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Answers
Step-by-step explanation:
Given :-
In the given figure, CD = 12 cm ,
BC = 9 cm , <A = 45° ,
<ABD = <BCD = 90°
To find :-
Find the following :
i) Area of ∆BCD
ii) Area of ∆ABD
iii) Area of the quadrilateral ABCD
Solution :-
Given that
In the given figure,
CD = 12 cm ,
BC = 9 cm
<A = 45° ,
<ABD = <BCD = 90°
In ∆BCD,
By Pythagoras Theorem,
=> BD² = BC² + CD²
=> BD² = 9²+12²
=> BD² = 81+144
=> BD² = 225
=> BD =√225
=> BD = 15 cm
We know that
Area of a right angled triangle
= (1/2)ab sq.units
Area of ∆ BCD = (1/2)×BC×CD sq.cm
=> Ar(∆BCD) = (1/2)×9×12 sq.cm
=> Ar(∆BCD) = 9×6 sq.cm
=> Ar(∆BCD) = 54 sq.cm
and
In ∆ ABD ,
< B = 90° ,
<A = 45°
We know that
The sum of the three angles in a triangle = 180°
=> <A + <B + <ADB = 180°
=> 45°+90°+<ADB = 180°
=> 135°+<ADB = 180°
=> <ADB = 180°-135°
=> <ADB = 45°
So, ∆ ADB is am Isosceles right angled triangle.
We have,
<A = <ADB = 45°
=> BD = AB
Since the sides opposite to the equal angles are equal.
=> AB = 15 cm
Now,
Area of ∆ ADB = (1/2)× AB×BD sq.cm
=> (1/2)×15×15 sq.cm
=> 225/2 sq.cm
=> 112.5 sq.cm
Area of ∆ ADB = 112.5 sq.cm
Now,
Area of the quadrilateral ABCD
=> Area(∆BCD)+Area(∆ADB)
=> 54+112.5
=> 166.5 sq.cm
Area of quadrilateral ABCD
= 166.5 sq.cm
Answer:-
i) Area of ∆BCD = 54 sq.cm
ii) Area of ∆ADB = 112.5 sq.cm
iii) Area of the quadrilateral ABCD
= 166.5 sq.cm
Used formulae:-
→ Area of a right angled triangle = (1/2)ab sq.units
→ Area of a triangle = (1/2)×base × height sq.units