Question

find area of triangle two sides are 16cm and 22cm and the perimeter is 64cm. ans.is
$32 \sqrt{30}$
cmsq.

Answer 1

Given,
The given two sides are 16cm and 22cm
And, Perimeter of Triangle is 64cm.

We know that,

$\bf \: Perimeter \: = \: a + b + c$

Let the first side ( i.e. 16cm) be a
and the second side ( i.e. 22cm) be b

Putting the values,

64cm = 16cm + 22cm + c

c = 64 - 38

c = 26cm.

Semi-perimeter (s) = perimeter / 2

S = 64 / 2

s = 32

Now,

Using the Heron's Formula,

$\bf \: \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{32(32 - 16)(32 - 22)(32 - 26)} \\ \\ = \sqrt{32(16)(10)(6)} \\ \\ = \sqrt{30720} \\ \\ = 32 \sqrt{30}$

Answer 2

Given : -
$\underline{two \:sides\: of\: triangle\: are,}$
(a) = 16 cm
(b) = 22 cm
(c) = ?

$\underline{Perimeter\: of\: triangle = 64 cm. }$

We know that,
$\bf{Perimeter\: of\: triangle = sum \:of \:all\: three \:sides \:of\: triangle.}$

$= > 64 = a + b + c \\ \\ = > 64 = 16 + 22 + c \\ \\ = > 64 = 38 + c \\ \\ = > 64 - 38 = c \\ \\ = > 26 = c$

Therefore, third side of triangle (c) = 26 cm.

Now,

$\bf{Semi\: Perimeter\: of \:triangle\:(s) }$ = $\frac{Perimeter}{2}$

= (s) = $\frac{a+b+c}{2}$

= (s) = $\frac{16+22+26}{2}$

= (s) = $\frac{64}{2}$

= (s) = 32

Now,
using Heron's Formula,

Area of triangle =
$\sqrt{s(s-a) (s-b) (s-c) }$ sq. unit
.

Area of triangle =
$\sqrt{32(32-16) (32-22) (32-26) }$cm²
.

Area of triangle =
$\sqrt{32(16) (10) (6) }$cm²
.

Area of triangle =
$\sqrt{2×2×2×2×2(4×4) (2×5) (2×3) }$cm²
.

Area of triangle =
2×2×2×4$\sqrt{2×5×3}$cm²
.

Area of triangle =
32$\sqrt{30}$cm²