S = 36 m 2 B K = 2 m B C : A D = 4 : 5 S=36m^2\\
BK=2m\\
BC:AD=4:5 S = 36 m 2 B K = 2 m BC : A D = 4 : 5
BK, CM - an altitude
Lets
B C = 4 x , A D = 5 x S = B C + A D 2 ⋅ B K S = 4 x + 5 x 2 ⋅ 2 = 36 9 x = 36 x = 4 B C = 16 m , A D = 20 m BC=4x, AD=5x\\
S=\frac{BC+AD}{2}\cdot BK\\
S=\frac{4x+5x}{2}\cdot2=36\\
9x=36\\
x=4\\
BC=16m, AD=20m BC = 4 x , A D = 5 x S = 2 BC + A D ⋅ B K S = 2 4 x + 5 x ⋅ 2 = 36 9 x = 36 x = 4 BC = 16 m , A D = 20 m
A K = M D = A D − B C 2 = 20 − 16 2 = 2 ( m ) AK=MD=\frac{AD-BC}{2}=\frac{20-16}{2}=2(m)\\ A K = M D = 2 A D − BC = 2 20 − 16 = 2 ( m )
△ A B K , ∠ K = 9 0 0 A B 2 = A K 2 + B K 2 A B 2 = 2 2 + 2 2 = 4 + 4 = 8 A B = 8 = 2 2 ( m ) A B = C D = 2 2 m P = A B + B C + C D + A D = = 2 2 + 16 + 2 2 + 20 = 36 + 4 2 ( m ) \triangle ABK, \angle K=90^0\\
AB^2=AK^2+BK^2\\
AB^2=2^2+2^2=4+4=8\\
AB=\sqrt{8}=2\sqrt{2}(m)\\
AB=CD=2\sqrt{2}m\\
P=AB+BC+CD+AD=\\
=2\sqrt{2}+16+2\sqrt{2}+20=36+4\sqrt{2}(m) △ A B K , ∠ K = 9 0 0 A B 2 = A K 2 + B K 2 A B 2 = 2 2 + 2 2 = 4 + 4 = 8 A B = 8 = 2 2 ( m ) A B = C D = 2 2 m P = A B + BC + C D + A D = = 2 2 + 16 + 2 2 + 20 = 36 + 4 2 ( m )
If AB and CD is increased by 1%=0.01:
A B ′ = 1.01 ( 2 2 ) = 2.02 2 ( m ) P ′ = 16 + 20 + 2.02 2 + 2.02 2 = 36 + 4.04 2 ( m ) P ′ ⋅ 100 % P = ( 36 + 4.04 2 ) ⋅ 100 % 36 + 4 2 ≈ 100.14 % AB'=1.01(2\sqrt{2})=2.02\sqrt{2}(m)\\
P'=16+20+2.02\sqrt{2}+2.02\sqrt{2}=36+4.04\sqrt{2}(m)\\
\frac{P'\cdot100\%}{P}=\frac{(36+4.04\sqrt{2})\cdot100\%}{36+4\sqrt{2}} \approx100.14\% A B ′ = 1.01 ( 2 2 ) = 2.02 2 ( m ) P ′ = 16 + 20 + 2.02 2 + 2.02 2 = 36 + 4.04 2 ( m ) P P ′ ⋅ 100% = 36 + 4 2 ( 36 + 4.04 2 ) ⋅ 100% ≈ 100.14%
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