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In trapezoid EFGH, \overline{EF} \parallel \overline{GH}, and P is the point on \overline{EH} such that EP:PH = 1:2. If the area of triangle PEF is 6, and the area of triangle PGH is 6, then find the area of trapezoid EFGH.

 
 Oct 29, 2024, 9:59:07 AM
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Let's analyze the problem step by step:

 

1. Understanding the Given Information:

 

EFGH is a trapezoid with EF parallel to GH.

 

P is a point on EH such that EP:PH = 1:2.

 

Area of triangle PEF = 6.

 

Area of triangle PGH = 6.

 

2. Visualizing the Situation:

 

trapezoid EFGH with point P dividing EH in a 1:2 ratio

 

3. Using the Ratio of Areas:

 

Since triangles PEF and PGH share the same height (the perpendicular distance between EF and GH), the ratio of their areas is equal to the ratio of their bases.

 

So, (Area of triangle PEF) / (Area of triangle PGH) = EP/PH = 1/2.

 

We know the area of both triangles, so we can confirm this ratio: 6/6 = 1/2.

 

4. Finding the Area of Triangle EGH:

 

Since EP:PH = 1:2, we can say that EH = 3 * EP.

 

Triangles EGH and PEF share the same height, so the ratio of their areas is equal to the ratio of their bases.

 

(Area of triangle EGH) / (Area of triangle PEF) = EH/EP = 3/1.

 

Therefore, the area of triangle EGH = 3 * (Area of triangle PEF) = 3 * 6 = 18.

 

5. Finding the Area of Trapezoid EFGH:

 

The area of trapezoid EFGH is the sum of the areas of triangles PEF and EGH.

 

Area of trapezoid EFGH = Area of triangle PEF + Area of triangle EGH = 6 + 18 = 24.

 

Therefore, the area of trapezoid EFGH is 24.

 11 hours ago

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