In the given figure, ABCD is a parallelogram. E is any point on AB. CB and DE are produced to meet at F. Prove that area of tria
1 answer:
You might be interested in
9514 1404 393
Answer:
(a) (-2, 4)
Step-by-step explanation:
The orthocenter is the point of concurrence of the altitudes of the triangle.
A plot of the triangle vertices shows the triangle to be acute. That means the orthocenter is inside. This eliminates choices B, C, D, which are all points outside the triangle.
__
Conveniently, side LM is horizontal. This means the orthocenter will have the x-coordinate of the vertical line through N, which is -2. Only one answer choice has an x-coordinate of -2:
a (-2, 4)
Answer:
B
Step-by-step explanation:
using the tangent ratio on the triangle on the right and the exact value
tan60° = , then letting the altitude be h
tan60° = =
=
( multiply both sides by h )
11 = h ×
( divide both sides by
)
11 = h
using the sine ratio in the triangle on the left and the exact value
sin45° = , then
sin45° = =
=
=
( cross- multiply )
× x = 22
( divide both sides by
)
x = 22