geometry question??
can someone explain how to do dis plz? (this is 10th grade geometry)
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If the vertices at the end of a side are (p,q) and (r,s), then from Pythagoras Theorem, the length of the side is:
sqrt((r – p)^2 + (s – q)^2).
Taking the points in pairs, the lengths are therefore:
AB = sqrt((4 – 2)^2 + (7 – 5)^2) = sqrt(8)
BC = sqrt((3 – 4)^2 + (-4 – 7)^2) = sqrt122)
CA = sqrt((3 – 2)^2 + (-4 – 5)^2) = sqrt(82)
The cosine rule enables you to find an angle from the sides.
If a, b, c are the sides opposite angles A, B, C respectively, then:
a^2 = b^2 + c^2 – 2bc cos(A)
cos(A) = (b^2 + c^2 – a^2) / 2bc …(1)
You need to find the sign of the cosine of the largest angle, to decide whether the triangle is acute- obtuse- or right-angled. If the cosine is positive, the angle is acute; if negative, it is obtuse; if zero, it is a right-angle.
The largest angle is opposite the largest side, which is:
BC = sqrt(122).
Using (1):
cos(A) = (82 + 8 – 122) (2 * 82 * 8) < 0. The triangle is obtuse-angled, and as no two sides are equal, it is also scalene.
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