Question

What is the area triangle with base 3.5 cm and height 1.2 cm​

Answer 1

Given:

• The base of a triangle is 3.5 cm
• The height of the triangle is 1.2 cm

To find:

• The area of the triangle

So,

The base and the height of the triangle are given so, we can use the formula

• The formula to get the area of the triangle when "base" and "height" are given = 1/2 * base * height unit sq.

Now,

Putting the respective values in the formula we get,

Area of the triangle = 1/2 * 3.5 * 1.2

Area of the triangle = 3.5 * 0.6

Area of the triangle = 2.1 cm sq.

Hence

• The area of the triangle is 2.1 cm sq.

More Information:

• For an equilateral triangle (all sides are equal in lengths)

Area = $\boxed{\frac{\sqrt{3}a^{2}}{4}}\ unit\ sq.$

Where 'a' is the side of the triangle.

• For a scalene triangle (all sides are different in lengths)

Area = $\sqrt{S(S-a)(S-b)(S-c)}$ unit sq.

Where, 'a', 'b', 'c' are the sides of the triangle and 'S' is the semi perimeter (i.e. Perimeter/2)

Answer 2

${\large{\bold{\rm{\underline{Given \; that}}}}}$

★ Base of triangle = 3.5 cm

★ Height of triangle = 1.2 cm

${\large{\bold{\rm{\underline{To \; find}}}}}$

★ Area of triangle

${\large{\bold{\rm{\underline{Solution}}}}}$

★ Area of triangle = 2.10 cm²

${\large{\bold{\rm{\underline{Using \; concept}}}}}$

★ Formula to find area of triangle.

${\large{\bold{\rm{\underline{Using \; formula}}}}}$

★ Area of triangle = ½ × Base × Height

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

➝ Base of triangle = 3.5 cm

➝ Height of triangle = 1.2 cm

➝ Area of triangle = ½ × B × H

• Area of triangle = ½ × 3.5 × 1.2

• Area of triangle = ½ × 4.20

• Area of triangle = 2.10 cm²

${\large{\bold{\rm{\underline{Additional \; information}}}}}$

Equilateral triangle

➜ Equilateral triangle is a triangle in which all the three sides have equal length.

➜ The sum of all three angles of an equilateral triangle is equal to 180°.

➜ Altitude of equilateral triangle = $\sf\dfrac{\sqrt{3}a}{2}$

➜ Perimeter of equilateral triangle = Sum of all sides (a + a + a)