In a square ABCD, T, U, V and W are points on side AB, AD, CD and BC such that AT: AB = AU: UD = CV: VD = CW: WB = 2:1. Further Z and X are midpoints of UT and UV respectively, while Y is a point on TW such that TY: YW = 1:2. Find the ratio of the area of the triangle XYZ to that of the square ABCD.
Answers
There is a small mistake in the given problem. That should be AT: TB = 2:1 rather than AT :AB.
see the diagram.
We use Pythagoras theorem repetitively to find the answer.
We find the sides of the triangle XYZ first. Then we apply Heron’s Formula for area of
triangle. Let us assume AB = BC = CD = DA = 3 units without loss of generality.
Area of square ABCD = 3² = 9
UV = √2 => UX = √2/2 = 1/√2
UT= 2√2 => UZ = TZ = √2
XZ = √[UX²+UZ²] = √5/√2 = 1.58
TW=√2 TY = TW/3 = √2/3
YZ = √[TZ²+TY²] = 2√5 / 3 = 1.49
YE = TE - TY = TW/2 - TY = 1/√2 - √2/3 = 1/(3√2)
XE = UT = 2√2
XY = √[XE²+ YE²] = √[8+1/18] = √29 *√5 / 3√2 = 2.83
Now semiperimeter s of triangle XYZ = s
s = [√5 *√29 / 3√2 + 2 √5 /
3 + √5 /√2 ] /2
= [ √29 + 2√2 + 3 ] * √5 / (6√2)
= 2.955
Area of triangle xyz = √[s(s-a)(s-b)(-c) = 0.8625
Ratio = 0.8625 / 9 = 0.0958
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Simpler way:
We find the areas of the triangles in the square. We subtract all those from Area of ABCD, to get that of XYZ.
Apply area of triangle = 1/2 * base * altitude
Area of Triangle UXZ = 1/2 * 1/√2 * √2 =
1/2
Area of Triangle TYZ = 1/2 * √2 * √2/3 =
1/3
Area of Triangle XYE = 1/2 * 2√2 * √2/ 6 = 1/3
Area of Triangle XYZ = Area UTEX - (area UXZ + area TYZ + area XYE)
= 2√2 * 1/√2 - (1/2+1/3+ 1/3)
= 2 - 7/6 = 5/6
Area of Square ABCD = 3² = 9
Ratio = (5/6) / 9 = 5 / 54
