Question

find the area of a triangle two sides of which are 28 cm and 10 cm and the perimeter is 42 cm

Answer 1

Correct Question:-

A triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

To FinD:-

The area.

SolutioN:-

• We know that measures of all 3 sides of a triangle sum up to its perimeter.
• Perimeter = 42 cm.
• 1st side = 18 cm.
• 2nd Side = 10 cm.

Let the third side be x.

According to the question,

$\large\implies{\sf{18+10+x=42}}$

$\large\implies{\sf{18+x=42}}$

$\large\implies{\sf{x=42-18}}$

$\large\therefore\boxed{\bf{x=14}}$

Therefore, the third side is 14 cm.

VerificatioN:-

$\large\implies{\sf{28+10+14=42}}$

$\large\implies{\sf{42=42}}$

$\large\therefore\boxed{\bf{LHS=RHS}}$

____________________________

Now the Area:-

By using Heron's Formula,

$\large{\green{\underline{\boxed{\bf{Area=\sqrt{s(s-a)(s-b)(s-c)}}}}}}$

where,

• s is Semiperimeter = Perimeter/2 = 42/2 = 21 cm
• a = 18 cm
• b = 10 cm
• c = 14 cm

Putting the values,

$\large\implies{\sf{Area=\sqrt{21(21-18)(21-10)(21-14)}}}$

$\large\implies{\sf{Area=\sqrt{21\times3\times11\times7}}}$

$\large\implies{\sf{Area=\sqrt{21\times21\times11}}}$

$\large\therefore\boxed{\bf{Area=21\sqrt{11}\:cm^2.}}$

The area is 21√11 cm².

Answer 2

$\implies \huge \sf\underline \red{Answer}$

$\sf\therefore \underline \purple{area = 21 \sqrt{11} {cm}^{2}}$

To find:

• Area of triangle

solution:

we know that area of triangle formula,

$\bf{ \boxed{ \underline{ \underline{ \red{ \sf{area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c) \: }}}}}}}$

$\sf \underline{so}$

• s is semi perimeter

• a,b,c is sides of triangle

$\sf \underline{Given}$

• a = 18 cm

• b = 10 cm

• perimeter = 42cm

$\sf \underline{we \: know \: that \: semi \: perimeter \: formula}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{semi \: perimeter = \dfrac{perimeter}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{s = \dfrac{42}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{semi = 21cm}$

we have to find c

$\sf \underline{now}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{perimeter = 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{a + b + c = 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{18cm + 10cm + c = 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{28cm + c= 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{c = 42 - 28cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{c = 14cm}$

$\sf \underline{take \: the \: formula \: again : }$

$\bf{ \boxed{ \underline{ \underline{ \red{ \sf{area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c) \: }}}}}}}$

• a = 18 cm

• b = 10 cm

• c = 14 cm

• s = 21 cm

$\bf{ \boxed{ \underline{ \underline{ \red{ \sf{area \: of \: triangle = \sqrt{s(s - a)(s - b)(s - c) \: }}}}}}}$

$\: \: \: \: \: \: \sf{Area \: of \: traingle = \sqrt{21(21 - 18)(21 - 10)(21 - 14)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21(3)(11)(7)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21(7 \times 3)(11)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21(21)(11)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21 \times 21 \times \sqrt{11}} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{ {21}^{2} \times \sqrt{11}} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 21 \times \sqrt{11} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 21 \sqrt{11} }$

$\sf \purple\therefore \underline{area = 21 \sqrt{11} {cm}^{2}}$