Question

if the ratio of the base and the hypotenuse of a right angled triangle is 4:5 , and the perimeter of the triangle is 48cm find its area​

Answer 1

GiveN:-

The ratio of the base and the hypotenuse of a right angled triangle is 4:5 , and the perimeter of the triangle is 48cm.

To FinD:-

The area.

SolutioN:-

Let the base be 4x and hypotenuse be 5x and side be s.

Now,

By Pythagoras Theorem,

$\large{\green{\underline{\boxed{\bf{(hypo)^2=(side)^2+(base)^2}}}}}$

where,

• Hypo is hypotenuse = 5x
• Base is 4x
• Side is s.

Putting the values,

$\large\implies{\sf{(5x)^2=(s)^2+(4x)^2}}$

$\large\implies{\sf{25x^2=s^2+16x^2}}$

$\large\implies{\sf{25x^2-16x^2=s^2}}$

$\large\implies{\sf{9x^2=s^2}}$

By square rooting both the sides,

$\large\implies{\sf{\sqrt{9x}=s}}$

$\large\implies{\sf{3x=s}}$

$\large\therefore\boxed{\bf{Side=3x.}}$

Therefore, the sides are 4x, 5x, 3x.

The perimeter is 48 cm.

We know that the measures of all the 3 sides of a triangle sum up to its Perimeter.

$\large\implies{\sf{4x+5x+3x=48}}$

$\large\implies{\sf{12x=48}}$

$\large\implies{\sf{x=\dfrac{48}{12}}}$

$\large\implies{\sf{x=\dfrac{\cancel{48}}{\cancel{12}}}}$

$\large\therefore\boxed{\bf{x=4}}$

The sides are:-

1. 4x = 4 × 4 = 16 cm
2. 5x = 5 × 4 = 20 cm
3. 3x = 3 × 4 = 12 cm

VerificatioN:-

$\large\implies{\sf{16+20+12=48}}$

$\large\implies{\sf{48=48}}$

$\large\therefore\boxed{\bf{LHS=RHS.}}$

Now the AreA:-

We know that area of triangle is,

$\large{\green{\underline{\boxed{\bf{Area=\dfrac{1}{2}\times\:Base\times\:Height}}}}}$

where,

• Base = 16 cm
• Height = 12 cm

Putting the values,

$\large\implies{\sf{Area=\dfrac{1}{2}\times16\times12}}$

$\large\implies{\sf{Area=\dfrac{1}{\cancel{2}}\times\cancel{16}\times12}}$

$\large\implies{\sf{Area=8\times12}}$

$\large\therefore\boxed{\bf{Area=96\:cm^2.}}$

The area of the triangle is 96 cm².

Answer 2

$\implies\huge \sf \underline \red{Answer}$

$\: \: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \pink{ \sf{ area = 96 {cm}^{2} }}}}}}$

__________________________________________________

$\implies \huge \sf \underline\blue{Given : }$

• base and hypotenuse = 4: 5
• perimeter = 48

$\implies \sf \huge \underline \pink{To \: find : }$

• Area of triangle

$\implies \sf \huge \underline \orange{solution : }$

• let us take base and hypotenuse be 4x and 5x

$\star \sf \underline{ \: now \: take \: phytagoras \: theorem : }$

$\: \: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \red{ \sf{ {h}^{2} = {p}^{2} + {b}^{2} \: }}}}}}$

so,

• hypotenuse is 4x
• base is 5x

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies\sf{ {p}^{2} = {5x}^{2} - {4x}^{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf{ {p}^{2} = {25x}^{2} - {16x}^{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf{ {p} = { \sqrt{ {9}^{2} } }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf{ {p =3}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \red{ \sf{p = 3 \: }}}}}}$

we know that,

$\sf \implies{perimeter = sum \: of \: all \: sides}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \blue{3x + 4x + 5x}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \blue{12x}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{given \: perimeter = 48}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \blue{12x = 48}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \blue{x = \dfrac{48}{12} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \blue{x = 4cm}$

Now,

$\star \sf \underline{putting \: the \: value \: of \: x}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{base = 4x}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{4(4) = 16cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{hypotenuse = 5x}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{3(4) = 12cm}$

$\star \implies \sf \underline{now \: formula}$

we know that area of triangle formula,

$\: \: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \red{ \sf{ \dfrac{1}{2} \times base \times height \: }}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{ \dfrac{1}{2} \times 16 \times 12 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{area = 96 {cm}^{2} }$

$\implies \sf \underline \purple{Area = 96 {cm}^{2}}$