Question

DEMONSTRATION

1. From the figure, AC^2 = AB^2 + BC = 12 + 12 = 2 or AC =√2
AD^2 = AC^2 + CD^2 = 2 + 1 = 3 or AD = √3​

Answer 1

Step-by-step explanation:

Given

ABCD is square

where AC=BD=rt4 2

cm (diagonals)

we know length of a diagonal of a square whose each side is 'a' cm=

2

a

∴4

2

=

2

a

∴a=4That is AB=BC=CD=AD=4cm

Area of the square, ABCD=4

2

=16cm

2

Next, ΔADE

AD=4cm, AE=2.5cm=DE

This is an isoceles triangle

using Heron's formula

Area of ΔADE=

s(s−a)(s−b)(s−c)

where s is the semi perimeter

Perimeter=AD+AE+DE=4+2.5+2.5=9cm

∴ semi perimeter=

2

9

=4.5cm

a=AD,b=AE,c=ED

Area of ΔADE=

4.5(4.5−4)(4.5−2.5)(4.5−2.5)

=

4.5×0.5×2×2

=

9

=3cm

2

∴ Area of ABCDE=Area of ABCD+Area of ΔADE

=16+3=19cm

2

.