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Sia ? 6 years, 4 months ago
Check solution here : https://mycbseguide.com/ncert-solutions.html
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Sia ? 6 years, 4 months ago
Given : Side AB and AC of {tex}\triangle{/tex}ABC are extended to points P and Q respectively. Also ∠PBC < ∠QCB.
To Prove : AC > AB.
Proof : ∠PBC < ∠QCB
{tex}\therefore{/tex} -∠PBC > - ∠QCB
{tex}\therefore{/tex} 180o - ∠PBC > 180o - ∠QCB
{tex}\therefore{/tex} ∠ABC > ∠ACB
{tex}\therefore{/tex} AC > AB . . . [Side opposite to greater angle is longer]
Posted by Disha Singh 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
- l = 1.5 m, b = 1.25 m, h = 65 cm = 0.65 m
∴ The area of the sheet required for making the box.
= lb + 2(bh + hl)
= (1.5) (1.25) + 2{(1.25)(0.65) + (0.65) (1.5)}
= 1.875 + 2{0.8125 + 0.975}
= 1.875 + 2(1.7875) = 1.875 + 3.575 = 5.45 m2 - The cost of sheet for it = Rs. 5.45 × 20 = Rs. 109
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Yogita Ingle 6 years, 4 months ago
Consider, a7 + ab6
= a(a6 + b6)
= a[(a2)3 + (b2)3]
= a[a2 + b2][a4 + a2b2 + b4]
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