Chapter 10 Locus Miscellaneous Exercise

Rajasthan Board RBSE Class 10 Maths Chapter 10 Locus Miscellaneous Exercise

Multiple Choice Questions (From 1 to 7)
Question 1.
A point equidistant from verties of triangle is called :
(A) Centroid
(B) Circumcentre
(C) Othrocentre
(D) Incentre
Solution :
(B) is correct.

Question 2.
Centroid of triangle is :
(A) Intersection point of ⊥ bisectors of sides of triangle.
(B) Intersection point of angle bisectors of triangle.
(C) Intersection point of medians of triangle.
(D) Intersection point of altitudes of triangle.
Solution :
(C) is correct.

Question 3.
Locus of center of circle rolling in plane is :
(A) Circle
(B) Curve
(C) Parallel line to plane
(D) Perpendicular line on plane
Solution :
(C) is correct.

Question 4.
If two medians of a triangle are same then triangle will be :
(A) Right triangle
(B) Isosceles triangle
(C) Equilateral triangle
(D) None of the above
Solution :
(B) is correct.

Question 5.
If AB and CD are two non-parallel lines, then locus of point p equidistant from these lines will be :
(A) line parallel to AB passing through point P.
(B) angle bisector of angle formed by lines AB and CD, passing through P.
(C) Line parallel to lines AB and CD, passing through P.
(D) line perpendicular to line AB and CD passing through point P.
Solution :
(B) is correct.

Question 6.
Triangle whose orthocentre, circumcentre and incentre are concurrent is called :
(A) Equilateral triangle
(B) Right angled triangle
(C) Isosceles triangle
(D) None of the above
Solution :
(A) is correct.

Question 7.
Triangle whose othocentre is vertices of triangle is called :
(A) Right angled triangle
(B) Equilateral triangle
(C) Isosceles triangle
(D) None of the above
Solution :
(A) is correct.

Question 8.
Write the locus of the end of pendulum of clock.
Solution :
According to figure, Pedulum of clock will be moving so by inspection, locus of end of pendulum will be an arc of circle.

Question 9.
If D, E and Fare mid-points of sides BC, CA and AB of triangle ABC respectively, then prove that EF, bisects AD.
Solution :
Given: In ΔABC, D, E and F are mid-points of sides BC, CA and AB.
Construction : Join DE and DF

Proof : In ΔABC, D and E are mid-points of sides BC and AC respectively.
∴ DE || AB
and DE = AB
Now, DE || AB ⇒ FA || DE …..(i)
Similarly, EA || DF ….(ii)
From equations (i) and (ii),
Thus ΔEAF will be a parallelogram.
Since we know that diagonals of parallelogram bisects each other.
Thus EF will bisects AD.