Find the area of quadrilateral ABCD in which <B = <C = 120 degrees and AB = 3 cm, BC = 4 cm and CD = 5 cm.

Question

The quadrilateral need not be a trapezium.

KJH, ABCD is NOT a trapezium! So you can't split ABCD into 2 right triangles and a rectangle the way you just described!

Answer 1

The figure could have been an isosceles trapezoid but it is not because side CD = 5 cm while AB = 3 cm. This means that quadrilateral ABCD is not a trapezoid nor a parallelogram.

If a line segment parallel to side BC is drawn from vertex A to intersect side CD, then we would have an isoscelestrapezoid,
The two base angles then would measure 60 degrees each.

Solving for h (the height of this isosceles trapezoid):
h = (AB)sin60deg = 3cmsin60 deg = 2.6 cm
x = (AB)cos60deg = 3cmcos60 deg = 1.5 cm

Area of isosceles trapezoid = (1/2)(B + b)h
A = (1/2)[(1.5 + 1.5 + 4)+ 4](2.6)
A = 14.3 cm^2

Solving for the area of small triangle:
altitutde = 7sin60 deg = 6.06 cm
As = (1/2)(2)(6.06)
As = 6.06 cm^2

Area of ABCD = A + As = 14.3 + 6.06 = 20.36 cm^2 ANS

Hope I help you.

teddy boy

Answer 2

Draw it, and then break the quadrilateral into triangles and rectangles.

The first triangle would have a hypotenuese of 3. The top angle would be 30, so do 3*cos(30) to find the height, and 3*sin(30) to find the base. A=.5bh Repeate the same for the other triangle (essentially, all you have to do is change the 3 to a 5 to find the b and h of that part. All that is left is the inner part of the quadrilateral, which should be simple to solve.

Answer 3

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Answer 4

Join diagonal AC.
Area of triangle ABC + area of triangle ACD = area of quadrilateral ABCD

Use Trigonometry to find the areas of these individual triangles.

For triangle ABC,
Area = (1/2) x AB x BC x sin(
Next, we have to find area of triangle ACD for which we need to know the length of side AC and measure of
To find side AC apply "cosine rule"
AC^2 = AB^2 + BC^2 - 2 x AB x BC x cos( = 3^2 + 4^2 - 2 x 3 x 4 x cos(120)
= 25 - 24x(-1/2)
= 37
So, AC = √37

To find In triangle ABC apply "sine rule"
AB / sin(ACB) = AC / sin(ABC)
3 / sin( 3 / sin( So,
For triangle ACD,
Area = (1/2) x AC x CD x sin(
Hence, area of quadrilateral ABCD = 5.196 + 15.155 = 20.35 sq units.

Answer 5

for triangle ABC
(AC)^2 = 3^2 + 4^2 - 2(3)(4)cos (120) = 37
AC = 6.083 cm
By sine law: 3/sin(BCA) = 6.083.04271/sin(120)
sin(BCA) = 0.4271
so, (AD)^2 = 6.083^2 + 5^2 - 2(6.083)(5) cos(94.717) = 67.0
AD = 8.18567
by Heron's formula
s[ABC] = (3+4+6.083)/2= 6.5415
area of ABC = sqrt{s-3)(s-4)(s-6.083)} = sqrt{3.5415x2.5415x0.4585} = 2.03146 sq cm
and s[ACD] = (6.083 + 5 + 8.18567)/2 = 9.6343
area of ACD = sqrt{(s- 6.083)(s - 5)(s - 8.1867)}
= sqrt{ 3.55134 x 4.63434 x 1.44867} =4.883 sq cm
Total area = 2.03146 + 4.883 = 6.9143 sq cm

Answer 6

i got 40