Question

The area of the quadrilateral ABCD in which AB = 6 cm, BC = 8 cm, CD = 8 cm, AD = 10 cm and
AC = 10 cm. (in cm) is
(a) 8 13
(b) 8(3+sqrt21 )
(c) 8sqrt21
(d) 24.

Answer 1

Answer:

(d) 24. answer is correct

Answer 2

Answer:

$\large{ \frak{ \green{ \underline{ \underline{☘ Question}}}}}$

The area of the quadrilateral ABCD in which AB = 6 cm, BC = 8 cm, CD = 8 cm, AD = 10 cm and AC = 10 cm is

$\large{ \frak{ \red{ \underline{ \underline{✿Answer}}}}}$

$\large{ \frak{ \underline{Given :}}}$

AB = 6 cm

BC = 8 cm

CD = 8 cm

AD = 10 cm

AC (diagonal of Quadrilateral) = 10 cm

$\large{ \frak{ \underline{To \: find :}}}$

Area of $\sf{ \square{ABCD}}$

$\large{ \frak{ \underline{Formula \: used :}}}$

Heron's formula

$\sf = \sqrt{s(s - a)(s - b)(s - c)}$

Where a, b & c are the sides of a triangle

And s is the semi perimeter

$\sf \: s = \frac{a + b + c}{2}$

$\large{ \frak{ \underline{Solⁿ :}}}$

$\sf{ \square{ABCD}}$ is divided into two triangles $\sf{ \triangle{ABC}}$ and $\sf{ \triangle{ADC}}$

∴ Area of $\sf{ \square{ABCD}}$ = Area of $\sf{ \triangle{ABC}}$ + Area of $\sf{ \triangle{ADC}}$

Now,

In $\sf{ \triangle{ABC}}$

$\sf \: a = 6 \: cm$

$\sf \: b = 8 \: cm$

$\sf \: c = 10 \: cm$

$\sf∴s = \frac{6 + 8 + 10}{2} cm$

$\sf \: \: \: \: \: \: \: = \frac{ \cancel{24}}{ \cancel{2}} \: cm$

$\sf \: \: \: \: \: \: \: = 12cm$

Area of $\sf{ \triangle{ABC}}$

$\sf = \sqrt{12(12 - 6)(12 - 8)(12 - 10)} {cm}^{2}$

$\sf = \sqrt{12 \times 6 \times 4 \times 2} {cm}^{2}$

$\sf = \sqrt{2 \times 2 \times 3 \times 2 \times 3 \times 2 \times 2 \times 2 \:} {cm}^{2}$

$\sf = \sqrt{ {2}^{2} \times {2}^{2} \times {2}^{2} \times {3}^{2}} \: {cm}^{2}$

$\sf = 2 \times 2 \times 2 \times 3 \: {cm}^{2}$

$\sf = 24 \: {cm}^{2}$

In $\sf{ \triangle{ADC}}$

$\sf \: a = 8 \: cm$

$\sf \: b = 10 \: cm$

$\sf \: c = 10 \: cm$

$\sf∴s = \frac{8 + 10 + 10}{2} \: cm$

$\sf \: \: \: \: \: \: \: = \frac{ \cancel{28}}{ \cancel{2}} cm$

$\sf \: \: \: \: \: \: \: = 14 \: cm$

Area of $\sf{ \triangle{ADC}}$

$\sf = \sqrt{14(14 - 8)(14 - 10)(14 - 10)} \: cm {}^{2}$

$\sf = \sqrt{14 \times 6 \times 4 \times 4} \: cm {}^{2}$

$\sf = \sqrt{2 \times 7 \times 2 \times 3 \times 2 \times 2 \times 2 \times 2} \: {cm}^{2}$

$\sf = \sqrt{ {2}^{2} \times {2}^{2} \times {2}^{2} \times 3 \times 7} \: {cm}^{2}$

$\sf = 8 \sqrt{21} cm {}^{2}$

∴ Area of $\sf{ \square{ABCD}}$

$\sf = 24 + 8 \sqrt{21} \: cm {}^{2}$

$\sf = 8(3 + \sqrt{21} )cm {}^{2}$

Hence, ⓑ $\sf 8(3 + \sqrt{21} )cm {}^{2}$ is correct.

$\large{ \frak{ \color{blue}{❅ \: hope \: it \: helps...}}}$

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