(1 point) let a=(2,4,−5)a=(2,4,−5), b=(−3,6,−5)b=(−3,6,−5), c=(−8,7,0)c=(−8,7,0), and d=(−3,5,0)d=(−3,5,0). find the area of the parallelogram determined by these four points, the area of the triangle abcabc, and the area of the triangle abdabd.

Areas and volumes of parallelograms and parallelepipeds in 3 dimensions are often easily found by making use of the cross product of the direction vectors of their edges. For edge vectors v1 and v2 of a triangle, the area is ...

... A = (1/2)║v1 × v2║

that is, half the norm of the cross-product vector. The area of a parallelogram with those edge vectors is simply ...

... A = ║v1 × v2║

Here, direction vectors are ...

We can see that ab = -cd and bc = -da, as required for a parallelogram.

The cross product ab × bc is (10, 25, 5), so the area of the parallelogram is

... ║(10, 25, 5)║ = √(10² +25² +5²) = √750

... Area = 5√30 ≈ 27.3861 . . . . square units (parallelogram area)

The areas of each of the mentioned triangles is half the area of the parallelogram, so is

... Area Δabc = Area ∆abd = (5/2)√30 ≈ 13.6931 . . . . square units (triangles)

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc