Answer:
Let P be the external point. O be the origin. join O and P get OP and nearest point on the circle from P be A.
Let Q be the point onthe circle in which, tangent make 90° with radius at Q.
PQ = 8 and OQ = 6
we get a right angled triangle PQO right angled at Q.
so, OP^2 = OQ^2 + PQ^2= 8^2 + 6^2 = 64 + 36 =1==
therefore OP =10cm
we need nearest point from P, which is PA
PA = OP - OA= 10 -6=4cm
Answer: y = 0.5, x = 50
Step-by-step explanation: Both triangles in the picture are isosceles, telling us that the 2 angles at the bottom are congruent. With this, we can find y by doing the following:
a triangle has 180 degrees so we subtract the given 50 which gives us 130
2(2y + 64) = 130
4y + 128 = 130
y = .5
This means that the bottom 2 angles are both 65. Since the top angle of the second triangle is supplementary to the bottom angle of the first one, the top angle of the second triangle is 115. So, we find x by:
2(45 - x/4) = 65
x = 50
This means the bottom 2 angles of the second triangle are both 32.5.
I don’t understand the question
Answer:
The answer to this question is x=3
Step-by-step explanation:
If you create the graph and then plot the points you will get a depiction of the trapazoids. using this you will find that they have reflected over the x=3 line.