Question

Find the area of the shaded region

Answer 1

Answer:

Area of the shaded region = 54 cm²

Step-by-step explanation:

i) In ∆ AOB,

<AOB = 90°,

and OA = 12 cm, OB = 5 cm,

/* By Pythagoras theorem :

AB² = OA² + OB²

= 12² + 5²

= 144 + 25

= 169

=> AB = √169 = √13² = 13 cm,

ii) In ∆AOB , Base (OA) = 12 cm,

Height (OB) = 5 cm

/* By Heron's formula:

$Area \: of \: \triangle AOB = \frac{1}{2} bh\\=\frac{1}{2}\times 12\times 5\\=6\times 5\\=30 \: cm^{2}\:---(1)$

iii ) In ∆ABC , BC = a = 15 cm ,

AC = b = 14 cm,

AB = c = 13 cm

$s = \frac{a+b+c}{3}\\=\frac{15+14+13}{3}\\=\frac{42}{2}\\=21$

s-a = 21 - 15 = 6 ,

s-b = 21 - 14 = 7 ,

s-c = 21 - 13 = 8 ,

$Area\:of \: \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{21\times 6 \times 7 \times 8 }\\=84\:--(2)$

Therefore,

Area of the shaded region = ∆ABC - ∆AOC

= 84 cm² - 30 cm²

= 54 cm²

•••♪

Answer 2

Step-by-step explanation:

Area of the shaded region = 54 cm²

Step-by-step explanation:

i) In ∆ AOB,

<AOB = 90°,

and OA = 12 cm, OB = 5 cm,

/* By Pythagoras theorem :

AB² = OA² + OB²

= 12² + 5²

= 144 + 25

= 169

=> AB = √169 = √13² = 13 cm,

ii) In ∆AOB , Base (OA) = 12 cm,

Height (OB) = 5 cm

/* By Heron's formula:

\begin{gathered}Area \: of \: \triangle AOB = \frac{1}{2} bh\\=\frac{1}{2}\times 12\times 5\\=6\times 5\\=30 \: cm^{2}\:---(1)\end{gathered}

Areaof△AOB=

2

1

bh

=

2

1

×12×5

=6×5

=30cm

2

−−−(1)

iii ) In ∆ABC , BC = a = 15 cm ,

AC = b = 14 cm,

AB = c = 13 cm

\begin{gathered}s = \frac{a+b+c}{3}\\=\frac{15+14+13}{3}\\=\frac{42}{2}\\=21\end{gathered}

s=

3

a+b+c

=

3

15+14+13

=

2

42

=21

s-a = 21 - 15 = 6 ,

s-b = 21 - 14 = 7 ,

s-c = 21 - 13 = 8 ,

\begin{gathered}Area\:of \: \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{21\times 6 \times 7 \times 8 }\\=84\:--(2)\end{gathered}

Areaof△ABC=

s(s−a)(s−b)(s−c)

=

21×6×7×8

=84−−(2)

Therefore,

Area of the shaded region = ∆ABC - ∆AOC

= 84 cm² - 30 cm²

= 54 cm²