Simple Triangle Question

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May 27th 2008, 01:07 PM
iwz

Simple Triangle Question

The triangle PQR isn't a square triangle.
PQ = 9.9cm
QR = 22.3cm
RP = 18.1cm

(Going around the triangle anti clockwise)

　　　　 P

Q 　　　　　　　　　　　　R

Calculate the area of the triangle.

Please help me and state how you did it :)
May 27th 2008, 01:50 PM
sean.1986

Cosine rule....

a^2 = b^2 + c^2 - 2bcCosA

2bcCosA = b^2 + c^2 - a^2

CosA = (b^2 + c^2 - a^2) / 2bc

CosA = (9.9^2 + 22.3^2 - 18.1^2) / 441.51

A = arccos((9.9^2 + 22.3^2 - 18.1^2) / 441.51) = 0.919389657 radians

Area = 0.5bcSinA = 87.8 cm^2
May 27th 2008, 02:08 PM
TheEmptySet

Quote:

Originally Posted by iwz

The triangle PQR isn't a square triangle.
PQ = 9.9cm
QR = 22.3cm
RP = 18.1cm

(Going around the triangle anti clockwise)

　　　　 P

Q 　　　　　　　　　　　　R

Calculate the area of the triangle.

Please help me and state how you did it :)

This problem can also be solved using Heron's formula

$A=\sqrt{s(s-a)(s-b)(s-c)}, \mbox{ where } s=\frac{a+b+c}{2}$

and a,b,c are the side lengths of the triangle.

$s=\frac{9.9+22.3+18.1}{2}=25.15$
so the Area is
$A=\sqrt{(25.15)(25.15-9.9)(25.15-22.3)(25.15-18.1)} \approx 87.79cm^2$
May 28th 2008, 11:15 AM
iwz

Quote:

Originally Posted by sean.1986

Cosine rule....

a^2 = b^2 + c^2 - 2bcCosA

2bcCosA = b^2 + c^2 - a^2

CosA = (b^2 + c^2 - a^2) / 2bc

CosA = (9.9^2 + 22.3^2 - 18.1^2) / 441.51

A = arccos((9.9^2 + 22.3^2 - 18.1^2) / 441.51) = 0.919389657 radians

Area = 0.5bcSinA = 87.8 cm^2

Thanks, that's what i was looking for :D