Answer:
The radius is:
Step-by-step explanation:
The equation of a circle in center-radius form is:
Where the center is at the point (h, k) and the radius is "r".
So, given the equation of the circle:
You can identify that:
Then, solving for "r", you get that the radius of this circle is:
Step-by-step explanation:
<em> </em><em>REFER</em><em> </em><em>TO</em><em> </em><em>THE</em><em> </em><em>ATTACHMENT</em><em> </em>
<em>HEY</em><em> </em><em>IN</em><em> </em><em>WHICH</em><em> </em><em>CLASS</em><em> </em><em>YOU</em><em> </em><em>ARE</em><em> </em><em>ROSA</em>
Answer:
Step-by-step explanation:
<em><u>Given</u></em><u>:</u> A line m is perpendicular to the angle bisector of ∠A. We call this
intersecting point as D. Hence, in figure ∠ADM=∠ADN =90°.
AD is angle bisector of ∠A. Hence, ∠MAD=∠NAD.
<u><em>To Prove</em></u>: <em><u>ΔAMN is an isosceles triangle. i.e any two sides in ΔAMN are</u></em>
<em> </em><em><u>equal. </u></em>
<em><u>Solution</u></em>: Now, In ΔADM and ΔADN
∠MAD=∠NAD ...(1) (∵Given)
AD=AD ...(2) (∵common side)
∠ADM=∠ADN ...(3) (∵Given)
<u><em> Hence, from equation (1),(2),(3) ΔADM ≅ ΔADN</em></u>
( ∵ ASA congruence rule)
⇒<u><em> AM=AN</em></u>
Now, In Δ AMN
AM=AN (∵ Proved)
Hence, ΔAMN is an isosceles triangle.
