The sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm (taken in order), respectively. The angle between the first two sides is a right angle. Its area is 24( √m +1)cm². The value of m is:
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The sides of a quadrilateral ABCD are 6cm,8cm,12cm and 14cm (taken in order).
i.e. ∠ABC=90°
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= √AB²+BC² =√ 6² +8² =√ 36+64 = 100 =10cm
So, the area of △ABC= 1/2 ×base×height
= 1/2×AB×BC
=1/2×6×8=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= (10+12+14)/2 = 36/2
=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)
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