+0  
 
0
89
6
avatar+91 

1) In triangle ABC, [ADE] = 4, [ABE] = 14, and [CDE] = 6. Compute [BCE].

2) In the diagram, WY = 12, XZ = 10, [AWX] = 15, and [AYZ] = 8. Find [AXY]

\(\text{In the diagram below, we have } \overline{AB}\parallel\overline{CD}, EF = FG, \angle AEG = x^\circ, \text{and } \angle BEF = 102^\circ + x^\circ. \text{Find the value of x.}\)

 

In triangle ABC, AB = AC. Find

 Oct 1, 2022
 #1
avatar+1005 
+2

1)

ADE and CDE share the same height, but different bases, given the ratio of areas being 4 : 6 (ADE : CDE), we can conclude that the ratio of AD:DC is 2:3

Also, the area of triangle ABD is 14 + 4 = 18 units squared. 

The triangles ABD and BCD share the same height, so the ratio of their areas is the ratio of AD:DC, which is 2:3, then we can conclude that triangle BCD has area of 27 units squared. Subtracting 6, the area of triangle BCE is 21 units squared.

 Oct 2, 2022
 #2
avatar+1005 
+2

2)

Area of triangle AXY now has area "x".

Since all the triangles in the image share the same height (the perpendicular from apex A to WZ), then the ratio of the triangles' areas are the ratio of their bases. WY/XZ = 12/10, so the areas of triangles AWY/AXZ = WY/XZ = 12/10. 

Substituting in, we have 15 + x / 8 + x = 12/10 

150 + 10x = 96 + 12x

54 = 2x

x = 27. Triangle AXY has area of 27 units squared.

 Oct 2, 2022
 #3
avatar+1005 
+2

3) In the diagram below, we have.....

Since AB || CD, by alternate interior angles, angle AEG = angle EGF = x*.

Also by alternate interior angles, angle BEF = angle EFG = 102 + x*.

Also EF = FG so the triangle within the two parallel lines is isosceles, by Isosceles Triangle Theorem, angle GEF = angle EGF = x.

The interior angle sum of a triangle is 180 degrees, so 102 + x + x + x = 180. 3x = 78, x = 26.

 Oct 2, 2022
 #4
avatar+1005 
0

4) You did not fully fill out the question --- please re-post/edit the bottom question since it got cut off after the "Find".

 Oct 2, 2022
 #5
avatar+91 
0

It was find bac im sorry (angle bac)

cosign  Oct 2, 2022
 #6
avatar+1005 
+2

4) If angle DCB is x, then angle BDC is also x.

Angle DBC = 180 - 2x

Angle DBE = x - (180 - 2x) = 3x - 180 degrees

Angle EDB = 3x - 180

Angle BED = 180 - 2(3x - 180) = 180 - 6x + 360 = 540 - 6x degrees

Angle AED = 180 - (540 - 6x) = 6x - 360 degrees

Angle A = 6x - 360 degrees.

Angle B + Angle C + Angle A = 180 

x + x + 6x - 360 = 180

8x = 540

Angle BAC = 67.5 degrees

 Oct 5, 2022

29 Online Users

avatar