1. For a triangle ABC, D and E are two points on AB and AC such that AD = ^{1}/_{4}AB, AE = ^{1}/_{4}AC. If BC = 12 cm then DE is :

3 cm

6 cm

5 cm

4 cm

Answer (a). X and Y are mid points of AB and BC respectively. As per mid-point theorem XY = BC/2 or 6 cm. Similarly, in triangle AXY, D and E are mid-points of AX and AY respectively. Therefore, as per mid-point theorem DE = XY/2 = 3 cm.

2. In an acute angled triangle ABC, if sin 2(A + B – C)= 1 and tan(B + C – A) = √3, then the value of angle B is

60°

30°

52 ½°

67 ½°

Answer (c). sin 2(A + B - C) = 1 2(A + B - C) = 90° (since sin 90° = 1) A + B - C = 45°......(1) tan (B + C - A) = √3 B + C - A = 60° (since tan 60° = √3) .......(2)
Adding (1) and (2) A + B - C + B + C - A = 45° + 60° = 105° 2B = 105° B = 52 ½°

3. If the in radius of a triangle with perimeter 32 cm is 6 cm, then the area of the triangle in sq. cm is

48

100

64

96

Answer (d). Area of the triangle = in radius x semi-perimeter Area = 6 x 16 = 96 sq. cm.

4. ABC is a right angled triangle, B being the right angle. Mid-points of BC and AC are respectively B’ and A’. The ratio of the area of the quadrilateral AA’ B’B to the area of the triangle ABC is

1 : 2

2 : 3

3 : 4

None of the above

Answer (c). Area of triangle ABC = bh/2 Since CB' is half of CB, area of triangle CA'B' = bh/8 The ratio of area of triangle CA'B' to ABC = 1 : 4. Therefore, the ratio of area of quadrilateral AA'BB' to the area of triangle ABC = 3 : 4.

5. In a triangle ABC, the side BC is extended up to D. Such that CD = AC, if angle BAD = 109° and angle ACB = 72° then the value of angle ABC is

35°

60°

40°

45°

Answer (a). Angle ACB = 72^{o}, hence angle ACD = 180^{o} - 72^{o} = 108^{o} Since CD = AC, triangle ADC is an isosceles triangle. Therefore angles CAD and CDA are both equal to 72^{o}/2 = 36^{o}. But angle BAD = 109^{o}, hence angle BAC = 109^{o} - 36^{o} = 73^{o} Finally angle ABC = 180^{o} - (73^{o} + 72^{o}) = 35^{o}

6. Side BC of triangle ABC is produced to D. If angle ACD = 140^{o} and angle ABC = 3 ⁄BAC, then find angle A.

45°

55°

35°

60°

Answer (c). Exterior angle is equal to the sum of 2 opposite interior angles. i.e. Angle A + angle B = 140^{o} Dividing 140^{o} in the ratio 1 : 3, we get angle A = 35^{o}

7. If O be the circum centre of a triangle PQR and angle QOR = 110°, angle OPR = 25°, then the measure of angle PRQ is

55°

60°

65°

50°

Answer (b). Since O is the circum centre, OP = OQ = OR Thus, triangle OPR is an isosceles triangle, hence angle PRO = 25^{o} Similarly, triangle OQR is also an isosceles triangle Hence, angle ORQ = (180^{o} - 110^{o})/2 = 35^{o} Therefore, angle PRQ = 35^{o} + 25^{o} = 60^{o}

8. D and E are mid-points of AB and AC of triangle ABC. If angle A = 80^{o}, angle C = 35^{o}, then angle EDB is equal to

100°

115°

120°

125°

Answer (b).Note: A line joining the mid-points of any two sides is parallel to the third side. In triangle ADE, angle E = 35^{o} (angle E and angle C are a pair of corresponding angles, hence equal) Therefore angle EDB = 80^{o} + 35^{o} = 115^{o} (Exterior angle is equal to the sum of two opposite interior angles)

9. In a right-angled triangle ABC, angle ABC = 90°, AB = 5 cm and BC = 12 cm. The radius of the circum circle of the triangle ABC is

6.5 cm

7 cm

7.5 cm

6 cm

Answer (a). The circumcentre of a right angled triangle is the mid-point of the hypotenuse. In the given diagram, AC^{2} = 12^{2} - 5^{2} AC = 13 cm which is also the diameter of circumcircle. Therefore the radius of the circumcircle = 6.5 cm.

10. If the circum radius of an equilateral triangle ABC be 8 cm, then the height of the triangle is

8 cm

12 cm

16 cm

6 cm

Answer (b). In an equilateral triangle, centroid and the circumcenter coincide. AD is thus the height as well as the median of the triangle. Since the centroid divides the median in the ratio 2 : 1, the height of the triangle will be 12 cm.

.......contd

11. 360 sq. cm and 250 sq. cm are the areas of two similar triangles. If the length of one of the sides of the first triangle be 8 cm, then the length of the correspoding side of the second triangle is

6 cm

6 ^{1}/_{5} cm

6 ^{1}/_{3} cm

6 ^{2}/_{3} cm

Answer (d). Property: Ratio of areas of two similar triangles is equal to the ratio of squares of the corresponding sides. Assuming the unknown side to be 'a' we have 360 : 250 = 8^{2} : a^{2} a^{2} = (250 x 64)/360 = 40/6 = 6 ^{2}/_{3} cm

12. If the perimeters of an equilateral triangle and that of a square are equal, then the ratio of their areas will be

4 : 3

4 : √3

4 : 3√3

4 : 2√3

Answer (c). Let the perimeter of square & the perimeter of triangle be A Side of a square = A/4 and its area = A^{2}/16 Side of the triangle = A/3 and its area = (√3 x A^{2})/4 x 9 = A^{2}√3/36 The ratio - A^{2}√3/36 : A^{2}/16 or 4 : 3√3

13. Length of each equal side of an isosceles triangle is 10 cm and the included angle between those two sides is 45^{o}. Find the area of the triangle.

25√2 cm^{2}

35√2 cm^{2}

5√2 cm^{2}

15√2 cm^{2}

Answer (a). Area of trianlge = abSin c/2 where a and b are sides and c is the included angle Area = 100 x Sin45^{o}/2 => 50/√2 (since Sin45 = 1/√2) => 25√2 cm^{2}

14. In a triangle ABC, the base BC is trisected at D and E. The line through D, parallel to AB, meets AC at F and the line through E parallel to AC meets AB at G. Let EG and DF intersect at H. What is the ratio of the sum of the area of parallelogram AGHF and the area of the triangle DHE to the area of the triangle ABC?

1 : 2

1 : 3

1 : 4

1 : 6

Answer (b). Ratio of areas of two similar triangles is equal to the ratio of squares of the corresponding sides. In the figure below, ABC, GBE, FDC and HDE are similar triangles. Sides of triangles GBE and FDC are 2/3 of the sides of triangle ABC. Hence if the area of the triangle ABC is assumed to be 1 sq. unit, the areas of triangles GBE and FDC would be (2/3)^{2} i.e. 4/9. Similarly sides of triangle HDE are 1/3 of the side of triangle ABC Therefore the area of triangle HDE would be 1/9 of the area of triangle ABC Area of trapezium GHDB would be 4/9 - 1/9 = 3/9 Area of parallelogram AGFH would be 1 - (3/9 + 4/9) = 2/9 Sum of the area of parallelogram AGHF and the area of the triangle DHE = 2/9 + 1/9 = 3/9 = 1/3. Therefore the required ratio = 1/3 : 1 or 1 : 3

15. PQR is an equilateral triangle. O is the point of intersection of altitudes PL, QM and RN. If OP = 8 cm, then what is the perimeter of the triangle PQR?

8√3 cm

12√3 cm

16√3 cm

24√3 cm

Answer (d). Altitude PL = 12 cm (see question 10 above) In triangle PLR, angle R = 60^{o} and angle LPR = 30^{o} , Therefore PL^{2} = 3LR^{2} =>144 = 3 x LR^{2} => LR = 12/√3 Therefore each side = 24/√3 and perimeter = 72/√3 = 24√3 cm.