Question

in triangle ABC angle B= 90° and AB : AC = 24 : 25. BC=14 find the perimeter of triangle ABC​

Answer 1

The perimeter of the triangle ABC is found to be 112.

Step-by-step explanation:

Given: i) △ ABC with ∠B = 90°

          ii) AB:AC = 24:25

          iii) BC = 14

To be found: Perimeter of △ ABC

Formulae to be used:

Pythagoras theorem $(AC^2 = AB^2 + BC^2)$ and  Perimeter of △ ABC  =  AB + AC + BCwhere,  

• AC is the hypotenuse, and
• AB and BC are the two sides of the △ ABC right- angled at B

Solution:

Let the sides be AB and AC be 24x and 25x

$(25x)^2 = (24x)^2 + (14)^2$

$625x^2 = 576x^2 +196$

$625x^2 - 576x^2 =196$

$49x^2 = 196$

$x^2 = \frac{196}{49}$

$x = \sqrt{\frac{196}{49}}$

$x = \frac{14}{7}}$

$x=2$

Thus, the sides would be

$AB = 24x = 24\times 2 = 48$ and $AC = 25x = 25\times 2 = 50$

Thus, Perimeter $= 48+50+14 = 112$

Answer 2

Answer:

112

Step-by-step explanation:

The perimeter of the triangle ABC is found to be 112.

Step-by-step explanation:

Given: (i) △ ABC with ∠B = 90°

(ii) AB:AC = 24:25

The perimeter of the triangle ABC is found to be 112.

Step-by-step explanation:

Given: i) △ ABC with ∠B = 90°

ii) AB:AC = 24:25

iii) BC = 14

To be found: Perimeter of △ ABC

Formulae to be used:

Pythagoras theorem (AC^2 = AB^2 + BC^2)(AC

2

=AB

2

+BC

2

) and Perimeter of △ ABC = AB + AC + BC where,

AC is the hypotenuse, and

AB and BC are the two sides of the △ ABC right- angled at B

Solution:

Let the sides be AB and AC be 24x and 25x

(25x)^2 = (24x)^2 + (14)^2(25x)

2

=(24x)

2

+(14)

2

625x^2 = 576x^2 +196625x

2

=576x

2

+196

625x^2 - 576x^2 =196625x

2

−576x

2

=196

49x^2 = 19649x

2

=196

x^2 = \frac{196}{49}x

2

=

49

196

x = \sqrt{\frac{196}{49}}x=

49

196

x = \frac{14}{7}}

x=2x=2

Thus, the sides would be

AB = 24x = 24\times 2 = 48AB=24x=24×2=48 and AC = 25x = 25\times 2 = 50AC=25x=25×2=50

Thus, Perimeter = 48+50+14 = 112=48+50+14=112 BC = 14

To be found: Perimeter of △ ABC

Formulae to be used:

Pythagoras theorem (AC^2 = AB^2 + BC^2)(AC

2

=AB

2

+BC

2

) and Perimeter of △ ABC = AB + AC + BC where,

AC is the hypotenuse, and

AB and BC are the two sides of the △ ABC right- angled at B

Solution:

Let the sides be AB and AC be 24x and 25x

(25x)^2 = (24x)^2 + (14)^2(25x)

2

=(24x)

2

+(14)

2

625x^2 = 576x^2 +196625x

2

=576x

2

+196

625x^2 - 576x^2 =196625x

2

−576x

2

=196

49x^2 = 19649x

2

=196

x^2 = \frac{196}{49}x

2

=

49

196

x = \sqrt{\frac{196}{49}}x=

49

196

x = \frac{14}{7}}

x=2x=2

Thus, the sides would be

AB = 24x = 24\times 2 = 48AB=24x=24×2=48 and AC = 25x = 25\times 2 = 50AC=25x=25×2=50

Thus, Perimeter = 48+50+14 = 112