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Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm.
Answer :

In the quadrilateral ABCD,
we have, ?ABC,

We have, AB = 3cm, BC = 4cm, CA = 5cm
image

We have, AC2=AB2+BC2
(By Pythagoras theorem)

= 32+42=9+16=25
? AC2=52
? AC = 5cm

Hence, ?ABC is a right triangle ...(i)

Now in ?ABD, we have,
AC = 5cm, CD = 4cm, DA = 5cm
We know that,
s=a+b+c2
? s=5+4+52=142=7cm

Now, Area of triangle
= 7(7?5)(7?4)(7?5)
(Since, Heron's formula [area = s(s?a)(s?b)(s?c)])

= 7×2×3×2
= 27×3cm2
= 221cm2 = 2 × 4.6cm2 = 9.2cm2(approx) ...(ii)

Since, ?ABC is an right angled triangle, ...(from(i))

Area of ?ABC = 12 × AB × BC
(Since, Area of triangle = 12 × Base × Height)

= 12×3×4=6cm2 ...(iii)

Area of quadrilateral ABCD = Area of ?ABC + Area of ?ACD

Hence from (ii) and (iii),

Area of quadrilateral ABCD = 9.2cm2+6m2=15.2cm2