BD² = ___ x ___
BD² = ___ x ___
BD² = ___ x ___
AB² = ___ x ___
AB = ______
Q.2. B) Solve.
Answers
Answer:
REF.Image.
In △ABC,∠A=90
AC=AB
D is an AB product
To prove :DC
2
−BD
2
=
2AB×AD
→ Acc to Pythagoras theorem
In △ACD
CD
2
=AC
2
+AD
.
..(1)
and In △ABC(∵AB+BD=AD)
BC
2
=AC
2
+AB
2
...(2)
from (1):
CD
2
=AC
2
+AD
2
CD
2
−BD
2
=AC
2
+AD
2
−BD
2
CD
2
−BD
2
=AC
2
+AD
2
−(AD−AB)
2
=AC
2
+AD
2
−AD
2
−AB
2
+2(AD/AB)
CD
2
−BD
2
=AC
2
+(−AB
2
)+2(AD)(AB)
=AB
2
−AB
2
+2(AB)(AD)
or (∵AB=AC)
CD
2
=BD
2
Answer:
REF.Image.
In △ABC,∠A=90
AC=AB
D is an AB product
To prove :DC
2
−BD
2
=
2AB×AD
→ Acc to Pythagoras theorem
In △ACD
CD
2
=AC
2
+AD
.
..(1)
and In △ABC(∵AB+BD=AD)
BC
2
=AC
2
+AB
2
...(2)
from (1):
CD
2
=AC
2
+AD
2
CD
2
−BD
2
=AC
2
+AD
2
−BD
2
CD
2
−BD
2
=AC
2
+AD
2
−(AD−AB)
2
=AC
2
+AD
2
−AD
2
−AB
2
+2(AD/AB)
CD
2
−BD
2
=AC
2
+(−AB
2
)+2(AD)(AB)
=AB
2
−AB
2
+2(AB)(AD)
or (∵AB=AC)
CD
2
=BD
2