Question

find the base of a triangle whose area is 0.48dm and altitude is 8 cm

Answer 1

The base of a triangle(b) = 12 cm or 1.2 dm

Step-by-step explanation:

Given,

The altitude of a triangle(h) = 8 cm = 0.1 × 8 dm = 0.8 dm and

The area of a triangle = 0.48 $dm^2$

To find, the base of a triangle(b) = ?

We know that,

The area of a triangle $=\dfrac{1}{2} bh$

⇒ $\dfrac{1}{2}\times b \times 0.8=0.48$

⇒ $b \times 0.4=0.48$

⇒ $b=\dfrac{.48}{0.4}=1.2dm$

⇒ b = 1.2 × 10 cm = 12 cm [ since, 1 dm = 10 cm]

Hence, the base of a triangle(b) = 12 cm or 1.2 dm

Answer 2

Answer:

Step-by-step explanation:

Given

The altitude of triangle =8cm

Area of triangle=0.48 decimete 0.48decimeters=4Khan8 CM

Given altitude or height of triangle=8cm

Area of triangle 1 upon 2 x base x height

Therefore 1 upon 2 x b x is equal to

1 upon 2 x b x h is equal to 48

1 upon 2 x b x 8 is equal to 48

B is equal to 48 upon 8 x 2

B is equal to 6 x 2

B is equal to 12

Therefore base is equal to 12 CM

May my this answer help you

From DAV duttnagar 7