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 →ABCD is a rectangle in which diagonal
Question: ABCD is a rectangle in which diagonal BD bisects B. Show that ABCD is a square. Solution: Given In a rectangle $A B C D$, diagonal $B D$ bisects $\angle B$. Construct Join AC. To show $A B C D$ is a square. Proof In $\triangle B A D$ and $\triangle B C D$, $\angle A B D=\angle C B D$ [given] $\angle A=\angle C$ $\left[\right.$ each $\left.90^{\circ}\right]$ and $B D=B D$ [common side] $\therefore \quad \triangle B A D \cong \triangle B C D \quad$ [by AAS congruence rule] $\therefore \q...
Read More →In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm,
Question: In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, AED = 90 and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region. Solution: We know that the opposite sides of a rectangle are equalAD = BC = 70 cmIn right triangle AEDAE2= AD2 DE2= (70)2 (42)2= 4900 1764= 3136 AE2= 3136⇒ AE = 56= Area of the shaded region = Area of rectangle (Area of triangle AED + Area of semicircle) $=\mathrm{AB} \times \mathrm{BC}-\left[\frac{1}{2} \tim...
Read More →If I3 denotes identity matrix of order 3 × 3,
Question: If $I_{3}$ denotes identity matrix of order $3 \times 3$, write the value of its determinant. Solution: In an identity matrix, all the diagonal elements are 1 and rest of the elements are 0.Here, $I_{3}=\mid \begin{array}{lll}1 0 0\end{array}$ $\begin{array}{lll}0 1 0\end{array}$ $\begin{array}{lll}0 0 1\end{array}$ $=1 \times \mid \begin{array}{ll}1 0\end{array}$ $\begin{array}{ll}0 1\end{array} \quad\left[\right.$ Expanding along $\left.\mathrm{C}_{1}\right]$ $=1$ $\Rightarrow \mathr...
Read More →Find the area of the shaded region in the given figure,
Question: Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made.$[$ Use $\sqrt{3}=1.73, \pi=3.14]$ Solution: In equilateral traingle all the angles are of 60 ABO = AOB = 60Area of the shaded region = (Area of triangle AOB Area of sector having central angle 60) + Area of sector having central angle (360 60) $=\frac{\sqrt{3...
Read More →P and O are points on opposite sides AD and BC
Question: P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O. Thinking Process Firstly, prove that ΔODP and ΔOBQ are congruent by ASA rule. Further show the required result by CPCT rule Solution: Given $A B C D$ is a parallelogram whose diagonals bisect each other at $O$. To show $P Q$ is bisected at $O$. In $\triangle O D P$ and $\triangle O B Q$, $\angle B O Q=\...
Read More →Solve the following
Question: Divide $9 x^{2} y-6 x y+12 x y^{2}$ by $-\frac{3}{2} x y$ Solution:...
Read More →If A = [aij] is a 3 × 3 scalar matrix
Question: If $A=\left[a_{i j}\right]$ is a $3 \times 3$ scalar matrix such that $a_{11}=2$, then write the value of $|A|$. Solution: A scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number. Given: $\mathrm{A}=\left[a_{i j}\right]$ is $3 \times 3$ matrix, where $a_{11}=2$ $\Rightarrow \mathrm{A}=\left[\begin{array}{lll}2 0 0\end{array}\right.$ $\begin{array}{lll}0 2 0\end{array}$ $\left.\begin{array}{lll}0 0 2\end{array}\right]$ $\Rightarrow|\ma...
Read More →ABCD is a field in the shape of a trapezium,
Question: ABCD is a field in the shape of a trapezium, AD || BC, ABC = 90 and ADC = 60. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m. Find the following: (i) total area of the four sectors, (ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m. Solution: (i) Area of fours sector = Area of sector having central angle 60 + Area of sector having central angle 90 + Area of sector having central angle 90 + Ar...
Read More →If A = [aij] is a 3 × 3 scalar matrix
Question: If $A=\left[a_{i j}\right]$ is a $3 \times 3$ scalar matrix such that $a_{11}=2$, then write the value of $|A|$. Solution: A scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number. Given: $\mathrm{A}=\left[a_{i j}\right]$ is $3 \times 3$ matrix, where $a_{11}=2$ $\Rightarrow \mathrm{A}=\left[\begin{array}{lll}2 0 0\end{array}\right.$ $\begin{array}{lll}0 2 0\end{array}$ $\left.\begin{array}{lll}0 0 2\end{array}\right]$ $\Rightarrow|\ma...
Read More →ABCD is a field in the shape of a trapezium,
Question: ABCD is a field in the shape of a trapezium, AD || BC, ABC = 90 and ADC = 60. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m. Find the following: (i) total area of the four sectors, (ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m. Solution: (i) Area of fours sector = Area of sector having central angle 60 + Area of sector having central angle 90 + Area of sector having central angle 90 + Ar...
Read More →If A = [aij] is a 3 × 3 scalar matrix
Question: If $A=\left[a_{i j}\right]$ is a $3 \times 3$ scalar matrix such that $a_{11}=2$, then write the value of $|A|$. Solution: A scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number. Given: $\mathrm{A}=\left[a_{i j}\right]$ is $3 \times 3$ matrix, where $a_{11}=2$ $\Rightarrow \mathrm{A}=\left[\begin{array}{lll}2 0 0\end{array}\right.$ $\begin{array}{lll}0 2 0\end{array}$ $\left.\begin{array}{lll}0 0 2\end{array}\right]$ $\Rightarrow|\ma...
Read More →Prove that the quadrilateral formed
Question: Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. Solution: Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of A, B, C and D, respectively. To prove Quadrilateral PQRS is a rectangle. Proof Since, ABCD is a parallelogram, then DC || AB and DA is a transversal. A+D= 180 [sum of cointerior angles of a parallelogram is 180] = A+ D = 90 [dividing both sides by 2] PAD + PDA = 90 APD = 90 [since,sum of all angles of a...
Read More →In the given figure, ABCD is a trapezium of area 24.5 cm2 ,
Question: In the given figure, ABCD is a trapezium of area 24.5 cm2, If AD || BC, DAB = 90, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region. Solution: Area of trapezium $=\frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AB}$ $\Rightarrow 24.5=\frac{1}{2}(10+4) \times \mathrm{AB}$ $\Rightarrow \mathrm{AB}=3.5 \mathrm{~cm}$ Area of shaded region = Area of trapezium ABCD Area of quadrant ABE $=24.5-\frac{1}{4} \pi(\mathrm{AB})^{2}$ $=24.5-\frac{1}{4...
Read More →E and F are respectively the mid-points
Question: E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = (AB + CD). Solution: Given $A B C D$ is a trapezium in which $A B \| C D$. Also, $E$ and $F$ are respectively the mid-points of sides $A D$ and $B C$. Construction Join $B E$ and produce it to meet $C D$ produced at $G$, also draw $B D$ which intersects $E F$ at $O$. To prove $E F \| A B$ and $E F=\frac{1}{2}(A B+C D)$. Proof In $\Delta G C B, E$ and $F$ are res...
Read More →In the given figure, ABCD is a trapezium of area 24.5 cm2 ,
Question: In the given figure, ABCD is a trapezium of area 24.5 cm2, If AD || BC, DAB = 90, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region. Solution: Area of trapezium $=\frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AB}$ $\Rightarrow 24.5=\frac{1}{2}(10+4) \times \mathrm{AB}$ $\Rightarrow \mathrm{AB}=3.5 \mathrm{~cm}$ Area of shaded region = Area of trapezium ABCD Area of quadrant ABE $=24.5-\frac{1}{4} \pi(\mathrm{AB})^{2}$ $=24.5-\frac{1}{4...
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