Difference between revisions of "1978 AHSME Problems/Problem 23"

Revision as of 11:48, 19 March 2023

Problem

Vertex $E$ of equilateral $\triangle ABE$ is in the interior of square $ABCD$ , and $F$ is the point of intersection of diagonal $BD$ and line segment $AE$ . If length $AB$ is $\sqrt{1+\sqrt{3}}$ then the area of $\triangle ABF$ is

$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$

Solution

C

Revision as of 16:41, 18 June 2021 (view source) Aopspandy (talk \| contribs) (→‎Problem) ← Older edit		Revision as of 11:48, 19 March 2023 (view source) Techno23 (talk \| contribs) m (→‎Solution) Newer edit →
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	==Solution==		==Solution==
−	~~No solutions yet!~~	+	C

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Difference between revisions of "1978 AHSME Problems/Problem 23"

Revision as of 11:48, 19 March 2023

Problem

Solution