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PLEASE SHARE WITH COMPLETE WRITTEN SOLUTION OR ELSE IT WILL BE REPORTED!!!
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Step-by-step explanation:
From the question it is given that,
∠ABC=90
∘
AB and DE is perpendicular to AC
(i) Consider the △ADE and △ACB,
∠A=∠A … [common angle for both triangle]
∠B=∠E … [both angles are equal to 90
∘
]
Therefore, △ADE∼△ACB
(ii) From (i) we proved that, △ADE∼△ACB
So,
AB
AE
=
AC
AD
=
BC
DE
… [equation(i)]
Consider the △ABC, is a right angle triangle
From Pythagoras theorem, we have:
AC
2
=AB
2
+BC
2
13
2
=AB
2
+5
2
169=AB
2
+25
AB
2
=169−25
AB
2
=144
AB=
144
AB=12 cm
Consider the equation (i),
AB
AE
=
AC
AD
=
BC
DE
Taking
AB
AE
=
AC
AD
12
4
=
13
AD
3
1
=
13
AD
AD=
3
(1×13)
cm
AD=4.33 cm
Now, take
AB
AE
=
BC
DE
12
4
=
5
DE
1/3=DE/5
DE=
3
(5×1)
DE=
3
5
DE=1.67 cm
(iii) Now, we have to find area of △ADE : area of quadrilateral BCED,
We know that, Area of △ADE=
2
1
×AE×DE
=
2
1
×4×
3
5
=
3
10
cm
2
Then, area of quadrilateral BCED= area of △ABC− area of △ADE
=
2
1
×BC×AB−
3
10
=
2
1
×5×12−
3
10
=1×5×6−
3
10
=30−
3
10
=
3
(90−10)
=
3
80
cm
2
So, the ratio of area of △ADE : area of quadrilateral BCED=
3
80
3
10
=
3
10
×
80
3
=
8
1
Therefore, area of △ADE : area of quadrilateral BCED is 1:8.
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