21. The diagonal of a right angle isosceles triangle is 5 cm. Its area will be
    

1. 1. 5 sq.cm
      

   
   
   

   2. 6.25 sq.cm
      

   
   
   

   3. 6.50 sq.cm
      

   
   
   

   4. 12.5 sq.cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   
   x² + x² = (5)² ⇒ 2x² = 25
   

   ⇒ x² =	25	⇒ x =	5
   	2		√2

   
   

   Area =	1	×	5	×	5
   	2		√2		√2

   
   

   =	25	= 6.25 sq.cm
   	4

   Correct Option: B

   Using Rule 1,
   
   x² + x² = (5)² ⇒ 2x² = 25
   

   ⇒ x² =	25	⇒ x =	5
   	2		√2

   
   

   Area =	1	×	5	×	5
   	2		√2		√2

   
   

   =	25	= 6.25 sq.cm
   	4

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22. The ratio of base of two triangles is x : y and that of their areas is a : b. Then the ratio of their corresponding altitudes will be:
    

1. 1. a	:	b
      x		y

   
   
   

   2. ax : by

   
   
   

   3. ay : bx
      

   
   
   

   4. x	:	a
      x		y

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   Let the respective altitudes be p₁ and p₂.
   

   ∴		=	1	× x × p1
   	a		2
   	b		1	× y × p2
   			2

   
   

   ⇒	p1	=	ay
   	p2		bx

   
   ⇒ ay : bx

   Correct Option: C

   Using Rule 1,
   Let the respective altitudes be p₁ and p₂.
   

   ∴		=	1	× x × p1
   	a		2
   	b		1	× y × p2
   			2

   
   

   ⇒	p1	=	ay
   	p2		bx

   
   ⇒ ay : bx

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23. If D and E are the mid-points of the side AB and AC respectively of the ∆ ABC in the figure given here, the shaded region of the triangle is what per cent of the whole triangular region?
    

1. 1. 50%
      

   
   
   

   2. 25%
      

   
   
   

   3. 75%
      

   
   
   

   4. 60%

   
   
   

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1. View Hint View Answer Discuss in Forum

   D is the mid-point of AB and E is the mid-point of AC.
   ∴ DE is parallel to BC.
   

   and DE =	1	BC
   	2

   
   ∆ ADE and ∆ ABC are similar, because
   D! = B! and E! = C!
   

   ∴	∆ ADE	=	DE²	=	1
   	∆ ABC		BC²		4

   
   ⇒ 4∆ ADE = ∆ ABC
   ∴ Area of trapezium DBCE
   = ∆ ABC – ∆ ADE
   4∆ADE – ∆ADE = 3.∆ ADE
   

   ∴ Required percentage =	3	× 100 = 75%
   	4

   Correct Option: C

   D is the mid-point of AB and E is the mid-point of AC.
   ∴ DE is parallel to BC.
   

   and DE =	1	BC
   	2

   
   ∆ ADE and ∆ ABC are similar, because
   D! = B! and E! = C!
   

   ∴	∆ ADE	=	DE²	=	1
   	∆ ABC		BC²		4

   
   ⇒ 4∆ ADE = ∆ ABC
   ∴ Area of trapezium DBCE
   = ∆ ABC – ∆ ADE
   4∆ADE – ∆ADE = 3.∆ ADE
   

   ∴ Required percentage =	3	× 100 = 75%
   	4

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24. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :
    

1. 1. 9 cm
      

   
   
   

   2. 18 cm
      

   
   
   

   3. 8 cm
      

   
   
   

   4. 12.5 cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   Area of the first triangle
   

   =	1	× base × height
   	2

   
   

   =	1	× 15 × 12 = 90 cm²
   	2

   
   According to the question Area of the another (second) triangle = 2 × 90 = 180 cm²
   ∴ Area of the new triangle = 180 cm²
   

   =	1	× 20 × height
   	2

   
   

   ∴ Height =	180 × 2	= 18 cm
   	20

   Correct Option: B

   Using Rule 1,
   Area of the first triangle
   

   =	1	× base × height
   	2

   
   

   =	1	× 15 × 12 = 90 cm²
   	2

   
   According to the question Area of the another (second) triangle = 2 × 90 = 180 cm²
   ∴ Area of the new triangle = 180 cm²
   

   =	1	× 20 × height
   	2

   
   

   ∴ Height =	180 × 2	= 18 cm
   	20

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25. If the area of a triangle is 1176 cm² and base : corresponding altitude is 3 : 4, then the altitude of the triangle is :
    

1. 1. 42 cm
      

   
   
   

   2. 52 cm
      

   
   
   

   3. 54 cm
      

   
   
   

   4. 56 cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Using Rule 1,
   Let the base and altitude be 3x and 4x respectively.
   ∴ According to question,
   

   1	base × altitude = 1176 cm²
   2

   
   

   or,	1	× 3x × 4x = 1176 cm
   	2

   
   12x² = 1176 × 2
   

   x² =	1176 × 2
   	2

   
   x² = 196
   ⇒ x = √196 = 14cm.
   ∴ Altitude of a triangle = 4x = 4 × 14 cm = 56 cm

   Correct Option: D

   Using Rule 1,
   Let the base and altitude be 3x and 4x respectively.
   ∴ According to question,
   

   1	base × altitude = 1176 cm²
   2

   
   

   or,	1	× 3x × 4x = 1176 cm
   	2

   
   12x² = 1176 × 2
   

   x² =	1176 × 2
   	2

   
   x² = 196
   ⇒ x = √196 = 14cm.
   ∴ Altitude of a triangle = 4x = 4 × 14 cm = 56 cm

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