# Intercept theorem in triangles

• Apr 11th 2010, 06:18 AM
AliceFisher
Intercept theorem in triangles
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).

The Question

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.

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)

The problem

How do I go about working out QR?
• Apr 11th 2010, 07:22 AM
sa-ri-ga-ma
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.
• Apr 11th 2010, 08:13 AM
Soroban
Hello, Alice!

You've done all the preliminary work.
. . The answers are waiting for you . . .

Quote:

Triangles $\displaystyle PQR$ and $\displaystyle XYZ$ are such that:

. . $\displaystyle \begin{array}{ccc}\angle P &=& \angle X \\ \angle Q &=& \angle Z \\ XY &=& 3\text{cm} \\ YZ &=& 4\text{cm}\\ PQ &=& 7\text{cm} \\ PR &=& 12\text{cm} \end{array}$

Find the lengths of $\displaystyle XZ$ and $\displaystyle QR$.

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 $\displaystyle PR = 12$ and $\displaystyle XY = 3$,

. . we see that $\displaystyle \Delta PQR$ is four times as large as $\displaystyle \Delta XYZ$

Then: .$\displaystyle QR \:=\:4\cdot YZ \quad\Rightarrow\quad QR \:=\:4(4) \quad\Rightarrow\quad QR\:=\:16$

And: .$\displaystyle PQ \:=\:4\cdot XZ \quad\Rightarrow\quad 7 \:=\:4\cdot XZ \quad\Rightarrow\quad XZ \:=\:\tfrac{7}{4}$

• Apr 11th 2010, 09:17 AM
AliceFisher
My thanks to both of you, it makes a lot more sense now =)
• Apr 11th 2010, 09:53 AM
razemsoft21
try to continue
Quote:

Originally Posted by AliceFisher
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).

The Question

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.

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)

The problem

How do I go about working out QR?

http://www.mathhelpforum.com/math-he...1&d=1271008329
• Apr 13th 2010, 11:14 AM
AliceFisher
So pretty