Question

find the area of the quadrilateral ABCD in which AB equal to 3 cm BC equal to 4 cm CD equal to 4 cm equal to 5 cm and ac equal to 5 cm. find the the answer of 12.2.​

Answer 1

Answer:

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

•••••• ::::::: SOLUTION ::::::::••••••

Area of Quadrilateral = Ar of ∆ABC + Ar of ∆ADC

◆ • • Area of ∆ ABC

by using Heron's formula :

\begin{lgathered}= \sqrt{s(s - a)(s - b)(s - c} \\ s = \frac{a + b + c}{2} = \frac{3 + 4 + 5}{2} = \frac{12}{2} \\ = 6 \\ \\ ar \: of \: triangle \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} {cm}^{2} \\ = \sqrt{6 \times 3 \times 2 \times 1} {cm}^{2} \\ = \sqrt{6 \times 6} = 6 {cm}^{2}\end{lgathered}

=

s(s−a)(s−b)(s−c

s=

2

a+b+c

=

2

3+4+5

=

2

12

=6

aroftriangle

=

s(s−a)(s−b)(s−c)

=

6(6−3)(6−4)(6−5)

cm

2

=

6×3×2×1

cm

2

=

6×6

=6cm

2

◆ • • Area of ∆ ADC

by using Heron's formula

\begin{lgathered}s = \frac{a + b + c}{2} \\ = \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7 \\ \\ ar \: of \: triangle \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} \\ = \sqrt{7 \times 2 \times 3 \times 2} \\ = 2 \sqrt{21} \ {cm}^{2} \\ = 2 \times 4.58 = 9.16 {cm}^{2}\end{lgathered}

s=

2

a+b+c

=

2

5+4+5

=

2

14

=7

aroftriangle

=

s(s−a)(s−b)(s−c)

=

7(7−5)(7−4)(7−5)

=

7×2×3×2

=2

21

cm

2

=2×4.58=9.16cm

2

now ,

AREA of Quadrilateral = ar of ∆ABC + ar of ∆ADC

\begin{lgathered}=( 6 + 9.16) {cm}^{2} \\ = 15.16 {cm}^{2} \\ = 15.2 {cm}^{2} (approx.)\end{lgathered}

=(6+9.16)cm

2

=15.16cm

2

=15.2cm

2

(approx.)

So , the area of Quadrilateral ABCD is 15.2cm sq.

Answer 2

Given :-

• ABCD is a quadrilateral in which AB = 3 cm , BC = 4 cm , CD = 4 cm , DA = 5 cm and AC = 5 cm.

To find :-

• The area of quadrilateral

Solution :-

In ∆ ABC

• a = 3 cm
• b = 4 cm
• c = 5 cm

$\rm\:s=\dfrac{a+b+c}{2}$

$\rm\longrightarrow\:s=\dfrac{3+4+5}{2}$

$\rm\longrightarrow\:s=\cancel\dfrac{12}{2}$

$\rm\longrightarrow\:s=6\:cm$

⭐ By using Heron's formula,

$\blue{\bigstar}\:\:{\underline{\boxed{\bf\red{Area=\sqrt{s(s-a)(s-b)(s-c)}}}}}$

$\rm\longrightarrow\:Area=\sqrt{6(6-3)(6-4)(6-5)}$

$\rm\longrightarrow\:Area=\sqrt{6\times\:3\times\:2\times\:1}$

$\rm\longrightarrow\:Area=\sqrt{36}$

$\rm\longrightarrow\:Area=6\:c{m}^{2}$

Hence,the area of ∆ ABC will be 6 cm².

Now,

In ∆ ACD

• a = 5 cm
• b = 4 cm
• c = 5 cm

$\rm\:s=\dfrac{a+b+c}{2}$

$\rm\longrightarrow\:s=\dfrac{5+4+5}{2}$

$\rm\longrightarrow\:s=\cancel\dfrac{14}{2}$

$\rm\longrightarrow\:s=7\:cm$

⭐ By using Heron's formula,

$\gray{\bigstar}\:\:{\underline{\boxed{\bf\purple{Area=\sqrt{s(s-a)(s-b)(s-c)}}}}}$

$\rm\longrightarrow\:Area=\sqrt{7(7-5)(7-4)(7-5)}$

$\rm\longrightarrow\:Area=\sqrt{7\times\:2\times\:3\times\:2}$

$\rm\longrightarrow\:Area=\sqrt{84}$

$\rm\longrightarrow\:Area=2\sqrt{21}$

$\rm\longrightarrow\:Area=2\times\:\sqrt{21}$

$\rm\longrightarrow\:Area=2\times\:4.60$

$\rm\longrightarrow\:Area=9.2\:c{m}^{2}$

Hence,the area of ∆ ACD will be 9.2 cm²

Then,

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

Area of quadrilateral ABCD = 6 + 9.2

Area of quadrilateral ABCD = 15.2 cm²

Hence,the area of quadrilateral will be 15.2 cm².