Answer:
It can be concluded that the intersection of a chord and the radius that bisects it is at right angle. The two are perpendicular.
Step-by-step explanation:
i. Construct the required circle of any radius as given in the question, then locate the chord. A chord joins two points on the circumference of a circle, but not passing through its center.
ii. Construct the radius to bisect the chord, dividing it into two equal parts.
Then it would be observed that the intersection of a chord and the radius that bisects it is at right angle. Thus, the chord and radius are are perpendicular to each other.
The construction to the question is herewith attached to this answer for more clarifications.
Answer:
It's true because plane geometry is 2 dimensional and solid geometry is 3 dimensional dont get them confused ok.
Answer:
ΔNAS≅ΔSEN by SSA axiom of congruency.
Step-by-step explanation:
Consider ΔNAS and ΔSEN,
NS=SN(Common ie . Both are the same side)
SA=NE( Given in the question that SA≅ NE)
∠SNA=∠NSE( Due to corresponding angle property where SE ║ NA)
Therefore, ΔNAS ≅ΔSEN by SSA axiom of congruency.
∴ NA≅SA by congruent parts of congruent Δ. Hence, proved.
Answer: 9 because of the (0,9)
Step-by-step explanation: