Q.5) Sides of a triangles are in the ratio of 12:17:25 and itsperimete

1.	Q.5) Sides of a triangles are in the ratio of 12:17:25 and itsperimeter is 540 cm. Find its area. OR Q.7)Prove that the sum of the angles of a triangles is 180.
Answer» Answer: 9000 cm^2 Step-by-step explanation: ⇒Let X be common ratio ∴ SIDES of triangle will be: 12x,17x and 25x ⇒Perimeter =540 cm(GIVEN) ⇒12x+17x+25x=540 cm, ⇒54x=540 cm ⇒x=10 cm ∴ Sides of triangle: a=120,b=170,c=250 cms ⇒2S=540 ⇒S=270 cm A= s(s−a)(s−b)(s−c) = 270(270−120)(270−170)(270−250) cm^2 = 270×150×100×20 cm^2 A=9000 cm^2

Q.5) Sides of a triangles are in the ratio of 12:17:25 and itsperimeter is 540 cm. Find its area. OR Q.7)Prove that the sum of the angles of a triangles is 180.

Answer»

Answer:

9000 cm^2

Step-by-step explanation:

⇒Let X be common ratio

∴ SIDES of triangle will be: 12x,17x and 25x

⇒Perimeter =540 cm(GIVEN)

⇒12x+17x+25x=540 cm,

⇒54x=540 cm

⇒x=10 cm

∴ Sides of triangle: a=120,b=170,c=250 cms

⇒2S=540

⇒S=270 cm

s(s−a)(s−b)(s−c)

270(270−120)(270−170)(270−250)

cm^2

270×150×100×20

cm^2

A=9000 cm^2

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Q.5) Sides of a triangles are in the ratio of 12:17:25 and itsperimeter is 540 cm. Find its area. OR Q.7)Prove that the sum of the angles of a triangles is 180.​

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Q.5) Sides of a triangles are in the ratio of 12:17:25 and itsperimeter is 540 cm. Find its area. OR Q.7)Prove that the sum of the angles of a triangles is 180.