wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, CAD = 80o and ACD = 40o .

Open in App
Solution

Draw line AB=5.5 cm
Draw an arc of radius 6.5 cm from point A and an other arc of radius 4.5 cm from point B both meet at point C
Join C to A and B
Draw an angle of 400 at point A and an other angle of 400 at point C. Both cut at point D
Join D to C and A
Hence ABCD is the required quadrilateral.
In triangle ABC,
s=6.5+5.5+4.52=8.25
Then area of ABC=8.25(8.256.5)(8.255.5)(8.254.5)=8.25×1.75×2.75×3.75=148.89=12.20
In triangle ACD
ADsin400=6.5sin1000AD=6.5×0.640.98=4.25cm
Then area triangle ACD =12absinθ=6.5×4.25sin400=27.63×0.64=17.68cm2
So area of ABCD quadrilateral =12.20+17.68=29.88 sq cm

685989_616675_ans_ec2f8d2fb37344b2baab6a00c8b092c5.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Quadrilaterals in General
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon