Let ABCD be a square having each side 4 cm. Then find the perimeter of the isosceles triangle ABD.
Answers
Answer:
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Answer:
⇒ Side of a square, ABCD=4cm
It is given that △CED is an equilateral triangle.
∴ EC=CD=DE=4cm
⇒ ∠ECD=60
o
. [ An angle of an equilateral triangle ]
AC is a diagonal of a square ABCD
∴ ∠ACD=45
o
.
⇒ ∠ECA=∠ECD−∠ACD
⇒ ∠ECA=60
o
−45
o
⇒ ∠ECA=15
o
In △ACE, draw perpendicular EM the base AC.
Now in △EMC,
⇒ sin15
o
=
H
P
=
EC
EM
⇒ sin15
o
=
4
EM
⇒
2
2
(
3
−1)
=
4
EM
⇒ (
3
−1)×
(2
2
×
2
)
2
=
4
EM
[ By rationalizing the denominator ]
⇒
4
2
(
3
−1)
=
4
EM
⇒ EM=
2
(
3
−1)
Diagonal of a square, (AC)=
2
a
AC=
2
×4=4
2
Diagonal of a square =4
2
cm
Now, in △AEC,
⇒ Area of △AEC=
2
1
×AC×EM
=
2
1
×4
2
×
2
(
3
−1)
=4(
3
−1)cm
2