Question

find the area of triangle whose sides are 4.5 cm and 10 cm and perimeter is equals to 10.5 CM

Answer 1

Answer:

The area of the triangle is $3\sqrt { 14 } cm^2$.

Step-by-step explanation:

Let a, b and c be the three sides of the triangle. Here, a=4.5 cm, b=10 cm and c=?

∴ Semi perimeter =$\frac { a+b+c }{ 2 }$ ⇒$\frac { (4.5+10+c) }{ 2 } =10.5$

⇒$14.5 +c = 2 \times 10.5 = 21$

⇒$c = 21 - 14.5 = 6.5 cm$

∴ Area of triangle (Using Heron's formula) $=\sqrt { (s(s-a)(s-b)(s-c)) }$

Square units $=\sqrt { 10.5 (10.5-10) (10.5-4.5) (10.5-6.5)} cm^2 \\=\sqrt{10.5 (0.5) (6) (4)} cm^2$

$=\sqrt {(10.5\times12)} cm^2 =\sqrt {126} cm^2 =\sqrt{ (14\times9)} cm^2 =3\sqrt{14} cm^2$

Hence, Area of triangle =$3\sqrt { 14 } cm^2$.

Answer 2

Answer:

∴ Semi perimeter =\frac { a+b+c }{ 2 }

2

a+b+c

⇒\frac { (4.5+10+c) }{ 2 } =10.5

2

(4.5+10+c)

=10.5

⇒14.5 +c = 2 \times 10.5 = 2114.5+c=2×10.5=21

⇒c = 21 - 14.5 = 6.5 cmc=21−14.5=6.5cm

∴ Area of triangle (Using Heron's formula) =\sqrt { (s(s-a)(s-b)(s-c)) }=

(s(s−a)(s−b)(s−c))

Square units \begin{lgathered}=\sqrt { 10.5 (10.5-10) (10.5-4.5) (10.5-6.5)} cm^2 \\=\sqrt{10.5 (0.5) (6) (4)} cm^2\end{lgathered}

=

10.5(10.5−10)(10.5−4.5)(10.5−6.5)

cm

2

=

10.5(0.5)(6)(4)

cm

2

=\sqrt {(10.5\times12)} cm^2 =\sqrt {126} cm^2 =\sqrt{ (14\times9)} cm^2 =3\sqrt{14} cm^2=

(10.5×12)

cm

2

=

126

cm

2

=

(14×9)

cm

2

=3

14

cm

2

Hence, Area of triangle =3\sqrt { 14 } cm^23

14

cm