Math, asked by anshubopchexyz, 7 months ago

owing:
1. In the adjoining figure, what is. the area of the
each triang
square ABCD?
A
B
45°
D.
Cf6cm →E
(a)
(b)
(a) 36 cm2
(c) 83 cm?
(b) 18 cm
?
(d) can't be determined
on the​

Answers

Answered by mukulsingh2871
1

Answer:

Given

ABCD is square

where AC=BD=42cm (diagonals)

we know length of a diagonal of a square whose each side is 'a' cm=2a

∴42=2a

∴a=4

That is AB=BC=CD=AD=4cm

Area of the square, ABCD=42=16cm2

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=29=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

=3cm2

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

=16+3=1

Answered by cyrilkalbande
0

Answer:

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