Download App

Squares and Square Roots — Maths STD 8 — Question

Maharashtra BoardEnglish MediumSTD 8MathsSquares and Square Roots2 Marks
Question
Show that the following numbers is a perfect square. In case, find the number whose square is the given number:
$2601$
✓

Answer

A perfect square is a product of two perfectly equal numbers.

Resolving into prime factors:

$2601$

$=9 \times 289$

$=3 \times 3 \times 17 \times 17$

$=3 \times 17 \times 3 \times 17$

$=51 \times 51$

$=(51)^2$

Thus, 2601 is the perfect square of 51.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "2 Mark Question" from Squares and Square Roots→Squares and Square Roots — all question types→Maths STD 8 question papers→Maharashtra Board STD 8 question papers→Maharashtra Board question paper generator→

Similar questions

Simplify:
$(4^{-1}-5^{-1})^{-1}\div3^{-1}$
→
How many edges are there in a:
Triangular prism.
→
Find the squares of the following numbers using the identity $(a-b)^2=a^2-2 a b+b^2$ :
498
→
What is the number of diagonals in a:
Octagon.
→
Verify the following:
$\frac{-5}{8}+\frac{-9}{13}=\frac{-9}{13}+\frac{-5}{8}$
→
Write the quotient and remainder when we divide:
$\left(x^3+1\right)$ by $(x+1)$
→
Express:
56% as a ratio.
→
Factorize : $343 a^3-512 b^3$
→
Factorise:
$x^2+15 x+56$
→
Factorise:
$63 a^2-112 b^2$
→