In the trapezium ABCD, AB || DC and angle C=60^ , angle D=45^ If AB = 6cr and BC = 4 cm , find
(1) the height of the trapezium
(ii) length of DC
Answers
Answer:
Correct option is B)
Through B, draw a straight line parallel to AD which meets CD at E.
Now, since AB∥DE and AD∥BE,
∴ABED is a parallelogram.
Thus, ED=AB=18cm
As, BE∥AD and CD is a transversal,
∴∠BED=∠D=60
o
(∵∠ABE=∠D).
Since, in an isosceles trapezium, the base angles are equal, ∠C=∠D=60
o
.
In ΔBEC,∠BEC+∠ECD+∠CBE=180
o
(Angle sum property of a triangle)
⇒60
o
+60
o
+∠CBE=180
o
⇒120
o
+∠CBE=180
o
⇒∠CBE=180
o
−120
o
⇒∠CBE=60
o
As the measure of ∠BEC=60
o
,∠ECB=60
o
and ∠CBE=60
o
,ΔCBE is an equilateral triangle, Thus
∴CE=BC=12cm(BC=AD=12cm)
Now, CE+ED=12+18
⇒DC=30cm
Answer:
(i) 2√3 cm (ii) 8+2√3
Step-by-step explanation:
DC=DE+EF+FC
find (i) FC
(ii) BF
(iii) AE
(iv) DE
solution:
In∆BFC
= cos 60°=FC/4
= 1/2=FC/4
= FC=4/2 = 2
= FC =2 cm
= sin 60°= BF/BC
= √3/2 = BF/4
= BF = 4*√3/2
= BF = 2√3
= AE = BF =2√3 cm.
ABFE is a parallelogram
In ∆ ADE
= tan 45°=2√3/DE
= 1= 2√3/DE
= DE = 2√3
= DC = DE+EF+FC
= DC= 2√3+6+2
= DC=8+2√3.
PLZ MAKE ME A BRIANLY
HOPE IT IS USEFUL
Answer:
(i) 2√3 cm (ii) 8+2√3
Step-by-step explanation:
DC=DE+EF+FC
find (i) FC
(ii) BF
(iii) AE
(iv) DE
solution:
In∆BFC
= cos 60°=FC/4
= 1/2=FC/4
= FC=4/2 = 2
= FC =2 cm
= sin 60°= BF/BC
= √3/2 = BF/4
= BF = 4*√3/2
= BF = 2√3
= AE = BF =2√3 cm.
ABFE is a parallelogram
In ∆ ADE
= tan 45°=2√3/DE
= 1= 2√3/DE
= DE = 2√3
= DC = DE+EF+FC
= DC= 2√3+6+2
= DC=8+2√3.
PLZ MAKE ME A BRIANLY
HOPE IT IS USEFUL