A few days ago
hmm …*

geometry question??

A triangle has vertex coordinates (2,5) , (4,7) and (3, -4). Find the length of each side, and decide what type of triangle it is.

can someone explain how to do dis plz? (this is 10th grade geometry)

A few days ago
Anonymous

Let the vertices be A(2,5) , B(4,7) and C(3, -4).

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.

0