Question

Find the area of a triangle two sides of which are 18 c.m. and 10 c.m. and perimeter is

42​

Answer 1

Given :-

First side of the triangle = 18 cm

Second side of the triangle = 10 cm

Perimeter of the triangle = 42 cm

To Find :-

The area of the triangle.

Analysis :-

Firstly, we've to find the third side of the triangle which should be considered as a variable.

Substitute the values given in the formula of perimeter of triangle.

Then find the third angle accordingly.

Next we've to find the semi perimeter by dividing the perimeter given by two.

Finally, substitute the values we got using Heron's formula and find the area accordingly.

Solution :-

We know that,

• p = Perimeter
• a = Area
• s = Semi perimeter

Let the third side of the triangle be 'x'.

By the formula,

$\underline{\boxed{\sf Perimeter \ of \ a \ triangle=a+b+c}}$

Given that,

Perimeter (p) = 42 cm

First side = 18 cm

Second side = 10 cm

Substituting their values,

42 = 18 + 10 + x

42 = 28 + x

By transposing,

x = 42 - 28

x = 14 cm

Therefore, the third side of the triangle is 14 cm.

Finding the semi perimeter,

$\underline{\boxed{\sf Semi \ perimeter=\dfrac{Perimeter}{2} }}$

Given that,

Perimeter (p) = 42 cm

Substituting them,

Semi perimeter = 42/2

Semi perimeter = 21 cm

Therefore, the semi perimeter of the triangle is 21 cm.

Using Heron's formula,

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

Given that,

Semi perimeter = 21 cm

First side = 18 cm

Second side = 10 cm

Third side = 14 cm

Substituting their values,

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

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

$\sf =\sqrt{4851}$

$\sf =21\sqrt{11} \ cm^2$

Therefore, the area of the triangle is 21√11 cm².

Answer 2

Answer:

★ǫᴜᴇsᴛɪᴏɴ★

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

★ANSWER★

Given :-

• First side of the triangle = 18 cm

• Second side of the triangle = 10 cm

• Perimeter of the triangle = 42 cm

To Find :-

• The area of the triangle.

Formula used:-

$\underline{\boxed{\sf Perimeter \ of \ a \ triangle=a+b+c}}$

where, a, b, c represents the 3 sides of a triangle.

$\underline{\boxed{\sf Semi \ perimeter \: of \: triangle=\dfrac{Perimeter}{2}}}$

Heron's Formula:-

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

Solution :-

Let the third side (c) of the triangle be 'x' cm.

★According to statement,

• Perimeter (2s) = 42 cm

• First side (a) = 18 cm

• Second side (b) = 10 cm

By the formula,

$\underline{\boxed{\sf Perimeter \ of \ a \ triangle=a+b+c}}$

Substituting thevalues of perimeter, a, b and c, we get

42 = 18 + 10 + x

⟹ 42 = 28 + x

⟹ x = 42 - 28

⟹ x = 14 cm

$\small \bf Therefore \: , the \: third \: side \: of \: the \: triangle \: is \: 14 \: cm.$

$\underline{\boxed{\sf Semi \ perimeter=\dfrac{Perimeter}{2}}}$

Perimeter (2s) = 42 cm

Substituting the value of perimeter in above formula, we get

Semi perimeter (s) = 42/2

Semi perimeter (s) = 21 cm

Therefore, the semi perimeter of the triangle is 21 cm.

Using Heron's formula,

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

Now,

• Semi perimeter (s) = 21 cm

• First side (a) = 18 cm

• Second side (b) = 10 cm

• Third side (c) = 14 cm

Substituting their values,

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

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

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

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

$\small \bf Therefore, \: the \: area \: of \: the \: triangle \: is \: 21 \sqrt{11} \: cm².$

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