| 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 |
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| 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 |
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| 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 |
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Difference of two sides < 3rd side |
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7 + 10 > 5 , 10 – 5 < 7 |
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7 + 10 > 7 , 10 –7 < 7 |
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7 + 10 > 11 , 11 –10 < 7 |
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7 + 10 > 13 , 13 –10 < 7 |
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| 4. |
From the above diagram express z in terms of x and y |
| Sol: |
 1 |
= |
x vertically opposite angles |
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x + y |
= |
z |
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Exterior angle |
= |
sum of interior opposite angles |
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| 5. |
If GF | | DE and CD | | AB find p,q |
| Sol: |
As GF | | DE |
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 JGF = GDE = 40° corresponding angles are equal |
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GDE + EDC + CDB |
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180° (straight angle) |
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40° +55° + p |
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180° |
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p |
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75° |
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p + q |
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180° (sum of cointerior angles = 180°) |
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75° + q |
= |
180° |
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q |
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180° – 75° |
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q |
= |
105° |
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p = 75° , q = 105° |
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| 6. |
From the given figure find p and q |
| Sol: |
In Δ ADC Exterior angle ADB |
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sum of interior opposite angles |
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p |
= |
DAC + ACD |
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20° + 50° |
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p |
= |
70° |
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In Δ ADB Exterior angle ABE |
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ADB + BAD |
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P + 60° |
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q |
= |
70° + 60° |
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q |
= |
130° |
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| 7. |
In a triangle ΔABC AC =  (12 + AC) Find AC? |
| Sol: |
4AC |
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12 + AC |
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3AC |
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12 cm |
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AC |
= |
4 cm |
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| 8. |
In Δ ABC, A = 30°, B = 90°, C = 60°. If AC = p + 6 and BC = 2p – 3 find the value of p |
| Sol: |
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when angles are 30°, 60° 90° |
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The ratio of sides will be 1 : √3 : 2 |
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4p – 6 |
= |
p + 6 |
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3p |
= |
12 |
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p |
= |
4cm |
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| 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 |
= |
 |
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Area of Δ PRS |
= |
 |
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The line joining the midpoints divides the triangle in the ratio 1 : 3 |
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Area of shaded portion |
= |
 |
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| 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 |
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From ΔDEF |
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1 < m < 7 (The range of 3rd side is difference of 2 sides < 3rd side <sum of two sides)––––(1) |
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From Δ DGF |
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3 < mc < 13–––––––––(2) |
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If EG is the diagonal (Δ DEG) |
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5 <m< 11––––––––––––(3) |
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From ΔGEF |
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1 < m < 9––––––––––––(4) |
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From (1),(2),(3),(4) 5 < m < 7 |
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The possible value of m is 5 < m < 7 |
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| 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 |
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a + 5 |
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5 + 5 |
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10 units |
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20 – a |
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20 – 5 |
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15units |
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a |
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5 units(10 + 5 > 15) |
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a |
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5(not correct option) |
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a |
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8, the sides will be 8 + 5, 20 – 8, 8 |
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a |
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8, is the correct option 13, 12, 8 |
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| 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 |
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Diameter of each circle |
= |
 cm |
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Radius of each circle |
= |
 |
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= |
 cm |
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unshaded portion area |
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Area of bigger circle – Area of 7 circles |
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| 13. |
Find the perimeter of the given figure |
| Sol: |
perimeter |
= |
AB + AD + CD + Arc  |
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2 + 1 + 2 + πr |
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where r |
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 unit |
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= |
5 + π × |
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| 14. |
If a,b,c are the sides of the triangle, then |
| Sol: |
sum of lengths of 2 sides > third side |
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b < c + a is true |
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| 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 |
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8 + 4 > 12 |
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8 – 4 < 12 |
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3rd side can be 4 cm |
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| 16. |
In ΔXYZ XY + YZ = 10 cm, YZ + ZX = 12 cm, ZX + XY = 16 cm. Find the sides of the triangle |
| Sol: |
XY + YZ |
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10 cm ––––(1) |
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YZ + ZX |
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12 cm–––––(2) |
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ZX + XY |
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16 cm––––––(3) |
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Adding (1)(2)(3) we get |
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2(XY + YZ + ZX) |
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38 cm |
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XY + YZ + ZX |
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19 cm |
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If we put XY + YZ |
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10 cm then |
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10 + ZX |
= |
19 cm |
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ZX |
= |
9 cm |
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If we put YZ + ZX |
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12cm then |
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XY + 12 |
= |
19cm |
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XY |
= |
7 cm |
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If we put ZX + XY |
= |
1 6cm then |
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YZ + 16 |
= |
19 cm |
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YZ |
= |
3 cm |
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