Question

The perimeter of a right triangle is 56 cm. If its hypotenuse is 25 cm. Find
other two sides. Find its area by using the formula for a right triangle
your answer by using Heron's formula.​

Answer 1

$\huge\bold\red{\underline{Solution:}}$

$\bold\green{\underline{Given:}}$

• The perimeter of Triangle=56cm
• The hypotenous=25cm

$\bold\green{\underline{Let:}}$

• The two sides be X and Y

$\bold\orange{\underline{A.T.Q}}$

• by using formula formula here

Sum of sides of Triangle=perimeter of Triangle

⟹X+Y+25=56

⟹X+Y=56-25

⟹X+Y=31

⟹X=31-Y

$\bold\purple{\boxed{X+Y=31-----(1)}}$

Than, using pathagoras theorem

⟹P²+b²=h²

⟹X²+Y²=25²

⟹X²+Y²=625

$\bold\purple{\boxed{X^2+Y^2=625----(2)}}$

• Solving the EQ by substituting method

Putting the value X=31-Y in EQ 2

⟹(31-Y)²+Y²=625

⟹961-62Y+Y²+Y²=625

⟹2Y²-62Y+961-625=0

⟹2Y²-62Y+336=0

• Dividing by 2 on both sides

⟹Y²-31Y+168=0

⟹Y²-24Y-7Y+168=0

⟹Y(Y-24)-7(Y-24)=0

SO, Y=24 and 7

• Putting the value of y=24 in Eq 1

⟹X+Y=31

⟹X+24=31

⟹X=7

Now, finding semi perimeter

$\bold\purple{\boxed{Semi\: perimeter=\frac{a+b+c}{2}}}$

Here, a=7 b=24 c=25

⟹Semi perimeter=7+24+25/2

⟹Semi perimeter=56/2

⟹Semi perimeter=28

As we know that,

Area of Triangle =1/2×base×height

Here, base=7cm height=24cm

Area=1/2×7×24

Area=12×7

Area=84cm²

Applying , heron's formula here to find the area of Triangle

⟹Area=√s(s-a)(s-b)(s-c)

⟹Area=√28(28-7)(28-24)(28-25)

⟹Area=√28x21x4x3

⟹Area=√7x2x2x7x3x2x2x3

⟹Area=7x2x2x3

⟹Aea=84 cm²

Hence,

The area of Triangle is 84cm²

$\large{\underline{\red{\rm{AnswEr:}}}}$

The two sides of Triangle=7cm and 24cm

The area of Triangle=84cm²

Answer 2

Answer:

hope this helps you