:
Explanation:
Given:
ABCDABCD is a Square,
DBDB is a diagonal of square,
\triangle DEB△DEB and \triangle CBF△CBF are Equilateral Triangles.
To Prove:
\dfrac{A(\triangle CBF)}{A(\triangle DEB)}=\dfrac{1}{2}
A(△DEB)
A(△CBF)
=
2
1
Proof:
Since, \triangle DEB△DEB and \triangle CBF△CBF are Equilateral Triangles.
\therefore∴ Their corresponding sides are in equal ratios.
In a Square ABCD, DB=BC\sqrt2DB=BC
2
.....(1).....(1)
\therefore∴ \dfrac{A(\triangle CBF)}{A(\triangle DEB)}=
A(△DEB)
A(△CBF)
=\dfrac{\dfrac{\sqrt3}{4}\times (BC)^2}{\dfrac{\sqrt3}{4}\times (DB)^2}
4
3
×(DB)
2
4
3
×(BC)
2
\therefore∴ \dfrac{A(\triangle CBF)}{A(\triangle DEB)}=
A(△DEB)
A(△CBF)
=\dfrac{\dfrac{\sqrt3}{4}\times (BC)^2}{\dfrac{\sqrt3}{4}\times (BC\sqrt2)^2}
4
3
×(BC
2
)
2
4
3
×(BC)
2
(From 1)
\therefore∴ \dfrac{A(\triangle CBF)}{A(\triangle DEB)}=\dfrac{1}{2}
A(△DEB)
A(△CBF)
= 1/2
plz refer to this attachment......□□
