Q.1) Find the area of the triangle whose sides are 13cm,14cm,15cm.
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Q.1) Find the area of the triangle whose sides are 13cm,14cm,15cm.
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❈ Given ❈
→ A triangle with sides 13 cm, 14 cm and 15 cm.
Let us assume a triangle ABC with,
❈ To Find ❈
The area of triangle ABC.
❈ Solution ❈
Let us suppose :-
• a = 13 cm
• b = 14 cm
• c = 15 cm
→ Perimeter of the triangle = 2s = a + b + c
⇒ 2s = 13 + 14 + 15
⇒ s = 42/2
⇒ s = 21 cm
______________
s - a = 21 - 13
⇒ s - a = 8
______________
s - b = 21 - 14
⇒ s - b = 7
______________
s - b = 21 - 15
⇒ s - c = 6
______________
Now, applying Heron's formula :-
Answer:
triangle with sides 13 cm, 14 cm and 15 cm.
Let us assume a triangle ABC with,
AB = 13 cm
BC = 14 cm
CA = 15 cm
❈ To Find ❈
The area of triangle ABC.
❈ Solution ❈
\bigstar\underline{\underline{\sf{DIAGRAM:-}}}★
DIAGRAM:−
\setlength{\unitlength}{1.6mm}\begin{picture}(50,20)\linethickness{0.1mm}\put(-3,-3){\line(1,1){20}}\put(36.6,-2.8){\line(-1,1){19.8}}\put(-3,-3){\line(1,0){39.5}}\put(30,5){15 cm}\put(-1.5,5){13 cm}\put(3,5){}\put(15,-5){14 cm}\put(15.5,17.5){A}\put(-4,-5){B}\put(35,-5){C}\end{picture}
Let us suppose :-
• a = 13 cm
• b = 14 cm
• c = 15 cm
→ Perimeter of the triangle = 2s = a + b + c
⇒ 2s = 13 + 14 + 15
⇒ s = 42/2
⇒ s = 21 cm
______________
s - a = 21 - 13
⇒ s - a = 8
______________
s - b = 21 - 14
⇒ s - b = 7
______________
s - b = 21 - 15
⇒ s - c = 6
______________
\bigstar★ Now, applying Heron's formula :-
\begin{gathered}\rm{A=\sqrt{s(s-a)(s-b)(s-c)} }\\\\\rm{\longrightarrow A= \sqrt{21(8)(7)(6)} }\\\\\rm{\longrightarrow A= \sqrt{3 \times 7 \times 7 \times 2 \times 2\times2\times2\times3} }\\\\\rm{\longrightarrow A= \sqrt{3^2\times7^2\times2^2\times2^2} }\\\\\rm{\longrightarrow A= 3\times7\times2\times2 }\\\\\rm{\longrightarrow A= 21\times4}\\\\\boxed{\boxed{\rm{\longrightarrow A= 84\ cm^2}}}\end{gathered}
A=
s(s−a)(s−b)(s−c)
⟶A=
21(8)(7)(6)
⟶A=
3×7×7×2×2×2×2×3
⟶A=
3
2
×7
2
×2
2
×2
2
⟶A=3×7×2×2
⟶A=21×4
⟶A=84 cm
2
Step-by-step explanation:
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