MathSolver74
A) Perhatikan bahwa AB/BD = AC/DE = BC/BE
b) BC = 10 cm CD : BD = 1 : 3 x + 3x = 10 4x = 10 x = 2,5 maka CD = 2,5 cm BD = 7,5 cm DE = AC = √(7,5² + 10²) = √(56,25 + 100) = √156,25 = 12, 5 cm
Luas segitiga BED = 1/2 x BD x BE = 1/2 x 7,5 cm x 10 cm = 37,5 cm²
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ichwanhandi
A) AB=BD, sdt α = sdt β, AE tegak lurus BC buktikan ΔABC≡ΔBED jwb: AB = AD ( diketahui ) sdt α =sdt β ( diketahui ) AE tegak lurus BC dan berpot. di titik B (diketahui) shg sdt ABC = sdt CBE sehingga ΔABC ≡ Δ BED ( sdt - sisi - sdt ) terbukti
B) BC = 10 CD = 1/3 BD BD = 2/3 BC = 2/3 . 10 =20/3 DE = √100+400/9 = √(900+400)/9 = √1300/9 = 10/3 √13 atau 3 1/3 √13
L ΔBED = 1/2 alas x tinggi = 1/2 . 20/3 . 10 = 100/3 atau 33 1/3
AB/BD = AC/DE = BC/BE
b) BC = 10 cm
CD : BD = 1 : 3
x + 3x = 10
4x = 10
x = 2,5
maka CD = 2,5 cm
BD = 7,5 cm
DE = AC = √(7,5² + 10²)
= √(56,25 + 100)
= √156,25
= 12, 5 cm
Luas segitiga BED = 1/2 x BD x BE
= 1/2 x 7,5 cm x 10 cm
= 37,5 cm²
AB=BD, sdt α = sdt β, AE tegak lurus BC
buktikan ΔABC≡ΔBED
jwb:
AB = AD ( diketahui )
sdt α =sdt β ( diketahui )
AE tegak lurus BC dan berpot. di titik B (diketahui)
shg sdt ABC = sdt CBE
sehingga ΔABC ≡ Δ BED ( sdt - sisi - sdt )
terbukti
B)
BC = 10
CD = 1/3 BD
BD = 2/3 BC
= 2/3 . 10 =20/3
DE = √100+400/9 = √(900+400)/9 = √1300/9 = 10/3 √13 atau 3 1/3 √13
L ΔBED = 1/2 alas x tinggi
= 1/2 . 20/3 . 10
= 100/3 atau 33 1/3