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Using the same identity, find the value of (205) square
To calculate this value shortcut using algebra, we separate 205 into 200 + 5. This fits our identity template perfectly with a as 200 and b as 5. The square of 200 is 40000, and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Summing 40000, 2000, and 25 resuRead more
To calculate this value shortcut using algebra, we separate 205 into 200 + 5. This fits our identity template perfectly with a as 200 and b as 5. The square of 200 is 40000, and the square of 5 is 25. The intermediate term is two times 200 times 5, which equals 2000. Summing 40000, 2000, and 25 results in the final answer of 42025.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessFactor completely 9x square + 24xy + 16y square
To factor this expression, we observe that it matches the standard identity template of a square + 2ab + b square. The first term 9x square can be written as (3x) square, and the last term 16y square can be written as (4y) square. The middle term 24xy perfectly represents two times 3x times 4y. TherRead more
To factor this expression, we observe that it matches the standard identity template of a square + 2ab + b square. The first term 9x square can be written as (3x) square, and the last term 16y square can be written as (4y) square. The middle term 24xy perfectly represents two times 3x times 4y. Therefore, we group these parts together to get the final factored form which is (3x + 4y) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessFind the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.
In a standard square root spiral, the first right triangle has perpendicular sides of length 1 unit each. By the Pythagorean theorem, its hypotenuse is the square root of 2. The next right triangle is built using this hypotenuse as a base and adding a perpendicular side of 1 unit, making its hypotenRead more
In a standard square root spiral, the first right triangle has perpendicular sides of length 1 unit each. By the Pythagorean theorem, its hypotenuse is the square root of 2. The next right triangle is built using this hypotenuse as a base and adding a perpendicular side of 1 unit, making its hypotenuse the square root of 3. This mathematical pattern continues sequentially for all successive right triangles, creating hypotenuse lengths of root 2, root 3, root 4, root 5, and so on.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/
See lessThree rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
We are given two equations: x + y + z = 0 and xy + yz + zx = 0. Using the algebraic identity, we know that (x + y + z) squared equals x2 + y2 + z2 + 2(xy + yz + zx). Substituting our given values into this identity results in 0 squared equals x2 + y2 + z2 + 2(0), which simplifies directly to x2 + y2Read more
We are given two equations: x + y + z = 0 and xy + yz + zx = 0. Using the algebraic identity, we know that (x + y + z) squared equals x2 + y2 + z2 + 2(xy + yz + zx). Substituting our given values into this identity results in 0 squared equals x2 + y2 + z2 + 2(0), which simplifies directly to x2 + y2 + z2 = 0. Since the square of any real rational number is always non-negative, their sum can only equal zero if x, y, and z are all simultaneously zero.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/
See lessA rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.
The denominator of the rational number in its lowest form is given as 2 cubed times 5, which equals 8 times 5, or 40. A fraction with a denominator of the form 2 raised to m times 5 raised to n will always have a terminating decimal expansion. The number of decimal places is determined by the higherRead more
The denominator of the rational number in its lowest form is given as 2 cubed times 5, which equals 8 times 5, or 40. A fraction with a denominator of the form 2 raised to m times 5 raised to n will always have a terminating decimal expansion. The number of decimal places is determined by the higher exponent of the prime factors 2 and 5. Since the exponent of 2 is 3, it will have exactly 3 decimal places.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 3 The world of numbers (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-3/
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