What is the perimeter of the smallest triangle that can be formed if all the sides are even integers and no two sides are congruent?
Answers
Answer:
REF.Image
Let ABC be isosceles is AB=AC
r is radius of circle
AF
2
+BF
2
=AB
2
⇒(3r)
2
+(y
2
)=x
2
________ (1)
From ΔAOD
(2r)
2
=r
2
+(AD)
2
=3r
2
=AD
2
=AD=
3
r
Now BD=BF & EC=FC (as tangents are equal from an external point)
AD+DB=x
=
3
r+y
2
=x
⇒y
2
=x=
3
r -(2)
From (1) & (2)
∴(3r)
2
+(x−
3
r)=x
2
9r
2
+x
2
+3r
2
−2
3
rx=x
2
=9r
2
+3r
2
−2
3
rx=0
=12r
2
=2
3
rx
=6r=
3
x
∴x=
3
6
r
From (2)
y/2=(6/
3
)r−
3
r
=y/2=(3
3
/3)r⇒y=2
3
r
perimeter = 2x+y=2(6/
3
)r+2
3
r
=(12r+6r)/
3
=(18/
3
)r
=6
3
r
Hence proved.
Answer:
Let the shortest side be x.
If the sides are consecutive odd integers, the other two sides will be (x+2)and(x+4)
The perimeter is the sum of the sides.
x+x+2+x+4=51
3x+6=51
3x=51−6
3x=45
x=15
The length of the shortest side.
The other sides are 17cmand19cm
Step-by-step explanation:
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