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

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 consider another side of Right angled triangle be x

$\underline{\frak{As~ we ~know~ that~:}}$

• $\boxed{\sf\pink{Pythagoras~Theorem~=~(Side_{(A)})^{2}~+~(Side_{(B)})^{2}~=~(Hypotenus)^2}}$

$~$

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

$~$

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

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

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

$~~~~~~~~~~{\sf:\implies{16~=~x^2}}$

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

$~~~~~~~~~~:\implies{\underline{\boxed{\frak{\pink{x~=~4~cm}}}}}$★

$~$

Therefore,

$\therefore\underline{\sf{Other~ Side~ of ~Right ~Angled~ Triangle ~is~\bf{4~cm}}}$

$~$

$~~~~~$ _____________________

❍ All Three sides of Triangle are :

• $\leadsto$First side = 5 cm

• $\leadsto$Second side = 3 cm

• $\leadsto$Third side = 4 cm

$~~~~~$ _____________________

$~$

• Finding Area of Triangle using Heron's Formula to Find Area of Triangle.

• Finding Semi-Perimeter of Triangle for Area of Triangle using the formula's given by :

$~$

• $\boxed{\sf\pink{Semi-Perimeter~=~\dfrac{(Side_{(A)})~+~(Side_{(B)})~+~(Side_{(C)})}{2}}}$

$~$

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

$~$

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

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

$~~~~~~~~~~:\implies{\underline{\boxed{\frak{\pink{Semi-Perimeter~=~6~cm}}}}}$

$~$

Therefore,

$\therefore\underline{\sf{Semi-Perimeter~of~Triangle~is~\bf{6~cm}}}$

$~$

____________________

$~$

$\underline{\frak{As~ we~ know ~that~:}}$

• $\boxed{\sf\pink{Area ~of~\triangle~=~\sqrt{s(s~-~a)(s~-~b)(s~-~c)}}}$

$~$

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

$~$

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

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

$~~~~~~~~~~{\sf:\implies{Area ~of ~\triangle~=~\sqrt{6~×~3~×~2}}}$

$~~~~~~~~~~{\sf:\implies{Area~ of ~\triangle~=~\sqrt{6~×~6}}}$

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

$~~~~~~~~~~:\implies\underset{\blue{\rm Required\ Answer}}{\underbrace{\boxed{\frak{\pink{Area ~of ~\triangle~=~6~cm^2}}}}}$

$~$

Hence,

$\therefore\underline{\sf{Area~of~Triangle(\triangle)~is~\bf{6~cm^2}}}$

$~~~~$$\qquad\quad\therefore\underline{\textsf{\textbf{Hence Verified!}}}$

Answer 2

Answer:

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

$\Large {\underline{\gray{\frak { Answer \:\:: \:}}}}$

• To Find : Area of Triangle.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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

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 .

As, We know that :

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

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

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

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

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

⠀⠀⠀⠀⠀$:\implies \sf{\:16 = x^{2} }$

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

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

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 :

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

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

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

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

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

____________________

$\underline{\frak{As~ we~ know ~that~:}}$

$\boxed{\sf\pink{Area ~of~\triangle~=~\sqrt{s(s~-~a)(s~-~b)(s~-~c)}}}$

~ 

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

$~~~~~~~~~~{\sf:\implies{Area~ of ~\triangle~=~\sqrt{6(6~-~3)(6~-~4) (6~-~5)}}}\\    :⟹Area of △ = 6(6 − 3)(6 − 4)(6 − 5)$

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

$~~~~~~~~~~{\sf:\implies{Area ~of ~\triangle~=~\sqrt{6~×~3~×~2}}}\\ :⟹Area of △ = 6 × 3 × 2$

$~~~~~~~~~~{\sf:\implies{Area~ of ~\triangle~=~\sqrt{6~×~6}}} \\         :⟹Area of △ = 6 × 6$

$~~~~~~~~~~{\sf:\implies{Area~ of ~\triangle~=~\sqrt{36}}}    \\       :⟹Area of △ = 36$

$~~~~~~~~~~:\implies\underset{\blue{\rm}}{\underbrace{\boxed{\frak{\pink{Area ~of ~\triangle~=~6~cm^2}}}}}       \\~$

$Hence, \underline{\sf{ \: Area~of~Triangle(\triangle)~is~\bf{6~cm^2}}}$