Tick (✓) the correct answer

Question: Tick (✓) the correct answer $\sqrt[3]{\frac{64}{343}}=?$ (a) $\frac{4}{9}$ (b) $\frac{4}{7}$ (c) $\frac{8}{7}$ (d) $\frac{8}{21}$ Solution: (b) $\frac{4}{7}$ $\sqrt[3]{\frac{64}{343}}=\frac{\sqrt[3]{64}}{\sqrt[3]{343}}=\frac{\sqrt[3]{4 \times 4 \times 4}}{\sqrt[3]{7 \times 7 \times 7}}=\frac{\sqrt[3]{(4)^{3}}}{\sqrt[3]{(7)^{3}}}$ $\sqrt[3]{\frac{64}{343}}=\frac{4}{7}$ $\therefore \sqrt[3]{\frac{64}{343}}=\frac{4}{7}$...

In figure, arcs have been drawn of radius 21 cm

Question: In figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region. Solution: Given that, radius of each arc (r) = 21 cm Area of sector with $\angle A=\frac{\angle A}{360^{\circ}} \times \pi r^{2}=\frac{\angle A}{360^{\circ}} \times \pi \times(21)^{2} \mathrm{~cm}^{2}$ $\left[\because\right.$ area of any sector with central angle $\theta$ and radius $\left.r=\frac{\pi r^{2}}{360^{\circ}} \times \theta\ri...

Tick (✓) the correct answer

Question: Tick (✓) the correct answer $\sqrt[3]{125 \times 64}=?$ (a) 100 (b) 40 (c) 20 (d) 30 Solution: (c) 20 $\sqrt[3]{125 \times 64}=\sqrt[3]{125} \times \sqrt[3]{64}=\sqrt[3]{5 \times 5 \times 5} \times \sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2}$ $\sqrt[3]{125 \times 64}=\sqrt[3]{(5)^{3}} \times \sqrt[3]{(2)^{3} \times(2)^{3}}=\sqrt[3]{(5)^{3}} \times \sqrt[3]{(4)^{3}}$ $\sqrt[3]{125 \times 64}=5 \times 4=20$ Hence, the cube root of $\sqrt[3]{125 \times 64}$ is 20 ....

Tick (✓) the correct answer

Question: Tick (✓) the correct answer $\sqrt[3]{512}=?$ (a) 6 (b) 7 (c) 8 (d) 9 Solution: (c) 8 $\sqrt[3]{512}=\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}=\sqrt[3]{(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times(2 \times 2 \times 2)}$ $\sqrt[3]{512}=\sqrt[3]{(2)^{3} \times(2)^{3} \times(2)^{3}}=8$ Hence, the cube root of 512 is 8....

A circular park is surrounded by a road 21 m wide.

Question: A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, then find the area of the road. Solution: Given that, a circuiar park is surrounded by a road. Width of the road $=21 \mathrm{~m}$ Radius of the park $(r)=105 \mathrm{~m}$ $\therefore$ Radius of whole circular portion (park + road), Now, area of road = Area of whole circular portion - Area of circular park $=\pi r_{8}^{2}-\pi r_{i}^{2} \quad\left[\therefore\right.$ area of circle $\left.=\pi r^{2}\ri...

Tick (✓) the correct answer

Question: Tick (✓) the correct answer Which of the following numbers is a perfect cube? (a) 1152 (b) 1331 (c) 2016 (d) 739 Solution: (a) $1152=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3=(2)^{3} \times(2)^{3} \times(2 \times 3 \times 3)$ Hence, 1152 is not a perfect cube. (b) (✓) $1331=11 \times 11 \times 11=(11)^{3}$ Hence, 1331 is a perfect cube. (c) $2016=2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7=(2)^{3} \times 2 \times 2 \times 3 \times 3 ...

Tick (✓) the correct answer

Question: Tick (✓) the correct answer Which of the following numbers is a perfect cube? (a) 141 (b) 294 (c) 216 (d) 496 Solution: (a)141 is not a perfect cube.(b)294 is not a perfect cube.(c) (✓)216 is a perfect cube. $216=(2 \times 2 \times 2) \times(3 \times 3 \times 3)=\left(2^{3}\right) \times\left(3^{3}\right)=6^{3}$ (d)496 is not a perfect cube....

In figure, arcs have been drawn with radii 14 cm

Question: In figure, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region. Solution: Given that, radii of each arc (r) = 14 cm Now, area of the sector with central $\angle P=\frac{\angle P}{360^{\circ}} \times \pi r^{2}$ $=\frac{\angle P}{360^{\circ}} \times \pi \times(14)^{2} \mathrm{~cm}^{2}$ $\left[\because\right.$ area of any sector with central angle $\theta$ and radius $\left.r=\frac{\pi r^{2}}{360^{\circ}} \times \theta\right]$ Area of...

Evaluate:

Question: Evaluate: $\sqrt[3]{\frac{-512}{343}}$ Solution: $\sqrt[3]{\frac{-512}{343}}$ By factorisation: $\sqrt[3]{\frac{512}{343}}=\frac{\sqrt[3]{8 \times 8 \times 8}}{\sqrt[3]{7 \times 7 \times 7}}$ $\sqrt[3]{\frac{-512}{343}}=\frac{-8}{7}$...

In figure arcs are drawn by taking vertices A, B

Question: In figure arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm, To intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region, (use = 3.14) Solution: Since, ABC is an equilateral triangle. $\angle A=\angle B=\angle C=60^{\circ}$ and $\quad A B=B C=A C=10 \mathrm{~cm}$ So, $E, F$ and $D$ are mid-points of the sides. $\therefore \quad A E=E C=C D=B D=B F=F A=5 \mathrm{~cm}$ Now, area of sector $C D E=\f...

Evaluate:

Question: Evaluate: $\sqrt[3]{\frac{729}{1000}}$ Solution: $\sqrt[3]{\frac{729}{1000}}$ On factorisation: $\sqrt[3]{\frac{729}{1000}}=\frac{\sqrt[3]{(3 \times 3 \times 3) \times(3 \times 3 \times 3)}}{\sqrt[3]{(2 \times 2 \times 2) \times(5 \times 5 \times 5)}}=\frac{\sqrt[3]{9 \times 9 \times 9}}{\sqrt[3]{10 \times 10 \times 10}}$ $\sqrt[3]{\frac{729}{1000}}=\frac{9}{10}$...

Find the area of the shaded region in figure,

Question: Find the area of the shaded region in figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-point P, Q, R and 5 of the sides AB, BC, CD and DA, respectively of a square ABCD. (use = 3.14) Solution: Given, side of a square BC = 12 cm Since, Q is a mid-point of BC $\therefore$Radius $=B Q=\frac{12}{2}=6 \mathrm{~cm}$ Now, $\quad$ area of quadrant $B P Q=\frac{\pi r^{2}}{4}=\frac{3.14 \times(6)^{2}}{4}=\frac{113.04}{4} \mathrm{~cm}^{2}$ Area of four quadrants $=\fr...

Evaluate:

Question: Evaluate: $\sqrt[3]{64 \times 729}$ Solution: $\sqrt[3]{64 \times 729}$ $\sqrt[3]{64 \times 729}=\sqrt[3]{64} \times \sqrt[3]{729}$ $=\sqrt[3]{4 \times 4 \times 4} \times \sqrt[3]{(3 \times 3 \times 3)} \times(3 \times 3 \times 3)$ $=\sqrt[3]{4 \times 4 \times 4} \times \sqrt[3]{(9 \times 9 \times 9)}$ $\sqrt[3]{64 \times 729}=(4) \times(9)=36$...

Question: Evaluate: $\sqrt[3]{\frac{-64}{343}}$ Solution: $\sqrt[3]{\frac{-64}{343}}$ On factorisation: $\sqrt[3]{\frac{64}{343}}=\sqrt[3]{\frac{2 \times 2 \times 2 \times 2 \times 2 \times 2}{7 \times 7 \times 7}}$ $\therefore \sqrt[3]{\frac{-64}{343}}=\frac{-4}{7}$...

Find the area of the minor segment

Question: Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60. Solution: Given that, radius of circle (r) = 14 cm and angle of the corresponding sector i.e., central angle () = 60 Since, in ΔAOB, OA = OB = Radius of circle i.e., ΔAOB is isosceles. $\Rightarrow \quad \angle O A B=\angle O B A=\theta$ Now, in $\triangle O A B \quad \angle A O B+\angle O A B+\angle O B A=180^{\circ}$ [since,sum of interior angles of any triangle is $180^{...

Evaluate:

Question: Evaluate: $\sqrt[3]{\frac{-27}{125}}$ Solution: $\sqrt[3]{\frac{-27}{125}}$ By factorisation: $\sqrt[3]{\frac{27}{125}}=\sqrt[3]{\frac{3 \times 3 \times 3}{5 \times 5 \times 5}}$ $\therefore \sqrt[3]{\frac{-27}{125}}=\frac{-3}{5}$...

Evaluate:

Question: Evaluate: $\sqrt[3]{\frac{125}{216}}$ Solution: $\sqrt[3]{\frac{125}{216}}$ By prime factorisation: $\sqrt[3]{\frac{125}{216}}=\frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{(2 \times 2 \times 2) \times(3 \times 3 \times 3)}}=\frac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{(6 \times 6 \times 6)}}=\frac{5}{6}$ $\therefore \sqrt[3]{\frac{125}{216}}=\frac{5}{6}$...

Find the area of the shaded region in figure.

Question: Find the area of the shaded region in figure. Solution: join GH and FE Here, breadth of the rectangle $B C=12 \mathrm{~m}$ $\therefore$ Breadth of the inner rectangle $E F G H=12-(4+4)=4 \mathrm{~cm}$ which is equal to the diameter of the semi-circle $E J F, d=4 \mathrm{~m}$ $\therefore \quad$ Radius of semi-circle $E J F, r=2 \mathrm{~m}$ $\therefore \quad$ Length of inner rectangle $E F G H=26-(5+5)=16 \mathrm{~m}$ $\therefore$ Area of two semi-circles EJF and $H / G=2\left(\frac{\pi...

Evaluate:

Question: Evaluate: $\sqrt[3]{\frac{27}{64}}$ Solution: $\sqrt[3]{\frac{27}{64}}$ By prime factorisation: $\sqrt[3]{\frac{27}{64}}=\frac{\sqrt[3]{27}}{\sqrt[3]{64}}=\frac{\sqrt[3]{(3 \times 3 \times 3)}}{\sqrt[3]{(2 \times 2 \times 2) \times(2 \times 2 \times 2)}}=\frac{\sqrt[3]{(3 \times 3 \times 3)}}{\sqrt[3]{(4 \times 4 \times 4)}}=\frac{3}{4}$ $\therefore \sqrt[3]{\frac{27}{64}}=\frac{3}{4}$...

Evaluate:

Question: Evaluate: $\sqrt[3]{-1331}$ Solution: $\sqrt[3]{-1331}$ By prime factorisation: $\sqrt[3]{1331}=\sqrt[3]{11 \times 11 \times 11}$ $\sqrt[3]{-1331}=-(11 \times 11 \times 11)^{\frac{1}{3}}=-11$ $\therefore \sqrt[3]{-1331}=-(\sqrt[3]{1331})=-11$...

Find the area of the shaded field shown in figure.

Question: Find the area of the shaded field shown in figure. Solution: In a figure, join ED From fiqure, radius of semi-circle $D F E, r=6-4=2 \mathrm{~m}$ Now, $\quad$ area of rectangle $A B C D=B C \times A B=8 \times 4=32 \mathrm{~m}^{2}$ and $\quad$ area of semi-circie $D F E=\frac{\pi r^{2}}{2}=\frac{\pi}{2}(2)^{2}=2 \pi \mathrm{m}^{2}$ $\therefore$ Area of shaded region $=$ Area of rectangle $A B C D+$ Area of semi-circle $D F E$ $=(32+2 \pi) m^{2}$...

Evaluate:

Question: Evaluate: $\sqrt[3]{-512}$ Solution: $\sqrt[3]{-512}$ By prime factorisation: $\sqrt[3]{512}=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ $\quad=(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times(2 \times 2 \times 2)$ $\sqrt[3]{-512}=-\sqrt[3]{(2 \times 2 \times 2)}=-8$ $\therefore \sqrt[3]{-512}=-(\sqrt[3]{512})=-8$...

In figure, AB is a diameter of the circle,

Question: In figure, AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region, (use = 3.14) Solution: Given, AC = 6 cm and BC = 8 cm We know that, triangle in a semi-circle with hypotenuse as diameter is right angled triangle. $\therefore \quad \angle C=90^{\circ}$ In right angled $\triangle A C B$, use Pythagoras theorem, $\therefore$ $A B^{2}=A C^{2}+C B^{2}$ $\Rightarrow \quad A B^{2}=100$ $\Rightarrow \quad A B=10 \mathrm{~cm} \quad$ [since, side cannot be ...

Evaluate:

Question: Evaluate: $\sqrt[3]{-216}$ Solution: $\sqrt[3]{-216}$ By prime factorisation: $216=2 \times 2 \times 2 \times 3 \times 3 \times 3$ $=(2 \times 2 \times 2) \times(3 \times 3 \times 3)$ $\sqrt[3]{-216}=-(2 \times 3)=-6$ $\therefore \sqrt[3]{-216}=-(\sqrt[3]{216})=-6$...

Evaluate:

Question: Evaluate: $\sqrt[3]{3375}$ Solution: $\sqrt[3]{3375}$ By prime factorisation: $3375=3 \times 3 \times 3 \times 5 \times 5 \times 5$ $=(3 \times 3 \times 3) \times(5 \times 5 \times 5)$ $\therefore \sqrt[3]{3375}=(3 \times 5)=15$...