Math, asked by radwintersidhu, 6 months ago

construct a quadrilateral ABCD with AB=5.5 cm , AD=6 CM, angle A =120° , angleD=90° and angle B =40°​




plzz answer me fast.....

Answers

Answered by shivasinghmohan629
1

Answer:

Step-by-step explanation:

ABmake ZBAX whose measure is 400

and at B on AB

make ZABY whose measure is 500. They

meet at C.

Step 4: With A and C as centres draw two arcs of radius 5cm and let them cut at D.

Step 5: Join ADand CD. ABCD is the required quadrilateral.

Step 6: From D draw

DEL ACand from B draw BCL AC Then measure the lengths of BC and DE. BC =

h1= 4.2 cm, DE = h

2= 4.3 cm and AC = d = 5 cm.

Calculation of area:

In the quadrilateral ABCD, d = 5 cm, h1= 4.2 cm and h2= 4.3 cm.

Area of the quadrilateral ABCD = 21d(h1+h2 )= 21

(57.8)(4.2+4.3)= 21×5x8.5.

Answered by amitnrw
2

Given :a quadrilateral ABCD with AB=5.5 cm , AD=6 CM, angle A =120° , angle D=90° and angle B =40°​

To Find : Construct quadrilateral ABCD

Solution:

Step 1 : Draw a line segment AB = 5.5 cm

Step 2 : Draw an angle of 120° using protector at point A on AB

Step 3 : Take a length of 6 cm from A at angle drawn in step 2 and label the other point as D

Step 4 : Draw an angle of 40° at B on AB using protector

Step 5 : Draw an angle of 90° at D on AD using protector such that it intersect angle drawn in step 4 at C

quadrilateral ABCD is constructed

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