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Science & Math for K6 - K12 Students

Cumulative Exercises SRQ
Exercise-I
1. Two sides of a triangle are 6 cm and 8 cm.If the triangle is isosceles, what are the possible values of the third side?
 Sol: The 3rd side can be 6 unit and 8 unit

2. Two sides of a triangle are 3 cm and 8 cm. If the triangle is isosceles what are the possible values of the 3rd side
 Sol: The length of the 3rd side can be 8 cm as 8 + 3 > 8, 8 + 8 > 3

3. Two sides of a triangle 7 units and 10 units which of the following can be the length of the 3rd side?
 Sol: In a triangle sum of two sides > 3rd side Difference of two sides < 3rd side 7 + 10 > 5 , 10 – 5 < 7 7 + 10 > 7 , 10 –7 < 7 7 + 10 > 11 , 11 –10 < 7 7 + 10 > 13 , 13 –10 < 7

4.
From the above diagram express z in terms of x and y
 Sol: 1 = x vertically opposite angles x + y = z Exterior angle = sum of interior opposite angles

5.
If GF | | DE and CD | | AB find p,q
 Sol: As GF | | DE JGF = GDE = 40° corresponding angles are equal GDE + EDC + CDB = 180° (straight angle) 40° +55° + p = 180° p = 75° p + q = 180° (sum of cointerior angles = 180°) 75° + q = 180° q = 180° – 75° q = 105° p = 75° , q = 105°

6.
From the given figure find p and q
 Sol: In Δ ADC Exterior angle ADB = sum of interior opposite angles p = DAC + ACD = 20° + 50° p = 70° In Δ ADB Exterior angle ABE = ADB + BAD = P + 60° q = 70° + 60° q = 130°

7. In a triangle ΔABC AC = (12 + AC) Find AC?
 Sol: 4AC = 12 + AC 3AC = 12 cm AC = 4 cm

8. In Δ ABC, A = 30°, B = 90°, C = 60°. If AC = p + 6 and BC = 2p – 3 find the value of p
 Sol: when angles are 30°, 60° 90° The ratio of sides will be 1 : √3 : 2 4p – 6 = p + 6 3p = 12 p = 4cm

9.
PQRS is a square in which x and y are mid points of PR and PS. If PR = m units. what is the area of the shaded portion
 Sol: Area of a square = Area of Δ PRS = The line joining the midpoints divides the triangle in the ratio 1 : 3 Area of shaded portion =

10.
From the given figure if the length of the diagonal DF is 'm' units then what can be possible value of(m)
 Sol: If DF is the diagonal From ΔDEF 1 < m < 7 (The range of 3rd side is difference of 2 sides < 3rd side

11.
In PQR, PQ = 20 – a, QR = a, PQ = a + 5 what is the possible value of 'a' (Hint a = 15, 17, 8, 5, 3)
 Sol: put a = 5 the sides will be a + 5 = 5 + 5 = 10 units 20 – a = 20 – 5 = 15units a = 5 units(10 + 5 > 15) a = 5(not correct option) a = 8, the sides will be 8 + 5, 20 – 8, 8 a = 8, is the correct option 13, 12, 8

12.
The radii of all the circles in the circle are equal.The radius of the larger circle is 'p' units. What is the area of unshaded region ?
 Sol: Total diameter of 3 circle = 2p Diameter of each circle = cm Radius of each circle = = cm unshaded portion area = Area of bigger circle – Area of 7 circles =

13.
Find the perimeter of the given figure
 Sol: perimeter = AB + AD + CD + Arc = 2 + 1 + 2 + πr where r = unit = 5 + π × =

14. If a,b,c are the sides of the triangle, then
 Sol: sum of lengths of 2 sides > third side b < c + a is true

15. Two sides of a triangle are 8 cm and 11 cm. Then a possible third side is from the given sides 1 cm; 2 cm; 3 cm; 4 cm; is
 Sol: If you take the side 4 cm 8 + 4 > 12 8 – 4 < 12 3rd side can be 4 cm

16. In ΔXYZ XY + YZ = 10 cm, YZ + ZX = 12 cm, ZX + XY = 16 cm. Find the sides of the triangle
 Sol: XY + YZ = 10 cm ––––(1) YZ + ZX = 12 cm–––––(2) ZX + XY = 16 cm––––––(3) Adding (1)(2)(3) we get 2(XY + YZ + ZX) = 38 cm XY + YZ + ZX = 19 cm If we put XY + YZ = 10 cm then 10 + ZX = 19 cm ZX = 9 cm If we put YZ + ZX = 12cm then XY + 12 = 19cm XY = 7 cm If we put ZX + XY = 1 6cm then YZ + 16 = 19 cm YZ = 3 cm

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