Question

Find the perimeter and area of the quadrilateral ABCD in which AB=8cm ,AD=10cm,BD=12cm,angle DBC=90degree ,DC=13cm.

Answer 1

Answer:

Step-by-step explanation:

Perimeter=sum of all side

=10+8+12+13=43

For area sorry your fig is wrong and missing some information

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Answer 2

Answer:

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Here is Your Answer...!!

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Step by step solution:

$Given \ AB=8 \ cm , AD=10 \ cm,BD=12 \ cm,DC=13 \ cm \ and \ DBC \ is \ right \ angled \ triangle.\\\\DBC \ is \ right \ angled \ triangle\\\\(DC)^2=(BD)^2+(BC)^2\\\\(13)^2=(12)^2+(BC)^2\\\\BC=\sqrt{169-144} \ cm\\ \\BC=\sqrt{25} \ cm =5 \ cm\\\\Now \ perimeter \ of \ quadrilateral \ ABCD=AB+BC+CD+AD\\\\Putting \ value \ here\\\\Perimeter \ of \ quadrilateral \ ABCD=10+8+5+13=36 \ cm$

$Area \ of \ quadrilateral \ ABCD=area \ of \ triangle \ ABD \ + \ area \ of \ triangle \ DBC\\\\Area \ of \ triangle \ DBC=\frac{1}{2} \times BC \times DC\\\\Area \ of \ triangle \ DBC=\frac{1}{2} \times12 \times5=\frac{60}{2}=30 \ cm^2$

$Area \ of \ triangle \ ABD=\sqrt{s(s-a)(s-b)(s-b)}\\\\s=\frac{a+b+c}{2}\\\\s=\frac{10+12+8}{2}=\frac{30}{2}=15 \ cm\\\\Area \ of \ triangle \ ABD= \sqrt{15(15-10)(15-12)(15-8)}\\\\Area \ of \ triangle \ ABD=\sqrt{3 \times5 \times5 \times3 \times7}\\\\Area \ of \ triangle \ ABD=15\sqrt{7} \ cm^2=39.68 \ cm^2$

$Area \ of \ quadrilateral \ ABCD=30 \ cm^2+39.68 \ cm^2\\\\Area \ of \ quadrilateral \ ABCD=69.18 \ cm^2$

Hope it is clear to you.