Question

The numerical value of the area of a right angled triangle is 9 more than three times of the length of its diagonal. If the difference of lengths of the other two sides of the triangle is 3 CM and the area is 54 sq CM, then find the lengths of the sides of the triangle​

Answer 1

Answer:

Step-by-step explanation:

In this right triangle, you are given the measurements for the hypotenuse, c, and one leg, b. The hypotenuse is always opposite the right angle and it is always the longest side of the triangle. To find the length of leg a, substitute the known values into the Pythagorean Theorem. Solve for a2

Answer 2

Answer:

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Step-by-step explanation:

Let ABC be the right angled triangle with right angle at A.

AB is the base, AC is the height and BC is the hypotenuse.

As the difference between the base and height is 3cm, let AB =xcm and AC =(x+3)cm

Area of triangle =21×base×height=54 sq cm

21×x×(x+3)=54

=>x2+3x=108

x2+12x−9x−(12×9)=0

Solving it x=9;−12

Since side cannot be negative, AB =x=9cm

AC =x+3=12cm

Since the third side  BC is the hypotenuse , BC2=92+122=225

Hence, BC =15cm

Perimeter of the triangle = Sum of all sides =15+9+12=36cm......

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