In the given figure, PQ = 24, PR = 7 cm and O is the centre of the circle.
Question: In the given figure,PQ= 24,PR= 7 cm andOis the centre of the circle. Find the area of the shaded region. Solution: In the rightΔRPQ, we have: $R Q=\sqrt{R P^{2}+P Q^{2}}$ $=\sqrt{7^{2}+24^{2}}$ $=\sqrt{49+576}$ $=25 \mathrm{~cm}$ OR=OQ= 12.5 cmNow, Area of the circle $=\pi r^{2}$ $=3.14 \times 12.5 \times 12.5$ $=490.625$ sq. $\mathrm{cm}$ Area of the semicircle $=\frac{490.625}{2}=245.31$ sq. $\mathrm{cm}$ Area of the triangle $=\frac{1}{2} \times b \times h=\frac{1}{2} \times 7 \time...
Read More →The figure obtained by joining the mid-points
Question: The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is (a)a rectangle of area 24 cm2 (b)a square of area 25 cm2 (c)a trapezium of area 24 cm2 (d)a rhombus of area 24 cm2 Solution: (d) Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus. Here, $\quad$ length of rectangle $A B C D=8 \mathrm{~cm}$ and breadth of rectangle...
Read More →Write the value of the determinant
Question: Write the value of the determinant $\left|\begin{array}{ccc}2 -3 5 \\ 4 -6 10 \\ 6 -9 15\end{array}\right|$. Solution: $=\mid 2 \quad-3 \quad 5$ $4-4-6+6 \quad 10-10$ $\begin{array}{lll}6 -9 15 \mid\end{array}$ [Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ ] $=\mid \begin{array}{lll}2 -3 5\end{array}$ $\begin{array}{lll}0 0 0\end{array}$ $\begin{array}{|lll|}6 -9 15\end{array}$ $=0$...
Read More →A round table cover has six equal designs as shown in the given figure.
Question: A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35 cm, then find the total area of the design. Solution: Join each vertex of the hexagon to the centre of the circle.The hexagon is made up of six triangles. Total area of the design = Area of the circle $-$ Area of six triangles $=\left[\frac{22}{7} \times 35 \times 35\right]-\left[6 \times\left(\frac{\sqrt{3}}{4} \times 35 \times 35\right)\right]$ $=(35 \times 35)\left[\frac{22}{7}-\...
Read More →In which of the following figures,
Question: In which of the following figures, you find two polygons on the same base and between the same parallels? Solution: (d)In figures (a), (b) and (c) there are two polygons on the same base but they are not between the same parallels. In figure (d), there are two polygons (PQRA and BQRS) on the same base and between the same parallels ....
Read More →A circular disc of radius 6 cm is divided into three sectors with central angles 90°,
Question: A circular disc of radius 6 cm is divided into three sectors with central angles 90,120 and 150. What part of the whole circle is the sector with central angle 150? Also, calculate the ratio of the areas of the three sectors. Solution: Area of sector having central angle $150^{\circ}==\frac{150^{\circ}}{360^{\circ}} \pi(6)^{2}=\frac{5}{12} \times$ Area of circular disc Now, Area of sector having central angle 90 : Area of sector having central angle 120 : Area of sector having central ...
Read More →Find the value of the determinant
Question: Find the value of the determinant $\left|\begin{array}{ccc}243 156 300 \\ 81 52 100 \\ -3 0 4\end{array}\right|$. Solution: $\begin{array}{ccc}\mid 243 156 300 \\ 81 52 100 \\ -3 0 4 \mid\end{array}$ $=\mid 243-(81 \times 3) 156-(52 \times 3) 300-(100 \times 3) 81$ $\begin{array}{llll}52 100-3 0 4 \mid\end{array}$ $\left[\right.$ Applying $\left.R_{1} \rightarrow R_{1}-3 R_{2}\right]$ $=\mid \begin{array}{lll}0 0 0\end{array}$ $\begin{array}{lll}81 52 100\end{array}$ $\begin{array}{lll...
Read More →The median of a triangle divides it into two
Question: The median of a triangle divides it into two (a)triangles of equal area (b)congruent triangles (c)right angled triangles (d)isosceles triangles Solution: (a)We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle divides it into two triangles of equal area....
Read More →A child draws the figure of an aeroplane as shown.
Question: A child draws the figure of an aeroplane as shown. Here, the wingsABCDandFGHIare parallelograms, the tailDEFis an isosceles triangle, the cockpitCKIis a semicircle andCDFIis a square. In the given figure,BPCD,HQFIandELDF.IfCD= 8 cm,BP=HQ= 4 cm andDE=EF= 5 cm, find the area of the whole figure. Solution: CD = 8 cmBP = HQ = 4 cmDE = EF = 5 cm Area of the parallelogram $\mathrm{ABCD}=B \times H$ $=8 \times 4$ $=32$ sq. $\mathrm{cm}$ Area of parallelogram $\mathrm{FGHI}=B \times H$ $=8 \ti...
Read More →If A is a square matrix such that
Question: If $A$ is a square matrix such that $|A|=2$, write the value of $\left|A A^{\top}\right|$. Solution: In a square matrix, $|\mathrm{A}|=\left|\mathrm{A}^{\mathrm{T}}\right|$. Since they are of same order, $\left|\mathrm{A} \mathrm{A}^{\mathrm{T}}\right|=|\mathrm{A}|\left|\mathrm{A}^{\mathrm{T}}\right|$. Given : $|\mathbf{A}|=2$ $\Rightarrow\left|\mathrm{A} \mathrm{A}^{\mathrm{T}}\right|=2 \times 2=4$...
Read More →Divide the following and find the quotient and remainder.
Question: Divide 9x4 4x2+ 4 by 3x2 4x+ 2 and find the quotient and remainder. Solution:...
Read More →P is the mid-point of the side CD of a parallelogram ABCD.
Question: P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR. Solution: Given in a parallelogram $A B C D, P$ is the mid-point of $D C$. To prove $D A=A R$ and $C Q=Q R$ Proof $A B C D$ is a parallelogram. $\therefore$ $B C=A D$ and $B C \| A D$ Also, $D C=A B$ and $D C \| A B$ Since, $P$ is the mid-point of $D C$. $\therefore$ $D P=P C=\frac{1}{2} D C$ Now, $Q C \| A P$ and $P C \| A ...
Read More →The area of an equilateral triangle is
Question: The area of an equilateral triangle is $49 \sqrt{3} \mathrm{~cm}^{2}$. Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of thetriangle. Find the area of the triangle not included in the circles. [Take = 1.73] Solution: Area of equilateral triangle $=49 \sqrt{3}$ $\Rightarrow \frac{\sqrt{3}}{4} \times(\text { Side })^{2}=49 \sqrt{3}$ $\Rightarrow(\text { Side })^{2}=7^{2} \times 2^{2}$ $\Rightarrow$ Side $=14 \mathrm{~cm}$ Radius of...
Read More →If A is a square matrix of order 3 with determinant 4, then write the value of |−A|.
Question: IfAis a square matrix of order 3 with determinant 4, then write the value of |A|. Solution: $|A|=4$ Here, Order of the matrix $(n)=3$ Using properties of matrices, we get $|\mathrm{kA}|=\mathrm{k}^{\mathrm{n}}|\mathrm{A}| \quad[$ For a square matrix of order $n$ and constant $k]$ $\Rightarrow|-\mathrm{A}|=(-1)^{3}|\mathrm{~A}|=(-1) \times 4=-4$...
Read More →Divide the following and find the quotient and remainder.
Question: Divide 30x4+ 11x3 82x2 12x+ 48 by 3x2+ 2x 4 and find the quotient and remainder. Solution:...
Read More →A is a skew-symmetric of order 3, write the value of |A|.
Question: Ais a skew-symmetric of order 3, write the value of |A|. Solution: We know that if a skew symmetric matrix $A$ is of odd order, then $|A|=0$ Since the order of the given matrix is $3,|A|=0$....
Read More →A is a skew-symmetric of order 3, write the value of |A|.
Question: Ais a skew-symmetric of order 3, write the value of |A|. Solution: We know that if a skew symmetric matrix $A$ is of odd order, then $|A|=0$ Since the order of the given matrix is $3,|A|=0$....
Read More →Divide following and find the quotient and remainder.
Question: Divide 6x3+ 11x2 39x 65 by 3x2+ 13x+ 13 and find the quotient and remainder. Solution: Quotient $=2 x-5$ Remainder $=0$...
Read More →The radius of a circular garden is 100 m.
Question: The radius of a circular garden is 100 m. There is a road 10 m wide, running all around it. Find the area of the road and the cost of levellingit at Rs 20 per m2. [Use = 3.14] Solution: Area of the road = Area of outer circle Area of inner circle $=\pi R^{2}-\pi r^{2}$ $=\pi\left[(110)^{2}-(100)^{2}\right]$ $=3.14 \times 2100$ $=6594 \mathrm{~m}^{2}$ Cost of levelling the road = Area of the road ⨯ Rate= 6594 ⨯ 20= Rs 131880...
Read More →If A and B are square matrices of the same order
Question: If $A$ and $B$ are square matrices of the same order such that $|A|=3$ and $A B=l$, then write the value of $|B|$. Solution: SinceABare square matrices of the same order, by the property of determinants we get $|A B|=|A| \times|B|$ $|A|=3, A B=I$ $\Rightarrow|A B|=1$ $\Rightarrow|A| \times|B|=1$ $3 \times|B|=1$ $|B|=\frac{1}{3}$...
Read More →If A and B are square matrices of the same order
Question: If $A$ and $B$ are square matrices of the same order such that $|A|=3$ and $A B=l$, then write the value of $|B|$. Solution: SinceABare square matrices of the same order, by the property of determinants we get $|A B|=|A| \times|B|$ $|A|=3, A B=I$ $\Rightarrow|A B|=1$ $\Rightarrow|A| \times|B|=1$ $3 \times|B|=1$ $|B|=\frac{1}{3}$...
Read More →A chord of a circle of radius 10 cm subtends a right angle at the centre.
Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use = 3.14.] Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{\theta}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{90^{\circ}}{360^{\circ}} \times 3.14(10)^{2}-\frac{1}{2} \times 10 \times 10$ $=78.5-50$ $=28.5 \mathrm{~cm}^{2}$ Hence, the area of minor segment is 28.5 cm2...
Read More →Divide the following 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 →A chord of a circle of radius 10 cm subtends a right angle at the centre.
Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use = 3.14.] Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{\theta}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{90^{\circ}}{360^{\circ}} \times 3.14(10)^{2}-\frac{1}{2} \times 10 \times 10$ $=78.5-50$ $=28.5 \mathrm{~cm}^{2}$ Hence, the area of minor segment is 28.5 cm2...
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...
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