Question

Find the area of a whose sides are 6cm, 8cm, 12cm respectively,​

Answer 1

Answer:

ok..but wjat is the shape of that

Answer 2

Answer:

247

Step-by-step explanation:

Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC.

We have, ABC is a right angled triangle at B.

Now, AC2=AB2+BC2 [by Pythagoras theorem]

=62+82=36+64=100

⇒ AC=10cm [taking positive square root]

∴Area of quadrilateral ABCD=Area of △ABC+Area of △ACD

Now, area of △ABC=12×AB×BC [∵area of triangle=12(base×height)]

=12×6×8=24cm2

In △ACD, AC=a=10cm,CD=b=12cm

and DA=c=14cm

Now,semi-perimeter of △ACD,s=a+b+c2=10+12+142=362=18cm

Area of △ACD=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√ [by Heron's formula]

=18(18−10)(18−12)(18−14)−−−−−−−−−−−−−−−−−−−−−−−−√

=18×8×6×4−−−−−−−−−−−−√=(3)2×2×4×2×3×2×4−−−−−−−−−−−−−−−−−−−−−−√

=3×4×23–√×2=246–√cm2

From Eq. (i)

Area of quadrilateralABCD=Area of △ABC+Area of △ACD

=24+246–√=24(1+6–√)cm2

Hence, the area of quadrilateral is 241+6–√−−−−−−√cm2.