56. If the length of each of two equal sides of an isosceles triangle is 10 cm. and the adjacent angle is 45°, then the area of the triangleis
    

1. 1. 20√2 square cm.
      

   
   
   

   2. 12√2 square cm.
      

   
   
   

   3. 25√2 square cm.
      

   
   
   

   4. 15√2 square cm.

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   AD = AB sin 45° = 10 ×	1	= 5√2 cm.
   	√2

   
   ∴ Area of ∆ABC
   

   =	1	× BC × AD
   	2

   
   

   =	1	× 10 × 5√2
   	2

   
   = 25√2 square cm.

   Correct Option: C

   
   

   AD = AB sin 45° = 10 ×	1	= 5√2 cm.
   	√2

   
   ∴ Area of ∆ABC
   

   =	1	× BC × AD
   	2

   
   

   =	1	× 10 × 5√2
   	2

   
   = 25√2 square cm.

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57. One of the angles of a right-angled triangle is 15°, and the hypotenuse is 1 metre. The area of the triangle (in square cm.) is
    

1. 1. 1220
      

   
   
   

   2. 1200
      

   
   
   

   3. 1250
      

   
   
   

   4. 1215

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   sin 15° = sin (45° – 30°)
   = sin 45° × cos 30° – cos 45° × sin 30°
   

   =	1	×	√3	-	1	×	1
   	√2		2		√2		2

   
   

   =	√3	-	1	=	√3 - 1
   	2√2		2√2		2√2

   
   and cos 15° = cos (45°– 30°)
   = cos 45°. cos 30° + sin 45°. sin 30°
   

   =	1	×	√3	+	1	×	1
   	√2		2		√2		2

   
   

   =	√3	+	1	=	√3 + 1
   	2√2		2√2		2√2

   
   ∴ AB = AC sin 15°
   

   =	√3 - 1	metre
   	2√2

   
   

   BC = AC cos 15° =	√3 + 1	metre
   	2√2

   
   

   ∴ Area of ∆ABC =	1	× AB × BC
   	2

   
   

   =		1	×	√3 - 1	×	√3 + 1		square metre
   		2		2√2		2√2

   
   

   =		3 - 1		square metre
   		16

   
   

   =	1	square metre
   	8

   
   

   =	10000	= 1250 square metre
   	8

   Correct Option: C

   
   sin 15° = sin (45° – 30°)
   = sin 45° × cos 30° – cos 45° × sin 30°
   

   =	1	×	√3	-	1	×	1
   	√2		2		√2		2

   
   

   =	√3	-	1	=	√3 - 1
   	2√2		2√2		2√2

   
   and cos 15° = cos (45°– 30°)
   = cos 45°. cos 30° + sin 45°. sin 30°
   

   =	1	×	√3	+	1	×	1
   	√2		2		√2		2

   
   

   =	√3	+	1	=	√3 + 1
   	2√2		2√2		2√2

   
   ∴ AB = AC sin 15°
   

   =	√3 - 1	metre
   	2√2

   
   

   BC = AC cos 15° =	√3 + 1	metre
   	2√2

   
   

   ∴ Area of ∆ABC =	1	× AB × BC
   	2

   
   

   =		1	×	√3 - 1	×	√3 + 1		square metre
   		2		2√2		2√2

   
   

   =		3 - 1		square metre
   		16

   
   

   =	1	square metre
   	8

   
   

   =	10000	= 1250 square metre
   	8

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58. If for an isosceles triangle the length of each equal side is ‘a’ units and that of the third side is ‘b’ units, then its area will be

1. 1. a	√4b² - a²
      4

   
   
   

   2. a	√2a² - b² square units
      2

      
      

   
   
   

   3. b	√4a² - b² square units
      4

   
   
   

   4. b	√a² - 2b² square units
      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   AD, is perpendicular on BC.
   BD = DC = b/2
   AD = √AB² - BD²
   

   = √a² -		b		²
   		2

   
   

   = √a² -	b²
   	2

   
   

   =	√4a² - b²
   	2

   
   

   ∴ Area of ∆ABC =	1	× BC × AD
   	2

   
   

   =	1	× b	√4a² - b²
   	2		2

   
   

   =	b	√4a² - b² square units.
   	4

   Correct Option: C

   
   AD, is perpendicular on BC.
   BD = DC = b/2
   AD = √AB² - BD²
   

   = √a² -		b		²
   		2

   
   

   = √a² -	b²
   	2

   
   

   =	√4a² - b²
   	2

   
   

   ∴ Area of ∆ABC =	1	× BC × AD
   	2

   
   

   =	1	× b	√4a² - b²
   	2		2

   
   

   =	b	√4a² - b² square units.
   	4

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59. The outer and inner diameter of a circular path be 728 metre and 700 metre respectively. The breadth of the path is
    

1. 1. 7 metre
      

   
   
   

   2. 28 metre
      

   
   
   

   3. 14 metre
      

   
   
   

   4. 20 metre

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   OA =	700	= 350 metre
   	2

   
   

   OB =	728	= 364 metre
   	2

   
   Width of path = OB – OA = 364 – 350 = 14 metre
   

   Correct Option: C

   
   

   OA =	700	= 350 metre
   	2

   
   

   OB =	728	= 364 metre
   	2

   
   Width of path = OB – OA = 364 – 350 = 14 metre
   

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60. The area of the parallelogram whose length is 30 cm, width is 20 cm and one diagonal is 40 cmis
    

1. 1. 200√15 cm²
      

   
   
   

   2. 100√15 cm²
      

   
   
   

   3. 300√15 cm²
      

   
   
   

   4. 150√15 cm²

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   
   In ∆ABD,
   AB = 20 cm. AD = 30 cm. BD = 40 cm.
   

   ∴ Semi–Perimeter (s) =	a + b + c	=	20 + 30 + 40
   	2		2

   
   = 45 cm
   ∴ Area of ∆ABD
   = √s(s - a)(s - b)(s - c)
   = √45(45 - 20)(45 - 30)(45 - 40)
   = √ 45 × 25 × 15 7times; 5
   = √5 × 3 × 5 × 5 × 5 × 3 × 5
   = = √5² × 5² × 5 × 3² × 3
   = 5 × 5 × 3√15 = 75√15 square cm.
   ∴ Area of parallelogram ABCD = 2 × 75√15
   = 150√15 square cm.

   Correct Option: D

   Using Rule 1,
   
   In ∆ABD,
   AB = 20 cm. AD = 30 cm. BD = 40 cm.
   

   ∴ Semi–Perimeter (s) =	a + b + c	=	20 + 30 + 40
   	2		2

   
   = 45 cm
   ∴ Area of ∆ABD
   = √s(s - a)(s - b)(s - c)
   = √45(45 - 20)(45 - 30)(45 - 40)
   = √ 45 × 25 × 15 7times; 5
   = √5 × 3 × 5 × 5 × 5 × 3 × 5
   = = √5² × 5² × 5 × 3² × 3
   = 5 × 5 × 3√15 = 75√15 square cm.
   ∴ Area of parallelogram ABCD = 2 × 75√15
   = 150√15 square cm.

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