What is the diameter of a circle
Question: What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm? Solution: Let the diameter of the required circle bed.Now, Area of required circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm $\Rightarrow \pi\left(\frac{d}{2}\right)^{2}=\pi\left(\frac{10}{2}\right)^{2}+\pi\left(\frac{24}{2}\right)^{2}$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=25+144$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=13^{2...
Read More →Write each of the following polynomials in the standard form. Also, write their degree.
Question: Write each of the following polynomials in the standard form. Also, write their degree. (i)x2+ 3 + 6x+ 5x4 (ii)a2+ 4 + 5a6 (iii) (x3 1)(x3 4) (iv) (y3 2)(y3+ 11) (v) $\left(a^{3}-\frac{3}{8}\right)\left(a^{3}+\frac{16}{17}\right)$ (vi) $\left(a+\frac{3}{4}\right)\left(a+\frac{4}{3}\right)$ Solution: (i) Standard form of the given polynomial can be expressed as: $\left(5 x^{4}+x^{2}+6 x+3\right)$ or $\left(3+6 x+x^{2}+5 x^{4}\right)$ The degree of the polynomial is 4 . (ii) Standard for...
Read More →What is the diameter of a circle
Question: What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm? Solution: Let the diameter of the required circle bed.Now, Area of required circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm $\Rightarrow \pi\left(\frac{d}{2}\right)^{2}=\pi\left(\frac{10}{2}\right)^{2}+\pi\left(\frac{24}{2}\right)^{2}$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=25+144$ $\Rightarrow\left(\frac{d}{2}\right)^{2}=13^{2...
Read More →Solve this
Question: If $\left|\begin{array}{ll}x+1 x-1 \\ x-3 x+2\end{array}\right|=\left|\begin{array}{cc}4 -1 \\ 1 3\end{array}\right|$, then write the value of $x$. Solution: $\left|\begin{array}{cc}x+1 x-1 \\ x-3 x+2\end{array}\right|=\left|\begin{array}{cc}4 -1 \\ 1 3\end{array}\right|$ $\Rightarrow(x+1)(x+2)-(x-1)(x-3)=12+1$ $\Rightarrow x^{2}+3 x+2-x^{2}+4 x-3=13$ $\Rightarrow 7 x-1=13$ $\Rightarrow 7 x=14$ $\Rightarrow x=2$ Hence, the value of $x$ is 2 ....
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Question: If $A=\left[\begin{array}{lll}5 3 8 \\ 2 0 1 \\ 1 2 3\end{array}\right]$. Write the cofactor of the element $a_{32}$. Solution: Minor of $a_{32}=M_{32}=\left|\begin{array}{ll}5 8 \\ 2 1\end{array}\right|=5-16=-11$ Cofactor of $a_{32}=A_{32}=(-1)^{3+2} M_{32}=11$ Hence, the cofactor of the element $a_{32}$ is 11 ....
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Question: If $A=\left[\begin{array}{lll}5 3 8 \\ 2 0 1 \\ 1 2 3\end{array}\right]$. Write the cofactor of the element $a_{32}$. Solution: Minor of $a_{32}=M_{32}=\left|\begin{array}{ll}5 8 \\ 2 1\end{array}\right|=5-16=-11$ Cofactor of $a_{32}=A_{32}=(-1)^{3+2} M_{32}=11$ Hence, the cofactor of the element $a_{32}$ is 11 ....
Read More →The circumference of a circle is 22 cm.
Question: The circumference of a circle is 22 cm. Find the area of its quadrant. Solution: Let the radius of the circle ber.Now, Circumference $=22$ $\Rightarrow 2 \pi r=22$ $\Rightarrow r=\frac{22 \times 7}{44}=\frac{7}{2} \mathrm{~cm}$ Now, Area of quadrant $=\frac{1}{4} \pi r^{2}=\frac{1}{4} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^{2}=\frac{77}{8} \mathrm{~cm}^{2}$ Hence, the area of the quadrant of the circle is $\frac{77}{8} \mathrm{~cm}^{2}$....
Read More →ABCD is a square. E and F are respectively
Question: ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF, prove that ar (ΔAER) = ar (ΔAFR). Solution: Given in square $A B C D, E$ and $F$ are the mid-points of $B C$ and $C D$ respectively. Also, $R$ is the mid-point of $E F$. To prove $\operatorname{ar}(A E R)=\operatorname{ar}(A F R)$. Construction Draw $A N \perp E F$ Proof $\because \quad$ ar $(\triangle A E R)=\frac{1}{2} \times$ Base $\times$ Height $=\frac{1}{2} \times E R \times A N$ ...
Read More →Evaluate:
Question: Evaluate: $\left|\begin{array}{ll}\cos 15^{\circ} \sin 15^{\circ} \\ \sin 75^{\circ} \cos 75^{\circ}\end{array}\right|$ Solution: $\begin{array}{ll}\mid \cos 15^{\circ} \sin 15^{\circ} \\ \sin 75^{\circ} \cos 75^{\circ} \mid\end{array}$ $=\cos 15^{\circ} \cos 75^{\circ}-\sin 15^{\circ} \sin 75^{\circ}$ $=\cos \left(15^{\circ}+75^{\circ}\right) \quad[\because \cos A \cos B-\sin A \sin B=\cos (A+B)]$ $=\cos 90^{\circ}$ $=0$ $\Rightarrow \mid \cos 15^{\circ} \sin 15^{\circ}$ $\sin 75^{\ci...
Read More →The difference between the circumference and radius of a circle is 37cm.
Question: The difference between the circumference and radius of a circle is $37 \mathrm{~cm}$. Using $\pi=\frac{22}{7}$, find the circumference of the circle. Solution: Let the radius of the circle berand circumferenceC. Now, $C-r=37$ $\Rightarrow 2 \pi r-r=37$ $\Rightarrow r\left(2 \times \frac{22}{7}-1\right)=37$ $\Rightarrow r=\frac{37 \times 7}{37}=7 \mathrm{~cm}$ Now, $C=2 \pi r=2 \times \frac{22}{7} \times 7=44 \mathrm{~cm}$ Hence, the circumference of the circle is 44 cm....
Read More →A matrix A of order 3 × 3 is such that
Question: A matrix $A$ of order $3 \times 3$ is such that $|A|=4$. Find the value of $|2 A|$. Solution: $|K A|=K^{n}|A|$ Here, $n$ is the order of $A$. Given: $|A|=4$ $\Rightarrow|2 A|=2^{3} \times 4=32$...
Read More →A matrix A of order 3 × 3 is such that
Question: A matrix $A$ of order $3 \times 3$ is such that $|A|=4$. Find the value of $|2 A|$. Solution: $|K A|=K^{n}|A|$ Here, $n$ is the order of $A$. Given: $|A|=4$ $\Rightarrow|2 A|=2^{3} \times 4=32$...
Read More →A matrix A of order 3 × 3 is such that
Question: A matrix $A$ of order $3 \times 3$ is such that $|A|=4$. Find the value of $|2 A|$. Solution: $|K A|=K^{n}|A|$ Here, $n$ is the order of $A$. Given: $|A|=4$ $\Rightarrow|2 A|=2^{3} \times 4=32$...
Read More →For what value
Question: For what value of $x$ is the matrix $\left[\begin{array}{ll}6-x 4 \\ 3-x 1\end{array}\right]$ singular? Solution: $\left|\begin{array}{ll}6-x 4 \\ 3-x 1\end{array}\right|$ is singular when its determinant is 0 . $\Rightarrow\left|\begin{array}{cc}6-x 4 \\ 3-x 1\end{array}\right|=0$ $\Rightarrow(6-x)-4(3-x)=0$ $\Rightarrow 6-x-12+4 x=0$ $\Rightarrow 3 x-6=0$ $\Rightarrow 3 x=6$ $\Rightarrow x=\frac{6}{3}=2$...
Read More →In ΔABC, D is the mid-point of AB and P is any
Question: In $\triangle A B C, D$ is the mid-point of $A B$ and $P$ is any point on $B C$. If $C Q \| P D$ meets $A B$ in $Q$ (shown in figure), then prove that ar $(\triangle B P Q)=1 / 2 \operatorname{ar}(\triangle A B C)$. Solution: Given in $\triangle A B C, D$ is the mid-point of $A B$ and $P$ is any point on $B C$. $C Q \| P D$ means $A B$ in $Q$. To prove $\operatorname{ar}(\triangle B P Q)=\frac{1}{2} \operatorname{ar}(\triangle A B C)$ Construction Join $P Q$ and $C D$. Proof Since, $D$...
Read More →What is the value of the determinant
Question: What is the value of the determinant $\left|\begin{array}{lll}0 2 0 \\ 2 3 4 \\ 4 5 6\end{array}\right| ?$ Solution: $\left|\begin{array}{lll}0 2 0 \\ 2 3 4 \\ 4 5 6\end{array}\right|$ $=0(18-20)-2(12-16)+0(10-12)$ $=8$...
Read More →If the diameters of the concentric circles shown in the figure below are in the ratio 1 : 2 : 3
Question: If the diameters of the concentric circles shown in the figure below are in the ratio 1 : 2 : 3 then find the ratio of the areas of three regions. Solution: The diameters of concentric circles are in ratio 1 : 2 : 3. let the diameters be $d_{1}=1 x_{1} d_{2}=2 x_{1} d_{3}=3 x$. So, the radii are $r_{1}=\frac{x}{2}, r_{2}=x, r_{3}=\frac{3 x}{2}$. The areas of three regions are given be $A_{1}=\pi r_{1}^{2}=\pi\left(\frac{x}{2}\right)^{2}=\frac{\pi x^{2}}{4}$ $\mathrm{~A}_{2}=\pi r_{2}^{...
Read More →If the diameters of the concentric circles shown in the figure below are in the ratio 1 : 2 : 3
Question: If the diameters of the concentric circles shown in the figure below are in the ratio 1 : 2 : 3 then find the ratio of the areas of three regions. Solution: The diameters of concentric circles are in ratio 1 : 2 : 3. let the diameters be $d_{1}=1 x_{1} d_{2}=2 x_{1} d_{3}=3 x$. So, the radii are $r_{1}=\frac{x}{2}, r_{2}=x, r_{3}=\frac{3 x}{2}$. The areas of three regions are given be $A_{1}=\pi r_{1}^{2}=\pi\left(\frac{x}{2}\right)^{2}=\frac{\pi x^{2}}{4}$ $\mathrm{~A}_{2}=\pi r_{2}^{...
Read More →Solve this
Question: If $|A|=2$, where $A$ is $2 \times 2$ matrix, find $|\operatorname{adj} A|$. Solution: For any square matrix $A$ of orde $r n,|\operatorname{adj} A|=|A|^{n-1}$. Given : $|A|=2$ Here, order is 2 . $\Rightarrow|\operatorname{adj} A|=|2|^{2-1}=2$...
Read More →Write the value of the determinant
Question: Write the value of the determinant $\left|\begin{array}{ccc}2 3 4 \\ 5 6 8 \\ 6 x 9 x 12 x\end{array}\right|$ Solution: $\left|\begin{array}{ccc}2 3 4 \\ 5 6 8 \\ 6 x 9 x 12 x\end{array}\right|$ $=\left|\begin{array}{lll}2 3 4 \\ 5 6 8 \\ 2 3 4\end{array}\right|$ [Taking $2 x$ common from $R_{3}$ ] $=0$ $\left[\because R_{1}\right.$ and $R_{3}$ are identical $]$...
Read More →Which of the following expressions are not polynomials?
Question: Which of the following expressions are not polynomials? (i)x2+ 2x2 (ii) $\sqrt{a x}+x^{2}-x^{3}$ (iv)ax1/2+ ax+ 9x2+ 4 (v) 3x2+ 2x1+ 4x+5 Solution: (i) $\mathrm{x}^{2}+2 \mathrm{x}^{-2}$ is not a polynomial because $-2$ is the power of variable $\mathrm{x}$ is not a non negative integer. (ii) $\sqrt{a x}+x^{2}-x^{3}$ is not a polynomial because $\frac{1}{2}$ is the power of variable $x$ is not a non negative integer. (iii) $3 \mathrm{y}^{3}-\sqrt{5} \mathrm{y}+9$ is a polynomial becaus...
Read More →With the vertices A, B and C of a triangle ABC as centres, arcs are drawn with radii 5 cm each as shown in the given figure.
Question: With the verticesA,BandCof a triangleABCas centres, arcs are drawn with radii 5 cm each as shown in the given figure. IfAB= 14 cm,BC= 48 cm andCA= 50 cm then find the area of the shaded region. [Use = 3.14] Solution: Given thatABCis a triangle with sidesAB= 14 cm, BC = 48 cm, CA = 50 cm.Clearly it is a right angled triangle. Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times 48 \times 14=336 \mathrm{~cm}^{2}$ Now we need to remove the area of ...
Read More →Find the value
Question: Find the value of $x$ from the following: $\left|\begin{array}{cc}x 4 \\ 2 2 x\end{array}\right|=0$ Solution: $\left|\begin{array}{cc}x 4 \\ 2 2 x\end{array}\right|=0$ $\Rightarrow 2 x^{2}-8=0$ $\Rightarrow 2 x^{2}=8$ $\Rightarrow x^{2}=\frac{8}{2}=4$ $\Rightarrow x=\sqrt{4}=\pm 2$...
Read More →Write the degree of each of the following polynomials.
Question: Write the degree of each of the following polynomials. (i) 2x2+ 5x2 7 (ii) 5x2 3x+ 2 (iii) 2x+x2 8 (iv) $\frac{1}{2} y^{7}-12 y^{6}+48 y^{5}-10$ (v) 3x3+ 1 (vi) 5 (vii) 20x3+ 12x2y2 10y2+ 20 Solution: (i) Correction : It is $2 \mathrm{x}^{3}+5 \mathrm{x}^{2}-7$ instead of $2 \mathrm{x}^{2}+5 \mathrm{x}^{2}-7$. The degree of the polymonial $2 \mathrm{x}^{3}+5 \mathrm{x}^{2}-7$ is 3 . (ii) The degree of the polymonial $5 x^{2}-35 x+2$ is 2 . (iii) The degree of the polymonial $2 \mathrm{...
Read More →Find the value
Question: Find the value of $x$ from the following: $\left|\begin{array}{cc}x 4 \\ 2 2 x\end{array}\right|=0$ Solution: $\left|\begin{array}{cc}x 4 \\ 2 2 x\end{array}\right|=0$ $\Rightarrow 2 x^{2}-8=0$ $\Rightarrow 2 x^{2}=8$ $\Rightarrow x^{2}=\frac{8}{2}=4$ $\Rightarrow x=\sqrt{4}=\pm 2$...
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