Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Friday, March 11, 2016

### Geometry Problem 1198: Right Triangle, Square, Rectangle, Quadrilateral, Altitude, Equal Areas

Labels:
altitude,
area,
quadrilateral,
rectangle,
right triangle,
square

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Observe that triangle AEF congruent to tri. ABC ….( case ASA)

ReplyDeleteSo AF= AC= AG

So S(AFDC)= S(ABDE)= AB^2

We have S(AHKG)=AG.AH=AC.AH= AB^2.. ( relation in right triangle ABC)

So S(AFDC)=S(AHKG)

S(AHKG)

ReplyDelete= 2×S(AHG)

= 2×S(ABG)

= 2×S(AEC)

= 2×S(AEB)

= S(AEDB)

= S(AFDB) + S(AEF)

= S(AFDB) + S(ABC)

= S(AFDC)

Since <EAF = < BAC, triangles AEF and ABC are congruent ASA and so AC = AF

ReplyDeleteHence S(AFDC) = S(ABDE) = AB^2 = AH.AC since triangles ABH and ABC are similar

Now AH.AC = AH.AG (since AC = AF = AG) = S(AHKG)

So S(AFDC)= S(AHKG)

Sumith Peiris

Moratuwa

Sri Lanka

Angles FAE and CAB are equal, as angles with orthogonal hands. So, the triangles AEF and ABC are congruent; both have the same area. Thus, area of the square ABDE is equal to area ABDF plus area ABC, that is area(ACBDF) = area(ABDE) = c^2, where the length of side AB = c. On the other side, area(AGKH) = bp, where p = length of AH.

ReplyDeleteHowever, in the right triangle ABC is valid c^2 = pb. It follows from the similar triangles ABC and AHB and c:p = b:c, that is c^2 = bp. Accordingly, area(AGKH) = area(ACDF).