Question

29. The sides of a triangle are 16cm,
30cm, and 34cm. Its area is
O (a) 225 sq.cm.
O (b) 240 sq.cm
O (c) 22572 sq.cm
O (d) 250 sq.cm​

Answer 1

Step-by-step explanation:

Given sides of triangles are a=16cm,b=30cm,c=34cm respectively

Semiperimeter s=

2

16+30+34

cm

=

2

80

=40cm

By heron's, area of the triangle is

s(s−a)(s−b)(s−c)

cm

2

=

40(40−16)(40−30)(40−34)

cm

2

=

40×24×10×6

cm

2

=

4×10×4×6×10×6

cm

2

=4×6×10cm

2

=240cm

2

Answer 2

GIVEN :-

• Sides of triangle are 16 cm , 30 cm , 34 cm.

TO FIND :-

• The area of the triangle .

SOLUTION :-

Sides or triangle are,

• a = 16 cm.
• b = 30 cm.
• c = 34 cm.

Now,

$\\ : \implies \displaystyle \sf \: perimeter \: of \: \triangle \: (s) = \frac{a + b + c}{2}$

$: \implies \displaystyle \sf \: perimeter \: of \: \triangle \: (s) = \frac{16 + 30 + 34}{2}$

$: \implies \displaystyle \sf \: perimeter \: of \: \triangle \: (s) = \frac{80}{2}$

$: \implies \underline{ \boxed{ \displaystyle \sf \: perimeter \: of \: \triangle \: (s) = 40 \: cm}}$

_____________________

$\\ \dashrightarrow \displaystyle \sf \: Area\: of \: \triangle \: = \sqrt{s(s - a)(s - b)(s - c)}$

$\dashrightarrow \displaystyle \sf \: Area\: of \: \triangle \: = \sqrt{40(40 - 16)(40 - 30)(40 - 34)}$

$\dashrightarrow \displaystyle \sf \: Area\: of \: \triangle \: = \sqrt{40 \times 24 \times 10 \times 6}$

$\dashrightarrow \displaystyle \sf \: Area\: of \: \triangle \: = \sqrt{57600}$

$\dashrightarrow \underline{ \boxed{\displaystyle \sf \: Area\: of \: \triangle \: = 240 \: cm ^{2} }}$