Draw triangle XYZ on PQR.
P coincides X, Z lies on PQ and Y lies on PR.
Since angle Z = angle Q and angle Y =angle R, ZY is parallel to QR
The two triangles a similar. So
XY/PR = XY/PQ = ZY/QR.
Now find the required sides.
Its straight line geometry time and I've reached another point that I cannot seem to get my head round (though I am sure it will be something obvious).
Triangles PQR and XYZ are such that:
1. Angle P = Angle X
2. Angle Q = Angle Z
3. XY = 3cm
4. YZ = 4cm
5. PQ = 7cm
6. PR = 12cm
I need to find the lengths of XZ and QR.
My answer so far
So I have a large PQR triangle drawn on my page with a line ZY which is parallel to QR, the symbols for angles are put in there to indicate that the two triangles are similar and that PQR is an enlargement of XYZ.
So to find XZ I knew the ratios of the intercepts would be the same so:
PZ:QR = PY:PR
so PZ:7 = 3:12 (or 1:4)
so PZ = (7*3)/12 (or 7/4)
How do I go about working out QR?
You've done all the preliminary work.
. . The answers are waiting for you . . .
Triangles and are such that:
Find the lengths of and .
Did you make a sketch like this?
Code:P * * * * * * * X 7 * * 12 * * * * * * * * * 3 * * * * Q * * * * * * * * * R Z * * * * * Y 4
The two triangles are similar; their sides are proportional.
Since and ,
. . we see that is four times as large as