Question
Find the square root of: $\frac{507}{4107}$
✓
Answer
$ \frac{507}{4107}$
$ =\frac{507+3}{4107+3}$
$ =\frac{169}{1369}$
Hence, square root of $\sqrt{\frac{169}{1369}}=\frac{13}{37}$
$ =\frac{507+3}{4107+3}$
$ =\frac{169}{1369}$
| $13$ | |
| $1$ | $169 1$ |
| $23$ | $69 69$ |
| $x$ |
| $37$ | |
| $3$ | $1369 9$ |
| $67$ | $469 469$ |
| $x$ |
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Factorise: $(a - 3b)^2 - 36 b^2$
→Find the perimeter and area of the adjoining figure, it being given that $AB \| CD , AB =12 cm, CD =15 cm$, $BD =4 cm$ and one end of the figure is a semi-circle with BD as diameter. (Take $\pi=3.14$.)
→Is it possible to have a regular polygon whose each exterior angle is: $80^\circ$
→Show that :
$\sqrt{x^{p-q}} \times \sqrt{x^{q-r}} \times \sqrt{x^{r-p}}=1$
→$\sqrt{x^{p-q}} \times \sqrt{x^{q-r}} \times \sqrt{x^{r-p}}=1$
In the adjoining figure, the diagonals AC and BD of parallelogram ABCD intersect at O . If $\angle OBC =36^{\circ}, \angle ODC$ $=28^{\circ}$ and $\angle AOD =64^{\circ}$; find the measure of
(i) $\angle OAD$
(ii) $\angle OCD$.
→(i) $\angle OAD$
(ii) $\angle OCD$.
Find the product : $\left(\frac{2}{x}+\frac{3}{y}\right)\left(\frac{2}{x}-\frac{3}{y}\right)$
→The length of a rectangle is 7 cm less than twice its breadth. If the length is decreased by 2 cm and breadth increased by 3 cm , the perimeter of the resulting rectangle is 66 cm . Find the length and breadth of the original rectangle.
Simplify:$a^5 ÷ a^3+ 3a \times 2a$
→Find the product : $\left(a-\frac{2}{3}\right)\left(a+\frac{2}{3}\right)$
→The two parallel sides and the distance between them are in the ratio $3: 4: 2$. If the area of the trapezium is $175\ cm^2$, find its height.
→