Q. 93.8( 8 Votes )

# Evaluate these using identity:

(i) 55 × 45 (ii) 33 × 27

(iii) 8.5 × 9.5 (iv) 102 × 98

Answer :

(i) 55 × 45

We can split 55 as (50+5)

And 45 as (50–5)

Now 55 × 45

= (50+5) (50–5)

Using the identity (a +b) (a–b) = a^{2} – b^{2}

Here a = 50 and b = 5

= 50^{2} – 5^{2}

= 2500 –25

= 2475

(ii) 33 × 27

= (30+3) (30–3)

Using the identity (a +b) (a–b) = a^{2} – b^{2}

Here a = 30 and b = 3

= (30)^{2} – 3^{2}

= 900 – 9

= 891

(iii) 8.5 × 9.5

= (9 – 0.5) (9 + 0.5)

Using the identity (a +b) (a–b) = a^{2} – b^{2}

Here a = 9 and b = 0.5

= 9^{2} – (0.5)^{2}

= 81 – 0.25

= 80.75

(iv) 102 × 98

= (100 + 2) (100 – 2)

Using the identity (a +b) (a–b) = a^{2} – b^{2}

Here a = 100 and b = 2

= (100)^{2} – 2^{2}

= 10000 – 4

= 9996

Rate this question :

Find the product:

(2x – 3y)(2x + 3y)(4x^{2} + 9y^{2})

If a, b are rational numbers such that a^{2} + b^{2} + c^{2} – ab – bc – ca = 0, prove that a = b = c.

Use the identity (a + b)(a – b) = a^{2} – b^{2} to find the products:

(i) (x – 6) (x + 6)

(ii) (3x + 5)(3x + 5)

(iii) (2a + 4b)(2a – 4b)

Karnataka Board - Mathematics Part I

Express the following as difference of two squares:

(i) (x + 2z)(2x + z);

(ii) 4(x + 2y)(2x + y);

(iii) (x + 98)(x + 102);

(iv) 505 × 495.

Karnataka Board - Mathematics Part IFind the product:

(2a + 3)(2a – 3)(4a^{2} + 9)

Simplify:

(i) (x + y)^{2} + (x – y)^{2};

(ii) (x + y)^{2} × (x – y)^{2}.

Find the product:

(p + 2) (p – 2)(p^{2} + 4)

Find the product:

(x – 3)(x + 3)(x^{2} + 9)

Evaluate these using identity:

(i) 55 × 45 (ii) 33 × 27

(iii) 8.5 × 9.5 (iv) 102 × 98

Karnataka Board - Mathematics Part IFind the product:

Karnataka Board - Mathematics Part I