the sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm respectively and the angle between first two sides is right angle. find its area.
Answers
angle B be right angle
now join diagonal AC
therefore quadrilateral ABCD now devides into two triangles ABC and ACD
The sum of areas of these two triangles is area of quadrilateral
now area of triangle ABC = 1/2 × 8 ×6
= 24 sq cm
to find area of triangle ACD we use Heron's formula
the sides of triangle ACD are as
AD =14 cm. and CD = 12 cm
the third side AC will be 10 cm can be find using Pythagoras theorem in triangle ABC
So ,
s = 10 +12 + 14 /2
= 36 /2
18
by Heron's formula ,
area of triangle ACD = √ s(s-a)(s-b)(s-c)
= √18 *(18-10)(18-12)(18-14)
= √18 ×8 ×6 ×4
= √576 × 6
= 24√6
= 24 × 2.45
= 58.8 sq cm
therefore area of rectangle is , 24 + 58.8= 82.8 sq cm
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(+1)