<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:
x+133= 130
maybe
Step-by-step explanation:
Answer:
Thus, option A is correct.
Explanation:
<u>match the proportions</u>
<u>set up a equation:</u>
- Please find the picture for your triangle.
- We know, area of a ∆ = <u>1/2 × base × height</u>
- Here, base = ( 1 + 7 ) units = <u>8 units</u>
- Height = ( 7 - 1 ) units = <u>6 units</u>
- Therefore, area of the ∆
- = <u>(1/2 × 8 × 6) square units</u>
- = <u>24 square units.</u>
<u>Answer</u><u>:</u>
<em><u>2</u><u>4</u><u> </u><u>square </u></em><em><u>units</u></em>
Hope you can get an idea from it.
Carry on learning.
Doubt clarification - use comment section.
We use the kinetic energy equation and substitute our data solving for the mass:
ke = (1/2)mv^2
m = 2*ke/v^2
m = 2*600/4^2
m = 1200/16
m = 75 kg
we can substitute directly because the units are homogeneus in the SI system: J, m/s, kg
