Question

In the given triangle PQR ,AB is parallel to QR and QP is parallel to CB and AR intersects CB at O.Using the given diagram answer the following questions.i)the triangle similar to triangle ARQ is ----ii)triangle PQR is similar to triangle BCR by axiom iii)If QC=6cm,CR=4cm,BR=3cm, the length of RP is -----iv)The ratio PQ:BC is​

Answer 1

Answer:

68 PQR is the correct answer okkk

Answer 2

Given:

ΔPQR with AB || QR and QP || CB

AR intersects CB at O

To Find:

i) The triangle that is similar to triangle ARQ

ii) The axiom by which triangle PQR is similar to triangle BCR

iii) The length of side RP

iv) The ratio PQ : BC

Solution:

(i) In ∆ARQ and ∆ORC we have,

AQ || OC  (Since given that, QP || CB)

⇒ ∠QAR = ∠COR  (Corresponding angles)

⇒∠ARQ = ∠ORC (Common angles)

and  ∠AQR = ∠OCR (Corresponding angles)

So, ∆ARQ ~ ∆ORC  By AAA similarity

(ii) In ∆PQR and ∆BCR we have,

QP || CB (given)

⇒ ∠QPR = ∠CBR (Corresponding angles)

⇒∠PRQ = ∠BRC (Common angles)

⇒ ∠PQR = ∠BCR (Corresponding angles)

So,  ∆PQR ~ ∆BCR  By AAA similarity

(iii) Since proven that ∆PQR ~ ∆BCR

⇒ PQ/BC = QR/CR = PR/BR  (when two triangles are similar the ratio of their corresponding sides are equal)

Thus, QR/CR = PR/BR

⇒(QC + CR)/CR = (PB + BR)/BR

Given that, QC =6 cm, CR = 4 cm, BR = 3 cm.

Substituting the values in the above equation we get:

(6 + 4)/4 = (PB + 3)/3

or 10/4 = (PB + 3)/3

or 5/2 = (PB + 3)/3

or 15 - 6 = 2 PB

or PB = 4.5

Hence, RP = RB + BP

⇒PR = BR + PB

or  PR = 3 + 4.5

or PR = 7.5 cm .

Thus, the length of RP is 7.5cm.

(iv) Since ∆PQR ~ ∆BCR

⇒ PQ/BC = QR/CR = PR/BR (when two triangles are similar the ratio of their corresponding sides are equal)

Thus, PQ/BC = PR/BR

⇒ PQ/BC = 7.5/3

⇒  PQ/BC = 75/30

⇒ PQ/BC = 5/2

or PQ : BC = 5 : 2

Therefore, the ratio of PQ : BC is 5 : 2.

(The figure from the question is attached below.)