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**

**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 **Δ**ABC. 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 **Δ**AEX and demonstrate the congruency. Show that XAY is a st. line.**

**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.**

**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**

**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

132 docs

### Value Based Questions- Triangles

- Doc | 2 pages

- Ex 7.5 NCERT Solutions- Triangles
- Doc | 2 pages
- Ex 7.4 NCERT Solutions- Triangles
- Doc | 3 pages
- Ex 7.3 NCERT Solutions- Triangles
- Doc | 4 pages
- Ex 7.2 NCERT Solutions- Triangles
- Doc | 5 pages