Question
In Fig. prove that:
✓
Answer
To prove,
We know that, in a triangle sum of any two sides is greater than the third side.
AB + BC > AC ....(i)
In $\triangle\text{ADC},$ we have
CD + DA > AC ....(ii)
Adding (i) and (ii), we get
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
And in $\triangle\text{ADC},$ we have
CD + DA > AC
Add AB on both sides
CD + DA + AB > AC + AB
From equation (iii) and (iv), we get,
CD + DA + AB > AC + AB > BC
CD + DA + AB > BC
Hence proved.
- CD + DA + AB + BC > 2AC
- CD + DA + AB > BC
We know that, in a triangle sum of any two sides is greater than the third side.
- So,
AB + BC > AC ....(i)
In $\triangle\text{ADC},$ we have
CD + DA > AC ....(ii)
Adding (i) and (ii), we get
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
- Now, In $\triangle\text{ABC},$ we have,
And in $\triangle\text{ADC},$ we have
CD + DA > AC
Add AB on both sides
CD + DA + AB > AC + AB
From equation (iii) and (iv), we get,
CD + DA + AB > AC + AB > BC
CD + DA + AB > BC
Hence proved.
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x:
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5
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6
|
7
|
8
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9
|
|
f:
|
4
|
8
|
14
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11
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3
|