Question

24. In the given triangle PQRAB II QR.QP II CB and AR intersects CB at O.
P
3cm
R
6cm
C.
4cm
Using the given diagram answer the following question​

Answer 1

Answer:i)

In ∆ARQ and ∆ORC we have ,

→ QA || CO { given that, QP || CB . }

So,

→ ∠QAR = ∠COR { Corresponding angles. }

→ ∠ARQ = ∠ORC { Common angles. }

then,

→ ∆ARQ ~ ∆ORC { By AA similarity. }

therefore, (a) ∆ORC is correct answer .

(ii)

since,

→ QP || CB,

In ∆PQR and ∆BCR we have,

→ ∠PQR = ∠BCR { Corresponding angles. }

→ ∠PRQ = ∠BRC { Common angles. }

→ ∠QPR = ∠CBR { Corresponding angles. }

So,

→ ∆PQR ~ ∆BCR { By AAA similarity. }

therefore, (b) AAA is correct answer .

(iii)

since,

→ ∆PQR ~ ∆BCR

→ QR/CR = RP/RB

→ (6+4)/4 = RP/3

→ 10/4 = RP/3

→ RP = 30/4

→ RP = 7.5 cm (C)

(iv)

also,

→ PQ/BC = QR/CR

→ PQ/BC = 10/4

→ PQ/BC = 5/2

→ PQ : BC = 5 : 2 (c)

Step-by-step explanation: