In geometry, it would be always helpful to draw a diagram to illustrate the given problem.
This will also help to identify solutions, or discover missing information.
A figure is drawn for right triangle ABC, right-angled at B.
The altitude is drawn from the right-angled vertex B to the hypotenuse AC, dividing AC into two segments of length x and 4x.
We will be using the first two of the three metric relations of right triangles.
(1) BC^2=CD*CA (similarly, AB^2=AD*AC)
(2) BD^2=CD*DA
(3) CB*BA = BD*AC
Part (A)
From relation (2), we know that
BD^2=CD*DA
substitute values
8^2=x*(4x) => 4x^2=64, x^2=16, x=4
so CD=4, DA=4*4=16 (and AC=16+4=20)
Part (B)
Using relation (1)
AB^2=AD*AC
again, substitute values
AB^2=16*20=320=8^2*5
=>
AB
=sqrt(8^2*5)
=8sqrt(5)
=17.89 (approximately)
Answer:
B = (- 4,
)
Step-by-step explanation:
Using the Section formula
=
=
=
= - 4
=
=
=
= 
Thus coordinates of B = (- 4,
)
Answer:
The coordinates of B are (-5, -1).
Step-by-step explanation:
You can make a linear relationship. Since AB is <em>1 unit down and 3 units to the left of</em> A, B must be <em>1 unit down and 3 units to the left of</em> AB.
A) x^2 + y^2 = 80
B) x * y = 12
What is x^2 -y^2?
B) x * y = 12
So y = 12 / x
A) x^2 + (12/x)^2 = 80
x^2 + (12/x)^2 = 80
x^2 + (144/x^2) = 80
Multiplying each side by x^2
x^4 + 144 = 80x^2
x^4 -80x^2 + 144 = 0
X = 1.355 (found by trial and error)
Y =
<span>
<span>
<span>
8.85608</span></span></span> (found by trial and error)