Question

sides of the triangle are 10 cm, 10 cm and 10 cm

find area of the triangle ?​

Answer 1

$\huge \bold \color{green}꧁ \: {\underbrace {\overbrace{\mathfrak \red{answer}}}} \: ꧂$

$\bold { \underline{ \underline \orange{answer}} \orange→}$

• 25√3 cm² or 43.30cm² ( approximately)

$\bold { \underline{ \underline \orange{solution}} \orange→}$

sides of triangle :-

• AB = 10 cm
• BC = 10 cm
• CD = 10 cm

that means triangle is a equilateral triangle

$\bold { \tt{ \underline{ \underline\color{blue}first \: method}}} :$

we know that:

$\bold{{ \tt \: area \: of \: equi \triangle = \frac{ \sqrt{3} }{4} {(side)}^{2} }}$

$\bold {{ \tt \implies \: \: ar( ABC) = \frac{ \sqrt{3} }{4} \times {(10)}^{2} }} \\ \\ \bold{{ \tt \implies \: \: ar(ABC) = \frac{ \sqrt{3} \times 100}{4} }} \\ \\ { \bold{ \tt \: \implies ar(ABC) = 25 \sqrt{3} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so area of triangle ABC is 25√3 cm²

$\bold { \tt { \underline{ \underline\color{blue}second\: method}}} :$

area of triangle ABC by heron's formula

$\bold{a = \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \bold{s = \frac{a + b + c}{2} } \\ \\ \bold{a , b \: and \: c \: are \: sides \: of \: triangle}$

here,

$\bold{s = \frac{10 + 10 + 10}{2} } \\ \\ \bold{ \implies \: s = \frac{30}{2} = 15 }$

so area of ∆ABC,

$\bold{ \tt \: a = \sqrt{15(15 - 10)(15 - 10)(15 - 10)} } \\ \\ \bold{{ \tt \implies \:a = \sqrt{15 \times 5 \times 5 \times 5} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \:a = 5 \times 5 \times \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \: \implies \: a = 25 \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

hence, area of triangle ABC is 25√3 cm²

Answer 2

Step-by-step explanation:

25√3 cm² or 43.30cm² ( approximately)

\bold { \underline{ \underline \orange{solution}} \orange→}

solution

→

sides of triangle :-

AB = 10 cm

BC = 10 cm

CD = 10 cm

that means triangle is a equilateral triangle

\bold { \tt{ \underline{ \underline\color{blue}first \: method}}} :

firstmethod

:

we know that:

\begin{gathered} \bold{{ \tt \: area \: of \: equi \triangle = \frac{ \sqrt{3} }{4} {(side)}^{2} }} \\ \end{gathered}

areaofequi△=

4

3

(side)

2

\begin{gathered} \bold {{ \tt \implies \: \: ar( ABC) = \frac{ \sqrt{3} }{4} \times {(10)}^{2} }} \\ \\ \bold{{ \tt \implies \: \: ar(ABC) = \frac{ \sqrt{3} \times 100}{4} }} \\ \\ { \bold{ \tt \: \implies ar(ABC) = 25 \sqrt{3} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}

⟹ar(ABC)=

4

3

×(10)

2

⟹ar(ABC)=

4

3

×100

⟹ar(ABC)=25

3

so area of triangle ABC is 25√3 cm²

\bold { \tt { \underline{ \underline\color{blue}second\: method}}} :

secondmethod

:

area of triangle ABC by heron's formula

\begin{gathered} \bold{a = \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \bold{s = \frac{a + b + c}{2} } \\ \\ \bold{a , b \: and \: c \: are \: sides \: of \: triangle}\end{gathered}

a=

s(s−a)(s−b)(s−c)

s=

2

a+b+c

a,bandcaresidesoftriangle

here,

\begin{gathered} \bold{s = \frac{10 + 10 + 10}{2} } \\ \\ \bold{ \implies \: s = \frac{30}{2} = 15 }\end{gathered}

s=

2

10+10+10

⟹s=

2

30

=15

so area of ∆ABC,

\begin{gathered} \bold{ \tt \: a = \sqrt{15(15 - 10)(15 - 10)(15 - 10)} } \\ \\ \bold{{ \tt \implies \:a = \sqrt{15 \times 5 \times 5 \times 5} }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \implies \:a = 5 \times 5 \times \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{ \tt \: \implies \: a = 25 \sqrt{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}

a=

15(15−10)(15−10)(15−10)

⟹a=

15×5×5×5

⟹a=5×5×

3

⟹a=25

3