ABC is a right angle triangle, in which
ABC = 90°. AB = 18 cms and BC = 24 cms.
D is a point which is equidistant from A
and C. If AD = 30 cms, then the area of the
quadrilateral ABCD in sq. cms is
Answers
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, AE/AB=AD/AC=DE/BC … [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=12cm
Consider the equation (i),
AE/AB=AD/AC=DE/BC
Take, AE/AB=AD/AC
4/12=AD/13
1/3=AD/13
(1×13)/3=AD
AD=4.33cm
Now, take AE/AB=DE/BC
4/12=DE/5
1/3=DE/5
DE=(5×1)/3
DE=5/3
DE=1.67cm
(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×(5/3)
=10/3cm
2
Then, area of quadrilateral BCED= area of △ABC− area of △ADE
=
2
1
×BC×AB−10/3
=
2
1
×5×12−10/3
=1×5×6−10/3
=30−10/3
=(90−10)/3
=80/3cm
2
So, the ratio of area of △ADE : area of quadrilateral BCED=(10/3)/(80/3)
=(10/3)×(3/80)
=(10×3)/(3×80)
=(1×1)/(1×8)
=1/8
Therefore, area of △ADE : area of quadrilateral BCED is 1:8.
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