Question

THE BASE OF A TRIANGULAR REGION IS THREE TIMES ITS ALTITUDE.IF THE AREA IS 27sq.m, find the base and altitude

Answer 1

Given :

The base of a triangular region is three times it's altitude.

Area of triangular region = 27m²

To find :

Base and altitude of triangular region.

Solution :

Let the altitude of triangular region be y m.

∴ Base of triangular region = 3y m -(1)

We know,

⇒ Area of a triangle = ½ × Base × Altitude

⇒ Area of triangular region = ½ × 3y × y

⇒ Area of triangular region = ½ × 3y²

Now, area of triangular region is given as 27 m²

According to condition,

⇒ ½ × 3y² = 27

⇒ 3y² = 27 × 2

⇒ 3y² = 54

⇒ y² = 54/3

⇒ y² = 18

⇒ y = √18 [Ignoring negative value as altitude can't be negative]

⇒ y = √(9 × 2)

⇒ y = 3√2 m

 ∴ Altitude of triangular region = 3√2 m

Now putting this value in (1) :

⇒ Base of triangular region = 3y m

⇒ Base of triangular region = 3 × 3√2 m

 ∴ Base of triangular region = 9√2 m

Therefore,

Base and altitude of triangular region are 9√2 m and 3√2 m respectively.

Answer 2

Step-by-step explanation:

topic :

• base and altitude

given :

• BASE OF A TRIANGULAR REGION = 3

• area = 27sq.m

to find :

• find the base and altitude = ?

• find the base and altitude = ?

knowledge about :

• area of traingle altitude formula = 1/2 ×ba

solution :

• traingular area = 1/2 × 3 × y

• traingular area = 1/2×3

• first we have to do area of 27 sq m

• area = 1/2 × 3y² = 27

• area = 3y² = 27 x 2

• area = 3y² = 54

• area y² = 54/3

• area = y² = 18

• area = y = √(9 × 2)

• area = y = 3√2 m

• Altitude = 3√2 m

• Base of traingular = 3y m

• Base of triangular = 3 × 3-√2 m

• Base of triangular = 9√2 m

hence, base and altitudes= 3√2 m and

9√2 m