Question

if one of the two equal side of an isosceles triangle is 10 cm and its perimeter is 32cm find area of a triangle​

Answer 1

Answer:

$\textsc{QUESTION:-}$

If one of the two equal side of an isosceles triangle is 10 cm and its perimeter is 32cm find area of a triangle?

Step-by-step explanation:

$\mathcal{Solution:-}$

We know 2 Sides of The Isosceles ∆le are Equal

i.e, a=10 and b=10

&a=b

We have to Find C in order to find Area✓

Perimeter=>a+b+c=32

10+10+c=32

=>20+c=32

$\orange{c=12[32-20]}$

Area of Isosceles ∆le

From Heron's Formula

S=↓

$\frac{a+a+c}{2}$

=>$\frac{10+10+12}{2}$

$\frac{20+12}{2}$

=>$\frac{32}{2}$

i.e,

$\red{16cm^2}$

now,

$\sqrt{s(s - 10)(s - 10)(s - 12)} \\ \\ \sqrt{16(16 - 10)(16 - 10)(16 - 12)} \\ \sqrt{ 16 \times 6 \times 6 \times 4} \\ \sqrt{2304} \\ = 48cm {}^{2}$

$\mathcal{BE \: BRAINLY}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}} \begin{cases} \sf{Perimeter \: = \: 32 \: cm} \\ \sf{a \: = \: b \: = \: 10 \: cm} \end{cases}$

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

As we know that :

$\large \star {\boxed{\sf{Perimeter \: = \: Sum \: of \: Sides}}} \\ \\ \implies {\sf{Perimeter \: = \: a \: + \: b \: + \: c}} \\ \\ \implies {\sf{36 \: = \: 10 \: + \: 10 \: + \: c}} \\ \\ \implies {\sf{32 \: = \: 20 \: + \: c}} \\ \\ \implies {\sf{c \: = \: 32 \: - \: 20}} \\ \\ \implies {\sf{c \: = \: 12}}$

∴ Third Side is 12 cm

$\rule{200}{1}$

$\star {\boxed{\sf{Semi \: Perimeter \: (S) \: = \: \dfrac{Perimeter}{2}}}} \\ \\ \implies {\sf{Semi \: Perimeter \: (S) \: = \: \dfrac{32}{2}}} \\ \\ \implies {\sf{Semi \: Perimeter \: (S) \: = \: 16 \: cm}}$

$\rule{200}{1}$

Now,

Use herons Formula

$\star {\boxed{\sf{Area \: = \: \sqrt{S(S \: - \: a) (S \: - \: b) (S \: - \: c) }}}} \\ \\ \small{\underline{\pink{\sf{\: \: \: \: \: \: \: \: \: \: Put \: Values \: \: \: \: \: \: \: \: \:}}}} \\ \\ \implies {\sf{Area \: = \: \sqrt{16(16 \: - \: 10)(16 \: - \: 10)(16 \: - \: 12)}}} \\ \\ \implies {\sf{Area \: = \: \sqrt{16(6)(6)(4)}}} \\ \\ \implies {\sf{Area \: = \: \sqrt{2304}}} \\ \\ \implies {\sf{Area \: = \: 48}} \\ \\ \large \leadsto {\boxed{\sf{Area \: = \: 48 \: cm^2}}}$