Question

Match the below figures with their respective areas in $c{m}^2$.

• A. $6+2\sqrt{2}1$
• B. 306
• C. 114

Answer 1

Case(1):
Quadrilateral ABCD can be divided into two triangles ABC and ADC
So, for triangle ABC
s = $\frac{a+b+c}{2}$
= $\frac{3+4+5}{2}$
= 6 cm
Area of ABC
$=\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$
$=\sqrt{6\times (6-3)\times (6-4)\times (6-5)}$
$=6c{m}^2$
Similarly,
for triangle ADC:
s $=\frac{a+b+c}{2}$
$=\frac{5+5+4}{2}$
$=7cm$

Area of ADC
= $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$
= $\sqrt{7\times (7-5)\times (7-5)\times (7-4)}$
= $2\sqrt{21}$ $c{m}^2$

So, area of quadrilateral ABCD
= ar(ABC) + ar(ADC)
= $6+2\sqrt{21}$ $c{m}^2$

Case(2):
Area of trepezium ABCD
= $\frac{1}{2}\times (bas{e}_1+bas{e}_2)\times height$
= $\frac{1}{2}\times (28+40)\times 9$= 306 $c{m}^2$

Case(3):
Quadrilateral ABCD can be divided into two right triangles ABC and ADC
area of right triangle = $\frac{1}{2}\times base\times height$

ar(ADC) = $\frac{1}{2}\times base\times height$
= $\frac{1}{2}\times 12\times 9$
= 54 $c{m}^2$
ar(ABC) = $\frac{1}{2}\times base\times height$
But, we dont know the value of BC
So, by applying the Pythagoreas theorem
$A{B}^2$ = $A{C}^2+B{C}^2$
${17}^2$ = ${15}^2+B{C}^2$
BC = 8 cm
So,
ar(ABC) = $\frac{1}{2}\times 8\times 15$
= 60 $c{m}^2$
So, area of quadrilateral ABCD = (60+54) $c{m}^2$
= 114 $c{m}^2$