Question

find the area of right triangle whose base and hypotenuse are 3 cm and 5 cm respectively. ​

Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$


$\underline{ \bf{GIVEN : }}$

ABC is a right angle triangle whose base is 3 cm and its hypotenuse is 5 cm respectively.


$\underline{ \bf{TO \: \: FIND : }}$

Area of the ∆ABC.


$\underline{ \bf{ PROCESS : \: }}$


As ∆ABC is a right angled triangle ,

Therefore, by Pythagoras Theorem

$\bf {(Hypotenuse)}^{2} = {(Base)}^{2} + {( Perpendicular )}^{2}$

$= > \: ( {5})^{2} = ( {3})^{2} \: + \: ( {h})^{2}$

$= > \: 25 \: = \: 9 \: + \: ( {h})^{2}$

$= > \: {h}^{2} \: = \: 25 \: - \: 9$

$= > {h }^{2} \: = 16$

$= > \: h \: = \: 4$


Now,

Area of the triangle = $\frac{1}{2} \times \: b \: \times \: h$

= $\frac{1}{2} \times \: 3\: \times \: 4$

= $3\: \times \: 2$

= $6\: {cm}^{2}$


$\boxed{ \bf{ \therefore \: AREA = 6 \: {cm}^{2} }}$

Answer 2

Answer:

ABC is a right angle triangle whose base is 3 cm and its hypotenuse is 5 cm respectively.

\underline{ \bf{TO \: \: FIND : }}

TOFIND:

Area of the ∆ABC.

\underline{ \bf{ PROCESS : \: }}

PROCESS:

As ∆ABC is a right angled triangle ,

Therefore, by Pythagoras Theorem

\bf {(Hypotenuse)}^{2} = {(Base)}^{2} + {( Perpendicular )}^{2}(Hypotenuse)

2

=(Base)

2

+(Perpendicular)

2

= > \: ( {5})^{2} = ( {3})^{2} \: + \: ( {h})^{2}=>(5)

2

=(3)

2

+(h)

2

= > \: 25 \: = \: 9 \: + \: ( {h})^{2}=>25=9+(h)

2

= > \: {h}^{2} \: = \: 25 \: - \: 9=>h

2

=25−9

= > {h }^{2} \: = 16=>h

2

=16

= > \: h \: = \: 4=>h=4

Now,

Area of the triangle = \frac{1}{2} \times \: b \: \times \: h

2

1

×b×h

= \frac{1}{2} \times \: 3\: \times \: 4

2

1

×3×4

= 3\: \times \: 23×2

= 6\: {cm}^{2}6cm

2

\boxed{ \bf{ \therefore \: AREA = 6 \: {cm}^{2} }}

∴AREA=6cm

2

Step-by-step explanation:

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