The sides AB, BC, CA of a triangle ABC have $3, 4$ and $5$ interior points respectively on them. The number of triangles that can be constructed using these points as vertices is-

205

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

210

No worries! We‘ve got your back. Try BYJU‘S free classes today!

315

No worries! We‘ve got your back. Try BYJU‘S free classes today!

216

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $205$
$\to$ Suppose we had $12$ points, none of them collinear, number of possible triangles $=^{12} C_{3}$
$\to$ If out of these $3, 4$ and $5$ points were chosen separately to lie on each side of a triangle ABC, then number of possible triangles
$=^{12} C_{3} -^{3} C_{3} -^{5} C_{3} -^{4} C_{3}$
$= 220 - 1 - 10 - 4$
$= 205$

Suggest Corrections

Similar questions

Q. The sides

A B, B C

and

C A

of a triangle

A B C

have

3, 4

and

5

interior points, respectively on them. The number of triangles that can be constructed using these interior points as vertices is _______.

Q. The sides AB, BC, CA of a triangle ABC have

3, 4

and

5

interior points respectively on them. Find the number of triangles that can be constructed using these points as vertices.

The sides BC, CA and AB of a triangle ABC have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these points as vertices is

Q. The sides

A B, B C

and

C A

△ A B C

are marked with

3, 4

and

5

interior points respectively. Number of triangles that can be constructed using these interior points as vertices is

Q. The sides AB, BC, CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The total number of triangles that can be constructed by using these points as vertices are

Join BYJU'S Learning Program

The sides AB, BC, CA of a triangle ABC have 3,4 and 5 interior points respectively on them. The number of triangles that can be constructed using these points as vertices is-

The correct option is A 205→Suppose we had 12 points, none of them collinear, number of possible triangles =12C3→If out of these 3,4 and 5 points were chosen separately to lie on each side of a triangle ABC, then number of possible triangles=12C3−3C3−5C3−4C3=220−1−10−4=205

The sides AB, BC, CA of a triangle ABC have $3, 4$ and $5$ interior points respectively on them. The number of triangles that can be constructed using these points as vertices is-