Question

Trapezul isoscel abcd are AB || CD, AB=24 cm, CD=18cm, iar m (《A)=60°. Determina: a) aria trapezului
b) lunigimile diagonalelor trapezului;
c) distanta de la punctul C la dreapta AD.

Answer 1

a) Am coborat inaltimile DM si CN (M,N ∈ AB)
MN=CD=18cm=>AM=NB=(24-18)/2=6/2=3cm
In ΔDMA dr. avem m(∡ADM)=90°-60°=30°
=>AM-cateta opusa unghiului de 30°=>AM=AD/2=>AD=6cm=CB (ipoteza, trapez isoscel)
Inaltimea DM se calculeaza cu t.Pit.
DM²=AD²-AM²
DM²=36-9
DM²=27=>DM=3√3cm
Aria=(B+b)xh/2
Aria=(24+18)x3√3/2=21x3√3=63√3cm²

b)CN=DM (inaltimi paralele)
In ΔCNA dr. AC²=CN²+AN²
AN=AM+MN=3+18=21
AC²=27+441
AC²=468
AC=6√13cm=BD