a) Daca DE =2√2
si EB= 4√2
⇒ BD = DE + EB =2√2 + 4√2 =6√2 (BD = diagonala)
In Δ ABD (dreptunghic in A) , AE = inaltime
AE² = DE ×EB (teorema inaltimii in Δ dreptunghic)
AE² = 2√2 × 4√2
AE² = 16
AE = √16
AE = 4
In Δ ADE (dreptunghic in E):
AD² = AE² +DE²
AD² = 16 +8
AD² = 24
AD =2√6
In Δ ABD :
BD² =AD² + AB²
AB² = BD² - AD²
AB² = (6√2)² - (2√6)²
AB² = 72 - 24
AB² =48
AB = 4√3
P (ABCD) = 2 (AB + AD) = 2 (4√3 + 2√6) =2(4√3 + 2√3 ×√2) =2×2√3(2+√2)
=4√3 (2+√2)
b) A (AED) = C1 × C2 / 2 (aria = cu semiprodusul catetelor)
= DE × AE / 2
=2√2 × 4 / 2
= 4√2 cm²
A (AEB) = AE × BE / 2
= 4 × 4√2 / 2 = 8√2 cm²
Raportul celor 2 arii :
A (AED) / A (AEB) = 4√2 / 8√2 = 1 / 2
Mulțumesc +coroana?
