Question

In triangle ABC; angle A=90degree, side AB=x cm, AC=(x+5) cm and area=150 cm square. Find the sides of a triangle.

Answer 1

REFER TO THE ATTACHMENT...

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Answer 2

Sides of triangle are 15 cm, 20 cm, 25 cm.

Step-by-step explanation:

Given:

AB = $'x' \ cm$

AC = $(x+5)\ cm$

Area of triangle = 150 sq. cm

m∠A =90°

We need to find the sides of the triangle.

Now we know that Area of triangle is half times base times height.

here base = AC and height = AB

Hence we can frame it as

Area of triangle = $\frac{1}{2} \times AB\times AC$

Substituting the given values we get;

$\frac{1}{2}\times x \times (x+5) = 150\\ \\x(x+5)=150\times 2\\\\x^2+5x=300\\\\x^2+5x-300=0$

Now finding the roots of 'x' we get;

$x^2+20x-15x-300=0\\\\x(x+20)-15(x+20)=0\\\\(x-15)(x+20)=0$

Now we will get 2 values of x

$x-15=0\\\\x=15\\\\OR\\\\x+20=0\\\\x=-20$

Now we get the value of x as 15 and -20.

Since side of triangle cannot be negative hence we can say that value of x is 15.

AB = $x =15\ cm$

AC = $x+5 =15+5=20\ cm$

Now By using Pythagoras theorem we get;

$BC^2=AB^2+AC^2 = 15^2+20^2 = 225+400 = 625\\\\BC^2=625$

Taking Square roots on both side we get;

$\sqrt{BC^2}=\sqrt{625} \\ \\BC = 25\ cm$

Hence side of triangle are 15 cm, 20 cm, 25 cm.