Tricky math help please! How can I solve these?

Question

1.If the ratio of the area of equilateral triangle A to the area of eq. triangle B is 100 to 25, then what is the ratio of their altitudes?

2. If the legs of an isos triangle are 13 inches long and the length of the altitude to the base is 12 in, what is the triangle's area?

3. A right triangle has an area of 100. If a segment is drawn though the midpoint of the triangle's legs, a new smaller triangle is formed. What is this triangle's area?

4.What is the altitude from the right angle of a triangle whose sides have lenghts 6, 8, 10?

5. The ratio of the lengths of the hypotenuse of 2 right triangles is 5 to 8. What is the ratio of the triangles' areas?

6. When 2 diagonals are drawn though a parallelogram, triangles are formed. If an area of one triangle is given, how can you find the area of a triangle adjacent to it?

7. How many right triangles have a hyp of lngth 10?

8. If a triangle has sides 6 and 8, what is the max area of this triangle?

Answer 1

1) the ratio of the altidues = sqrt ratio of areas = sqrt (4) = 2
8) the largest area = 24 = 1/2 (6)(8), when the included angle is right,
The rest is for you to do.

Answer 2

Height of an equlateral triangle = s*sqrt(3)/2 = h
base of equilateral triangle = s
Area of equilateral triangle = hs/2
Area of eqilateral triangle 4X as big is 2hs
This means both h and s doubled
So A/B=2h/h=2

The base must be 10 to because 5^2+12^2 = 13^2, so
Area = 12*10/2 = 60 in^2

The altitude is 1/2 of original triangle's altitude as is the base. Thus the area is 1/4 *100 = 25

altitude = 8sin(arcsin(.8)) = 6.4

The legs must also be in the ratio of 5to 8
So areas have ratio of 25/64

The area of an adjacent triangle is equal to one-half the area of the parallellogram - the area of the given triangle.

1. it is the 6-8-10 triangle.

8*6/2 = 24, since it is a rt triangle which has max area.

Answer 3

#8:
Area of triangle, K, knowing two sides a and b:

K=0.5ab·sin C

where C is angle between sides a and b.

So the maximum of the area would be when sin C is at its maximum. The maximum of the sine funtion is 1, when C is 90°.

So a right triangle, given the sides adjacent to the right angle, would have the maximum area. K = 0.5·6·8·sin 90° = 3·8 = 24.