ICOME
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Unit 3

Unit-3:Geometry

(a) Triangles

(b)Circles


Q1. Geometric figures having the same shape but different sizes are known as _________ figures.

(a) Similar (b) Congruent (c) Both(S&C) (d) None of these

Solution.
(a) Similar
Solution

Q2. Two polygons are said to be similar to each other if.
(i) Their corresponding angles are equal.
(ii) The lengths of their corresponding sides are proportional.

(a) Only (i) (b) Only (ii) (c) Both (d) None of these

Solution.
(c) Both
Solution

Q3. In a triangle ABC ED ∥ CB and
AD/DB
=
5/7
. AE = 5 cm find EC.


(a) 8 (b) 9 (c) 6 (d) 7

Solution.
By thale's theorem
let EC = x
AD/DB
=
AE/EC

5/7
=
5/x

x = 7 cm
Solution

Q4. In a triangle ABC , DE ∥ BC find x, if AD = x, BD = x + 2, AE = x - 1, EC = x - 2.


(a) 2 (b)
3/2
(c)
2/3
(d) 3

Solution.
AD/DB
=
AE/EC

x/x + 2
=
x - 1/x - 2

x2 - 2x = x2 - x + 2x - 2
x =
2/3
Solution

Q5. In a triangle ABC, AD is the bisector of ∠A, AB = 8 cm, AC = 6 cm , BD = 4 cm, Aand DC = x cm


(a) 5 (b) 3 (c) 2 (d) 4

Solution.
AB/AC
=
BD/DC

8/6
=
4/x

x = 3 cm
Solution

Q6. The diagonals of a quadrilateral divide each other proportionally, then it is a ____________.

(a) Rectangle (b) Parallelogram (c) Trapezium (d) None of these

Solution.
(c) Trapezium
Solution

Q7. The length of a tangent from a point A, at a distance 5 cm from the centre of the circle is 4 cm, find the radius of the circle .

(a) 5 cm (b) 4 cm (c) 2 cm (d) 3 cm

Solution.
(d) 3 cm
Solution

Q8. In a circle of radius 21 cm, an are subtends an angle of 600 at the centre. Find the lenth of the arc.

(a) 30 cm (b) 22 cm (c) 10 cm (d) 11 cm

Solution.
L = rθ
L = 21 x
π/3

= 21 x
22/7 x 3

L = 22 cm
Solution

Q9. Find the distance between two parallel tangents of a circle of radius 10 cm.

(a) 9 cm (b) 5 cm (c) 10 cm (d) 20 cm

Solution.
(d) 20 cm
Solution

Q10. The sides AB,BC and CA of triangle ABC touch a circle with centre o and radius r at P,Q and R respectively, then which relation is true .

(a) Area of triangle ABC =
1/2
(Perimetre of triangle ABC) x r
(b)Area of triangle ABC = (Perimetre of triangle ABC) x r
(c)Area of triangle ABC =
1/2
(Perimetre of triangle ABC)
(d)Area of triangle ABC =
1/2
(Perimetre of triangle ABC) - r

Solution.
(a) Area of triangle ABC =
1/2
(Perimetre of triangle ABC) x r
Solution