Question

Sa se afle in fiecare caz in parte aria unui trape dreptunghic ABCD, AB paralel cu CD, AD perpendicular pe DC, daca AB, CD, BC sunt direct proporționale cu numerele 3,7,5 si a) distanta de la D la latura BC este 42 cm b) DB= 12 cm.​

Answer 1

Daca {AB,CD,BC} d.p {3,7,5} vom avea:

$\frac{AB}{3} =\frac{CD}{7} =\frac{BC}{5} =k$

AB=3k

CD=7k

BC=5k

Ducem BE⊥DC

DE=AB=3k

EC=DC-DE=7k-3k=4k

In ΔBCE dreptunghic in E, aplicam Pitagora (suma catetelor la patrat este egala cu ipotenuza la patrat)

BC²=BE²+EC²

25k²=BE²+16k²

BE²=9k²

BE=3k

a) Din ipoteza stim ca d(D,BC)=42 cm

Notam x=d(D,BC)=42 cm

Aplicam Aria in doua moduri

$A_{DBC}=\frac{BE\times DC}{2} =\frac{x\times BC}{2}$

3k×7k=42×5k

21k=42×5

k=10

AB=30 cm

CD=70 cm

BC=50 cm

BE=30 cm

Aria unui trapez

$A=\frac{(b+B)\times h}{2} \\\\\A_{ABCD}=\frac{(AB+CD)\times BE}{2}$

A=(30+70)×30:2=1500 cm²

b) BD=12 cm

Aplicam Pitagora in ΔBAD

BD²=AD²+AB²

144=9k²+9k²

18k²=144

k²=8

k=2√2

AB=6√2 cm

CD=14√2 cm

BC=10√2 cm

BE=6√2 cm

A=(6√2+14√2)×6√2:2=120 cm²