in figure AB = 11 cm, BC = 8cm, AD = 3 cm, AC 5cm and A DAC is a right angle.
the area of the quadrilateral ABCD
In "and A DAC is a right angle." I will assume you have a typo and I will just ignore the A in front of DAC
the area of the quadrilateral ABCD = triangle ACD + triangle ABC
= (1/2)(5)(3) + triangle ABC
In triangle ABC, use the cosine law to find angle B
5^2 = 11^2 + 8^2 - 2(11)(8)cos B
..... etc ....
then the area of triangle ABC = (1/2)(11)(8)sin B.
sub back in above and you got it
To find the area of quadrilateral ABCD, we need to break it down into two smaller triangles and then add their areas together.
First, let's find the area of triangle ABC:
We can use the formula for the area of a triangle:
Area = (1/2) * base * height
In triangle ABC, we have:
Base AB = 11 cm
Height BC = 8 cm
So, the area of triangle ABC is:
Area(ABC) = (1/2) * AB * BC
= (1/2) * 11 cm * 8 cm
= 44 cm^2
Next, let's find the area of triangle ACD:
Since triangle ACD is a right-angled triangle, we can use the formula for the area of a right-angled triangle:
Area = (1/2) * base * height
In triangle ACD, we have:
Base AC = 5 cm
Height AD = 3 cm
So, the area of triangle ACD is:
Area(ACD) = (1/2) * AC * AD
= (1/2) * 5 cm * 3 cm
= 7.5 cm^2
Finally, to find the area of quadrilateral ABCD, we add the areas of triangle ABC and triangle ACD:
Area(ABCD) = Area(ABC) + Area(ACD)
= 44 cm^2 + 7.5 cm^2
= 51.5 cm^2
Therefore, the area of quadrilateral ABCD is 51.5 cm^2.