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Sir mathematics ma ex 6.3 nahi ha muja ncrt nahi chai

Posted by Mahesh Gujjar 3 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 3 years, 11 months ago

6. In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ½ QPR.

Solution:

Consider the ΔPQR. PRS is the exterior angle and QPR and PQR are interior angles.

So, PRS = QPR+PQR (According to triangle property)

Or, PRS -PQR = QPR ———–(i)

Now, consider the ΔQRT,

TRS = TQR+QTR

Or, QTR = TRS-TQR

We know that QT and RT bisect PQR and PRS respectively.

So, PRS = 2 TRS and PQR = 2TQR

Now, QTR = ½ PRS – ½PQR

Or, QTR = ½ (PRS -PQR)

From (i) we know that PRS -PQR = QPR

So, QTR = ½ QPR (hence proved).

0Thank You

ANSWER

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