Question

Question 1
The area of triangle with given two sides 18cm and 10cm respectively and perimeter equal to 42 cm
is:

Plz

Answer 1

Answer:

The area of triangle is 21√11 cm².

Step-by-step explanation:

Sides of triangle = 18 cm and 10 cm

Perimeter = 42 cm

Let the third side be y.

★ Perimeter of triangle = Sum of all sides

$\longrightarrow$ 42 = 18 + 10 + y

$\longrightarrow$ 42 = 28 + y

$\longrightarrow$ y = 42 – 28

$\longrightarrow$ y = 14 cm

★ Area of triangle =

By heron's formula,

• Side a = 18 cm
• Side b = 10 cm
• Side c = 14 cm

$\boxed{\rm{Semi \: perimeter = \frac{Perimeter}{2}}}$

$\rm{\longrightarrow} \: s = \dfrac{42}{2}$

$\rm{\longrightarrow} \: s = 21$

Area =

$\boxed{\rm{A=\sqrt{s(s - a)(s - b)(s - c)}}}$

$\rm{\longrightarrow} \: A=\sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

$\rm{\longrightarrow} \: A=\sqrt{21 \times 3 \times 11 \times 7}$

$\rm{\longrightarrow} \: A=\sqrt{7 \times 3 \times 3 \times 11 \times 7}$

$\rm{\longrightarrow} \: A=7 \times 3 \sqrt{11}$

$\rm{\longrightarrow} \: A=21 \sqrt{11} \: {cm}^{2}$

Area of triangle = 21√11 cm²

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

Answer 2

Answer:

‣ Area of triangle $\mapsto\:{\boxed{\tt{20\sqrt{11}\:cm^2}}}$

Explanation:

Given information,

We have a triangle with given two sides 18 cm and 10 cm respectively and perimeter equal to 42 cm. Find it's area.

• a = 18 cm
• b = 10 cm
• c = ?
• Perimeter of ∆ = 42 cm

⚘ [a, b, c are sides of ∆]

★ Finding third side (c) ::

• $\boxed{\underline{\underline{\bf{\red{Perimeter_{\triangle} = a + b + c}}}}}\:\bf{\dag}$

Putting all values,

$\\ \longrightarrow\:\tt 42 = 18 + 10 + c$

$\\ \longrightarrow\:\tt 42 = 28 + c$

$\\ \longrightarrow\:\tt 42 - 28 = c$

$\\ \longrightarrow\:{\underline{\boxed{\tt{c = 14\:cm}}}}\:\bigstar$

• Hence, third side of ∆ (c) is 14 cm.

★ Finding semi perimeter (s) ::

• $\boxed{\underline{\underline{\bf{\pink{s = \dfrac{a + b + c}{2}}}}}}\:\bf{\dag}$

Putting all values,

$\\ \longrightarrow\:\tt s = \dfrac{18 + 10 + 14}{2}$

$\\ \longrightarrow\:\tt s = {\cancel{\dfrac{42}{2}}}$

$\\ \longrightarrow\:{\underline{\boxed{\tt{s = 21\:cm}}}}\:\bigstar$

• Hence, semi perimeter of ∆ (s) is 21 cm.

⚘ Now, we have all required values. So, let's find area of triangle by using heron's formula;

★ Finding area of ∆ ::

• $\boxed{\underline{\underline{\bf{\purple{Area_{\triangle} = \sqrt{s(s - a)(s - b)(s - c)}}}}}}\:\bf{\dag}$

Putting all values,

$\\ \longrightarrow\:\tt Area_{\triangle} = \sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

$\\ \longrightarrow\:\tt Area_{\triangle} = \sqrt{21(3)(11)(7)}$

$\\ \longrightarrow\:\tt Area_{\triangle} = \sqrt{21\:\times\:3\:\times\:11\:\times\:7}$

$\\ \longrightarrow\:\tt Area_{\triangle} = \sqrt{3\:\times\:7\:\times\:3\:\times\:11\:\times\:7}$

$\\ \longrightarrow\:\tt Area_{\triangle} = \sqrt{\underline{3\:\times\:3}\:\times\:\underline{7\:\times\:7}\:\times\:11}$

$\\ \longrightarrow\:\tt Area_{\triangle} = 3\:\times\:7\:\times\:\sqrt{11}$

$\\ \longrightarrow\:\tt Area_{\triangle} = 21\:\times\:\sqrt{11}$

$\\ \longrightarrow\:{\underline{\boxed{\tt{Area_{\triangle} = 21\sqrt{11}\:cm^2}}}}\:\bigstar$

• Hence, area of ∆ is 21√11 cm².

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