Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm.
Answers
Step-by-step explanation:
Given:
\text{In quardilateral ABCD,}In quardilateral ABCD,
\text{AB=8 cm, BC=6 cm, CD=8 cm, AD=10 cm and AC= 10 cm}AB=8 cm, BC=6 cm, CD=8 cm, AD=10 cm and AC= 10 cm
\textbf{To find:}To find:
\text{Area of quadrilateral ABCD}Area of quadrilateral ABCD
\textbf{Solution:}Solution:
\textbf{Area of triangle ABC:}Area of triangle ABC:
\text{AB= 8 cm, BC= 6 cm and AC=10 cm}AB= 8 cm, BC= 6 cm and AC=10 cm
\text{Here,}\;AB^2+BC^2=AC^2Here,AB
2
+BC
2
=AC
2
\implies\,\triangle\,ABC\;\text{right angled triangle}⟹△ABCright angled triangle
\text{Area of triangle ABC=}\dfrac{1}{2}{\times}b{\times}hArea of triangle ABC=
2
1
×b×h
\text{Area of triangle ABC=}\dfrac{1}{2}{\times}8{\times}6Area of triangle ABC=
2
1
×8×6
\text{Area of triangle ABC=}4{\times}6=24\,\text{square cm.}Area of triangle ABC=4×6=24square cm.
\textbf{Area of triangle ACD:}Area of triangle ACD:
\text{AC=10 cm, CD=8 cm, AD=10 cm}AC=10 cm, CD=8 cm, AD=10 cm
s=\dfrac{a+b+c}{2}s=
2
a+b+c
s=\dfrac{10+8+10}{2}=14s=
2
10+8+10
=14
\text{By using Heron's formula}By using Heron’s formula
\textbf{Area of triangle=}Area of triangle=
=\bf\sqrt{s(s-a)(s-b)(s-c)}=
s(s−a)(s−b)(s−c)
=\sqrt{14(14-10)(14-8)(14-10)}=
14(14−10)(14−8)(14−10)
=\sqrt{14(4)(6)(4)}=
14(4)(6)(4)
=4\sqrt{14(6)}=4
14(6)
=8\sqrt{21}\;\text{square cm.}=8
21
square cm.
\textbf{Area of quadrilateral ABCD}=\textbf{Area of triangle ABC}+\text{Area of triangle ACD}Area of quadrilateral ABCD=Area of triangle ABC+Area of triangle ACD
=24+8\sqrt{21}\;\text{square cm.}=24+8
21
square cm.
\therefore\textbf{Area of quadrilateral ABCD is $\bf24+8\sqrt{21}\;\text{square cm.}$}∴Area of quadrilateral ABCD is 24+8
21
square cm.
Find more:
In quadrilateral ABCD, angleB = angleD= 90°,
AB = 7 cm, AC = 25 cm, CD = 20 cm.
Find the area of ABCD