Triangle Problems

Triangle problems with 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

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 to triangles and geometry.


Parallel Lines and Angles Problems

Triangles

Solve Right Triangle Problems

Cosine Law Problems

Geometry Tutorials, Problems and Interactive Applets.


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Updated: 2 April 2013

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