Question

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

Answer 1

Step-by-step explanation:

A=√s(s﹣a)(s﹣b)(s﹣c)

P=a+b+c

s=a+b+c÷2

Solving forA

A=1

4﹣P4+4P3a+4P3b﹣4(Pa)2﹣12P2ab﹣4(Pb)2+8Pa2b+8Pab2=1

4·﹣424+4·423·18+4·423·10﹣4·(42·18)2﹣12·422·18·10﹣4·(42·10)2+8·42·182·10+8·42·18·102≈69.64912cm²

Answer 2

Given :-

• One side of triangle = 18 cm.

• Other side of triangle = 10 cm.

• Perimeter of triangle = 42 cm.

To Find :-

• Area of the triangle.

Solution :-

We know that,

$\bullet\:\: \underline{\boxed{\bf{Perimeter=a+b+c}}}$

Here,

→ a = 18 cm

→ b = 10 cm

→ c = ?

Now,

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

$:\implies\:\sf{42=28+c}$

$:\implies\:\sf{c=42-28}$

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

∴ Third side of the triangle = 14 cm.

Semi perimeter of triangle,

$:\implies\:\sf{\dfrac{18+10+14}{2} }$

$:\implies\:\sf{\dfrac{42}{2} }$

$:\implies\:\sf{21\:cm }$

Now,

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

Here,

→ s = 21 cm

→ a = 18 cm

→ b = 10 cm

→ c = 14 cm

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

$:\implies\:\sf{\sqrt{21 \times 3\times 11 \times 7} }$

$:\implies\:\sf{\sqrt{3 \times 7 \times 3\times 11\times 7} }$

$\sf:\implies \underline{\boxed{\pink{\mathfrak{21\sqrt{11}\:cm}}}}\:\:\bigstar$