Question 198416: In rectangle ABCD, AB=6 and the length of diagonal AC is 10. The area of ABCD is
1.60
2.48
3.32
4.28 Answer by anantha(86) (Show Source):
You can put this solution on YOUR website! sol:
in rectangle ABCD in which AC and BD are diagonals
then AC^2+BD^2=AB^2+BC^2+CD^2+DA^2
PROOF:
IN RIGHT ANGLED TRIANGLE ABC
AC^2=AB^2+BC^2 (PYTHAGORAS THEOREM)
IN RIGHT ANGLED TRIANGLEBDC
BD^2=BC^2+CD^2
ADDING ABOVE EQUATIONS
AC^2+BD^2=AB^2+BC^2+BC^2+CD^2
AC^2+BD=AB^2+BC^2+DA^2+CD^2 (BC=DA)
in rectangle AB=DC=6 given,AD=BC=x taken,AC=BD=10 GIVEN
substitute these values in the above formula
10^2+10^2=6^2+x^2+x^2+6^2
100+100=36+x^2+x^2+36
200=72+2x^2
2x^2=200-72
2x^2=128
x^2=64
x=sqrt(64)=8
AB=6 AND BC=8
Area of the rectangle=AB*BC=6*8=48
(