The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively. and The angle between the first two sides is a right angle. Find its area..
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Answers
Answer:
The sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm (taken in order).
i.e. ∠ABC=90
o
Now, let us join the points A & C.
So, we get two triangles namely △ABC (a right angled triangle) and △ACD.
Applying Pythagoras theorem in △ABC, we get
AC=
AB
2
+BC
2
=
6
2
+8
2
=
36+64
=
100
=10cm
So, the area of △ABC=
2
1
×base×height
=
2
1
×AB×BC
=
2
1
×6×8=24
Now, in △ACD, we have
AC=10cm,CD=12cm,AD=14cm.
According to Heron's formula the area of triangle (A)=
[s(s−a)(s−b)(s−c)]
where, 2s=(a+b+c).
Here, a=10cm,b=12cm,c=14cm
s=
2
(10+12+14)
=
2
36
=18
Area of △ACD=
[18×(18−10)(18−12)(18−14)]
=
(18×8×6×4)
=
(2×3×3×2×2×2×2×3×2×2)
=
[(2×2×2×2×2×2×3×3)×2×3]
=2×2×2×3×
6
=24
6
So, total area of quadrilateral ABCD =△ABC+△ACD
=24+24
6
=24(
6
+1)
Step-by-step explanation:
I hope this will help you
Answer:
to solve are of a right angled triangle we have to use pythogeros theorem so according to it base x hight which is base will be always the longest side of the figure so the base is 14 cm now u have to multiply 14 with the height I and orry after this u have to solve cause without seeing picture we can't figure out which is the height kindly add the picture of the question