Question

The lengths of the sides of a right triangle are a and b, and the hypotenuse is c. Find the area of the triangle. b = 2 in.; c = 6 in. A = sq. in.

Answer 1

Answer:

$\sqrt{32}sq. in$  

Step-by-step explanation:

To find the area of a triangle, we need to know the base and perpendicular to the base (height) of the triangle.

For a right triangle,

(Hypotenuse)² = (Base)² + (Perpendicular)²

We can find the perpendicular first from the above relation.

Given, Hypotenuse, c = 6 in

Base, b = 2 in

Perpendicular, $a = \sqrt{c^2-b^2} = \sqrt {6^2-2^2}=\sqrt{36-4}=\sqrt{32}$

The area of triangle:

$A = \frac{1}{2}{b}{a} = \frac{1}{2}\times 2 \times \sqrt{32} =\sqrt{32}sq. in.$

Answer 2

Answer:

$a = \sqrt{32}\text{ inch}$

Rea of triangle = $A = \sqrt{32}\text{ square inches}$

Step-by-step explanation:

We are given the following information:

The lengths of the sides of a right angled triangle are a and b, and the hypotenuse is c.

b = 2 inch

c = 6 inch

We have to find the area of the triangle.

Since, the given triangle is a right angles triangle, it satisfies the Pythagoras theorem.

The Pythagoras statement states that:

• In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Thus,

$a^2 + b^2 = c^2\\\text{Putting all the values}\\a^2 + (2)^2 = (6)^2\\a^2 = 36 -4 = 32\\a = \sqrt{32}\text{ inch}$

Area of triangle =

$\displaystyle\frac{1}{2}\times \text{Base} \times \text{Height}\\\\= \frac{1}{2}\times a \times b\\\\=\frac{1}{2}\times 2\times \sqrt{32}\\=\sqrt{32}\text{ square inches}$