Prvo rješenje.
Neka je ∠ E D F = φ \angle EDF=\varphi ∠ E D F = φ D E DE D E A B AB A B C D CD C D ∠ F D C = 90 ∘ − φ \angle FDC=90^\circ-\varphi ∠ F D C = 9 0 ∘ − φ
Tada je ∠ D C B = ∠ B A D = φ \angle DCB=\angle BAD=\varphi ∠ D C B = ∠ B A D = φ
Iz trokuta D F C DFC D F C
cos φ = ∣ C F ∣ ∣ C D ∣ = 1 3 , \cos\varphi=\frac{|CF|}{|CD|}=\frac13, cos φ = ∣ C D ∣ ∣ C F ∣ = 3 1 , odnosno
∣ C F ∣ = 1 3 ∣ C D ∣ , |CF|=\frac13|CD|, ∣ C F ∣ = 3 1 ∣ C D ∣ , a primjenom Pitagorinog poučka je
32 2 + ( 1 3 ∣ C D ∣ ) 2 = ∣ C D ∣ 2 , 32^2+\left(\frac13|CD|\right)^2=|CD|^2, 3 2 2 + ( 3 1 ∣ C D ∣ ) 2 = ∣ C D ∣ 2 , 8 9 ∣ C D ∣ 2 = 32 2 , \frac89|CD|^2=32^2, 9 8 ∣ C D ∣ 2 = 3 2 2 , odnosno
∣ C D ∣ = 24 2 . |CD|=24\sqrt2. ∣ C D ∣ = 24 2 . Analogno, iz trokuta D E A DEA D E A
cos φ = ∣ A E ∣ ∣ A D ∣ = 1 3 , \cos\varphi=\frac{|AE|}{|AD|}=\frac13, cos φ = ∣ A D ∣ ∣ A E ∣ = 3 1 , odnosno
∣ A E ∣ = 1 3 ∣ A D ∣ , |AE|=\frac13|AD|, ∣ A E ∣ = 3 1 ∣ A D ∣ , 20 2 + ( 1 3 ∣ A D ∣ ) 2 = ∣ A D ∣ 2 , 20^2+\left(\frac13|AD|\right)^2=|AD|^2, 2 0 2 + ( 3 1 ∣ A D ∣ ) 2 = ∣ A D ∣ 2 , 8 9 ∣ A D ∣ 2 = 20 2 , \frac89|AD|^2=20^2, 9 8 ∣ A D ∣ 2 = 2 0 2 , odnosno
∣ A D ∣ = 15 2 . |AD|=15\sqrt2. ∣ A D ∣ = 15 2 . Slijedi:
∣ C F ∣ = 1 3 ∣ C D ∣ = 8 2 , |CF|=\frac13|CD|=8\sqrt2, ∣ C F ∣ = 3 1 ∣ C D ∣ = 8 2 , ∣ B F ∣ = 15 2 − 8 2 = 7 2 , |BF|=15\sqrt2-8\sqrt2=7\sqrt2, ∣ B F ∣ = 15 2 − 8 2 = 7 2 , ∣ A E ∣ = 1 3 ∣ A D ∣ = 5 2 , |AE|=\frac13|AD|=5\sqrt2, ∣ A E ∣ = 3 1 ∣ A D ∣ = 5 2 , ∣ B E ∣ = 24 2 − 5 2 = 19 2 . |BE|=24\sqrt2-5\sqrt2=19\sqrt2. ∣ B E ∣ = 24 2 − 5 2 = 19 2 . Traženu površinu četverokuta D E B F DEBF D E B F D E B DEB D E B B F D BFD B F D
P = ∣ D E ∣ ⋅ ∣ B E ∣ 2 + ∣ D F ∣ ⋅ ∣ B F ∣ 2 P=\frac{|DE|\cdot|BE|}{2}+\frac{|DF|\cdot|BF|}{2} P = 2 ∣ D E ∣ ⋅ ∣ B E ∣ + 2 ∣ D F ∣ ⋅ ∣ B F ∣ = 20 ⋅ 19 2 2 + 32 ⋅ 7 2 2 = 302 2 . =\frac{20\cdot19\sqrt2}{2}+\frac{32\cdot7\sqrt2}{2}=302\sqrt2. = 2 20 ⋅ 19 2 + 2 32 ⋅ 7 2 = 302 2 . Nadalje, u četverokutu D E B F DEBF D E B F ∠ F B E = 180 ∘ − φ \angle FBE=180^\circ-\varphi ∠ F B E = 18 0 ∘ − φ
P D E B F = P E F D + P E B F = ∣ D E ∣ ⋅ ∣ D F ∣ 2 sin φ + ∣ E B ∣ ⋅ ∣ B F ∣ 2 sin ( 180 ∘ − φ ) P_{DEBF}=P_{EFD}+P_{EBF}=\frac{|DE|\cdot|DF|}{2}\sin\varphi+\frac{|EB|\cdot|BF|}{2}\sin(180^\circ-\varphi) P D E B F = P E F D + P E B F = 2 ∣ D E ∣ ⋅ ∣ D F ∣ sin φ + 2 ∣ E B ∣ ⋅ ∣ B F ∣ sin ( 18 0 ∘ − φ ) = 20 ⋅ 32 2 sin φ + 19 2 ⋅ 7 2 2 sin φ = 302 2 . =\frac{20\cdot32}{2}\sin\varphi+\frac{19\sqrt2\cdot7\sqrt2}{2}\sin\varphi=302\sqrt2. = 2 20 ⋅ 32 sin φ + 2 19 2 ⋅ 7 2 sin φ = 302 2 . Drugo rješenje.
Slika je ista kao u prvom rješenju.
Neka je ∠ E D F = φ \angle EDF=\varphi ∠ E D F = φ D E DE D E A B AB A B C D CD C D ∠ F D C = 90 ∘ − φ \angle FDC=90^\circ-\varphi ∠ F D C = 9 0 ∘ − φ
Tada je ∠ D C B = ∠ B A D = φ \angle DCB=\angle BAD=\varphi ∠ D C B = ∠ B A D = φ
Kako je φ \varphi φ
sin φ = 1 − cos 2 φ = 1 − 1 9 = 2 2 3 . \sin\varphi=\sqrt{1-\cos^2\varphi}=\sqrt{1-\frac19}=\frac{2\sqrt2}{3}. sin φ = 1 − cos 2 φ = 1 − 9 1 = 3 2 2 . Iz trokuta D F C DFC D F C sin φ = ∣ D F ∣ ∣ C D ∣ \sin\varphi=\dfrac{|DF|}{|CD|} sin φ = ∣ C D ∣ ∣ D F ∣
∣ C D ∣ = ∣ D F ∣ sin φ = 32 2 2 3 = 24 2 . |CD|=\frac{|DF|}{\sin\varphi}=\frac{32}{\frac{2\sqrt2}{3}}=24\sqrt2. ∣ C D ∣ = sin φ ∣ D F ∣ = 3 2 2 32 = 24 2 . Analogno, iz trokuta D E A DEA D E A sin φ = ∣ D E ∣ ∣ A D ∣ \sin\varphi=\dfrac{|DE|}{|AD|} sin φ = ∣ A D ∣ ∣ D E ∣
∣ A D ∣ = ∣ D E ∣ sin φ = 20 2 2 3 = 15 2 . |AD|=\frac{|DE|}{\sin\varphi}=\frac{20}{\frac{2\sqrt2}{3}}=15\sqrt2. ∣ A D ∣ = sin φ ∣ D E ∣ = 3 2 2 20 = 15 2 . Sada je površina paralelograma A B C D ABCD A B C D
P A B C D = ∣ C D ∣ ⋅ ∣ D E ∣ = 24 2 ⋅ 20 = 480 2 . P_{ABCD}=|CD|\cdot|DE|=24\sqrt2\cdot20=480\sqrt2. P A B C D = ∣ C D ∣ ⋅ ∣ D E ∣ = 24 2 ⋅ 20 = 480 2 . Površina trokuta D E A DEA D E A
P D E A = 1 2 ∣ A D ∣ ⋅ ∣ D E ∣ ⋅ sin ( 90 ∘ − φ ) P_{DEA}=\frac12|AD|\cdot|DE|\cdot\sin(90^\circ-\varphi) P D E A = 2 1 ∣ A D ∣ ⋅ ∣ D E ∣ ⋅ sin ( 9 0 ∘ − φ ) = 1 2 ∣ A D ∣ ⋅ ∣ D E ∣ ⋅ cos φ = 1 2 ⋅ 15 2 ⋅ 20 ⋅ 1 3 = 50 2 . =\frac12|AD|\cdot|DE|\cdot\cos\varphi=\frac12\cdot15\sqrt2\cdot20\cdot\frac13=50\sqrt2. = 2 1 ∣ A D ∣ ⋅ ∣ D E ∣ ⋅ cos φ = 2 1 ⋅ 15 2 ⋅ 20 ⋅ 3 1 = 50 2 . Nadalje, površina trokuta D F C DFC D F C
P D F C = 1 2 ∣ D F ∣ ⋅ ∣ C D ∣ ⋅ sin ( 90 ∘ − φ ) P_{DFC}=\frac12|DF|\cdot|CD|\cdot\sin(90^\circ-\varphi) P D F C = 2 1 ∣ D F ∣ ⋅ ∣ C D ∣ ⋅ sin ( 9 0 ∘ − φ ) = 1 2 ∣ D F ∣ ⋅ ∣ C D ∣ ⋅ cos φ = 1 2 ⋅ 32 ⋅ 24 2 ⋅ 1 3 = 128 2 . =\frac12|DF|\cdot|CD|\cdot\cos\varphi=\frac12\cdot32\cdot24\sqrt2\cdot\frac13=128\sqrt2. = 2 1 ∣ D F ∣ ⋅ ∣ C D ∣ ⋅ cos φ = 2 1 ⋅ 32 ⋅ 24 2 ⋅ 3 1 = 128 2 . Konačno površina četverokuta D E B F DEBF D E B F
P = P A B C D − P D E A − P D F C = 480 2 − 50 2 − 128 2 = 302 2 . P=P_{ABCD}-P_{DEA}-P_{DFC}=480\sqrt2-50\sqrt2-128\sqrt2=302\sqrt2. P = P A B C D − P D E A − P D F C = 480 2 − 50 2 − 128 2 = 302 2 . 
