Question

Find the area of a triangle two sides of which are 14 cm and 8 cm and the perimeter is 34 cm do it in rough note​

Answer 1

Step-by-step explanation:

$\rm Given, \: two\: sides\: of\: the \: triangle$

$\rm First \: side \: (a) = 14cm$

$\rm Second \: side \: (b) = 8cm$

$\rm Let \: the \: third \: side \: be \: " c "$

$\rm Perimeter \: of \: the \: triangle = 34cm$

$\rm \dashrightarrow a + b + c = 34$

$\rm \dashrightarrow 14 + 8 + c = 34$

$\rm \dashrightarrow c = 34 - 22$

$\rm \dashrightarrow c = 12cm$

$\rm \therefore The \: third \: side \: (c) = 12cm$

$\rm We \: have :-$

• a = 14cm

• b = 8cm

• c = 12cm

$\rm Semi- perimeter\: (s)= \dfrac{Perimeter}{2}$

$\rm \dashrightarrow s = \dfrac{34}{2} = 17$

$\rm \leadsto Area \: of \: the \: triangle = \sqrt{s(s-a)(s-b)(s-c)}$

$\rm = \sqrt{17(17-14)(17-8)(17-12)}$

$\rm = \sqrt{17 \times 3 \times 9 \times 5}$

$\rm = \sqrt{2295} \approx 47.91$

$\underline{\boxed{\rm \therefore Area \: of \: the \: triangle = 47.91cm^{2}}}$

Answer 2

$\huge \mathbb{\underbrace{\red{Question\:}}}$

Find the area of a triangle two sides of which are 14 cm and 8 cm and the perimeter is 34 cm

$\huge{\boxed{\sf Solution}}$

Side(a) = 14cm

breadth (b) = 8cm

Perimeter (P) = 34cm

$A \: = \sqrt{s(s - a)(s - b)(s - c)} \\ p \: = a + b + c \\ s \: = \frac{a + b + c}{2} \\ A = \frac{1}{4} \sqrt{ { - p}^{2} + {4p}^{3}a + {4p}^{3}b - 4 {pa}^{2} - 4 {pb}^{2} + 8 {pa}^{2} b + {8pab}^{2} } \\ = \frac{1}{4} \sqrt{ { - 34}^{4} + {4.34}^{3}.14 + {4.34}^{3} .8 - 4. {34.14}^{2} - {12.24}^{2}.14.8 - 4. {34.8}^{2} + 8.34. {14}^{2} .8 + 8.34.14. {8}^{2} } </strong></p><p></p><p><strong>[tex]A \: = \sqrt{s(s - a)(s - b)(s - c)} \\ p \: = a + b + c \\ s \: = \frac{a + b + c}{2} \\ A = \frac{1}{4} \sqrt{ { - p}^{2} + {4p}^{3}a + {4p}^{3}b - 4 {pa}^{2} - 4 {pb}^{2} + 8 {pa}^{2} b + {8pab}^{2} } \\ = \frac{1}{4} \sqrt{ { - 34}^{4} + {4.34}^{3}.14 + {4.34}^{3} .8 - 4. {34.14}^{2} - {12.24}^{2}.14.8 - 4. {34.8}^{2} + 8.34. {14}^{2} .8 + 8.34.14. {8}^{2} }$

≈ 47.90616cm²