Triangle problems are presented along with their detailed solutions.

## Problem 1The right triangle shown below has an area of 25. Find its hypotenuse.Solution to Problem 1:
- Since the x coordinates of points A and B are equal, segment AB is parallel to the y axis. Since BC is perpendicular to AB then BC is parallel to the x axis and therefore y, the y coordinate of point C is equal to 3. We now need to find the x coordinate x of point C using the area as follows
area = 25 = (1/2) d(A,B) * d(B,C) d(A,B) = 5 d(B,C) = |x - 2|
- We now substitute d(A,B) and d(B,C) in the area formula above to obtain.
25 = (1/2) (5) |x - 2|
- We solve the above as follows
|x - 2| = 10 x = 12 and x = - 8
- We select x = 12 since point C is to the left of point B and therefore its x coordinate is greater than 2.
- We have the coordinates of point A and C and we can find the hypotenuse using the distance formula.
hypotenuse = d(A,C) = sqrt[ (12 - 2)^{ 2}+ (3 - 8)^{ 2}] = sqrt(125) = 5 sqrt(5)
## Problem 2Triangle ABC shown below is inscribed inside a square of side 20 cm. Find the area of the triangleSolution to Problem 2:
- The area is given by
area of triangle = (1/2) base * height = (1/2)(20)(20) = 200 cm^{ 2}
## Problem 3Find the area of an equilateral triangle that has sides equal to 10 cm.Solution to Problem 3:
- Let A,B and C be the vertices of the equilateral triangle and M the midpoint of segment BC. Since the triangle is equilateral, AMC is a right triangle. Let us find h the height of the triangle using Pythagorean theorem.
h^{ 2}+ 5^{ 2}= 10^{ 2}
- Solve the above equation for h.
h = 5 sqrt(3) cm - We now find the area using the formula.
area = (1/2)* base * height = (1/2)(10)(5 sqrt(3)) = 25 sqrt(3) cm^{ 2}= 43.3 cm^{ 2}
## Problem 4An isosceles triangle has angle A 30 degrees greater than angle B. Find all angles of the triangle.Solution to Problem 4:
- An isosceles triangle has two angles equal in size. In this problem A is greater than B therefore angles B and C are equal in size. Since angle A is 30 greater than angle B then A = B + 30
^{ o}. The sum of all angles in a triangle is equal to 180^{ o}. (B+30) + B + B = 180
- Solve the above equation for B.
B = 50^{ o}
- The sizes of the three angles are
A = B + 30 = 80^{ o} C = B = 50^{ o}
## Problem 5Triangle ABC, shown below, has an area of 15 mm^{ 2}. Side AC has a length of 6 mm and side AB has a length of 8 mm and angle BAC is obtuse. Find angle BAC to the and find length of side BC.
Solution to Problem 5:
- Let the size of angle BAC = t. One of the many formulas for the area triangle
is.
area = 15 = (1/2) (AC)(AB) sin(t)
- Solve for sin(t) to obtain.
sin(t) = 30 / (8*6) = 0.625
- Solve for t above and take the solution that gives t obtuse
t = Pi - arcsin(0.625)
- Convert t to degrees to obtain
t (approximately) = 141.3^{ o}
- We now use the cosine rule
to calculate the length of side BC
BC^{ 2}= AB^{ 2}+ AC^{ 2}- 2(AB)(AC)cos(t) = 64 + 36 - 2(8)(6)cos(141.3^{ o}) BC = 13.23 mm.
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