## Saturday, May 13, 2023

### Geometry Problem 1538: Solve for the Area of a Quadrilateral Using External Squares and a Segment Length

Geometry Problem 1538. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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1. Trigonometry Solution

S(ACGD) = C^2 /2 + a^2 /2 + (ac/2)SinB + (ac/2)Sin (180 - B)
S(ACGD) = (a^2 + c^2)/2 + (ac/2)sinB........(1)

EF^2 = 2c^2 + 2a^2 - 2(V2a)(V2c)cos(90+B)
EF^2 = 4{(a^2 + c^2) + (ac/2)sinB}......(2)

From (1) & (2),
S(ACGD) = EF^2 / 4 = 10^2 / 4 = 25

Sumith Peiris
Moratuwa
Sri Lanka

1. Some clarifications and correcting of typos

S(AGCD) is the sum of areas of Triangles ABD, DGB, CBG & ABC
Equation (1) should be corrected to read as
S(ACGD) = (a^2 + c^2)/2 + ac.sinB......(1)

The length of EF is obtained by applying the Cosine Rule to Triangle EBF
Equation (2) should be corrected to read as
4{(a^2 + c^2)/2 + ac.sinB}................(2)

RHS of (2) is 4 times RHS of (1)

2. A slightly different version of Stan's excellent geometry proof.

Let AG,CD meet at P
Triangles ABG & CBD are congruent SAS, so AG = CD......(1)

Also since < AGB = < BCD, BPCG is concyclic and so EF^2 =2. AG^2 = 2. 2. S(ACGD) from (1)
So S(AGCD) = EF^2 / 4 = 10^2 / 4 = 25

Sumith Peiris
Moratuwa
Sri Lanka