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Question : 29 of 160
Marks:
+1,
-0
Solution:
r1+r2+r3−r=4R ∴‌‌r1+r2+r3=4. . . (i)
Let
‌=‌=‌=k ‌=‌=‌=k Then,
a=4k,b=5k,c=6k s=‌=‌ Area
∆=√s(s−a)(s−b)(s−c) ‌‌=√‌(‌−4k)(‌− ‌‌=15√7‌ r‌‌=‌=‌=√7‌ ∴‌‌r=‌=‌=√7‌‌...‌ (ii) ‌ ‌ and ‌‌‌R=‌=‌| (4k)⋅(5k)(6k) |
| 4×15√7‌ |
=‌‌...‌ (iii) ‌ Now,
‌[r1+r2+r3]=‌ [from Eq. (i)]
‌‌=1+‌ ‌‌=1+‌ ‌‌=1+‌=‌
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