If A is a square matrix satisfying
Question: If $A$ is a square matrix satisfying $A^{\top} A=l$, write the value of $|A|$. Solution: Let $A=\left[a_{i j}\right]$ be a square matrix of order $\mathrm{n}$. Here, $|A|=\left|A^{T}\right|$ [By property of determinants] Given: $\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I}$ $\Rightarrow\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}\right|=1$ Then, $\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}\right|=\left|\mathrm{A}^{\mathrm{T}}\right||\mathrm{A}| \quad$ [Since the determinants are of the same ...
Read More →Divide the following and find the quotient and remainder.
Question: Divide 6x3x2 10x 3 by 2x 3 and find the quotient and remainder. Solution: Quotient $=3 x^{2}+4 x+1$ Remainder $=0$...
Read More →The perimeter of the quadrant of a circle is 25 cm.
Question: The perimeter of the quadrant of a circle is 25 cm. Find its area. Solution: Let the radius of the circle ber Now, Perimeter of quadrant $=\frac{1}{4}(2 \pi r)+2 r$ $\Rightarrow 25=\frac{1}{2} \times \frac{22}{7} \times r+2 r$ $\Rightarrow 25=\frac{25 r}{7}$ $\Rightarrow r=7 \mathrm{~cm}$ Area of quadrant $=\frac{1}{4} \pi r^{2}=\frac{1}{4} \times \frac{22}{7} \times 7 \times 7=38.5 \mathrm{~cm}^{2}$ Hence, the area of quadrant is 38.5 cm2...
Read More →Prove that the line joining the mid-points
Question: Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. Solution: Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively. To prove $M N\|A B\| C D$ Construction Join $C N$ and produce it to meet $A B$ at $E$. In $\triangle C D N$ and $\triangle E B N$, we have $D N=B N$ [since, $N$ is the mid-point of $B D$ ] and $\angle D C N=\angle B E N$ [alternat...
Read More →Divide the sum of and find the quotient and remainder.
Question: Divide 6x3x2 10x 3 by 2x 3 and find the quotient and remainder. Solution: Quotient $=3 x^{2}+4 x+1$ Remainder $=0$...
Read More →On a circular table cover of radius 42 cm,
Question: On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design.[Use $\sqrt{3}=1.73, \pi=\frac{22}{7}$ ] Solution: Construction: Join AO and extend it to D on BC. Radius of the circle,r= 42 cmOCD= 30 $\cos 30^{\circ}=\frac{\mathrm{DC}}{\mathrm{OC}}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{DC}}{42}$ $\Rightarrow \mathrm{DC}=21 \sqrt{3}$ $\Rightarrow \mathrm{BC}=2 ...
Read More →Write the value
Question: Write the value of $\left|\begin{array}{cc}\sin 20^{\circ} -\cos 20^{\circ} \\ \sin 70^{\circ} \cos 70^{\circ}\end{array}\right|$. Solution: Let $\Delta=\mid \sin 20^{\circ} \quad-\cos 20^{\circ}$ $\sin 70^{\circ} \quad \cos 70^{\circ}$ $=\sin 20^{\circ} \cos 70^{\circ}+\cos 20^{\circ} \sin 70^{\circ}$ $=\sin \left(20^{\circ}+70^{\circ}\right) \quad[$ trignometric identity $]$ $=\sin 90^{\circ}$ $=1$...
Read More →On a circular table cover of radius 42 cm,
Question: On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design.[Use $\sqrt{3}=1.73, \pi=\frac{22}{7}$ ] Solution: Construction: Join AO and extend it to D on BC. Radius of the circle,r= 42 cmOCD= 30 $\cos 30^{\circ}=\frac{\mathrm{DC}}{\mathrm{OC}}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{DC}}{42}$ $\Rightarrow \mathrm{DC}=21 \sqrt{3}$ $\Rightarrow \mathrm{BC}=2 ...
Read More →In question 18, write the value
Question: In question 18, write the value of $a_{11} C_{21}+a_{12} C_{22}+a_{13} C_{23}$. Solution: We know that in a square matrix of order n, the sum of the products of elements of a row (or a column) with the cofactors of the corresponding elements of some other row (or column ) is zero. Therefore, $A=\left[a_{i j}\right]$ is a square matrix of order $n$. $\Rightarrow \sum_{j=1}^{n} a_{i j} C_{k j}=0$ and $\sum_{i=1}^{n} a_{i j} C_{i k}=0$ $\Rightarrow a_{11} C_{21}+a_{12} C_{22}+a_{13} C_{23...
Read More →Divide the sum of and find the quotient and remainder
Question: Divide 14x3 5x2+ 9x 1 by 2x 1 and find the quotient and remainder Solution: Quotient $=7 x^{2}+x+5$ Remainder $=4$...
Read More →In question 18, write the value of a_{11} C_{21}+a_{12} C_{22}+a_{13} C_{23}.
Question: In question 18, write the value of $a_{11} C_{21}+a_{12} C_{22}+a_{13} C_{23}$. Solution: We know that in a square matrix of order n, the sum of the products of elements of a row (or a column) with the cofactors of the corresponding elements of some other row (or column ) is zero. Therefore, $A=\left[a_{i j}\right]$ is a square matrix of order $n$. $\Rightarrow \sum_{j=1}^{n} a_{i j} C_{k j}=0$ and $\sum_{i=1}^{n} a_{i j} C_{i k}=0$ $\Rightarrow a_{11} C_{21}+a_{12} C_{22}+a_{13} C_{23...
Read More →D, E and F are respectively the mid-points
Question: D, E and F are respectively the mid-points of the sides AB, BC and CA of a ΔABC. Prove that by joining these mid-points D, E and F, the ΔABC is divided into four congruent triangles. Solution: Given In a ΔABC, D, E and F are respectively the mid-points of the sides AB, BC and CA. To prove ΔABC is divided into four congruent triangles. Proof Since, ABC is a triangle and D, E and F are the mid-points of sides AB, BC and CA, respectively. Then, $\quad A D=B D=\frac{1}{2} A B, B E=E C=\fra...
Read More →Let A = [aij] be a square matrix of order 3 × 3 and Cij
Question: Let $A=\left[a_{i j}\right]$ be a square matrix of order $3 \times 3$ and $C_{i j}$ denote cofactor of $a_{i j}$ in $A$. If $|A|=5$, write the value of $a_{31} C_{31}+a_{32} C_{32} a_{33} C_{33}$. Solution: If $A=\left[\mathrm{a}_{\mathrm{i}}\right]$ is a square matrix of order $n$ and $\mathrm{C}_{\mathrm{i} \mathrm{j}}$ is a cofactor of $\mathrm{a}_{\mathrm{i} \mathrm{j}}$, then $\sum_{i=1}^{n} a_{i j} C_{i j}=|A|$ and $\sum_{j=1}^{n} a_{i j} C_{i j}=|A|$ Given: $|\mathrm{A}|=5$ and ...
Read More →In the given figure, a circle is inscribed in an equilateral triangle ABC of side 12 cm.
Question: In the given figure, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region. $[$ Use $\sqrt{3}=1.73, \pi=3.14]$ Solution: We can find the radius of the incircle by using the formula $r=\frac{2 \times \text { Area of triangle }}{\text { Perimeter of triangle }}=\frac{2 \times \frac{\sqrt{3}}{4} \times(12)^{2}}{3 \times 12}=2 \sqrt{3} \mathrm{~cm}$ Now, area of shaded region = Area of triangle Area of circ...
Read More →On expanding by first row, the value of the determinant of 3 × 3 square matrix
Question: On expanding by first row, the value of the determinant of 3 3 square matrix $A=\left[a_{i j}\right]$ is $a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13}$, where $\left[C_{i j}\right]$ is the cofactor of $a_{i j}$ in $A$. Write the expression for its value on expanding by second column. Solution: If $A=\left[\mathrm{a}_{\mathrm{i}}\right]$ is a square matrix of order $n$, then the sum of the products of elements of a row (or a column) with their cofactors is always equal to det (A). Therefor...
Read More →On expanding by first row, the value of the determinant of 3 × 3 square matrix
Question: On expanding by first row, the value of the determinant of 3 3 square matrix $A=\left[a_{i j}\right]$ is $a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13}$, where $\left[C_{i j}\right]$ is the cofactor of $a_{i j}$ in $A$. Write the expression for its value on expanding by second column. Solution: If $A=\left[\mathrm{a}_{\mathrm{i}}\right]$ is a square matrix of order $n$, then the sum of the products of elements of a row (or a column) with their cofactors is always equal to det (A). Therefor...
Read More →In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°.
Question: In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and BOD = 90. Find the area of shaded region.[Use = 3.14] Solution: In right triangle ABCBC2= AB2+ AC2= (7)2+ (24)2= 49 + 576= 625 BC2= 625⇒ BC = 25Now, COD + BOD = 180 (Linear pair angles)⇒COD = 180 90 = 90Now, Area of the shaded region = Area of sector having central angle (360 90) Area of triangle ABC $=\frac{270^{\circ}}{360^{\circ}} \pi\left(\frac{\mathrm{BC}}{2}\right)^{2}-\frac{1}{2} \mathrm{AB} \times...
Read More →On expanding by first row, the value of the determinant of 3 × 3 square matrix
Question: On expanding by first row, the value of the determinant of 3 3 square matrixA=[aij]isa11C11+a12C12+a13C13A=aijisa11C11+a12C12+a13C13, where [Cij] is the cofactor ofaijinA. Write the expression for its value on expanding by second column. Solution: IfA=[aij]A=aijis a square matrix of ordern, then the sum of the products of elements of a row (or a column) with their cofactors is always equal to det (A). Therefore, $\sum_{i=1}^{n} a_{i j} C_{i j}=|A|$ and $\sum_{j=1}^{n} a_{i j} C_{i j}=|...
Read More →In the given figure, from a rectangular region ABCD with AB
Question: In the given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside' the region. Find the area of the shaded region. [Use = 3.14] Solution: In right triangle AEDAD2= AE2+ DE2= (9)2+ (12)2= 81 + 144= 225 AD2= 225⇒ AD = 15 cmWe know that the opposite sides of a rectangle are equalAD = BC = 15 cm= Area of the shaded region = Area of rectangle Area of t...
Read More →A matrix A of order 3 × 3 has determinant 5.
Question: A matrix $A$ of order $3 \times 3$ has determinant 5 . What is the value of $|3 A| ?$ Solution: If $A=\left[a_{i j}\right]$ is a square matrix of order $n$ and $k$ is a constant, then $|\mathrm{kA}|=\mathrm{k}^{\mathrm{n}}|\mathrm{A}|$ Here, Number of rows $=n$ $k$ is a common factor from each row of $k$ $|3 \mathrm{~A}|=3^{3}|\mathrm{~A}|=27 \times 5=135$ [Given matrix is $3 \times 3$ such that $|\mathrm{A}|=5$ ] Thus, $|3 \mathrm{~A}|=135$...
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