Question

If the base of a right angled triangle is 6 units and the hypotenuse is 10 units, find its area

Answer 1

Given :

• Base of the triangle = 6 units

• Hypotenuse of the triangle = 10 units

To Find :

The area of the triangle .

Solution :

To find the area of the triangle first we need to find the height of the triangle.

We can find the height of the triangle by using the Pythagoras theorem .

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Pythagoras theorem :

$\underline{\bf{H^{2} = P^{2} + B^{2}}}$

Where :-

• H = Hypotenuse
• P = Height
• B = Base

Now, using the formula and substituting the values in it, we get :

Let the height of the triangle be x units.

$:\implies \bf{H^{2} = P^{2} + B^{2}}$

$:\implies \bf{10^{2} = x^{2} + 6^{2}}$

$:\implies \bf{10^{2} - 6^{2} = x^{2}}$

$:\implies \bf{\sqrt{10^{2} - 6^{2}} = x}$

$:\implies \bf{\sqrt{100 - 36} = x}$

$:\implies \bf{\sqrt{64} = x}$

$:\implies \bf{8 = x}$

$\therefore \bf{Height = 8\:units}$

Hence, the height of the triangle is 8 units.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Area of the triangle :

We know the formula for area of a triangle :-

$\underline{\bf{A = \sqrt{s(s - a)(s - b)(s - c)}}}$

Where :-

• A = Area of the triangle
• s = Semi-perimeter
• a , b and c = Side of the triangle

Semi-perimeter = $\rm{\dfrac{a + b + c}{2}}$

Here , Semi-perimeter =

$:\implies \rm{s = \dfrac{a + b + c}{2}}$

$:\implies \rm{s = \dfrac{6 + 8 + 10}{2}}$

$:\implies \rm{s = \dfrac{24}{2}}$

$:\implies \rm{s = 12\:units}$

Hence, the semi-perimeter is 12 units.

Now , Using the formula and substituting the values in it , we get :

$:\implies \bf{A = \sqrt{s(s - a)(s - b)(s - c)}}$

$:\implies \bf{A = \sqrt{12(12 - 6)(12 - 8)(12 - 10)}}$

$:\implies \bf{A = \sqrt{12 \times 6 \times 4 \times 2}}$

$:\implies \bf{A = \sqrt{12 \times 48}}$

$:\implies \bf{A = \sqrt{576}}$

$:\implies \bf{A = 24}$

$\therefore \bf{Area = 24\:unit^{2}}$

Hence, the area of the triangle is 24 unit².

Answer 2

$\pink{\large{\underline{\underline{ \rm{Given: }}}}}$

◍ Base of a right angle triangle = 6 units

◍ Hypotenuse of a right angle triangle = 10 units

$\pink{\large{\underline{\underline{ \rm{To \: Find: }}}}}$

◉ Area of Triangle.

$\pink{\large{\underline{\underline{ \rm{Solution: }}}}}$

Let,

◎ ABC be the right angle triangle.

◎ $\sf{ \angle{B = 90 \degree}}$

◎ BC i.e., base = 6 units

◎ AC i.e., hypotenuse = 10 units

$\sf{ \blue{How \: to \: solve?}}$

For finding the area of triangle first we have to find AB i.e., height and then by using simple formula of area of triangle, we will find the area of a right angled triangle.

In ∆ ABC,

By using Pythagoras theorem,

$\sf{ {H}^{2} = {B}^{2} + {P}^{2}}$

Here,

★ H = Hypotenuse

★ B = Base

★ P = Perpendicular

$\sf{ {AC}^{2} = {BC}^{2} + {AB}^{2} }$

By Substituting the values, we have:

➯ $\sf{ {10}^{2} = {6}^{2} + {AB}^{2} }$

➯ $\sf{ {AB}^{2} = {10}^{2} - {6}^{2}}$

➯ $\sf{ {AB}^{2} = 100 - 36}$

➯ $\sf {AB}^{2} = 64$

➯ $\sf{ {AB} = \sqrt{64} }$

➯ $\sf{AB = 8 \: units}$

Now let's find out the area of triangle.

If one side (base) and the corresponding height (altitude) of the triangle are known, its

$\underline{ \boxed{ \sf{Area = \dfrac{1}{2} \times base \times height}}}$

We know,

Base = 6 units

Height = 8 units

By substituting the values in the formula, we have:

$\sf{Area \: of \: \triangle \: ABC = (\dfrac{1}{2} \times 6 \times 8) \: sq. \: units}$

$\sf = (\dfrac{1}{2} \times 48) \: sq. \: units$

$\sf= 24 \: sq. \: units$

∴ Area of Triangle = $\blue{ \underline{ \boxed{ \sf{24 \: sq. \: units}}}}$

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