In the following diagram AD bisects angle BAC. The area of the triangle ABD is 30 sq. cm and AC is 3 times AB. Find the area of triangle ABC.
ans is 120(90+30) use 1/2absinC formula for triangle areas.
120 is ans y is this msg not being displayed?
i think the correct answer is 300 sq cm(30+30*3^2)
n/a
120
the ans is 120
how have you solved it?explain it please
As it says in the question itself, AD bisects the angle BAC. Moreover, AC is 3times AB.Area of traingle ABD = 30 = 0.5 * AD *BDBD = 60/AD
Area of ABC = 0.5 * AD * BC (You can prove triangle ABC is similar to ABD, Thus when AC = 3AB, then DC = 3BD. AC = BD +DC.AC = 4BC)
Area of ABC = 0.5* AD * BC= 0.5*AD*4BD=0.5*AD*4*(60/AD)=120
This is how the answer can be done
AD is the bisector. Hence it divides the opposite side in the ratio as the sides. AC is 3 times AB. Hence DC is 3 times BD. Hence the area of ADC is 3 times that of ABD. AD is the common base. Hence 120.
More information about formatting options
ans is 120(90+30) use 1/2absinC formula for triangle areas.
120 is ans y is this msg not being displayed?
i think the correct answer is 300 sq cm(30+30*3^2)
n/a
120
the ans is 120
n/a
how have you solved it?explain it please
As it says in the question itself, AD bisects the angle BAC. Moreover, AC is 3times AB.
Area of traingle ABD = 30 = 0.5 * AD *BD
BD = 60/AD
Area of ABC = 0.5 * AD * BC
(You can prove triangle ABC is similar to ABD, Thus when AC = 3AB, then DC = 3BD. AC = BD +DC.
AC = 4BC)
Area of ABC = 0.5* AD * BC
= 0.5*AD*4BD
=0.5*AD*4*(60/AD)
=120
This is how the answer can be done
AD is the bisector. Hence it divides the opposite side in the ratio as the sides. AC is 3 times AB. Hence DC is 3 times BD. Hence the area of ADC is 3 times that of ABD. AD is the common base. Hence 120.
Post new comment