Algebraic Expressions Solution of TS & AP Board Class 8 Mathematics

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Exercise 11.1

Question 1.

Find the product of the following pairs:

6, 7k

Answer:

Given:6 and 7k

The product of 6 and 7k is = 6×7k = 42k

Hence the product of 6 and 7k is 42k.

Question 2.

Find the product of the following pairs:

−3l, −2m

Answer:

Given: -3l and -2m

The product of -3l and -2m is = -3l × -2m = 6lm

Hence the product of -3l and -2m is 6lm.

Question 3.

Find the product of the following pairs:

−5t² −3t²

Answer:

Given: -5t² and -3t²

The product of -5t² and -3t²

⇒ -5t²×-3t² = 15(t².t²)

⇒15t⁴

Hence the product of -5t² and -3t² is 15t⁴.

Question 4.

Find the product of the following pairs:

6n, 3m

Answer:

Given: 6n and 3m

The product of 6n and 3m

⇒ 6n × 3m = 18nm

Hence the product of 6n and 3m is 18nm.

Question 5.

Find the product of the following pairs:

−5p², −2p

Answer:

Given: -5p² and -2p

The product of -5p² and -2p

⇒-5p²×-2p = 10(p².p)

⇒10p³

Hence the product of -5p² and -2p is 10p³.

Question 6.

Complete the table of the products.

Answer:

Here the product is calculated in the following manner:

We multiply the terms on the x axis (horizontal)and y axis(vertical).

For example, when -2y² is on the X axis (observe the 1^st row 3^rd column) and 3x is on the Y axis(2^nd row 1^st column) then ,the product

becomes -2y²×3x = -6xy²(observe the 2^nd row 3^rd column).

Question 7.

Find the volumes of rectangular boxes with given length, breadth and height in the following table.

Answer:

Here we have calculated the Volume V using the Length, Breadth and Height.

V = Length×Breadth×Height

Question 8.

Find the product of the following monomials

xy, x²y, xy, x

Answer:

Here we have xy,x²y,xy,x

Hence the product of the above terms is (xy)×(x²y)×(x) =

(x.x².x) (y.y)

= x⁴y²

Question 9.

Find the product of the following monomials

a, b, ab, a³b, ab³

Answer:

Here we have a,b,ab,a³b,ab³

Hence the product of the above terms is (a)×(b)×(ab)×(a³b)×(ab³)

= (a.a.a³.a)(b.b.b.b³)

= a⁶b⁶

Question 10.

Find the product of the following monomials

kl, lm, km, klm

Answer:

Here we have kl,lm,km,klm

Hence the product of the above terms is (kl)×(lm)×(km)×(klm)

= (k.k.k)(l.l.l)(m.m.m)

= k³l³m³

Question 11.

Find the product of the following monomials

pq ,pqr, r

Answer:

Here we have pq,pqr,r

Hence the product of the above terms is (pq)×(pqr)×(r)

= (p.p)(q.q)(r.r)

= p²q²r²

Question 12.

Find the product of the following monomials

−3a, 4ab, −6c, d

Answer:

Here we have -3a,4ab,-6c,d

Hence the product of the above terms is (-3a)×(4ab)×(-6c)×(d)

= (-3×4×-6)(a.a)(b)(c)(d)

= 72a²bcd

Question 13.

If A = xy, B = yz and C = zx, then find ABC = ...........

Answer:

Here we have A = xy ,B = yz and C = zx.

Therefore the product of A,B and C will be ABC = (xy)×(yz)×(zx) = x²y²z².

Question 14.

If P = 4x², T = 5x and R = 5y, then  = ..........

Answer:

Here we have P = 4x² ,T = 5x and R = 5y.

Therefore the product of P,T and R will be PTR = (4x²)×(5x)×(5y) = 100x³y

Hence  is equal to  i.e. x³y.

Question 15.

Write some monomials of your own and find their products.

Answer:

The monomials are (i)5x,6y,7z and (ii)3x²y,4xy²,7x³y³

(i) The given monomial is 5x,6y,7z .

Therefore the product of above terms is (5x)×(6y)×(7z) = 210xyz

(ii) The given monomial is 3x²y,4xy²,7x³y³ .

Therefore the product of above terms is (3x²y)×(4xy²)×(7x³y³) = 84x⁶y⁶

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Exercise 11.2

Question 1.

Complete the table:

Answer:

Here we have the First and Second expressions as 5q and p + q-2r respectively.

The process involved in the multiplication is :

Step1.Write the product of the expressions using multiplication symbol: 5q×(p + q-2r)

Step2.Use distributive law :Multiply the monomial by the first term of the polynomial then multiply the monomial by the second term and then multiply the monomial by the third term of the polynomial and add their products: 5pq + 5q²-10qr

Step3.Simplify the terms: 5q² + 5pq-10qr

2. Similarly ,following the above process we calculate

(kl + lm + mn)(3k)

= (3k)(kl) + (3k)(lm) + (3k)(mn)

= 3k²l + 3klm + 3kmn

3. Similarly ,following the above process we calculate

(ab²)( a + b² + c³) = ab²(a) + ab² (b²) + ab²(c³) = a²b² + ab⁴ + ab²c³

4.Similarly ,following the above process we calculate

(x-2y + 3z)(xyz) = ( x-2y + 3z)×(xyz) = x²yz-2xy²z + 3xyz²

5. Similarly ,following the above process we calculate

(a²bc + b²cd-abd²)(a²b²c²) = (a²bc + b²cd- abd²)×(a²b²c²) = a⁴b³c³ + a²b⁴c³d-a³b³c²d²

Question 2.

Simplify: 4y(3y + 4)

Answer:

Here we have 4y and 3y + 4.

Now the product of these terms is as follows:

The procedure involved in the multiplication is :

Step1.Write the product of the expressions using multiplication symbol: 4y(3y + 4).

Step2.Use distributive law :Multiply the monomial by the first term of the binomial then multiply the monomial by the second term of the binomial and then add their products: 4y(3y) + 4y(4)

Step3.Simplify the terms: 12y² + 16y

Hence 4y(3y + 4) = 12y² + 16y

Question 3.

Simplify x(2x²−7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0

Answer:

Here we have the monomial x and polynomial 2x²-7x + 3

Now the product of these terms is as follows:

The process involved in the multiplication is :

Step1.Write the product of the expressions using multiplication symbol:

x(2x²-7x + 3).

Step2.Use distributive law :Multiply the monomial by the first term of the trinomial then multiply the monomial by the second term of the trinomial and then the monomial by the third term of the trinomial and then add their products: x(2x²)-x(7x) + x(3)

Step3.Simplify the terms: 2x³-7x² + 3x

Hence x(2x²−7x + 3) = 2x³-7x² + 3x

(i)When x = 1, 2x³-7x² + 3x = 2(1)³-7(1)² + 3(1) = -2

(ii)When x = 0, 2x³-7x² + 3x = 2(0)³-7(0)² + 3(0) = 0

Question 4.

Add the product: a(a−b), b(b−c), c(c−a)

Answer:

Here we have the expressions a(a-b),b(b-c) and c(c-a)

Using the distributive law a(a-b) = a²-ab,b(b-c) = b²-bc and c(c-a) = c²-ca

Now we add the terms (a²-ab) + (b²-bc) + (c²-ca)

= a² + b² + c²-ab-bc-ca

Hence addition of the product of the above expressions is a² + b² + c²-ab-bc-ca.

Question 5.

Add the product: x(x + y−r), y(x−y + r), z(x−y−z)

Answer:

Here we have the expressions x(x + y-r),y(x-y + r) and z(x-y-z)

Using the distributive law x(x + y-r)

= xy-y² + yr and

z(x-y-z) = xz-yz-z²

Now we add these terms (x² + xy-xr) + (xy-y² + yr) + (xz-yz-z²)

= x²-y²-z² + 2xy-xr + yr + xz-yz

Hence addition of the product of the above expressions is x²-y²-z² + 2xy-xr + yr + xz-yz.

Question 6.

Subtract the product of 2x(5x−y) from product of 3x(x + 2y)

Answer:

Here we have 2x(5x-y) and 3x(x + 2y)

Using the distributive law 2x(5x-y) = 2x(5x)-2x(y)

= 10x²-2xy

and 3x(x + 2y)

= 3x(x) + 3x(2y)

= 3x² + 6xy.

Now , we have to subtract 10x²-2xy from 3x² + 6xy.

i.e. (3x² + 6xy)-(10x²-2xy)

= 3x² + 6xy-10x² + 2xy

= -7x² + 8xy

Hence after subtraction of the product of 2x(5x−y) from the product of 3x(x + 2y) is

-7x² + 8xy.

Question 7.

Subtract 3k(5k−l + 3m) from 6k(2k + 3l−2m)

Answer:

Here we have 3k(5k−l + 3m) and 6k(2k + 3l−2m)

Using the distributive law 3k(5k−l + 3m) = 15k²-3kl + 9km

and 6k(2k + 3l-2m) = 12k² + 18kl-12km

Now , we have to subtract 15k²-3kl + 9km from 12k² + 18kl-12km.

i.e. (12k² + 18kl-12km)-(15k²-3kl + 9km) = -3k² + 21kl-21k²m²

Hence after subtraction of the product of 3k(5k−l + 3m) from the product of 6k(2k + 3l−2m) is = -3k² + 21kl-21k²m²

Question 8.

Simplify: a²(a−b + c) + b²(a + b−c)−c²(a−b−c)

Answer:

Here we have the polynomials as a²(a−b + c), b²(a + b−c) and c²(a−b−c)

Now using the distributive law, a²(a−b + c) = a²(a)-a²(b) + a²(c) = a³-a²b + a²c

b²(a + b−c) = b²(a) + b²(b)-b²(c) = ab² + b³-b²c and c²(a−b−c) = c²a-c²b-c³

Now we will simplify it, a²(a−b + c) + b²(a + b−c)−c²(a−b−c) = (a³-a²b + a²c) + (ab² + b³-b²c)-(c²a-c²b-c³) = a³-a²b + a²c + ab² + b³-b²c-c²a + c²b + c³ = a³ + b³ + c³-a²b + a²c + ab²-b²c-c²a + c²b

Hence , a²(a−b + c) + b²(a + b−c)−c²(a−b−c) = a²b + a²c + ab²-b²c-c²a + c²b

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Exercise 11.3

Question 1.

Multiply the binomials:

2a−9 and 3a + 4

Answer:

Consider two binomials as 2a−9 and 3a + 4

Now,to get the product of two binomials 2a−9 and 3a + 4 we use the

distributive law i.e. (2a−9)×(3a + 4) = 2a(3a) + 2a(4)-9(3a)-9(4) = 6a² + 8a-27a-36 = 6a²-19a-36

Therefore ,the product of 2a−9 and 3a + 4 is 6a²-19a-36.

Question 2.

Multiply the binomials:

x−2y and 2x−y

Answer:

Consider two binomials as x−2y and 2x−y

Now,to get the product of two binomials x−2y and 2x−y we use the

distributive law i.e. (x−2y)×( 2x−y) = x(2x)-x(y)-2y(2x)-2y(-y) = 2x²-xy-4xy + 4y²

Therefore ,the product of x−2y and 2x−y is 2x² + 4y²-5xy.

Question 3.

Multiply the binomials:

kl + lm and k−l

Answer:

Consider two binomials as kl + lm and k-l

Now,to get the product of two binomials kl + lm and k-l we use the

distributive law i.e. (kl + lm)×( k-l) = kl(k) + kl(-l) + lm(k) + lm(-l) = k²l-kl² + klm-l²m

Therefore , the product of (kl + lm) and ( k-l) is k²l-kl² + klm-l²m

Question 4.

Multiply the binomials:

m²−n² and m + n

Answer:

Consider two binomials as m²−n² and m + n

Now, to get the product of two binomials ) m²−n² and m + n we use the

distributive law i.e. (m²−n²)×( m + n) = m²(m) + m²(n)-n²(m)-n²(n) =

m³ + m²n-mn²-n³

Therefore , the product of m²−n² and m + n is m³ + m²n-mn²-n³.

Question 5.

Find the product:

(x + y)(2x−5y + 3xy)

Answer:

Consider the binomial x + y and the trinomial 2x−5y + 3xy.

Now, to get the product of above expressions x + y and 2x−5y + 3xy,we use the

distributive law i.e (x + y)×(2x−5y + 3xy) = x(2x-5y + 3xy) + y(2x-5y + 3xy)

= x(2x) + x(-5y) + x(3xy) + y(-5y) + y(3xy) = 2x²-5xy + 3x²y-5y² + 3xy²

Therefore , the product of x + y and 2x−5y + 3xy is 2x²-5y² + 3x²y + 3xy²-5xy

Question 6.

Find the product:

(mn−kl + km) (kl−lm)

Answer:

Consider the trinomial mn−kl + km and the binomial kl−lm.

Now, to get the product of mn−kl + km and kl−lm,we use the

distributive law i.e. (mn−kl + km)×( kl−lm) = mn(kl-lm)-kl(kl-lm) + km(-lm)

= mnkl-m²nl-k²l² + kl²m + kml-klm²

Therefore , the product of mn−kl + km and kl−lm is mnkl-m²nl-k²l² + kl²m + kml-klm²

Question 7.

Find the product:

(a−2b + 3c)(ab²−a²b)

Answer:

Consider the trinomial a−2b + 3c and the binomial ab²−a²b.

Now, to get the product of a−2b + 3c and ab²−a²b,we use the

distributive law i.e. (a−2b + 3c)×( ab²−a²b) = a(ab²-a²b)-2b(ab²-a²b) + 3c(ab²-a²b)

= a²b²-a³b-2ab³ + 2a²b² + 3ab²c-3a²bc = 3a²b²-a³b-2ab³ + 3ab²c-3a²bc

Therefore , the product of a−2b + 3c and ab²−a²b is 3a²b²-a³b-2ab³ + 3ab²c-3a²bc

Question 8.

Find the product:

(p³ + q³)(p−5q + 6r)

Answer:

Consider the binomial p³ + q³ and the trinomial p−5q + 6r.

Now, to get the product of p³ + q³ and p−5q + 6r,we use the

distributive law i.e (p³ + q³)(p−5q + 6r) = p³(p-5q + 6r) + q³(p-5q + 6r) = (p³.p-5p³q + p³.6r) + (p.q³-5q.q³ + 6q³r) = p⁴-5p³q + 6p³r + pq³-5q⁴ + 6q³r

Therefore , the product of p³ + q³and p−5q + 6r is p⁴-5p³q + 6p³r + pq³-5q⁴ + 6q³r

Question 9.

Simplify the following:

(x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y)

Answer:

Here we have (x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y)

Re-arranging, we get,

(x - 2y)(y - 3x) - (y - 3x)(4x - 5y) + (x + y)(x - 3y)

= (y - 3x)[(x - 2y) - (4x - 5y)] + (x + y)(x - 3y)

= (y - 3x)(-3x + 3y) + (x + y)(x - 3y)

= - 3 (y - 3x)(x - y) + (x + y)(x - 3y)

= -3 (xy - y² - 3x² + 3xy) + x² - 3xy + xy - 3y²= 9x² + 3y² - 12xy + x² - 3y² - 2xy

= 10x² - 14xy

Question 10.

Simplify the following:

(m + n)(m²−mn + n²)

Answer:

We have the binomial m + n and trinomial m²−mn + n²

Here we will use the distributive law as follows:

(m + n)(m²−mn + n²) = m(m²−mn + n²) + n(m²−mn + n²) = (m.m²-m.mn + m.n²) + (n.m²-mn.n + n.n²) = m³-m²n + mn² + m²n-mn² + n³ = m³ + n³

Hence , (m + n)(m²−mn + n²) = m³ + n³

Question 11.

Simplify the following:

(a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b)

Answer:

We have (a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b)

Here, (a−2b + 5c)(a−b) = (a-b)(a-2b + 5c)(using Commutative law)

= a(a-2b + 5c)-b(a-2b + 5c)(using distributive law )

= a²-2ab + 5ac-ab + 2b²-5bc = a² + 2b²-3ab + 5ac-5bc

(a−b−c)(2a + 3c) = (2a + 3c)(a-b-c) (using Commutative law)

= 2a(a-b-c) + 3c(a-b-c) (using distributive law )

= 2a²-2ab-2ac + 3ac-3bc-3c²

= 2a² + ac-2ab-3bc-3c²

and (6a + b)(2c−3a−5b) = 6a(2c−3a−5b) + b(2c−3a−5b) = 12ac-18a²-30ab + 2bc-3ab-5b² = -18a² + 5b²-33ab + 2bc + 12ac

Therefore, a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b) = a² + 2b²-3ab + 5ac-5bc-(2a² + ac-2ab-3bc-3c²) + ( -18a² + 5b²-33ab + 2bc + 12ac) =

a² + 2b²-3ab + 5ac-5bc-2a²-2ac + 2ab + 3bc + 3c²-18a² + 5b²-33ab + 2bc + 12ac =

= -19a² + 7b² + 3c²-34ab + 15ac

Question 12.

Simplify the following:

(pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r)

Answer:

We have , (pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r)

(pq-qr + pr)(pq + qr) = (pq + qr)( pq-qr + pr) (using Commutative law)

= pq(pq-qr + pr) + qr( pq-qr + pr) (using distributive law )

= p²q²-pq²r + p²qr + pq²r-q²r² + pqr² = p²q² + p²qr + pqr²-q²r²

(pr + pq)(p + q−r) = pr(p + q-r) + pq(p + q−r) (using distributive law )

= p²r + prq-pr² + p²q + pq²-prq

= p²r-pr² + p²q + pq²

(pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r) = p²q² + p²qr + pqr²-q²r²-( p²r-pr² + p²q + pq²)

= p²q² + p²qr + pqr²-q²r²-p²r + pr²-p²q-pq²

Hence, (pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r) = p²q² + p²qr + pqr²-q²r²-p²r + pr²-p²q-pq²

Question 13.

If a, b, c are positive real numbers such that, find the value of 

Answer:

_Given:

 , where a, b, c are positive real numbers

Here, we have 

⇒b(a + b-c) = c(a-b + c)

⇒ab + b²-bc = ac-bc + c²(bc is on both the sides and they get subtracted)

⇒ab + b² = ac + c²

⇒ab-ac = c²-b²

⇒a(b-c) = (c-b)(c + b)

⇒-a(c-b) = (c + b)(c-b)

⇒-a = (c + b)[Here (c-b) is on both the sides,so when they get divided and the result is 1]

⇒(b + c) = -a(commutative property)

We have , 

⇒a(a-b + c) = b(-a + b + c)

⇒a²-ab + ac = -ab + b² + bc

⇒a² + ac = b² + bc

⇒a²-b² = bc-ac

⇒(a + b)(a-b) = -c(a-b)

⇒a + b = -c

We have ,

⇒a(a + b-c) = c(-a + b + c)

⇒a² + ab-ac = -ac + bc + c²

⇒a² + ab = bc + c²

⇒a²-c² = bc-ab

⇒(a + c)(a-c) = b(c-a)

⇒(a + c)(a-c) = -b(a-c)

⇒a + c = -b

Therefore, = -1(Here we have substituted the values of (a + b),(b + c),(c + a) from above evaluation)

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Exercise 11.4

Question 1.

Select a suitable identity and find the following products

(3k + 4l) (3k + 4l)

Answer:

Given: (3k + 4l) (3k + 4l),it is a product of 2 binomial expressions which have the same terms 3k + 4l and 3k + 4l.

Now when we compare these expressions with the identities, we

find it in the form of (a + b)²,where a = 3k and b = 4l,

The identity (a + b)² = a² + 2ab + b²,

Hence, (3k + 4l) (3k + 4l) = (3k + 4l)² = (3k)² + 2(3k)(4l) + (4l)² = 9k² + 24kl + 16l²

Question 2.

Select a suitable identity and find the following products

(ax² + by²) (ax² + by²)

Answer:

Given: (ax² + by²) (ax² + by²),it is a product of 2 binomial expressions which have the same terms ax² + by² and ax² + by².

Now when we compare these expressions with the identities, we

find it in the form of (a + b)²,where a = ax² and b = by²,

The identity (a + b)² = a² + 2ab + b²,

Hence, (ax² + by²) (ax² + by²) = (ax² + by²)² = (ax²)² + 2(ax²)(by²) + (by²)²

= a²x⁴ + 2abx²y² + b²y⁴

Question 3.

Select a suitable identity and find the following products

(7d – 9e) (7d – 9e)

Answer:

Given: (7d – 9e) (7d – 9e),it is a product of 2 binomial expressions which have the same terms 7d – 9e and 7d – 9e.

Now when we compare these expressions with the identities, we

find it in the form of (a-b)²,where a = 7d and b = 9e,

The identity (a-b)² = a²-2ab + b²,

Hence, (7d – 9e) (7d – 9e) = (7d-9e)² = (7d)²-2(7d)(9e) + (9e)² = 49d²-126de + 81e²

Question 4.

Select a suitable identity and find the following products

(m² – n²) (m² + n²)

Answer:

Given:( m² – n²) (m² + n²),it is a product of 2 binomial expressions which have the terms (m² – n²) and (m² + n²).

Now when we compare these expressions with the identities, we

find it in the form of (a-b)(a + b),where a = m² and b = n²,

The identity (a-b)(a + b) = a²-b²,

Hence, (m² – n²) (m² + n²) = (m²)²-(n²)² = m⁴-n⁴

Question 5.

Select a suitable identity and find the following products

(3t + 9s) (3t – 9s)

Answer:

Given:(3t + 9s) (3t – 9s),it is a product of 2 binomial expressions which have the terms :(3t + 9s) and (3t – 9s).

Now when we compare these expressions with the identities, we

find it in the form of (a-b)(a + b),where a = 3t and b = 9s,

The identity (a-b)(a + b) = a²-b²,

Hence, (3t + 9s) (3t – 9s) = (3t)²-(9s)² = 9t²-81s²

Question 6.

Select a suitable identity and find the following products

(kl – mn) (kl + mn)

Answer:

Given:( kl – mn) (kl + mn),it is a product of 2 binomial expressions which have the terms :( kl – mn) and (kl + mn).

Now when we compare these expressions with the identities, we

find it in the form of (a-b)(a + b),where a = kl and b = mn,

The identity (a-b)(a + b) = a²-b²,

Hence, (kl – mn) (kl + mn) = (kl)²-(mn)²

Question 7.

Select a suitable identity and find the following products

(6x + 5) (6x + 6)

Answer:

Given:(6x + 5) (6x + 6),it is a product of 2 binomial expressions which have the terms (6x + 5) and (6x + 6).

Now when we compare these expressions with the identities, we

find it in the form of (x + a)(x + b),where x = 6x and a = 5 and b = 6,

Thus,(x + a)(x + b) = x² + x(b + a) + ab

Hence, (6x + 5) (6x + 6) = (6x)² + 6x(5 + 6) + 30 = 12x² + 66x + 30

Question 8.

Select a suitable identity and find the following products

(2b – a) (2b + c)

Answer:

Given: (2b – a) (2b + c),it is a product of 2 binomial expressions which have the terms (2b – a) (2b + c).

Now when we compare these expressions with the identities, we

find it in the form of (x-a)(x + b),where x = 2b and a = a and b = c,

Thus,(x-a)(x + b) = x² + bx-ax-ab = x² + x(b-a)-ab

Hence, (2b – a) (2b + c) = (2b)² + 2b(c-a)-ac = 4b² + 2b(c-a)-ac

Question 9.

Evaluate the following by using suitable identities:

304²

Answer:

Given:304² = (300 + 4)² = (300)² + 2(300)(4) + (4)²

= 90000 + 2400 + 16, where a = 300 and b = 4 in the identity (a + b)² = a² + 2ab + b²

= 92416

Question 10.

Evaluate the following by using suitable identities:

509²

Answer:

Given: 509² = (500 + 9)² = (500)² + 2(500)(9) + (9)²

= 2500 + 9000 + 81, where a = 500 and b = 9 in the identity (a + b)² = a² + 2ab + b²

= 11581

Question 11.

Evaluate the following by using suitable identities:

992²

Answer:

Given: 992² = (900 + 2)² = (900)² + 2(900)(2) + (2)²

= 8100 + 3600 + 4, where a = 900 and b = 2 in the identity (a + b)² = a² + 2ab + b²

= 11704

Question 12.

Evaluate the following by using suitable identities:

799²

Answer:

Given: 799² = (800-1)² = (800)²-2(800)(1) + (1)²

= 6400 + 1600 + 1, where a = 800 and b = 1 in the identity (a-b)² = a²-2ab + b²

= 8001

Question 13.

Evaluate the following by using suitable identities:

304 × 296

Answer:

Given: 304×296 = (300 + 4)(300-4)

= (300)²-(4)², where a = 300 and b = 4 in identity (a + b)(a-b) = a²-b²

= 90000-16 = 89984

Question 14.

Evaluate the following by using suitable identities:

83 × 77

Answer:

Given:83×77 = (80 + 3)(80-3)

= (80)²-(3)²,where a = 80 and b = 3 in identity (a + b)(a-b) = a²-b²

= 6400-9 = 6391

Question 15.

Evaluate the following by using suitable identities:

109×108

Answer:

Given:109×108 = (100 + 9)(100 + 8)

= (100)² + (9 + 8)100 + (9×8),

where x = 100,a = 9 and b = 8 in identity (x + a)(x + b) = x² + (a + b)x + ab

= 10000 + 1700 + 72

= 11772

Question 16.

Evaluate the following by using suitable identities:

204×206

Answer:

Given: (200 + 4)(200 + 6)

= (200)² + (4 + 6)200 + (4×6),

where x = 200,a = 4,b = 6 in identity (x + a)(x + b) = x² + (a + b)x + ab

= 40000 + 2000 + 24

= 42024

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Exercise 11.5

Question 1.

Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking

a = 2 units, b = 4 units

Answer:

(a + b)²Ξa² + 2ab + b²

Draw a square with the side a + b i.e.,2 + 4

L.H.S of the whole square = (2 + 4)² = (6)² = 36

R.H.S = Area of the square with 2 units + Area of the square with 4 units +

Area of the 2,4 units + Area of the square with 4 ,2 units = 2² + 4² + 2×4 + 2×4 = 4 + 16 + 8 + 8 = 36

L.H.S = R.H.S

Hence,the identity is verified.

Question 2.

Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking

a = 3 units, b = 1 unit

Answer:

(a + b)²Ξa² + 2ab + b²

Draw a square with the side a + b i.e.,3 + 1

L.H.S of the whole square = (3 + 1)² = (4)² = 16

R.H.S = Area of the square with 3 units + Area of the square with 1 unit +

Area of the 3,1 unit + Area of the square with 1 ,3 units = 3² + 1² + 3×1 + 1×3 = 9 + 1 + 3 + 3 = 16

L.H.S = R.H.S

Hence, the identity is verified.

Question 3.

Verify the identity (a + b)² ≡ a² + 2ab + b² geometrically by taking

a = 5 units, b = 2 unit

Answer:

(a + b)²Ξa² + 2ab + b²

Draw a square with the side a + b i.e.,5 + 2

L.H.S of the whole square = (5 + 2)² = (7)² = 49

R.H.S = Area of the square with 5 units + Area of the square with 2 units +

Area of the 5,2 units + Area of the square with 2 ,5 units = 5² + 2² + 5×2 + 2×5 = 49

L.H.S = R.H.S

Hence,the identity is verified.

Question 4.

Verify the identity (a – b)² ≡ a² - 2ab + b²geometrically by taking

a = 3 units, b = 1 unit

Answer:

(a – b)² ≡ a² - 2ab + b²

Consider a square with side a.i.e.a = 3

The square is divided into 4 regions.

It consists of 2 squares with sides a-b and b respectively and 2 rectangles with length and breadth as ‘a-b’ and ‘b’ respectively.

Here a = 3 and b = 1.Therefore the 2 squares consist of sides ‘3-1’ and ‘1’ respectively and 2 rectangles with length and breadth as ‘3-1’ and ‘1’ respectively.

Now area of figure I = Area of whole square with side ‘a’ i.e.3 units-Area of figure II

-Area of figure III –Area of figure IV

L.H.S of area of figure I = (3-1)(3-1) = 2(2) = 4 units

R.H.S = Area of whole square with side 3 units-Area of figure II with 1,(3-1)units

-Area of figure III with 1,(3-1) units –Area of figure IV with 1,1 unit = 3²-(1×(3-1))-

(1×(3-1))-(1×1) = 4 units

L.H.S = R.H.S

Hence,the identity is verified.

Question 5.

Verify the identity (a – b)² ≡ a² - 2ab + b²geometrically by taking

a = 5 units, b = 2 units

Answer:

(a – b)² ≡ a² - 2ab + b²

Consider a square with side a.i.e.a = 5

The square is divided into 4 regions.

It consists of 2 squares with sides a-b and b respectively and 2 rectangles with length and breadth as ‘a-b’ and ‘b’ respectively.

Here a = 5 and b = 2.Therefore the 2 squares consist of sides ‘5-2’ and ‘2’ respectively and 2 rectangles with length and breadth as ‘5-2’ and ‘2’ respectively.

Now area of figure I = Area of whole square with side ‘a’ i.e.5 units-Area of figure II

-Area of figure III –Area of figure IV

L.H.S of area of figure I = (5-2)(5-2) = 3(3) = 9 units

R.H.S = Area of whole square with side 5 units-Area of figure II with 2,(5-2)units

-Area of figure III with 2,(5-2) units –Area of figure IV with 2,2 units = 25-(2×3)-

( 2×3)-2² = 25-6-6-4 = 9 units

L.H.S = R.H.S

Hence,the identity is verified.

Question 6.

Verify the identity (a + b) (a – b) ≡ a² – b² geometrically by taking

a = 3 units, b = 2 units

Answer:

Remove square from this whose side is ‘b’.(b<a)

Consider a square with side ‘a’.

We get the above figure by removing the square with side ‘b’.It consists of 2 parts

I and II.

So a²-b² = Area of figure I + Area of figure II = a(a-b) + b(a-b) = (a-b)(a + b)

Thus a²-b² = (a-b)(a + b).

Here a = 3 units and b = 2 units

So,L.H.S = a²-b² = 3²-2² = 5 units

R.H.S = (a-b)(a + b) = (3-2)(3 + 2) = 1(5) = 5 units

Therefore L.H.S = R.H.S

Hence,the identity is verified.

Question 7.

Verify the identity (a + b) (a – b) ≡ a² – b² geometrically by taking

a = 2 units, b = 1 unit

Answer:

Refer the above figure.

Here a²-b² = Area of figure I + Area of figure II = a(a-b) + b(a-b) = (a-b)(a + b)

Thus a²-b² = (a-b)(a + b).

Here a = 2 units and b = 1 units

So, L.H.S = a²-b² = 2²-1² = 3 units

R.H.S = (a-b)(a + b) = (2-1)(2 + 1) = 3 units

Therefore L.H.S = R.H.S

Hence,the identity is verified.