Question

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

Answer 1

Answer:Answer:4851^1/2 cm^2

Step-by-step explanation:perimeter=42cm

1st side=18cm

2nd side=10cm

3rd side=42cm-(18+10)cm

=42cm-28cm

=14cm

Area of scalene triangle=Answer:4851^1/2 cm^2

Here S = Perimeter/2

Answer 2

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

$\huge{\underline{\tt{\orange{Sloution-}}}}$

Given: $\sf{a = 18}$

$\sf{b = 10}$

$\sf{Perimeter = 42}$

$\longrightarrow \sf{a+b+c=42}$

$\longrightarrow \sf{18+10+c=42}$

$\longrightarrow \sf{28+c=42}$

$\longrightarrow \sf{c=48-28}$

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

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$\longrightarrow\sf{2s=42}$

$\longrightarrow\sf \frac{s = 42}{2}$

$\longrightarrow \sf\blue{s = 21}$

____________________________________________

$\longrightarrow\sf\green{Area\:of\:triangle = \sqrt{s(s-a)(s-b)(s-c)}}$

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

$\longrightarrow\sf { =\sqrt{21\times3\times11\times7}}$

$\longrightarrow\sf{=\sqrt{7\times3\times3\times11\times7}}$

$\longrightarrow \sf{=\sqrt{49\times9\times11}}$

$\longrightarrow\sf{=7\times3\sqrt{11 }}$

$\huge\longrightarrow\tt\green{=21\sqrt{11}}$