Hots Questions- Triangles Class 9 Notes | EduRev

Class 9 Mathematics by Full Circle

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Class 9 : Hots Questions- Triangles Class 9 Notes | EduRev

The document Hots Questions- Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
All you need of Class 9 at this link: Class 9

Question 1. ABCD is a square. P is any point inside it, such that ΔPQR is another square. Prove that AP = CR

Hots Questions- Triangles Class 9 Notes | EduRev

Hint: Join AP and CR.
In ΔADP and ΔCDR, we have :
AD = CD                      [sides of a square]
∠ADP = ∠CDR                     [each = 90° - ∠PDC]
DP = DR                    [side of a square]
⇒ ΔADP ≌ ΔCDR                     [SAS congruence]
⇒ AP = CR                     [C.P.C.T]

 

Question 2. E and F are the mid points of sides AB, AC of ΔABC. CE and BF are produced to X and Y respectively, such that EX = CE and FY = BF. AX and AY are joined. Find in your figure, a triangle congruent to ΔAEX and demonstrate the congruency. Show that XAY is a st. line.

Hots Questions- Triangles Class 9 Notes | EduRev

Hint: Prove ΔAEX ≌ ΔBEC                    [By SAS congruency]
⇒ ∠ XAE = ∠ CBE                    [c.p.c.t.]
⇒ ∠XAB = ∠CBA
But they form a pair of co-interior angles.
⇒ XA || BC                    ...(1)
Similarly, ΔAFY ≌ ΔCFB
⇒ AY || BC                    ...(2)
from (1) and (2) XAY is a st. line.

 

Question 3. In the adjacent figure, BA || DF and CA || EG. If BD = EC then prove that BG = DF and EG = CF.

Hots Questions- Triangles Class 9 Notes | EduRev

Hint: In ΔGBE and ΔFDC ∠ABC = ∠FDE and ∠DED = ∠ACB
also BE = DC
∴ ΔGBE ≌ ΔFDC                    [ASA congruency]
⇒ BG = DF and EG = CF

 

Question 4. ABCD is a square. M is the mid point of AB and PQ ⊥ CM meets AD at P. CB produced meet at Q. Prove that (i) PA = BQ and (ii) CP = AB + PA

Hots Questions- Triangles Class 9 Notes | EduRev

Hint: Prove, ΔAMP ≌ ΔBMQ                    [ASA cong.]
⇒ MP = MQ and PA = QB                    [c.p.c.t.]
⇒ PA = BQ
Again, prove, ∆CMP ≌ ΔCMQ                    [SAS cong.]
⇒ CP = CQ                     [c.p.c.t.]
⇒ CP = CB + BQ = AB + PA

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