Tillbaka
A — 1020
1 p
Givet triangeln
A
B
C
ABC
A
B
C
med sidlängder
∣
A
B
∣
=
c
|AB| = c
∣
A
B
∣
=
c
,
∣
B
C
∣
=
a
|BC| = a
∣
B
C
∣
=
a
,
∣
C
A
∣
=
b
|CA| = b
∣
C
A
∣
=
b
, låt
h
c
h_c
h
c
vara längden av höjden mot sidan
A
B
AB
A
B
. Då gäller
A
h
c
=
(
2
a
+
2
b
−
c
)
(
2
a
−
2
b
+
c
)
(
2
b
+
2
c
−
a
)
(
a
+
b
+
c
)
2
c
h_c = \dfrac{\sqrt{(2a+2b-c)(2a-2b+c)(2b+2c-a)(a+b+c)}}{2c}
h
c
=
2
c
(
2
a
+
2
b
−
c
)
(
2
a
−
2
b
+
c
)
(
2
b
+
2
c
−
a
)
(
a
+
b
+
c
)
B
h
c
=
a
b
c
(
a
+
b
+
c
)
2
c
h_c = \dfrac{\sqrt{abc\,(a+b+c)}}{2c}
h
c
=
2
c
ab
c
(
a
+
b
+
c
)
C
h
c
=
(
a
+
b
+
c
)
(
a
+
b
−
c
)
(
a
−
b
+
c
)
(
b
+
c
−
a
)
2
c
h_c = \dfrac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(b+c-a)}}{2c}
h
c
=
2
c
(
a
+
b
+
c
)
(
a
+
b
−
c
)
(
a
−
b
+
c
)
(
b
+
c
−
a
)
D
h
c
=
a
b
(
a
+
b
+
c
)
2
c
h_c = \dfrac{\sqrt{ab\,(a+b+c)}}{2c}
h
c
=
2
c
ab
(
a
+
b
+
c
)
Svara