From the given data find perimeter of □ABCD, AD = 6√2cm.
Answers
→Consider Δ BAC
Using the Pythagoras theorem
BC
2
=AC
2
+AB
2
By substituting the values
BC
2
=20
2
+21
2
On further calculation
BC
2
=400+441
By addition
BC
2
=841
By taking out the square root
BC=
841
So we get
BC = 29cm
We know that
Perimeter of quadrilateral ABCD = AB + BC + CD + AD
By substituting the values
Perimeter = 21 + 29 + 42 + 34
By addition
Perimeter = 126cm
We know that area of Δ ABC =
2
1
×b×h
It can be written as
Area of Δ ABC =
2
1
×AB×AC
By substituting the values
Area of Δ ABC =
2
1
×21×20
On further calculation
Area of Δ ABC = 210cm
2
Consider Δ ACD
We know that AC = 20cm, CD = 42cm and AD = 34cm
It can be written as a = 20cm, b = 42cm and c = 34cm
So we get
s=
2
a+b+c
s=
2
20+42+34
By division
s=48cm
We know that
Area=
s(s−a)(s−b)(s−c)
By substituting the values
Area=
48(48−20)(48−42)(48−34
So we get
Area=
48×28×6×14
It can be written as
Area=
16×3×14×2×3×2×14
On further calculation
Area=4×3×14×2
By multiplication
Area=336cm
2
So the area of quadrilateral ABCD =Area of ΔABC+ Area of ΔACD
By substituting the values
Area of the quadrilateral ABCD=210+336=546cm
2
Therefore, the perimeter is 210cm and the area is $$546cm^2