Question

Find the area of the triangles field with the length of the sides 9 cm, 10 cm and 17 cm.

Answer 1

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Find the area of the triangles field with the length of the sides 9 cm, 10 cm and 17 cm.

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Area of Triangle is 36cm²

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• Sides of triangle are 9cm, 10cm and 17cm

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• Area of Triangle

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In this question, all sides of triangle are given

So, we will find area of Triangle by Heron's Formula

Heron's Formula

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

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Let,

a = 9cm

b = 10cm

c = 17cm

Semi perimeter = a + b + c /2 = 36/2 = 18cm

Area of Triangle by Heron's Formula

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

$\bf \sqrt{18(18 - 9)(18 - 10)(18 - 17)}$

$\bf \sqrt{18 \times 9 \times 8 \times 1}$

$\bf \sqrt{1296}$

$\bf 36 {cm}^{2}$

Therefore, Area of triangle is 36cm².

$\bf\blue{hope \: } \pink{it \: } \purple{helps \: } \red{uh...!!}$

Answer 2

$\huge\fbox\red{\fbox\blue{Solution:-}}$

$\fbox\red{\fbox\pink{Given:-}}$

$\red{Sides = 9cm }\\ = \orange{ 10cm} \\ = \green{17cm}$

$\fbox\pink{\fbox\red{By \: the \: problem:-}}$

$\blue{perimeter = {(a + b + c)} } \\ = > \green{ semi \: perimeter \:(s) = \blue{ \frac{(a + b + c)}{2}}} \\ = > \red{ \frac{(9+ 10 + 17)}{2} }\\ = > \pink{ \frac{36}{2 } } \\ = > \green{18 \: cm}$

$\fbox\red{\fbox\green{Formula:-}}$

$\underline \pink{Heron's\: formula }\\ \green{ Area \: of \: the \: triangle} \\ = \red{ \sqrt{s(s - a)(s - b)(s - c)} } -\\ \pink{where \: a, \: b, \: c \: are \: sides \: } \\ \green{and \: s \: is \: semi \: perimeter}.$

$\fbox\pink{\fbox\red{by \: the \: problem:-}}$

$\blue{Area \: of \: that \: field } \\ = \red{\sqrt{s(s - a)(s - b)(s - c)}} \\ = \orange{ \sqrt{18(18 - 9)(18 - 10)(18 - 17)} }\\ = \green{ \sqrt{18 \times 9 \times 8 \times 1} } \\ = \blue{ \sqrt{2 \times 3 \times 3 \times 3 \times 3 \times 2 \times 2 \times 2 \times 1}} \\ = \red{\sqrt{ {2}^{4} \times {3}^{4} } } \\ = > \green{{2}^{2} \times {3}^{2} = 36 {cm}^{2} }$

$\fbox\red{\fbox\green{Answer:-}}$

$\pink{ \: 36 {cm}^{2} }$