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**kempos123****Guest**

The area of the triangle ABC = 24.

Point X lies on AB and divide AB into two equal parts. Point Z lies on BC and diviede BC in the ratio: BZ:ZC = 1:3

Point Y lies on AC and divide AC in the ratio: AY:YC=1:2

Find the area of the triangle XYZ

**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

is this a right angled triangle, isoceles? what

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**kempos123****Guest**

no info about the type of the triangle. i started with finding area of the small one on the left = 4 and the big on the right = 12. i am stuck with the one on the top. once we have them three we can find XYZ.

as i said no info about the kind

**pi man****Guest**

The area of a triangle can be computed as: Area = 1/2sin(C)ab. (capital letters signify the angle; lowercase letters signify the length of the size opposite the angle; i.e. a is the length of the side opposite of angle A.

So the area of ABC can be computed in 3 ways, all equal to 24:

Area = 1/2 sin(A)bc

Area = 1/2 sin(B)ac

Area = 1/2 sin(C)ab

The area of XYZ = area of ABC - area of AYX - area of BXZ - area of CYZ

The area of AYX = 1/2 sin(A) (1/3 b)(1/2 c) = 1/2 sin(A) 1/6 (bc). That's 1/6 of the area of ABC, or 4.

Likewise the area of BXZ = 1/2 sin(B) (1/2 c) (1/4 a) = 1/8 of ABC or 3

Area of CYZ = 1/2 sin(C) (2/3 b) (3/4 a) = 1/2 of ABC or 12

So the area of XYZ = 24 - 4 - 3 - 12 = 5

**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

i see it like this, looking at pic

we dont know any of the angles, or any of the sides

let a be the length BC

let b be the length AC

let c be the length AB

and change ABC to be the opposite angles to a,b,c

*Last edited by luca-deltodesco (2006-07-05 06:55:23)*

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