P is external points to a circle having center O. and pa, PB are tangents. PO=17 cm, AO = 8 cm, then find the perimeter of PAOB quadrilateral
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Answered by
2
from the the phythgoras theorum we can get the length of the tangent
PO² = AO² + PA²
PO² = BO² + PB²
17² = 8² + PA²
PA² = 289 - 64 = 225
PA = PB = 15 CMS
so the perimeter of the quadrilateral is 15 + 15 + 8 + 8 = 30 + 16 = 46 cms
ans = perimeter is 46 cms
PO² = AO² + PA²
PO² = BO² + PB²
17² = 8² + PA²
PA² = 289 - 64 = 225
PA = PB = 15 CMS
so the perimeter of the quadrilateral is 15 + 15 + 8 + 8 = 30 + 16 = 46 cms
ans = perimeter is 46 cms
Answered by
1
First of all,use pythagores thorem=Hsqr=Psqr+Bsqr, then we find the sides.we already given that OA=8 and OP=17,by pg throem we find AP=15,now we know that OA is radius,so OA=OB mean ,if OA=8,SO OB=8 also,,and AP=PB,by thoem tangents are equal,so,AP=PB,means AP=15,so also PB=15,,,now add all the sides of quadrilateral,,8+8+15+15=46,,so the perimeter of PAOB is 46cm*,,hence proof
mandeepkaurR:
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