Find all the partial products. then add to find the final product draw area model as needed. 32
1 answer:
You might be interested in
Answer:
Transitive property of equality is not a justification for the proof.
Step-by-step explanation:
We draw a right angle ΔACB. CD is perpendicular to AB.
Let AC = a , BC = b , AB = c and CD = h
Now in ΔABC and ΔACD
∠C = ∠D and ∠A = ∠A
from AA similarity postulate
ΔABC similar to ΔACD.
Hence,
=
= c × x ·····················(1)
Now in ΔABC and ΔCBD
∠C = ∠D and ∠B = ∠B
from AA similarity postulates
ΔABC similar to ΔCBD
Hence,
= c × y······················(2)
Add equation (1) and (2)
+
= cx + cy
+
= c(x+y)
+
=
[because x+y=c]
Transitive property is not useful for this proof.
