The length of the line segment AB is 40. The point C is the midpoint of the line segment AB. The triangles ACD and CBG are isosceles. The length of the height of the triangle on the side AC from the point D is 15, and the length of the height of the triangle on the side CB from the point G is 11. The perimeter of the triangle GDC is:
Answers
Step-by-step explanation:
11+15+40
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The perimeter of ΔGDC ≅ 53 units
Given: AB = 40, AC = 20 = CB,
To Find: The perimeter of triangle GDC
Solution:
1. Since ΔACD and ΔCBG are isosceles therefore their respective altitudes are perpendicular bisectors of the sides AC and CB.
2. Let's call the points where the perpendicular bisectors meet to be M and N respectively. Thus according to the information given DM = 15 units and GN = 11 units
3. Another thing to note is that, MN = DG = 20 units
Now coming to the numerical part,
perimeter of ΔGDC = DC + CG + DG ..... (1)
Using pythagorus theorem,
DC = √( 15^2 + 10^2 ) = √325 = 18.02 ≅ 18 units
CG = √( 11^2 + 10^2 ) = √221 = 14.86 units
DG = 20 units
Putting the respective values in the (1) , we get
perimeter of ΔGDC = DC + CG + DG
= 18 + 14.86 + 20
= 52.86
≅ 53 units
∴ the perimeter of ΔGDC ≅ 53 units
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