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Explanation:
3. 2+√3/2-√3 = 2+√3/2-√3 × 2+√3/2+√3
= (2+√3)²/(2)²-(√3)² = 4+3+4√3/4-3
= 7+4√3/1 = 7+4√3
4.i)Let ABCD be a trapezium with,
AB∥CD
AB=25m
CD=10m
BC=14m
AD=13m
Draw CE∥DA. So, ADCE is a parallelogram with,
CD=AE=10m
CE=AD=13m
BE=AB−AE=25−10=15m
In ΔBCE, the semi perimeter will be,
s= a+b+c/2
s=14+13+15/2
s=21m
Area of ΔBCE,
A=√s(s-a)(s-b)(s-c)
=√21(21-14)(21-13)(21-15)
=√21(7)(8)(6)
=√7056
=84m²
Area of ΔBCE is,
A=1/2 × base × height
84=1/2×15×CL
CL=56/5m
Are of trapezium
A=1/2×(25+10)(56/5)
A = 196m²
Hence,the area of the trapezium is 196m²
ii)4:1
Hope it helps
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