Bunuel wrote:
mailnavin1 wrote:
Triangle PQR is right angled at Q. QT is perpendicular to PR, where T is the point on PR. If PQ = 3cm and QR = 6cm. Find QT
1. 3/5
2. 6\sqrt{5}
3. 6/\sqrt{5}
4. 4
5. None
Look at the diagram:
Attachment:
untitled.PNG
You can solve this question using the fact that:
perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle, which means that all 3 triangles PQT, QRT and PRQ are similar. Now, as
in similar triangles, corresponding sides are all in the same proportion then QT/QR=PQ/PR, as \(PR=\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5}\) then \(\frac{QT}{6}=\frac{3}{3\sqrt{5}}\) --> \(QT=\frac{6}{\sqrt{5}}\).
Or you can equate the area: \(area_{PQR}=\frac{1}{2}*PQ*QR=\frac{1}{2}*QT*PR\) --> the same here: \(PR=3\sqrt{5}\) --> \(QT=\frac{6}{\sqrt{5}}\).
Answer: C.
Hi Bunuel,
I'm a little confused here. let me try to explain:
If I equate the ratios:
I get: QT/6 = 5/3Root5
I'm equating the ratios as such: the height of the QTR over the height of the the main triangle which equals, the hypotenuse of QTR to the hypotenuse of the main, which i 3 Root 5.
Can you please clarify what the ratios actually are(height of one vs. base of another etc). It's a little confusing since the variables are repeated between the 3 triangles. My worry comes in because if I did this same ratio and I equated it to QTP, i would've gotten a completely different answer. I'm surely doing something wrong -- perhaps in how i'm setting up the ratios?
Corresponding sides are opposite the corresponding (equal) angles. Corresponding (equal) angles are marked in the diagram below: