In the given figure if PA=20 cm, what is the perimeter of ΔPQR.
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From P we have tangents PA and PB
Hence PA = PB …tangents from same point are equal …(a)
Point Q is on PA
From Q we have tangents QA and QC
⇒ QA = QC …tangents from same point are equal …(i)
Point R is on PB
From R we have two tangents RC and RB
⇒ RC = RB … tangents from same point are equal …(ii)
Consider ΔPQR
⇒ perimeter of ΔPQR = PQ + QR + PR
From figure QR = QC + CR
⇒ perimeter of ΔPQR = PQ + QC + CR + PR
Using (i) and (ii)
⇒ perimeter of ΔPQR = PQ + QA + RB + PR
From figure we have
PQ + QA = PA and RB + PR = PB
⇒ perimeter of ΔPQR = PA + PB
Using (a)
⇒ perimeter of ΔPQR = PA + PA
⇒ perimeter of ΔPQR = 2(PA)
PA is 20 cm given
⇒ perimeter of ΔPQR = 2 × 20
⇒ perimeter of ΔPQR = 40 cm
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