Question

Find the area of a quadrilateral ABCD in which ab=42,CM bc=21cm ,cd=29cm,da=34cm,and diagonal bd=20cm

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{In quadrilateral ABCD,}$

$\mathsf{AB=42cm,\,BC=21cm,\,CD=29cm,\,AD=34m,\,BD=20cm}$

$\underline{\textsf{To find:}}$

$\textsf{Area of the quadrilateral ABCD}$

$\underline{\textsf{Solution:}}$

$\textsf{Area of quadrilateral ABCD}$

$\textsf{=Area of triangle ABD+Area of triangle BCD}$

$\textsf{In triangle ABD,}$

$\textsf{AB=42cm,\,BD=20cm,\,AD=34cm}$

$\mathsf{s=\dfrac{a+b+c}{2}=\dfrac{42+20+34}{2}=48}$

$\textsf{Area of triangle ABD}$

$\mathsf{=\sqrt{s(s-a)(s-b)(s-c)}}$

$\mathsf{=\sqrt{48(48-42)(48-20)(48-34)}}$

$\mathsf{=\sqrt{48(6)(28)(14)}}$

$\mathsf{=\sqrt{(16{\times}3)(2{\times}3)(4{\times}7)(2{\times}7)}}$

$\mathsf{=\sqrt{4^2{\times}3^2{\times}4^2{\times}7^2}}$

$\mathsf{=4{\times}3{\times}4{\times}7}$

$\mathsf{=336}\,\textsf{square units}$

$\textsf{In triangle BCD,}$

$\textsf{BC=21cm,\,CD=29cm,\,BD=20cm}$

$\mathsf{s=\dfrac{a+b+c}{2}=\dfrac{21+29+20}{2}=35}$

$\textsf{Area of triangle BCD}$

$\mathsf{=\sqrt{s(s-a)(s-b)(s-c)}}$

$\mathsf{=\sqrt{35(35-21)(35-29)(35-20)}}$

$\mathsf{=\sqrt{35(14)(6)(15)}}$

$\mathsf{=\sqrt{(7{\times}5)(2{\times}7)(2{\times}3)(3{\times}5)}}$

$\mathsf{=\sqrt{2^2{\times}3^2{\times}5^2{\times}7^2}}$

$\mathsf{=2{\times}3{\times}5{\times}7}$

$\mathsf{=210}\,\textsf{square units}$

$\textsf{Now,}$

$\textsf{Area of quadrilateral ABCD}\;\mathsf{=336+210}$

$\textsf{Area of quadrilateral ABCD}\;\mathsf{=546}\;\textsf{square units}$

Find more:

In quadrilateral ABCD, angleB = angleD= 90°,

AB = 7 cm, AC = 25 cm, CD = 20 cm.

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