Question

find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.​

Answer 1

${\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}$

• $\bf\:perimeter=42\:cm$

• $\bf\:First\:side\:(a)=18\:cm$

• $\bf\:second\:side\:(b)=10\:cm$

$\bf\:Third\:side\:(c)=perimeter-(a+b)$

$\bf\:Third\:side=42-(18+10)\:cm$

$\bf\:Third\:side=42-28\:cm$

• $\bf\:Third\:side=14\:cm$

${\bold{\blue{\underline{\red{To}\:\pink{Fi}\green{nd}\purple{:-}}}}}$

• $\bf\green{{Area\:of\:triangle=?}}$

${\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$

Now we have,

a = 18 cm ,b = 10 cm and c = 14 cm

$\bf\:Semi\:perimeter=\dfrac{a+b+c}{2}$

$\bf\:Semi\:perimeter=\dfrac{18+10+14}{2}$

$\bf\:Semi\:perimeter=\dfrac{42}{2}$

$\bf\red{{Semi\:perimeter=21\:cm}}$

$\bf\boxed\star\pink{\underline{\underline{{Using\:heron's\:formula:-}}}}$

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

$\bf\:Area=\sqrt{21(21-18)(21-10)(21-14)}$

$\bf\:Area=\sqrt{21\times\:3\times\:11\times\:7}$

$\bf\:Area=\sqrt{3\times\:7\times\:3\times\:11\times\:7}$

$\bf\:Area=3\times\:7\sqrt{11}\:c{m}^{2}$

$\bf\orange{{Area=21\sqrt{11}\:c{m}^{2}}}$

Hence,area of triangle will be 21√11 cm²

Answer 2

{\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}

Given:−

\bf\:perimeter=42\:cmperimeter=42cm

\bf\:First\:side\:(a)=18\:cmFirstside(a)=18cm

\bf\:second\:side\:(b)=10\:cmsecondside(b)=10cm

\bf\:Third\:side\:(c)=perimeter-(a+b)Thirdside(c)=perimeter−(a+b)

\bf\:Third\:side=42-(18+10)\:cmThirdside=42−(18+10)cm

\bf\:Third\:side=42-28\:cmThirdside=42−28cm

\bf\:Third\:side=14\:cmThirdside=14cm

{\bold{\blue{\underline{\red{To}\:\pink{Fi}\green{nd}\purple{:-}}}}}

ToFind:−

\bf\green{{Area\:of\:triangle=?}}Areaoftriangle=?

{\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}

Solution:−

Now we have,

a = 18 cm ,b = 10 cm and c = 14 cm

\bf\:Semi\:perimeter=\dfrac{a+b+c}{2}Semiperimeter=

2

a+b+c

\bf\:Semi\:perimeter=\dfrac{18+10+14}{2}Semiperimeter=

2

18+10+14

\bf\:Semi\:perimeter=\dfrac{42}{2}Semiperimeter=

2

42

\bf\red{{Semi\:perimeter=21\:cm}}Semiperimeter=21cm

\bf\boxed\star\pink{\underline{\underline{{Using\:heron's\:formula:-}}}}

⋆

Usingheron

′

sformula:−

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

s(s−a)(s−b)(s−c)

\bf\:Area=\sqrt{21(21-18)(21-10)(21-14)}Area=

21(21−18)(21−10)(21−14)

\bf\:Area=\sqrt{21\times\:3\times\:11\times\:7}Area=

21×3×11×7

\bf\:Area=\sqrt{3\times\:7\times\:3\times\:11\times\:7}Area=

3×7×3×11×7

\bf\:Area=3\times\:7\sqrt{11}\:c{m}^{2}Area=3×7

11

cm

2

\bf\orange{{Area=21\sqrt{11}\:c{m}^{2}}}Area=21

11

cm

2

Hence,area of triangle will be 21√11 cm²