Simplify:
Question: $\frac{16 m^{3} y^{2}}{4 m^{2} y}$ Solution: $\frac{16 m^{3} y^{2}}{4 m^{2} y}$ $=\frac{16 \times m \times m \times m \times y \times y}{4 \times m \times m \times y}$ $=4 m^{(3-2)} y^{(2-1)}$ $=4 m y$...
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....
Read More →ABCD is a parallelogram in which BC
Question: ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F If ar (ΔDFB) = 3 cm2, then find the area of the parallelogram ABCD. Solution: Given, $A B C D$ is a parallelogram and $C E=B C$ i.e., $C$ is the mid-point of $B E$. Also, $\operatorname{ar}(\Delta D F B)=3 \mathrm{~cm}^{2}$ Now, $\triangle A D F$ and $\triangle D F B$ are on the same base $D F$ and between parallels $C D$ and $A B$. Then, ar $(\triangle A D F)=\operatorname{ar}(\triangle D F B...
Read More →Write the value of the determinant
Question: Write the value of the determinant $\left|\begin{array}{cc}p p+1 \\ p-1 p\end{array}\right|$. Solution: $\left|\begin{array}{cc}p p+1 \\ p-1 p\end{array}\right|=p^{2}-(p+1)(p-1)$ $=p^{2}-\left(p^{2}-1\right)$ $=p^{2}-p^{2}+1$ $=1$ Hence, the value of the determinant $\left|\begin{array}{cc}p p+1 \\ p-1 p\end{array}\right|$ is 1 ....
Read More →Define the following
Question: Divide 72a4b5c8by 9a2b2c3. Solution: $\frac{-72 a^{4} b^{5} c^{8}}{-9 a^{2} b^{2} c^{3}}$ $=\frac{-72 \times a \times a \times a \times a \times b \times b \times b \times b \times b \times c \times c \times c \times c \times c \times c \times c \times c}{-9 \times a \times a \times b \times b \times c \times c \times c}$ $=8 a^{(4-2)} b^{(5-2)} c^{(8-3)}$ $=8 a^{2} b^{3} c^{5}$...
Read More →Find the area of a circle whose circumference is 8π.
Question: Find the area of a circle whose circumference is 8. Solution: Let the radius of the circle ber.Now, Circumference $=8 \pi$ $\Rightarrow 2 \pi r=8 \pi$ $\Rightarrow r=4 \mathrm{~cm}$ Now, Area of circle $=\pi r^{2}=\pi \times 4^{2}=16 \pi \mathrm{cm}^{2}$ Hence, the area of the circle is 16 cm2....
Read More →Define the following
Question: Divide 72xyz2by 9xz. Solution: $\frac{72 x y z^{2}}{-9 x z}$ $=\frac{72 \times x \times y \times z \times z}{-9 \times x \times z}$ $=-8 x^{(1-1)} y z^{(2-1)}$ $=-8 y z$...
Read More →Solve this
Question: If $A$ is a $3 \times 3$ matrix, $|A| \neq 0$ and $|3 A|=k|A|$ then write the value of $k$. Solution: Let $A=\left[\begin{array}{lll}a_{1} a_{2} a_{3} \\ b_{1} b_{2} b_{3} \\ c_{1} c_{2} c_{3}\end{array}\right]$. then, $3 A=\left[\begin{array}{lll}3 a_{1} 3 a_{2} 3 a_{3} \\ 3 b_{1} 3 b_{2} 3 b_{3} \\ 3 c_{1} 3 c_{2} 3 c_{3}\end{array}\right]$. $|3 A|=\left|\begin{array}{lll}3 a_{1} 3 a_{2} 3 a_{3} \\ 3 b_{1} 3 b_{2} 3 b_{3} \\ 3 c_{1} 3 c_{2} 3 c_{3}\end{array}\right|$ $=3^{3}\left|\be...
Read More →Solve this
Question: If $A$ is a $3 \times 3$ matrix, $|A| \neq 0$ and $|3 A|=k|A|$ then write the value of $k$. Solution: Let $A=\left[\begin{array}{lll}a_{1} a_{2} a_{3} \\ b_{1} b_{2} b_{3} \\ c_{1} c_{2} c_{3}\end{array}\right]$. then, $3 A=\left[\begin{array}{lll}3 a_{1} 3 a_{2} 3 a_{3} \\ 3 b_{1} 3 b_{2} 3 b_{3} \\ 3 c_{1} 3 c_{2} 3 c_{3}\end{array}\right]$. $|3 A|=\left|\begin{array}{lll}3 a_{1} 3 a_{2} 3 a_{3} \\ 3 b_{1} 3 b_{2} 3 b_{3} \\ 3 c_{1} 3 c_{2} 3 c_{3}\end{array}\right|$ $=3^{3}\left|\be...
Read More →Solve this
Question: If $A$ is a $3 \times 3$ matrix, $|A| \neq 0$ and $|3 A|=k|A|$ then write the value of $k$. Solution: Let $A=\left[\begin{array}{lll}a_{1} a_{2} a_{3} \\ b_{1} b_{2} b_{3} \\ c_{1} c_{2} c_{3}\end{array}\right]$. then, $3 A=\left[\begin{array}{lll}3 a_{1} 3 a_{2} 3 a_{3} \\ 3 b_{1} 3 b_{2} 3 b_{3} \\ 3 c_{1} 3 c_{2} 3 c_{3}\end{array}\right]$. $|3 A|=\left|\begin{array}{lll}3 a_{1} 3 a_{2} 3 a_{3} \\ 3 b_{1} 3 b_{2} 3 b_{3} \\ 3 c_{1} 3 c_{2} 3 c_{3}\end{array}\right|$ $=3^{3}\left|\be...
Read More →Find the diameter of the circle
Question: Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm. Solution: Let the diameter of the required circle bed.Now, Area of the required circle = Area of circle having radius 4 cm + Area of circle having radius 3 cm $\Rightarrow \pi\left(\frac{d}{2}\right)^{2}=\pi(4)^{2}+\pi(3)^{2}$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=16+9$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=25=(5)^{2}$ $\Rightarrow \frac{d}{2}=5$ $\Rightarrow d=1...
Read More →Define the following
Question: Divide 21abc2by 7abc. Solution: $\frac{-21 a b c^{2}}{7 a b c}$ $=\frac{-21 \times a \times b \times c \times c}{7 \times a \times b \times c}$ $=-3 a^{(1-1)} b^{(1-1)} c^{(2-1)}$ $=-3 c$...
Read More →Find the diameter of the circle
Question: Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm. Solution: Let the diameter of the required circle bed.Now, Area of the required circle = Area of circle having radius 4 cm + Area of circle having radius 3 cm $\Rightarrow \pi\left(\frac{d}{2}\right)^{2}=\pi(4)^{2}+\pi(3)^{2}$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=16+9$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=25=(5)^{2}$ $\Rightarrow \frac{d}{2}=5$ $\Rightarrow d=1...
Read More →Divide the following
Question: Divide 24a3b3by 8ab. Solution: $\frac{24 a^{3} b^{3}}{-8 a b}$ $=\frac{24 \times a \times a \times a \times b \times b \times b}{-8 \times a \times b}$ $=-3 a^{(3-1)} b^{(3-1)}$ $=-3 a^{2} b^{2}$...
Read More →Solve this
Question: If $\left|\begin{array}{cc}2 x 5 \\ 8 x\end{array}\right|=\left|\begin{array}{cc}6 -2 \\ 7 3\end{array}\right|$, write the value of $x$ Solution: $\left|\begin{array}{cc}2 x 5 \\ 8 x\end{array}\right|=\left|\begin{array}{cc}6 -2 \\ 7 3\end{array}\right|$ $\Rightarrow 2 x^{2}-40=18+14$ $\Rightarrow 2 x^{2}-40=32$ $\Rightarrow 2 x^{2}=72$ $\Rightarrow x^{2}=36$ $\Rightarrow x=\pm 6$ Hence, the value of $x$ is $\pm 6$....
Read More →Find the length of the arc of a circle of diameter 42 cm
Question: Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60 at the centre. Solution: We have $r=\frac{\text { diameter }}{2}=\frac{42}{2}=21 \mathrm{~cm}$ and $\theta=60^{\circ}$ Length of 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 cm....
Read More →Solve this
Question: If $\left|\begin{array}{cc}3 x 7 \\ -2 4\end{array}\right|=\left|\begin{array}{ll}8 7 \\ 6 4\end{array}\right|$, find the value of $x .$ Solution: $\left|\begin{array}{ll}3 x 7 \\ -2 4\end{array}\right|=\left|\begin{array}{ll}8 7 \\ 6 4\end{array}\right|$ $\Rightarrow 12 x+14=32-42$ $\Rightarrow 12 x+14=-10$ $\Rightarrow 12 x=-24$ $\Rightarrow x=-2$ $\therefore x=-2 .$...
Read More →Solve this
Question: If $\left|\begin{array}{cc}3 x 7 \\ -2 4\end{array}\right|=\left|\begin{array}{ll}8 7 \\ 6 4\end{array}\right|$, find the value of $x .$ Solution: $\left|\begin{array}{ll}3 x 7 \\ -2 4\end{array}\right|=\left|\begin{array}{ll}8 7 \\ 6 4\end{array}\right|$ $\Rightarrow 12 x+14=32-42$ $\Rightarrow 12 x+14=-10$ $\Rightarrow 12 x=-24$ $\Rightarrow x=-2$ $\therefore x=-2 .$...
Read More →Divide the sum of
Question: Divide 15m2n3by 5m2n2. Solution: $\frac{15 m^{2} n^{3}}{5 m^{2} n^{2}}$ $=\frac{15 \times m \times m \times n \times n \times n}{5 \times m \times m \times n \times n}$ $=3 m^{(2-2)} n^{(3-2)}$ $=3 m^{0} n^{1}$ $=3 n$...
Read More →What is the perimeter of a square which circumscribes a circle of radius a cm?
Question: What is the perimeter of a square which circumscribes a circle of radiusacm? Solution: We know that if a square circumscribes a circle, then the side of the square is equal to the diameter of the circle. Side of Square = 2aNow, Perimeter of the square = 4 ⨯ Side of square = 4 ⨯ 2a= 8acmHence, the perimeter of the square is 8acm....
Read More →Solve the following equations
Question: If $\left|\begin{array}{cc}2 x x+3 \\ 2(x+1) x+1\end{array}\right|=\left|\begin{array}{ll}1 5 \\ 3 3\end{array}\right|$, then write the value of $x$ Solution: $\left|\begin{array}{cc}2 x x+3 \\ 2(x+1) x+1\end{array}\right|=\left|\begin{array}{cc}1 5 \\ 3 3\end{array}\right|$ $\Rightarrow(2 x)(x+1)-2(x+1)(x+3)=3-15$ $\Rightarrow(x+1)(2 x-2 x-6)=-12$ $\Rightarrow-6 x-6=-12$ $\Rightarrow-6 x=-6$ $\Rightarrow x=1$ Hence, the value of $x$ is 1 ....
Read More →If the area of the circle is numerically
Question: If the area of the circle is numerically equal to twice its circumference then what is the diameter of the circle? Solution: Let the diameter of the required circle bed.Now, Area of circle = 2 ⨯ Circumference of the circle $\Rightarrow \pi\left(\frac{d}{2}\right)^{2}=2\left(2 \pi \times \frac{d}{2}\right)$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=2 d$ $\Rightarrow d^{2}=8 d$ $\Rightarrow d^{2}-8 d=0$ $\Rightarrow d(d-8)=0$ $\Rightarrow d=8 \mathrm{~cm} \quad[\because d \neq 0]$ Hence, ...
Read More →Divide the sum of
Question: Divide 6x3y2z2by 3x2yz. Solution: $\frac{6 x^{3} y^{2} z^{2}}{3 x^{2} y z}$ $=\frac{6 \times x \times x \times x \times y \times y \times z \times z}{3 \times x \times x \times y \times z}$ $=2 x^{(3-2)} y^{(2-1)} z^{(2-1)}$ $=2 x y z$...
Read More →If the area of the circle is numerically
Question: If the area of the circle is numerically equal to twice its circumference then what is the diameter of the circle? Solution: Let the diameter of the required circle bed.Now, Area of circle = 2 ⨯ Circumference of the circle $\Rightarrow \pi\left(\frac{d}{2}\right)^{2}=2\left(2 \pi \times \frac{d}{2}\right)$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=2 d$ $\Rightarrow d^{2}=8 d$ $\Rightarrow d^{2}-8 d=0$ $\Rightarrow d(d-8)=0$ $\Rightarrow d=8 \mathrm{~cm} \quad[\because d \neq 0]$ Hence, ...
Read More →O is any point on the diagonal PR
Question: O is any point on the diagonal PR of a parallelogram PQRS (figure). Prove that ar(ΔPSO) = ar(ΔPQO). Solution: Given in a parallelogram $P Q R S, O$ is any point on the diagonal $P R$. To prove $\operatorname{ar}(\Delta P S O)=\operatorname{ar}(\triangle P Q O)$ Construction Join $S Q$ which intersects $P R$ at $B$. Proof We know that, diagonals of a parallelogram bisect each other, so $B$ is the mid-point of $S Q$. Here, $P B$ is a median of $\triangle Q P S$ and we know that, a median...
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