Question

Using integration, find the area of the triangle PQR, whose vertices are at P (2, 5), Q (4, 7) and R (6, 2).

Answer 1

To solve the length of PR first find how far over (x), and up (y) R is from P.

x = xR-xP = 5 - 1 = 4

y = yR=yP = 15-3 = 12

Use pythagorean theorm to solve for PR (the hypotenuse):

PR = sqrt(x^2 + y^2) = sqrt(4^2 + 12^2) = 12.6

To solve the area we first need the height, which is the perpendicular line from Q to PR.  First we need the lengths of the other two sides PQ, and RQ.  We solve the same way as we did to find PR:

xPQ = xQ-xP = 4

yPQ = yQ-yP = 1

PQ = sqrt(4^2 + 1^2) = 4.1

xRQ = 0

yRQ = yR = yQ = 15-4 = 11

Using the cosine law of a non-right triangle (c^2 = a^2 + b^2 -2bc(cos[theta])) we can solve for angle PRQ:

4.1^2 = 12.6^2 + 11^2 - 2(12.6)(11)cos[theta]

theta = 18.45 deg

Using sine law in a right triangle we can find the height (h):

sin(18.45) = h/11 --> h = 11*sin(18.45) = 3.5

Finally, to find the area of the triangle:

A = 0.5*b*h = 0.5*PR*h = 0.5*12.6*3.5 = 22

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Answer 2

To solve the length of PR first find how far over (x), and up (y) R is from P.

x = xR-xP = 5 - 1 = 4

y = yR=yP = 15-3 = 12

Use pythagorean theorm to solve for PR (the hypotenuse):

PR = sqrt(x^2 + y^2) = sqrt(4^2 + 12^2) = 12.6

To solve the area we first need the height, which is the perpendicular line from Q to PR.  First we need the lengths of the other two sides PQ, and RQ.  We solve the same way as we did to find PR:

xPQ = xQ-xP = 4

yPQ = yQ-yP = 1

PQ = sqrt(4^2 + 1^2) = 4.1

xRQ = 0

yRQ = yR = yQ = 15-4 = 11

Using the cosine law of a non-right triangle (c^2 = a^2 + b^2 -2bc(cos[theta])) we can solve for angle PRQ:

4.1^2 = 12.6^2 + 11^2 - 2(12.6)(11)cos[theta]

theta = 18.45 deg

Using sine law in a right triangle we can find the height (h):

sin(18.45) = h/11 --> h = 11*sin(18.45) = 3.5

Finally, to find the area of the triangle:

A = 0.5*b*h = 0.5*PR*h = 0.5*12.6*3.5 = 22

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