Triangle Problems

Triangle problems are presented along with their detailed solutions.

Problem 1

The right triangle shown below has an area of 25. Find its hypotenuse.
right triangle problem 1

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 2

Triangle ABC shown below is inscribed inside a square of side 20 cm. Find the area of the triangle
triangle inscribed in square

Solution to Problem 2:

  • The area is given by
    area of triangle = (1/2) base * height
    = (1/2)(20)(20) = 200 cm 2

Problem 3

Find 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.
    triangle inscribed in square

    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 4

An 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 5

Triangle 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.
triangle inscribed in square

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.

More References and Links to

Triangles
Solve Right Triangle Problems
Cosine Law Problems
Geometry Tutorials, Problems and Interactive Applets.

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