Question

The vertices of a triangle are A(a,0), B(0,b) and C(a,b), then the foot of altitude from C is

a) (a/2,b/2)

b) (a,b)

c) (a³/a²+b²,b³/a²+b²)

d) (2a/3,2b/3)​

Answer 1

Answer:

ANSWER

Steps of construction

Draw a line segment OA.

Taking O as center and any radius, draw an arc cutting OA at B.

Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.

With C as center and the same radius, draw an arc cutting the arc at D.

With C and D as center and radius more than

2

1

CD, draw two arc intersecting at P.

Join OP.

Thus, ∠AOP=90

o

Justification

Join OC and BC

Thus,

OB=BC=OC [Radius of equal arcs]

∴△OCB is an equilateral triangle

∴∠BOC=60

o

Join OD,OC and CD

Thus,

OD=OC=DC [Radius of equal arcs]

∴△DOC is an equilateral triangle

∴∠DOC=60

o

Join PD and PC

Now,

In △ODP and △OCP

OD=OC [Radius of same arcs]

DP=CP [Arc of same radii]

OP=OP [Common]

∴△ODP≅△OCP [SSS congruency]

∴∠DOP=∠COP [CPCT]

So, we can say that

∠DOP=∠COP=

2

1

∠DOC

∠DOP=∠COP=

2

1

×60=30

o

Now,

∠AOP=∠BOC+∠COP

∠AOP=60+30

∠AOP=90

o

Hence justified.