The areas of two circles are in the ratio 4: 9.
Question: The areas of two circles are in the ratio 4: 9. What is the ratio between their circumferences? Solution: Let the radii of the two circles berandR, the circumferences of the circles becandCand the areas of the two circles beaandA.Now, $\frac{a}{A}=\frac{4}{9}$ $\Rightarrow \frac{\pi r^{2}}{\pi R^{2}}=\left(\frac{2}{3}\right)^{2}$ $\Rightarrow \frac{r}{R}=\frac{2}{3}$ Now, the ratio between their circumferences is given by $\frac{c}{C}=\frac{2 \pi r}{2 \pi R}$ $=\frac{r}{R}$ $=\frac{2}{...
Read More →Solve this
Question: If $x \in \mathrm{N}$ and $\left|\begin{array}{cc}x+3 -2 \\ -3 x 2 x\end{array}\right|=8$, then find the value of $x$. Solution: $\left|\begin{array}{cc}x+3 -2 \\ -3 x 2 x\end{array}\right|=8$ $\Rightarrow(x+3) 2 x-(-2)(-3 x)=8$ $\Rightarrow 2 x^{2}+6 x-6 x=8$ $\Rightarrow 2 x^{2}=8$ $\Rightarrow x^{2}-4=0$ $\Rightarrow x^{2}=4$ $\Rightarrow x=2 \quad[x \neq-2 \because x \in \mathrm{N}]$...
Read More →If the mid-points of the sides of a quadrilateral
Question: If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure). Solution: Given Let $A B C D$ is a quadrilateral and $P, F, R$ and $S$ are the mid-points of the sides $B C, C D$, $A D$ and $A B$ respectively and $P F R S$ is a parallelogram. To prove ar (parallelogram PFRS) $=\frac{1}{2}$ ar (quadrilateral $A B C D$ ) Construction Join $B D$ and $B R$. Proof Median $...
Read More →The circumferences of two circles are in the ratio 2: 3.
Question: The circumferences of two circles are in the ratio 2: 3. What is the ratio between their areas? Solution: Let the the radii of the two circles berandR, the circumferences of the circles becandCand the areas of the two circles beaandA.Now, $\frac{c}{C}=\frac{2}{3}$ $\Rightarrow \frac{2 \pi r}{2 \pi R}=\frac{2}{3}$ $\Rightarrow \frac{r}{R}=\frac{2}{3}$ Now, the ratio between their areas is given by $\frac{a}{A}=\frac{\pi r^{2}}{\pi R^{2}}$ $=\left(\frac{r}{R}\right)^{2}$ $=\left(\frac{2}...
Read More →The circumferences of two circles are in the ratio 2: 3.
Question: The circumferences of two circles are in the ratio 2: 3. What is the ratio between their areas? Solution: Let the the radii of the two circles berandR, the circumferences of the circles becandCand the areas of the two circles beaandA.Now, $\frac{c}{C}=\frac{2}{3}$ $\Rightarrow \frac{2 \pi r}{2 \pi R}=\frac{2}{3}$ $\Rightarrow \frac{r}{R}=\frac{2}{3}$ Now, the ratio between their areas is given by $\frac{a}{A}=\frac{\pi r^{2}}{\pi R^{2}}$ $=\left(\frac{r}{R}\right)^{2}$ $=\left(\frac{2}...
Read More →Find the maximum value
Question: Find the maximum value of $\left|\begin{array}{ccc}1 1 1 \\ 1 1+\sin \theta 1 \\ 1 1 1+\cos \theta\end{array}\right|$ Solution: Let $\Delta=\left|\begin{array}{ccc}1 1 1 \\ 1 1+\sin \theta 1 \\ 1 1 1+\cos \theta\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we get $\Delta=\left|\begin{array}{ccc}1 1 1 \\ 0 \sin \theta 0 \\ 0 0 \cos \theta\end{array}\right|$ $=\sin \theta \cos \theta$ $=\frac{\sin 2 \theta}{2}$ We know that $-1 \leq \si...
Read More →Define the following
Question: Divide $\sqrt{3} a^{4}+2 \sqrt{3} a^{3}+3 a^{2}-6 a$ by $3 a$. Solution: $\frac{\sqrt{3} a^{4}+2 \sqrt{3} a^{3}+3 a^{2}-6 a}{3 a}$ $=\frac{\sqrt{3} a^{4}}{3 a}+\frac{2 \sqrt{3} a^{3}}{3 a}+\frac{3 a^{2}}{3 a}-\frac{6 a}{3 a}$ $=\frac{1}{\sqrt{3}} a^{(4-1)}+\frac{2}{\sqrt{3}} a^{(3-1)}+a^{(2-1)}-2$ $=\frac{1}{\sqrt{3}} a^{3}+\frac{2}{\sqrt{3}} a^{2}+a-2$...
Read More →In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.
Question: In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre. Find the length of the arc. Solution: We have $r=21 \mathrm{~cm}$ and $\theta=60^{\circ}$ Length of $\operatorname{arc}=\frac{\theta}{360^{\circ}} \times 2 \pi r=\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21=22 \mathrm{~cm}$ Hence, the length of the arc of the circle is $22 \mathrm{~cm}$....
Read More →Find the area of the sector of a circle having radius 6 cm and of angle 30°.
Question: Find the area of the sector of a circle having radius 6 cm and of angle 30. Solution: We have $r=6 \mathrm{~cm}$ and $\theta=30^{\circ}$ Now, Area of sector $=\frac{\theta}{360^{\circ}} \times \pi r^{2}=\frac{30^{\circ}}{360^{\circ}} \times 3.14 \times 36=9.42 \mathrm{~cm}^{2}$ Hence, the area of the sector of the circle is $9.42 \mathrm{~cm}^{2}$....
Read More →Define the following
Question: Divide 5z3 6z2+ 7zby 2z. Solution: $\frac{5 z^{3}-6 z^{2}+7 z}{2 z}$ $=\frac{5 z^{3}}{2 z}-\frac{6 z^{2}}{2 z}+\frac{7 z}{2 z}$ $=\frac{5}{2} z^{(3-1)}-3 z^{(2-1)}+\frac{7}{2}$ $=\frac{5}{2} z^{2}-3 z+\frac{7}{2}$...
Read More →Solve this
Question: If $A=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$, then for any natural number, find the value of $\operatorname{Det}\left(A^{n}\right)$. Solution: Let $A=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$. Then, $A^{2}=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}...
Read More →Define the following
Question: Divide $-x^{6}+2 x^{4}+4 x^{3}+2 x^{2}$ by $\sqrt{2} x^{2}$ Solution: $\frac{-x^{6}+2 x^{4}+4 x^{3}+2 x^{2}}{\sqrt{2} x^{2}}$ $=\frac{-x^{6}}{\sqrt{2} x^{2}}+\frac{2 x^{4}}{\sqrt{2} x^{2}}+\frac{4 x^{3}}{\sqrt{2} x^{2}}+\frac{2 x^{2}}{\sqrt{2} x^{2}}$ $=\frac{-1}{\sqrt{2}} x^{(6-2)}+\sqrt{2} x^{(4-2)}+2 \sqrt{2} x^{(3-2)}+\sqrt{2} x^{(2-2)}$ $=\frac{-1}{\sqrt{2}} x^{4}+\sqrt{2} x^{2}+2 \sqrt{2} x+\sqrt{2}$...
Read More →The radii of two circles are 8 cm and 6 cm.
Question: The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of the areas of the two circles. Solution: Let the radius of the required circle ber.Now, Area of the required circle = Area of circle having radius 8 cm + Area of circle having radius 6 cm $\Rightarrow \pi r^{2}=\pi(8)^{2}+\pi(6)^{2}$ $\Rightarrow r^{2}=64+36$ $\Rightarrow r^{2}=10^{2}$ $\Rightarrow r=10 \mathrm{~cm}$ Hence, the radius of the circle is 10 cm....
Read More →In trapezium ABCD, AB || DC and L is the mid-point of BC.
Question: In trapezium $A B C D, A B \| D C$ and $L$ is the mid-point of $B C$. Through $L$, a line $P Q \| A D$ has been drawn which meets $A B$ in $P$ and $D C$ produced in $Q$. Prove that $\operatorname{ar}(A B C D)=\operatorname{ar}(A P Q D)$. $\therefore \mathrm{BL}=\mathrm{CL}$ Solution: Given in trapezium $A B C D, A B \| D C, D C$ produced in $Q$ and $L$ is the mid-point of $B C$. $\therefore \quad B L=C L$ To prove $\operatorname{ar}(A B C D)=\operatorname{ar}(A P Q D)$ Proof Since, $D ...
Read More →The radii of two circles are 8 cm and 6 cm.
Question: The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of the areas of the two circles. Solution: Let the radius of the required circle ber.Now, Area of the required circle = Area of circle having radius 8 cm + Area of circle having radius 6 cm $\Rightarrow \pi r^{2}=\pi(8)^{2}+\pi(6)^{2}$ $\Rightarrow r^{2}=64+36$ $\Rightarrow r^{2}=10^{2}$ $\Rightarrow r=10 \mathrm{~cm}$ Hence, the radius of the circle is 10 cm....
Read More →Define the following
Question: Divide 4a3+ 4a2+aby 2a. Solution: $\frac{-4 a^{3}+4 a^{2}+a}{2 a}$ $=\frac{-4 a^{3}}{2 a}+\frac{4 a^{2}}{2 a}+\frac{a}{2 a}$ $=-2 a^{(3-1)}+2 a^{(2-1)}+\frac{1}{2}$ $=-2 a^{2}+2 a+\frac{1}{2}$...
Read More →The radii of two circles are 19 cm and 9 cm,
Question: The radii of two circles are 19 cm and 9 cm, Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. Solution: Let the radius of the required circle ber.Now, Circumference of the required circle = Circumference of circle having radius 19 cm + Circumference of circle having radius 9 cm $\Rightarrow 2 \pi r=2 \pi \times 19+2 \pi \times 9$ $\Rightarrow r=19+9$ $\Rightarrow r=28 \mathrm{~cm}$ Hence, the radius of the required circle ...
Read More →The radii of two circles are 19 cm and 9 cm,
Question: The radii of two circles are 19 cm and 9 cm, Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. Solution: Let the radius of the required circle ber.Now, Circumference of the required circle = Circumference of circle having radius 19 cm + Circumference of circle having radius 9 cm $\Rightarrow 2 \pi r=2 \pi \times 19+2 \pi \times 9$ $\Rightarrow r=19+9$ $\Rightarrow r=28 \mathrm{~cm}$ Hence, the radius of the required circle ...
Read More →Divide the following
Question: Divide $y^{4}-3 y^{3}+\frac{1}{2} y^{2}$ by $3 y$. Solution: $\frac{y^{4}-3 y^{3}+\frac{1}{2} y^{2}}{3 y}$ $=\frac{y^{4}}{3 y}-\frac{3 y^{3}}{3 y}+\frac{\frac{1}{2} y^{2}}{3 y}$ $=\frac{1}{3} y^{(4-1)}-y^{(3-1)}+\frac{1}{6} y^{(2-1)}$ $=\frac{1}{3} y^{3}-y^{2}+\frac{1}{6} y$...
Read More →Write the value of the determinant
Question: Write the value of the determinant $\left|\begin{array}{ccc}x+y y+z z+x \\ z x y \\ -3 -3 -3\end{array}\right|$. Solution: $\left|\begin{array}{ccc}x+y y+z z+x \\ z x y \\ -3 -3 -3\end{array}\right|$ $=\left|\begin{array}{ccc}x+y+z x+y+z z+x+y \\ z x y \\ -3 -3 -3\end{array}\right|$ $\left[\right.$ Applying $\left.R_{1} \rightarrow R_{1}+R_{2}\right]$ $=(x+y+z)\left|\begin{array}{ccc}1 1 1 \\ z x y \\ -3 -3 -3\end{array}\right|$ [Taking $(x+y+z)$ common from $R_{1}$ ] $=(x+y+z)\left|\b...
Read More →Define the following
Question: Dividex+ 2x2+ 3x4x5by 2x. Solution: $\frac{x+2 x^{2}+3 x^{4}-x^{5}}{2 x}$ $=\frac{x}{2 x}+\frac{2 x^{2}}{2 x}+\frac{3 x^{4}}{2 x}-\frac{x^{5}}{2 x}$ $=\frac{1}{2}+x+\frac{3}{2} x^{3}-\frac{1}{2} x^{4}$...
Read More →Find the radius of a circle whose perimeter and area are numerically equal.
Question: Find the radius of a circle whose perimeter and area are numerically equal. Solution: Let the radius of the required circle ber.Now, Area of circle = Perimeter of the circle $\Rightarrow \pi r^{2}=2 \pi \times r$ $\Rightarrow r^{2}=2 r$ $\Rightarrow r^{2}-2 r=0$ $\Rightarrow r(r-2)=0$ $\Rightarrow r-2=0 \quad[\because r \neq 0]$ $\Rightarrow r=2$ units Hence, the radius of the circle is 2 units....
Read More →Find the radius of a circle whose perimeter and area are numerically equal.
Question: Find the radius of a circle whose perimeter and area are numerically equal. Solution: Let the radius of the required circle ber.Now, Area of circle = Perimeter of the circle $\Rightarrow \pi r^{2}=2 \pi \times r$ $\Rightarrow r^{2}=2 r$ $\Rightarrow r^{2}-2 r=0$ $\Rightarrow r(r-2)=0$ $\Rightarrow r-2=0 \quad[\because r \neq 0]$ $\Rightarrow r=2$ units Hence, the radius of the circle is 2 units....
Read More →Simplify:
Question: Simplify: $\frac{32 m^{2} n^{3} p^{2}}{4 m n p}$ Solution: $\frac{32 m^{2} n^{3} p^{2}}{4 m n p}$ $=\frac{32 \times m \times m \times n \times n \times n \times p \times p}{4 \times m \times n \times p}$ $=8 m^{(2-1)} n^{(3-1)} p^{(2-1)}$ $=8 m n^{2} p$...
Read More →Find the perimeter of a semicircular protractor whose diameter is 14 cm.
Question: Find the perimeter of a semicircular protractor whose diameter is 14 cm. Solution: Perimeter of a semicircular protractor = Circumference of semicircular protractor + diameter of semicircular protractor $=\frac{1}{2}(2 \pi r)+d$ $=\pi r+d$ $=\pi \frac{d}{2}+d$ $=\frac{22}{7} \times 7+14$ $=22+14$ $=36 \mathrm{~cm}$ Hence, the perimeter of a semicircular protractor is 36 cm....
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