Question

The hypotenuse of a right triangle is 5 cm and one of its side is 3 cm find the area of

triangle​

Answer 1

Answer:

4cm

Step-by-step explanation:

The lengths of the side of the triangle are 4 cm , 3 cm and 5 cm

Step-by-step explanation:

Hypotenuse = 5 cm

Let the perpendicular and base of right angled triangle be x and y

We are given that the other 2 sides differ by 1cm

So, x-y=1

y = x-1

So, Base = x-1

Using Pythagoras theorem

\begin{gathered}Hypotenuse^2 = Perpendicular^2+Base^2\\5^2=x^2+(x-1)^2\\25 = x^2+x^2+1-2x\\2x^2-2x-24=0\\x^2-x-12=0\\x^2-4x+3x-12=0\\x(x-4)+3(x-4)=0\\(x+3)(x-4)=0\\x=-3,4\end{gathered}Hypotenuse2=Perpendicular2+Base252=x2+(x−1)225=x2+x2+1−2x2x2−2x−24=0x2−x−12=0x2−4x+3x−12=0x(x−4)+3(x−4)=0(x+3)(x−4)=0x=−3,4

Since the side cannot be negative

So, Perpendicular = 4 cm

Base = x-1 = 4-1 = 3 cm

Hence The lengths of the side of the triangle are 4 cm , 3 cm and 5 cm

Answer 2

Given : The hypotenuse of a right triangle is 5 cm and one of its side is 3 cm .

To Find : Area of Triangle.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's Assume another side of Right angled triangle be x .

$\frak {\underline { As,\:We\:Know\:that\::}}\\\\{\star{\bf{ By \:Pythagoras \:Theorem \::}}}$

⠀⠀⠀⠀⠀ $\underline {\boxed {\sf{ Pythagoras \:Theorem\:=\: (Side_{(A)})^{2} + (Side\:_{(B)})^{2} = (Hypotenuse)^{2}.}}}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀$:\implies \sf{\:5^{2} = 3^{2} + x^{2} }$

⠀⠀⠀⠀⠀$:\implies \sf{\:25 = 9 + x^{2} }$

⠀⠀⠀⠀⠀$:\implies \sf{\:25- 9 = x^{2} }$

⠀⠀⠀⠀⠀$:\implies \sf{\:16 = x^{2} }$

⠀⠀⠀⠀⠀$:\implies \sf{\:\sqrt {16} = x }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  x = 4\: cm}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{ \mathrm {  Other \:Side\:of\:\:Right\:Angled\:Triangle \:is\:4\: cm}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ All Three sides of Triangle are :

⠀⠀⠀⠀⠀➢ First Side = 5 cm

⠀⠀⠀⠀⠀➢ Second Side = 3 cm

⠀⠀⠀⠀⠀➢Third Side = 4 cm

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Findig Area of Triangle using Heron's Formula to Find Area of Triangle:

⠀⠀Finding Semi-Perimeter of Triangle for Area of Triangle using the formulais given by :

⠀⠀⠀⠀⠀ $\underline {\boxed {\sf{ Semi-Perimeter \:=\dfrac{\: (Side_{(A)}) + (Side\:_{(B)}) + (Side_{(C)})}{2}}}}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀ $: \implies \sf{Semi-Perimeter \:= \dfrac{5 + 4 + 3 }{2}}$

⠀⠀⠀⠀⠀ $: \implies \sf{ Semi-Perimeter \: =\dfrac{\cancel {12} }{\cancel{2}}}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Semi-Perimeter \:= 6\: cm}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Semi\:-Perimeter \:of\:Triangle \:is\:6\: cm}}}$

$\frak {\underline { As,\:We\:Know\:that\::}}\\\\{\star{\bf{ By \:Heron's \:Formula \::}}}$

⠀⠀⠀⠀⠀$\underline {\boxed {\sf{\star Area\:of\:\triangle = \sqrt { s ( s - a ) (s -b)(s-c)}}}}$

⠀⠀⠀⠀⠀ Here s is the Semi-Perimeter of Triangle in cm , a , b and c are three sides of Triangle in cm .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀$:\implies \sf{ Area\:of\:\triangle = \sqrt { 6 ( 6 - 3 ) (6 -4)(6-5)}}$

⠀⠀⠀⠀⠀$:\implies \sf{ Area\:of\:\triangle = \sqrt { 6 ( 3 ) (2)(1)}}$

⠀⠀⠀⠀⠀$:\implies \sf{ Area\:of\:\triangle = \sqrt { 6 \times 3 \times2}}$

⠀⠀⠀⠀⠀$:\implies \sf{ Area\:of\:\triangle = \sqrt { 6 \times 6}}$

⠀⠀⠀⠀⠀$:\implies \sf{ Area\:of\:\triangle = \sqrt { 36}}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  Area\:of\:\triangle = 6\: cm^{2}}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore{ \mathrm {  Area \:of\:Triangle \:is\:6\: cm^{2}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀