CBSE Exam Paper

Ganita Prakash | Part 2 | Class 8 | Ch 7 Area  Area Solutions Question Answer 7.1 Rectangle and Squares Figure It Out (Pages 150-152) Question 1. Identify the missing sidelengths. Solution: (i) After naming the figure In rectangle ABCD, Area of rectangle = Length × Breadth 7 × BC = 21 ⇒ BC = 3 in ∴ AF = AD + DF = 3 in + 4 in = 7 in In the rectangle EFAG, EF × AF = 28 in 2 ⇒ EF × 7 in = 28 in 2 ⇒ EF = 4 in ∴ HA = HG + GA = 3 in + 4 in = 7 in In rectangle HIJA, Area = HA × AJ ⇒ 35 in 2  = 7 in × AJ ⇒ AJ = 5 in ∴ AK = AJ + JK = 5 in + 2 in = 7 in In rectangle KLMA, Area = KL × LM ⇒ x in × 7 in = 14 in 2 ⇒ x = 2 in Thus, the missing sidelength = 2 in (ii) After naming the figure, In rectangle ABGH, Area = AB × AH AB × 4 m = 29 m² AB =  29 4  m or 7.25 m Area of rectangle HGDC = Area of rectangle ABDC – Area of rectangle ABGH = 50 m 2  – 29 m 2 = 21 m 2 In rectangle HGDC, CD × GD = 21 m 2 ⇒  29 4 m × GD = 21 m 2 ⇒ GD =  84 29  m or 2.9 m In rectangle B...

Ganita Prakash | Part 2 | Class 8 | Ch 6. Algebra Play 6.1 Algebra Play & 6.2 Thinking about ‘Think of a Number’ Tricks NCERT Intext Questions (Page 136) Question 1. How would you change this game to make the final answer 3? What about 5? Solution: (a) We can understand such tricks through algebra. Step 1: Think of a number = x Step 2: Triple it = 3x Step 3: Add 9 = 3x + 9 Step 4: Divide by 3 = \(\frac{3 x+9}{3}\) = x + 3 Step 5: Subtract the original number you thought of (x + 3) – x = 3. For Example: Consider a number 23. Triple it = 3 × 23 = 69 Add 9 = 69 + 9 = 78 Divide by 3 = 78 ÷ 3 = 26 ∴ 26 – 23 = 3 (b) To make the final answer 5. Step 1: Think of a number = x Step 2: Double it = 2x Step 3: Add 10; 2x + 10 Step 4: Divide by 2 = \(\frac{2 x+10}{2}\) = x + 5 Step 5: Subtract the original number = x + 5 – x = 5 Question 2. Can you come up with more complicated steps that always lead to the same final value? Solution: Yes. Here is an example. We can understand such tricks throug...