Gujarat BoardEnglish MediumSTD 9MathsCongruent Triangles4 Marks
Question
In Fig. prove that:
✓
Answer
To prove,
$i. CD + DA + AB + BC > 2AC$
$ii.CD + DA + AB > BC$
From the given figure, We know that, in a triangle sum of any two sides is greater than the third side.
$i.$ So,
In $\triangle\text{ABC},$ we have
$AB + BC > AC ....(i)$
In $\triangle\text{ADC},$ we have
$CD + DA > AC ....(ii)$
Adding $(i)$ and $(ii)$, we get
$AB + BC + CD + DA > AC + AC$
$AB + BC + CD + DA > 2AC$
$ii.$ Now, In $\triangle\text{ABC},$ we have,
$AB + AC > BC ...(iii)$
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.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
