In a quadrilateral ABCD ,which is not a trapezium.It is known that <DAB=<ABC=60 DEGREE.moreover , <CAB=<CBD.Then A) AB=BC+CD B) AB=AD+CD C) AB=BC+AD D) AB=AC+AD ? Choose correct option.
Option( c)is correct
i.e.AB=BC+AD
Given

#"In quadrilateral "ABCD, /_DAB=/_ABC=60^@# 
#/_CAB=/_CBD# Construction

# BC and AD " are produced to intersect at E"# 
# In Delta ABE,/_EBA=/_EAB=60^@ =>/_BEA=60^@# 
#:. Delta ABE " is equilateral">AB=BE=EA# Now
#In DeltaABC and Delta BED, /_BED=/_CBA=60^@#
#/_EBD=/_CAB (given)and BE= AB (proved)# # DeltaABC and Delta BED " are congruent by ASA Rule"# #:.DE = BC " corresponding sides"#
#"of congruent triangles"# # "Finally", AB=AE=AD+DE=AD+BC#
So option (c) is correct

By signing up, you agree to our Terms of Service and Privacy Policy
C)
To ease the comprehension of the problem refer to the figure below:
If the sides BC and AD are extended the
First, in addition to the angles informed it was possible to determine other interior angles of the triangles ABC and ABD, as shown in the figure.
Then using the Law of Sines on these triangles:
#triangle_(ABC)#
#"AB"/sin (120^@alpha)="BC"/sin alpha="AC"/sin 60^@#
#triangle_(ABD)#
#"AB"/sin (60^@+alpha)="AD"/sin (60^@alpha)="BD"/sin 60^@#
To focus on viable hypotheses I suggest to begin trying to prove a specific case.
For instance
From
From
Note that
Finally,
#AD+BC=5*(sin 15^@+sin 45^@)/(sin 75^@)=5*1=AB#
So in this case hypothesis (C) is true (I tried the others and verified that they are false.)
Now for the general proof of the hypothesis (C), following the same steps that worked for the specific proof:
From
Observation:
#sin (120^@alpha)=sin (180^@(120^@alpha))=sin(60^@+alpha)#
Therefore,
# "BC" ="AB" * (sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#
From
This will get you
#AD="AB"*sin(60^@alpha)/sin (60^@+alpha)#
#=AB((sqrt(3)/2)*cos alpha(1/2)*sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)# Finally,
#BC+AD="AB"(sin alpha+(sqrt(3)/2)*cos alpha(1/2)*sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#
#BC + AD ="AB" * 1# Therefore,
#AB=BC+AD# and the hypothesis (C) is true.**
By signing up, you agree to our Terms of Service and Privacy Policy
Option D) AB = AC + AD
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
 How do you prove this theorem on trapezoids and its median? The median (or midsegment) of a trapezoid is parallel to each base and its length is one half the sum of the lengths of the bases.
 Two rhombuses have sides with lengths of #2 #. If one rhombus has a corner with an angle of #pi/3 # and the other has a corner with an angle of #(5pi)/12 #, what is the difference between the areas of the rhombuses?
 What is the area of a rhombus?
 Two rhombuses have sides with lengths of #4 #. If one rhombus has a corner with an angle of #pi/12 # and the other has a corner with an angle of #(5pi)/8 #, what is the difference between the areas of the rhombuses?
 A parallelogram has sides with lengths of #9 # and #8 #. If the parallelogram's area is #63 #, what is the length of its longest diagonal?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7