Question

Let p be an interior point of a right-angled isosceles triangle abc with hypotenuse ab. If the perpendicular distance of p from each of ab, bc, and ca is 4 (2 - l) cm, then the area, in sq cm, of the triangle abc is

Answer 1

Answer:

Step-by-step explanation:

From CAT 2017 – Quantitative Aptitude – Geometry – Triangles, we can see that,

Quantitative Aptitude - Geometry - Triangles - Let P be an interior point  

PQ = PR = PS = 4(√2-1)

CS = PR

(PC)^2 = (PS)^2 + (CS)^2

On solving, we get, PC = 4√2(√2-1)

So, QC = PC + PQ = 4

Area of a right angled triangle = ½ * Base * Height

So, ½ * AC * BC = ½ * QC * AB

On solving, we get a = 4√2

Area of triangle = ½ * a^2 = 16

Answer: 16

Answer 2

Answer:

If the perpendicular distance of P from each of AB,BC,and CA is 4 ( 2 − 1 ) 4(\sqrt{2}-1) 4(2 −1) cm,then the area, in sq cm, of the triangle ABC is. P is equidistant from all the side of the triangle. Hence P is the incenter and the perpendicular distance is the inradius.