Question

In APQR, T and S are the points on PQ and PR respectively such that QT : PT = 2:3. If U is the intersection point of QS and RT such that U is the mid- point of RT, then SR: SP is Р T S R​

Answer 1

Answer:

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.(I) Prove ΔPQR∼Δ SPR

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.(I) Prove ΔPQR∼Δ SPR(II) Find the length of QR and PS

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.(I) Prove ΔPQR∼Δ SPR(II) Find the length of QR and PS(III)

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.(I) Prove ΔPQR∼Δ SPR(II) Find the length of QR and PS(III) areaofΔSPR

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.(I) Prove ΔPQR∼Δ SPR(II) Find the length of QR and PS(III) areaofΔSPRareaofΔPQR

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.(I) Prove ΔPQR∼Δ SPR(II) Find the length of QR and PS(III) areaofΔSPRareaofΔPQR

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR=∠QPR/ Given QP=8cm, PR=6cm and SR=3cm.(I) Prove ΔPQR∼Δ SPR(II) Find the length of QR and PS(III) areaofΔSPRareaofΔPQR .

Step-by-step explanation:

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