Question

the perimeter of right angle triangle is 60 cm and its area is 120 CM square find the length of its sides

Answer 1

Step-by-step explanation:

Given :-

The perimeter of a right angle triangle is 60 cm and its area is 120 cm²

To find :-

Find the length of its sides ?

Solution :-

Let the sides of a right angled triangle be

a cm , b cm and c cm

Let c cm be the hypotenuse

We know that

Perimeter of a right angled triangle

= Sum of the all sides

Perimeter of the given right angled triangle = ( a+b+c ) cm

According to the given problem

Perimeter of the given right angled triangle = 60 cm

=> a+b+c = 60 ------------------(1)

=> c = 60-(a+b) ------------------(2)

and

Area of a right angled triangle

= (1/2) ab sq.units

According to the given problem

Area of the right angled triangle = 120 cm

=> (1/2) ab = 120

=> ab = 120×2

=> ab = 240 ------------------------(3)

We know that

By Pythagoras Theorem ,

c² = a²+b² --------------------------(4)

On substituting the value of c from (2) in (4)

=> [60-(a+b)]² = a²+b²

=>[60-(a+b)]² = (a+b)² -2ab

=> (60)² -2(60)(a+b)+(a+b)² = (a+b)² -2ab

=> 3600-120(a+b)+(a+b)² = (a+b)²-2(240)

=> 3600-120(a+b)+(a+b)² = (a+b)² -480

=> 3600-120(a+b)+480 = (a+b)²-(a+b)²

=> 4080-120(a+b) = 0

=> 4080 = 120(a+b)

=> (a+b) = 4080/120

=> (a+b) = 408/12

=> (a+b) = 34 --------------------(5)

Substituting this value in (2) then

c = 60-34

=> c = 26 cm

We know that

(a-b)² = (a+b)²-4ab

=> (a-b)² = (34)²-4(240) (from (3))

=> (a-b)² = 1156-960

=> (a-b)² = 196

=> a-b = ±√196

=> a-b = ±14

The lengths of the sides can't be negative .

a-b = 14 -----------------------------(6)

On adding (5) and (6)

a + b = 34

a - b = 14

(+)

_________

2a + 0 = 48

_________

=> 2a = 48

=> a = 48/2

=> a = 24 cm

From (5)

=> 24+b = 34

=> b = 34-24

=> b = 10 cm

Therefore, a = 24 cm, b= 10 cm and

c = 26 cm

The three sides of the triangle are 24 cm, 10 cm and 26 cm

Answer:-

The lengths of the three sides of the right angled triangle are 24 cm , 10 cm and 26 cm

Check :-

a = 24 cm , b = 10 cm and c = 26 cm

a²+b² = 24² + 10²

=> a²+b² = 576+100

=> a²+b² = 676

=> a²+b² = 26²

=> a²+b² = c²

a,b and c are the sides of the right angled triangle.

and

The perimeter = 24+10+26 = 60 cm

Area of the triangle = (1/2)(24×10)

=> (24×10)/2

=> 240/2

=> 120 sq.cm

Verified the given relations in the given problem.

Used formulae:-

→ Perimeter of a right angled triangle

= ( a+b+c ) units

→ Area of a right angled triangle = (1/2) ab sq.units

Where, a , b and c are the three sides

→ In right angled triangle, c² = a²+b²,

Where c is the hypotenuse.

______________________________

→ (a+b)² = a²+2ab+b²

→ (a-b)² = a²-2ab+b²

→ (a-b)² = (a+b)²-4ab

Used Theorem:-

Pythagoras Theorem:-

" In a right angled triangle, The square of the hypotenuse is equal to the sum of the squares of the other two sides".

Answer 2

Answer:

Given :-

• The perimeter of a right angle triangle is 60 cm and its area is 120 cm².

To Find :-

• What is the length of its sides.

Solution :-

Let,

$\mapsto \bf First\: Side =\: x\: cm$

$\mapsto \bf Second\: Side =\: y\: cm$

$\mapsto \bf Third\: Side =\: z\: cm$

According to the question,

$\implies \sf x + y + z =\: 60$

$\implies \sf\bold{\purple{z =\: 60 - x - y\: ------\: (Equation\: No\: 1)}}$

Now,

Given :

• Area = 120 cm²
• Height = x cm
• Base = y cm

As we know that :

$\longrightarrow \sf\boxed{\bold{\pink{Area_{(Triangle)} =\: \dfrac{1}{2} \times Height \times Base}}}$

According to the question by using the formula we get,

$\implies \sf 120 =\: \dfrac{1}{2} \times x \times y$

$\implies \sf 120 =\: \dfrac{xy}{2}$

By doing cross multiplication we get,

$\implies \sf xy =\: 120(2)$

$\implies \sf\bold{\purple{xy =\: 240\: ------\: (Equation\: No\: 2)}}$

Again, by using Pythagoras theorem we get,

$\implies \sf\bold{\purple{(z)^2 =\: (x)^2 + (y)^2\: ------\: (Equation\: No\: 3)}}$

By putting the value of z in the equation no 3 we get,

$\implies \sf z^2 =\: x^2 + y^2$

$\implies \sf (60 - x - y)^2 =\: x^2 + y^2$

$\implies \sf 3600 + x^2 + y^2 - 120(x + y) + 2xy =\: x^2 + y^2$

$\implies \sf 3600 + 2xy - 120(x + y) =\: 0$

$\implies \sf\bold{\purple{3600 + 2xy =\: 120(x + y)\: ------\: (Equation\: No\: 4)}}$

Again, by putting the value of xy in the equation no 4 we get,

$\implies \sf 3600 + 2(240) =\: 120(x + y)$

$\implies \sf 3600 + 480 =\: 120(x + y)$

$\implies \sf 4080 =\: 120(x + y)$

$\implies \sf \dfrac{\cancel{4080}}{\cancel{120}} =\: x + y$

$\implies \sf 34 =\: x + y$

$\implies \sf\bold{\purple{x + y =\: 34\: ------\: (Equation\: No\: 5)}}$

Again,

$\implies \sf (x + y)^2 =\: (x - y)^2 + 4xy$

$\implies \sf (34)^2 =\: (x - y)^2 + 4(240)$

$\implies \sf 1156 =\: (x - y)^2 + 960$

$\implies \sf 1156 - 960 =\: (x - y)^2$

$\implies \sf 196 =\: (x - y)^2$

$\implies \sf\bold{\purple{x - y =\: 14\: ------\: (Equation\: No\: 6)}}$

Again, by adding the equation no 5 and 6 we get,

$\implies \sf x {\cancel{+ y}} + x {\cancel{- y}} =\: 34 + 14$

$\implies \sf x + x =\: 48$

$\implies \sf 2x =\: 48$

$\implies \sf x =\: \dfrac{\cancel{48}}{\cancel{2}}$

$\implies \sf\bold{\red{x =\: 24\: cm}}$

Again, by putting x = 24 in the equation no 5 we get,

$\implies \sf x + y =\: 34$

$\implies \sf 24 + y =\: 34$

$\implies \sf y =\: 34 - 24$

$\implies \sf\bold{\red{y =\: 10\: cm}}$

Again, by putting x = 24 and y = 10 in the equation no 1 we get,

$\implies \sf z =\: 60 - x - y$

$\implies \sf z =\: 60 - 24 - 10$

$\implies \sf\bold{\red{z =\: 26\: cm}}$

$\therefore$ The length of its sides is 24 cm, 10 cm and 26 cm respectively.

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