In the given figure, ABCD is a trapezium in which
(1) AB II CD
10
12
(III) AB - 5 cm
Find the value of CD,
Answers
Answer:
@PRAVEE14
============= ••
(( ★ INSTAGRAM → @FARKAT__ ★ ))
__________________
#STAYHOME
#STAYSAFE
__________________
In a trapezium ABCD with bases AB and CD where AB=52,BC=12,CD=39 and DA=5. The area of the trapezium ABCD is
EASY
Help best friend
Study later
ANSWER
Given,
The parallel sides of a trapezium ABCD are52 m and 39 m and the remaining two sides are 12 m and 5 m .
AB=52cm
AE+FB=52−39
=13
Let AE=x, therefore FB=(13−x)
Now,
△DEA is a right angled triangle.
DA
2
=AE
2
+DE
2
=>DE
2
=DA
2
−AE
2
=>DE
2
=12
2
−x
2
=>DE
2
=144−x
2
Similarly,
△CBF is a right angled triangle.
CB
2
=CF
2
+BF
2
=>CF
2
=CB
2
−BF
2
=>CF
2
=5
2
−(13−x)
2
=>CF
2
=25−[(13)
2
+(x
2
)−2.13.x]
=>CF
2
=25−169−x
2
+26x
=>CF
2
=−x
2
+26x−144
Since DE = CF,
144−x
2
=−x
2
+26x−144
=>−26x=−288
=>x=11
AE=11 and
FB=13−11=2
∴DE=CF=
144−11
2
=4.8
Area of trapezum ABCD=Area of △DEA+Area of △CFB+Area of rectangle ABEF
=(
2
1
×11×4.8+
2
1
×2×4.8+39×4.8)m
2
=(26.4+4.8+187.2)m
2
=218.4m
2