Question

A park in the shape of a quadrilateral ABCD has angle c=90 degree, AB=9m , BC = 12m , CD=5m and AD =8m . How much area does it occupy

Answer 1

$\huge{\bf{\underline{\purple{Answer :}}}}$

Given,

→ABCD is a quadrilateral

→AB = 9 m , BC = 12 m , CD = 5 m , AD = 8 m and $\bold{\angle C = 90}$

To Find :

• Area of quadrilateral = ?

Solution :

• Construction : Join BD , its help to quadrilateral divide into two triangle

⇒ In ΔBCD ,

By applying Pythagoras theorem to find Hypotenuse [ BD] :

$\implies \bold{BD^{2} = BC^{2} \ + \ CD^{2}}$

$\implies \bold{BD^{2} = [ 12]^{2} \ + \ [5]^{2}}$

$\implies \bold{BD^{2} = 169}$

$\implies \bold{BD = 13 \ m }$

So, Now we have Hypotenuse so we find area of ΔBCD :

Area of ΔBCD = $\bold{\frac{1}{2} \ x \ 12 \ x \ 5 }$

Area of ΔBCD = 30 $\bold{m^{2}}$

Now , we find the area of ΔABD ,

We have ,

• a = 13 cm
• b = 9 cm
• c = 8 cm

So , we find firstly Semi perimeter :

→Semi perimeter of ΔABD = $\bold{\frac{8 \ + \ 9 \ + \ 13 \ }{2}}$

Semi perimeter of ΔABD =    $\bold{\frac{30}{2}} = \bold{ 15 \ m }$

Using Heron's Formula :

Area of ΔABD = $\bold{\sqrt{s[s-a] \ [s-b] \ [s-c] }}$

Area of ΔABD = $\bold{\sqrt{15[15 - 13] \ [15 - 13] \ [15-13] }}$

Area of ΔABD = $\bold{\sqrt{15 \ x \ 2 \ x \ 6 \ x 7 }}$

Area of ΔABD = $\bold{\sqrt[6]{35}}$

Area of ΔABD = 35.5 $\bold{m^{2}}$[Approx}

Now, we find Area of Quadrilateral ABCD :

Area of ΔBCD + Area of ΔABD = Area of quadrilateral ABCD

$\implies \bold{30 m^{2}} \ + \bold{ 35.5 m^{2}} = \bold{65.5 m^{2}}$

So Area of quadrilateral = 65.5 m²................Answer

Answer 2

Answer:

${\boxed{ \boxed {\sf{65.5 {m}^{2} }}}}$

Step-by-step explanation:

Given:

ABCD is a Quadrilateral Park , In which

$\angle \: C = 90 \degree \\ AB \: = 9m \\ BC = 12m \\ CD = 5m \\ AD = 8m$

To Find:

$\sf{area \: occupied \: by \: quadrilateral \: field}$

Solution:-

$\sf \red{total \: area \: of \: park = area \: \triangle \: ABD \: + \triangle \: BCD} \\ \\ \rm{according \: to \: question : } \\ \\ \sf {\green{area \: of \triangle \: BCD}} \\ \\ \rm \green{given \: that \: \: CD \: = 5m \: and \: BC = 12m \: and \angle \: C = 90 \degree} \\ \\ \rm \green{ \triangle \: BCD \: is \: right \: angled \: triangle} \\ \\ \rm \green{ \therefore \: area \: of \: \triangle \: BCD = \frac{1}{2} \times base \times \: height} \\ \rm \green{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times 12 \times 5 {m}^{2} } \\ \rm \green{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 30 {m}^{2}}$

$\rm \pink{area \: of \triangle \: ABD} \\ \\ \rm \pink{area \: of \triangle \: = \: \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \rm \pink{s, \: is \: semi \: perimeter} \\ \rm \pink{ \: and \: also \: a , \: b, \: c \: are \: sides \: of \triangle} \\ \\ \rm \pink{given \: that, \: } \\ \rm{a = 8m \: , b = 9m \: and \: c = BD} \\ \rm{also \:( c )\: is \: 90 \degree} \\ \\ \rm \pink{by \: applying \: pythagoras \: theorem} \\ \\ \rm{B{D}^{2} =BC^{2} + CD^{2} } \\ B {D}^{2} = ({12})^{2} + ( {5})^{2} \\ BD ^{2} = 144 + 25 \\ BD ^{2} = 169 \\ BD = \sqrt{169 } \\ BD = 13m$

$\rm \pink{c = BD = 13m} \\ \\ \rm{s = \frac{a + b + c}{2} = \frac{8 + 9 + 13}{2} = \frac{30}{2} = 15m}$

$\rm \blue{area \: of \triangle \: ABD = \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \rm \blue{putting \: value \: of \: a, = 8m \: b = 9m , c = 13m\: and \: s = 15m} \\ \\ \rm \blue{area \: of \triangle \: ABD = \sqrt{15(15 - 8)( 15 - 9)(15 - 13 )}} \\ \rm \blue { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt {15 \times 7 \times 6 \times 2} } \\ \: \: \: \: \rm \blue{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{(3 \times 5) \times 7 \times 6 \times 2}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \blue{ = \sqrt{(3 \times 2) \times 6 \times (7 \times 5)}} \\ \rm \blue{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{( {6}^{2}) \times 35 } } \\ \rm \blue{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6 \sqrt{35}} \\ \rm \blue{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (6 \times 5.91) {m}^{2} } \\ \rm \blue{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = (35.46) {m}^{2} }$

$\rm {\red{ \: area \: of \: park = \: area \: of \: \triangle \: ABD + \: area \: of \triangle \: BCD \: }} \\ \\ \rm \red { \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 35.46 + 30 {m}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \red {\boxed{ = 65.46 {m}^{2} }}$