Algebraic Expressions and Identities
(NCERT Ex 9.2 Answers)
(i) 4 , 7p
Answer:
4 ☓ 7p = 4 ☓ 7 ☓ p = 28p
(ii) -4p , 7p
Answer:
-4p ☓ 7p = -4 ☓ p ☓ 7 ☓ p = (-4☓7)☓ (p☓ p) = -28p2
(iii) 4p , 7pq
Answer:
-4p ☓ 7p = -4 ☓ p ☓7 ☓ p ☓ q = (-4☓ 7)☓ (p☓ p☓ q) = -28p2q
(iv) 4p3 , -3p
Answer:
4p3 ☓ -3p = 4 ☓(-3) ☓ p ☓ p☓ p☓ p = -12p4
(v) 4p , 0
Answer:
4p ☓ 0 = 4☓ p☓ 0 = 0
Q2: Find the areas of rectangles with the following pairs of monomials as their lengths
and breadths respectively.
(p , q);(10m , 5n);(20x2 , 5y2);(4x , 3x2);(3mn ,4np)
Answer:
We know that,
Area of rectangle = Length ☓ Breadth
Area of 1st rectangle = p ☓ q = pq
Area of 2nd rectangle = 10m ☓ 5n = 10☓ 5 ☓ m ☓ n = 50mn
Area of 3rd rectangle = 20x2 ☓ 5y2 = 20☓ 5 ☓ x2 ☓ y2 = 100x2y2
Area of 4th rectangle = 4x ☓ 3x2 = 4☓3 ☓ x ☓ x2 = 12x3
Area of 5th rectangle = 3mn ☓4np = 3☓ 4 ☓ m ☓ n ☓ n ☓ p = 12mn2p
Q3: Complete the table of products.
First Monomial (⇨) Second Monomial(⇩) |
2x | -5y | 3x2 | -4xy | 7x2y | -9x2y2 |
2x | 4x2 | ... | ... | ... | ... | ... |
-5y | ... | ... | -15x2y | ... | ... | ... |
3x2 | ... | ... | ... | ... | ... | ... |
-4xy | ... | ... | ... | ... | ... | ... |
7x2y | ... | ... | ... | ... | ... | ... |
-9x2y2 | ... | ... | ... | ... | ... | ... |
Answer:
The table can be completed as follows.
First Monomial (⇨) Second Monomial(⇩) |
2x | -5y | 3x2 | -4xy | 7x2y | -9x2y2 |
2x | 4x2 | -10xy | 6x3 | -8x2y | 14x3y | -18x3y2 |
-5y | -10xy | 25y2 | -15x2y | 20xy2 | -35x2y2 | 45x2y3 |
3x2 | 6x3 | -15x2y | 9x4 | -12x3y | 21x4y | -27x4y2 |
-4xy | -8x2y | 20xy2 | -12x3y | 16x2y2 | -28x3y2 | 36x3y3 |
7x2y | 14x3y | -35x2y2 | 21x4y | -28x3y2 | 49x4y2 | -63x4y3 |
-9x2y2 | -18x3y2 | 45x2y3 | -27x4y2 | 36x3y3 | -63x4y3 | 81x4y4 |
Q4: Obtain the volume of rectangular boxes with the following length,breadth
and height respectivly.
(i) 5a , 3a2 , 7a4
(ii) 2p , 4q , 8r
(iii) xy , 2x2y , 2xy2
(iv) a , 2b , 3c
Answer:
We know that,
Volume = Length ☓ Breadth ☓ Height
(i) Volume= 5a ☓ 3a2☓ 7a4= 5☓ 3☓ 7☓a☓ a2☓a4 = 105a7
(ii) Volume=2p☓ 4q ☓ 8r= 2☓ 4☓ 8☓ p☓ q☓ r = 64pqr
(iii) Volume=xy ☓ 2x2y☓ 2xy2= 2☓2☓ xy☓ x2y☓ xy2 = 4x4y4
(iv) Volume= a ☓ 2b ☓ 3c= 2☓ 3☓ a☓ b☓ c = 6abc
Q5: Obtain the product of
(i) xy , yz , zx
Answer:
xy ☓ yz ☓ zx = x2y2z2
(ii) a , -a2 , a3
Answer:
a☓ -a2☓ a3= -a1 + 2 + 3 = -a6
(iii) 2 , 4y , 8y2 , 16y3
Answer:
2☓ 4y☓ 8y2☓16y3= 2☓ 4☓ 8☓ 16☓y ☓y2☓y3 = 1024y6
(iv) a , 2b , 3c , 6abc
Answer:
a☓ 2b☓ 3c☓ 6abc= 2☓ 3☓ 6☓ a☓ b ☓ c ☓ abc = 36a2b2c2
(v) m , -mn , mnp
Answer:
m ☓ (-mn)☓ mnp= -m3n2p
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