टोपी शुक्ला ने कब, कौन-सी कसम खाई इसका कया कारण था? 5 mark questions
Answers
Explanation:
Join oa <br /><br /><br /><br />
1. Pa = (pn-an) , pb = (pn+bn ) <br /><br /><br /><br />
pa.Pb = (pn-an)(pn+bn) <br /><br /><br /><br />
on perpendicular ab so an = bn <br /><br /><br /><br />
pa.Pb = (pn-an)(pn+an) = pn^2-an^2 <br /><br /><br /><br />
2. Pn^2-an^2 = <br /><br /><br /><br />
in ∆ona <br /><br /><br /><br />
oa^2 = on^2+an^2 <br /><br /><br /><br />
an^2 = oa^2-on^2 <br /><br /><br /><br />
pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 <br /><br /><br /><br />
in ∆ onp <br /><br /><br /><br />
op^2 = on^2+pn^2 <br /><br /><br /><br />
so. Op^2- oa^2 <br /><br /><br /><br />
oa= ot <br /><br /><br /><br />
pn^2-an^2= op^2-ot^2 <br /><br /><br /><br />
3. From 1 & 2 <br /><br /><br /><br />
pa.Pb = op^2-ot^2 <br /><br /><br /><br />
in ∆ otp <br /><br /><br /><br />
op^2 = pt^2+ot^2 <br /><br /><br /><br />
op^2-ot^2= pt^2 <br /><br /><br /><br />
so pa.Pb = pt^2 <br /><br /><br /><br />
log in to add aJoin oa <br /><br /><br /><br />
1. Pa = (pn-an) , pb = (pn+bn ) <br /><br /><br /><br />
pa.Pb = (pn-an)(pn+bn) <br /><br /><br /><br />
on perpendicular ab so an = bn <br /><br /><br /><br />
pa.Pb = (pn-an)(pn+an) = pn^2-an^2 <br /><br /><br /><br />
2. Pn^2-an^2 = <br /><br /><br /><br />
in ∆ona <br /><br /><br /><br />
oa^2 = on^2+an^2 <br /><br /><br /><br />
an^2 = oa^2-on^2 <br /><br /><br /><br />
pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 <br /><br /><br /><br />
in ∆ onp <br /><br /><br /><br />
op^2 = on^2+pn^2 <br /><br /><br /><br />
so. Op^2- oa^2 <br /><br /><br /><br />
oa= ot <br /><br /><br /><br />
pn^2-an^2= op^2-ot^2 <br /><br /><br /><br />
3. From 1 & 2 <br /><br /><br /><br />
pa.Pb = op^2-ot^2 <br /><br /><br /><br />
in ∆ otp <br /><br /><br /><br />
op^2 = pt^2+ot^2 <br /><br /><br /><br />
op^2-ot^2= pt^2 <br /><br /><br /><br />
so pa.Pb = pt^2 <br /><br /><br /><br />
log in to add aJoin oa <br /><br /><br /><br />
1. Pa = (pn-an) , pb = (pn+bn ) <br /><br /><br /><br />
pa.Pb = (pn-an)(pn+bn) <br /><br /><br /><br />
on perpendicular ab so an = bn <br /><br /><br /><br />
pa.Pb = (pn-an)(pn+an) = pn^2-an^2 <br /><br /><br /><br />
2. Pn^2-an^2 = <br /><br /><br /><br />
in ∆ona <br /><br /><br /><br />
oa^2 = on^2+an^2 <br /><br /><br /><br />
an^2 = oa^2-on^2 <br /><br /><br /><br />
pn^2-(oa^2-on^2)= pn^2+on^2-oa^2 <br /><br /><br /><br />
in ∆ onp <br /><br /><br /><br />
op^2 = on^2+pn^2 <br /><br /><br /><br />
so. Op^2- oa^2 <br /><br /><br /><br />
oa= ot <br /><br /><br /><br />
pn^2-an^2= op^2-ot^2 <br /><br /><br /><br />
3. From 1 & 2 <br /><br /><br /><br />
pa.Pb = op^2-ot^2 <br /><br /><br /><br />
in ∆ otp <br /><br /><br /><br />
op^2 = pt^2+ot^2 <br /><br /><br /><br />
op^2-ot^2= pt^2 <br /><br /><br /><br />
so pa.Pb = pt^2 <br /><br /><br /><br />
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