Star - Part 1

Hmmm... All these solutions seem so complicated. Just mark off what the answer represents on the diagram, and you realize you have marked the perimeter of the star and pentagon. 50+30=80.

Star perimeter+pentagon perimeter

Perimeter of a pentagon( in the centre) is 30 units. Perimeter of pentagon is the sum of all sides of pentagon, That implies that length of each side is permineter divided by number of sides, Which is 30/5 =6. Perimeter of star is 50 units. The star consists of ten sides. (Two sides per arm) Each side therefore will be 5 units. AB will be equal to HA+GH+BG that is 5+6+5 units (16 units). There are in totality 5 such long sides ( like AB,BC,CD...). 16 times 5 gives 80 units and therefore the answer.

doesn't make sense. because Pentagon wont give out any perimeter regularly. you know Pentagon, everything classified

If HI is 6 ,AH and AI cannot be 5 and have line AHGB be a straight line. For that line to be straight, AH would be 9.708, unless I am missing something.

Star Perimeter Line Segments = (Star Perimeter / Total Line Segments) = (50 / 10) = 5

AI = IC = CJ = JE = EF = FB = BG = GD = DH = HA = 5

Pentagon Line Segments = (Pentagon Perimeter / Total Line Segments) = (30 / 5) = 6

HI = IJ = JF = FG = GH = 6

AB = (HA + GH + BG) = (5 + 6 + 5) = 16

BC = (FB + JF + CJ) = (5 + 6 + 5) = 16

CD = (IC + HI + DH) = (5 + 6 + 5) = 16

DE = (GD + FG + EF) = (5 + 6 + 5) = 16

EA = (JE + IJ + AI) = (5 + 6 + 5) = 16

(AB + BC + CD + DE + EA) = (16 + 16 + 16 + 16 + 16) = 80

(5+6+5)*5=80

Unfortunately I believe your answer is incorrect. AHGB is not a straight line and AB must be interpreted as the straight line from A to B and its length is 15.46 given the dimensions you have given. Therefore the total of AB+BC+CD+DE+EA = 5x15.46 = 77.31 (to 2dp). Using the cosine rule angle HAI = 73.74 degrees and given the angle at the point of a 5-pointed star is 36 degrees, AB = 2x5cos18.87 + 6 = 15.46

AB + BC + CD + DE + EA = (AH + HG + GB) + (BF + FJ + JC) + (CI + IH + HD) + (DG + GF + FE) + (EJ + JI + IA)

Rearranging the RHS:

AB + BC + CD + DE + EA = (AH + IA + CI + JC + EJ + FE + BF + GB + HD + DG) + (GF + FJ + JI + IH + HG)

AB + BC + CD + DE + EA = (Outside perimeter of the star) + (Perimeter of pentagon)

AB + BC + CD + DE + EA = 50 + 30 = 80

I know how to solve this problem, but I dont think the input is correct. AH = AI = 5 <HI = 6 however HAI is a simmetric triangel.p, AH & AI must be > HI

Length $AB$ covers 2 of the smaller triangle lengths and the side of the pentagon. $AB=BC=CD=DE=EA$ and we want to add all 5 so let’s just compute the $\text{pentagon perimeter} + \text{star perimeter} = 30+50=\boxed{80}$

They don't say if the inside polygon is regular (i.e. all sides are equal) or not. But we don't care about that. The sum AB + BC + CD + DE + EA will be the sum of the two perimeters. That is 50 + 30 = 80 !

Divide outside perimeter by 10 identical triangle sides, you get 5 for the blue segments. Then divide perimeter of pentagon by 5, you get 6. One big line is 2 blue + 1 green => 5 + 6 + 5 => 16.

16 x 5 = 80, done deal.

Easy, $50+30=\boxed{\large{80}}$ because $AB+BC+CD+DE+EA$ contains both shapes perimeter

Something is wrong here! The internal angles of a regular pentagon are 108 degrees. AGFJ is a quadrilateral with 3 angles in the pentagon. So the angle GAJ is 36 degrees. The triangle AHI has sides 5,5,6. You can't have the longest side (6) opposite the smallest angle (36).

AH+HD+DG+GB+BF+FE+EJ+JC+CI+IA=50 HG+GF+FJ+JI+IH=30

AH+GB+BF+JC+CI+HD+DG+FE+EJ+IA+ (50) +HG+FJ+IH+GF+IJ (30) = ? S=50+30=80

I would like to suggest that any problem that is presented must be meaningful. In the current problem, we apparently have a regular pentagon, each interior angle being 108 degrees, with a supplementary angle of 72 degrees, which is the base angle of any of the isosceles triangles forming an external blue star. Its vertex angle is 36 degrees with opposite side equal to 6. We are forced to have by the law of cosines: 6^2 = 5^2 + 5^2 - (2)(5)(5)cos(36), or 36 = 25 + 25 -50cos(36) = 50(1 - cos(36) = 50 (1 - .80901699) = 50 .19098301 = 9.5491505, a long way from 36. Ed Gray

The question is wrong!!! AB,BC,CD,DE,or EA can't be straight lines as the question suggests, just because for AB to be straight angle<AHI has to be 72 degrees(pentagon's interior angle is 108 deg). So the length of AH is already fixed and has to be greater than HI, which in this case(AH=5 and HI=6) is not true. Cheers!

While the correct answer to the question is 80, GIVEN the conditions, the conditions as stated are impossible.

There is no way to draw a pentagram on a plane such that the length of AH = 5 and length of HG = 6.

It is an exact ratio. If AH = 5 then HG will always be equal to approximately 3.09.

Just add the two figures together: 50+30=80. But note this won't produce a <perfect> star since the ratios along AB should be 1 : 0.618... : 1 (Golden Ratio) not 1 : 0.6 : 1

HG = 30/5 = 6

AH = 50/10 = 5

AB = BC = CD = DE = EA = (2*5) = 6 = 16

AB + BC + CD + DE + EA = 16+16+16+16+16 = 80

Given the data, 80 is the correct answer. However, if HA = 6 and HA < AI then AI cannot be 5. In fact, this triangle is known as the Golden triangle. The correct length of the lateral side is 6*phi = 9.71 and the perimeter of the star is 97.082. Given that the perimeter of the hexagon is 30, the correct answer to the question is 127.082. Phi, the Golden ratio is 1.618 ... which is the solution to phi-squared - phi - 1 = 0.

No need for calculation:

AB + BC + CD + DE +EA

= (AH + HG + GB) + (BF + FJ + JC) + (CI + IH + HD) + (DG + GF + FE) + (EJ + JI + IA)

= (AH + GB + BF + JC + CI + HD + DG + FE + EJ + IA) + (HG + FJ + IH + GF + JI)

= (Perimeter of Star) + (Perimeter of Pentagon)

= (50) + (30)

= $\boxed{80}$

Perimeter of Pentagon + Perimeter of the Star

will complete Each Segment (AB , BC ,CD , DE and EA)

The outside goes around ten sides and the pentagon has five sides; therefore:

50/10= 5

30/5=6

Each line segment passes through two of the outer lines and one of the inner line. This repeats five times since each line segment given is regular:

5(5+5+6)=x

5(16)=x

x=80

Each of the star's line segments is 5 long

(total perimeter is 50 and there are 10 star segments so 50/10 = 5)..

Each of the pentagon's is 6 long,

(total perimeter is 30 and there are 5 pentagon segments so 30/5 = 6)..

Since AB, BC, CD, DE and EA all contain 2 star segments and 1 pentagon segment, then

each of them is = 2(5) + 6 = 16

the sum of the five segments = 16*5 = [80]

If the outside perimeter is 50, then each of the external sides of the triangle has a length of 5 (50/10). Then, applying the same method, each side of the pentagon has a length of 6 (30/5). Finally, each segment length is the sume of two triangle sides and the pentagon side. Thus the total is eight times that result, which is: ((5x2)+6)x5)=80

50+30= 80 All the side are coming in calculation atleast once. And the perimeter is sum of sides.

Perimeter of a pentagon( in the centre) is 30 units. Perimeter of pentagon is the sum of all sides of pentagon, That implies that length of each side is permineter divided by number of sides, Which is 30/5 =6. Perimeter of star is 50 units. The star consists of ten sides. (Two sides per arm) Each side therefore will be 5 units. AB will be equal to HA+GH+BG that is 5+6+5 units (16 units). There are in totality 5 such long sides ( like AB,BC,CD...). 16 times 5 gives 80 units and therefore the answer.

50+30= 80 All the side are coming in calculation atleast once. And the perimeter is sum of sides.

Inside perimeter+outside perimeter, 50+30. AB, BC, CD and DE lines cross all the points in the outside perimeter and the intern perimeter. For example, AB=AH+HG+GB, the you just have to plus the perimeters, 50+30=80. If I can not make you understand, draw the outside and the inside apart and write each perimeter value, then overlap the two drawings and you will notice you get the complete star. Have a nice math day :).

It is basically the sum of the sides of thestar amd pentagon so 50+30= 80

If perimeter of the regular pentagon is 30 units, then each side of the pentagon will be 6 units. Assuming the triangles formed at the ends of the star are isosceles, the equal sides of the triangle will be equal to 5 units as the perimeter is 50 units and there are ten line segments bordering the star. Then we can simply add the values as per given

Perimeter of star is 50 so length of one arm will be 5, and for hexagon it will be 6, so total length will be for ab is 5+5+6 =16, total length 16*5=80, or simply requirement is completely covering inner and outer perimeter so it will be total = 50+30= 80

Stat perimeter = 50= 10 sections of 5

Pentagon Perimeter =30= 5 sections of 6

AB=BC=CD=DE=EA= 16

16x5=80

Perimeter of star + perimeter of pentagon

As we know that :- AH+DH+DG+GB+BF+FE+EJ+JC+CI+IA=50----------(Eqn. 1)

AS WELL AS :- HI+IJ+JF+FG+GH=30---------(Eqn 2)

ADDING 1 AND 2, WE GET,

(AH+HG+GB)+(BF+FJ+JC)+(DH+HI+IC)+(DG+GF+FE)+(AI+IJ+JE)=80

THEREFORE AB+BC+CD+DE+EA=80

Calculate each side of star 10/2=5 Pentagon 30/5=6 AB =5+5+6=16 16*5=80