<span>Consider a angle â BAC and the point D on its defector
Assume that DB is perpendicular to AB and DC is perpendicular to AC.
Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC
Proof
Consider triangles ΔADB and ΔADC
Both are right angle, â ABD= â ACD=90 degree
They have congruent acute angle â BAD and â CAD( since AD is angle bisector)
They share hypotenuse AD
therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
Answer:
2½
Step-by-step explanation:
33÷15 you got this don't give up
Hey there! I'm happy to help!
First, let's find the radius. The radius is half of the diameter.
8.4/2=4.2
Here's how to find the area of a circle.
Square the radius.
4.2²=17.64
And multiply by pi (3.14 in our case)
17.64(3.14)=55.3896
We round to the nearest tenth, giving us an area of 55.4 ft²
Have a wonderful day! :D
Answer:
A THE LOCATION OF THE LIGHT BULB
I got this right on the assignment
Step-by-step explanation: