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# Perimeter

+9
608
17
+841

?

Jul 9, 2014

#12
+97586
+16

I think the answer is 15*sqrt3+15

Now let me try and explain how I got this.

< KJA = 60°     Ext < of a triangle is equal sum of opposite interior angles

triangle JKF is isosceles (2 equal angles) therefore  KJ=JF=10

consider triangle AJK

Sin30=1/2=AJ/10   so AJ=5

Using Pythagoras' Theorum

$$AK^2=10^2-5^2 AK^2=100-25 AK^2=75 AK=5\sqrt3$$

AF=5+10=15

$$\\KF^2=15^2+(5\sqrt3)^2\\\\ KF^2=225+75\\\\ KF^2=300\\\\ KF=10\sqrt3\\\\$$

$$\\Perimeter = 5\sqrt3+10\sqrt3+15\\\\ Perimeter = 15\sqrt3+15\;\; units\\\\$$

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Jul 10, 2014

#1
+8263
0

A triangle has 180 degrees, right. So you add 30+30=60. So, the perimeter is 180 degrees.

Jul 9, 2014
#2
+841
+6

Perimiter is the lengths of the sides not the angles

Jul 9, 2014
#3
+146
+3

well we know that the length of KJ is 10 heres a start

Jul 9, 2014
#4
+841
+6

Here is the hint -

Jul 9, 2014
#5
+8263
0

Well, I don't know how to find the perimeter of a triangle using degrees.

Jul 9, 2014
#6
+841
+6

Converse of the base angle theorem - If two angles of a triangle are congruent, then the sides opposite those angles are congruent

Jul 9, 2014
#7
+8263
0

What grade level is this?

Jul 9, 2014
#8
+841
+6

Geometry - 10th grade

Jul 9, 2014
#9
+146
+8

does mit give you options for the correct answer? beacuse i believe i found it

Jul 9, 2014
#10
+841
+6

I have two chances to get it right so yes :)

Jul 9, 2014
#11
+146
+8

sorry i took so long ive been away but i believe the answer is 41.03 only because if you look at the formula or a diagram for a 30,60,90 triangle the smaller leg is X the second leg is x square root3 and the hypotenuse is 2x using these if you fine the perimeter for KAJ you can also use the same formulas on the larger triangle that you need hope im right lol

Jul 10, 2014
#12
+97586
+16

I think the answer is 15*sqrt3+15

Now let me try and explain how I got this.

< KJA = 60°     Ext < of a triangle is equal sum of opposite interior angles

triangle JKF is isosceles (2 equal angles) therefore  KJ=JF=10

consider triangle AJK

Sin30=1/2=AJ/10   so AJ=5

Using Pythagoras' Theorum

$$AK^2=10^2-5^2 AK^2=100-25 AK^2=75 AK=5\sqrt3$$

AF=5+10=15

$$\\KF^2=15^2+(5\sqrt3)^2\\\\ KF^2=225+75\\\\ KF^2=300\\\\ KF=10\sqrt3\\\\$$

$$\\Perimeter = 5\sqrt3+10\sqrt3+15\\\\ Perimeter = 15\sqrt3+15\;\; units\\\\$$

Melody Jul 10, 2014
#13
+841
+8

Thanks everybody!  Melody your amazing <3

Jul 10, 2014
#14
+96302
+13

JF = KJ  = 10     Sides opposite equal angles are themselves equal

Therefore, m<KJF = 120

m<KJF = 120   Therefore m< KJA = 60   (The sum of any two angles comprising a strainght angle = 180)

Therefore, triangle KAJ  is 30 - 60 - 90

And AJ = 5     And KA = 5√3

Then, by the Pytagorean theorem,  KF = √[ (AJ + JF)^2 + (KA)^2] = √[ (5 + 10)^2 + (5√3)^2] =

√[ (15)^2 + (5√3)^2] =  √[ 15^2 + 75] =  √[ 225 + 75]  = √[ 300] = 10√3

So....the perimeter =  AJ + JF + KF + KA = 5 + 10 + 10√3  + 5√3 =

15 + 15√3

Jul 10, 2014
#15
+97586
+11

Thanks Rose98.

You are pretty COOL too!

Jul 11, 2014
#16
+394
+5

Yes she is.

Rose98 is very polite too.

This is like finding a lily in a Russian snow storm.

7up

Jul 11, 2014
#17
+841
0

That just made my day :D thanks SevenUP and Melody :)

Jul 11, 2014