(4)
(2) 66 cm
(3) 96 cm
Direction (Q.36 & Q.37): Read the passage and answer the following questions.
In the given figure ABCD is a square. AC
BD - 42 cm. AE DE - 2.5 cm.
E
A
D
F332
2.5
S
༡༠
C
B В
Read the above passage and answers the questions below.
EF =
(1) 1.5 cm
(2) 2.5 cm
(3) 3 cm
4) 4 cm
Answers
Answered by
0
Answer:
Given
ABCD is square
where AC=BD=4
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=4
That 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
.
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