In a circle of radius 7 cm, a square ABCD is inscribed.
Question: In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the circle which is outside the square. Solution: Let the diagonal of the square bed.We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square. d= 2 ⨯ 7 = 14 cmNow,Area of required region = Area of circle Area of square $=\pi r^{2}-\frac{1}{2} d^{2}$ $=\frac{22}{7} \times(7)^{2}-\frac{1}{2} \times(14)^{2}$ $=56 \mathrm{~cm}^{2}$ Hence, the required area...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $6.5 x+\frac{19.5 x-32.5}{2}=6.5 x+13+\left(\frac{13 x-26}{2}\right)$ Solution: $6.5 \mathrm{x}+\frac{19.5 \mathrm{x}-32.5}{2}=6.5 \mathrm{x}+13+\frac{13 \mathrm{x}-26}{2}$ or $\frac{19.5 x-32.5}{2}-\frac{13 x-26}{2}=13$ or $\frac{19.5 \mathrm{x}-32.5-13 \mathrm{x}+26}{2}=13$ or $6.5 \mathrm{x}-6.5=26[$ After $c$ ross multiplication $]$ or $6.5 \mathrm{x}=26+6.5$ or $\mathrm{x}=\frac{32.5}{6.5}=5$ Thus, $x=5$...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $6.5 x+\frac{19.5 x-32.5}{2}=6.5 x+13+\left(\frac{13 x-26}{2}\right)$ Solution: $6.5 \mathrm{x}+\frac{19.5 \mathrm{x}-32.5}{2}=6.5 \mathrm{x}+13+\frac{13 \mathrm{x}-26}{2}$ or $\frac{19.5 x-32.5}{2}-\frac{13 x-26}{2}=13$ or $\frac{19.5 \mathrm{x}-32.5-13 \mathrm{x}+26}{2}=13$ or $6.5 \mathrm{x}-6.5=26[$ After $c$ ross multiplication $]$ or $6.5 \mathrm{x}=26+6.5$ or $\mathrm{x}=\frac{32.5}{6.5}=5$ Thus, $x=5$...
Read More →Find the perimeter of the shaded region in the figure,
Question: Find the perimeter of the shaded region in the figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. Solution: Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC $=\frac{1}{2} \times 2 \pi r+\frac{1}{2} \times 2 \pi r+14+14$ $=2 \pi r+28$ $=2 \times \frac{22}{7} \times \frac{14}{2}+28$ $=72 \mathrm{~cm}$ Hence, the perimeter of the shaded region is 72 cm....
Read More →Find the perimeter of the shaded region in the figure,
Question: Find the perimeter of the shaded region in the figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. Solution: Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC $=\frac{1}{2} \times 2 \pi r+\frac{1}{2} \times 2 \pi r+14+14$ $=2 \pi r+28$ $=2 \times \frac{22}{7} \times \frac{14}{2}+28$ $=72 \mathrm{~cm}$ Hence, the perimeter of the shaded region is 72 cm....
Read More →AD is a diameter of a circle and AB is a chord.
Question: AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is (a)17 cm (b)15 cm (c)4 cm (d)8 cm Solution: (d)Given, AD = 34 cm and AB = 30 cm In figure, draw OL AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord. $\therefore$$A L=L B=\frac{1}{2} A B=15 \mathrm{~cm}$ In right angled $\triangle O L A, \quad O A^{2}=O L^{2}+A L^{2}$ [by Pythagoras theorem] $\therefore$ $(17)^{2}=O L^{2}+(15...
Read More →In the given figure, OABC is a quadrant of a circle of radius 3.5 cm with centre O.
Question: In the given figure,OABCis a quadrant of a circle of radius 3.5 cm with centreO. If OD = 2 cm, find the area of the shaded portion. Solution: Area of the right-angled $\Delta \mathrm{COD}=\frac{1}{2} \times b \times h$ $=\frac{1}{2} \times 3.5 \times 2=3.5 \mathrm{~cm}^{2}$ Area of the sector $\mathrm{AOC}=\frac{\theta}{360} \times \pi \times r^{2}$ $=\frac{90}{360} \times \frac{22}{7} \times(3.5)^{2}$ $=9.625 \mathrm{~cm}^{2}$ Area of the shaded reaion = Area of the $\triangle \mathrm...
Read More →find the problem
Question: Find $A(\operatorname{adj} A)$ for the matrix $A=\left[\begin{array}{ccc}1 -2 3 \\ 0 2 -1 \\ -4 5 2\end{array}\right]$ Solution: $A=\left[\begin{array}{ccc}1 -2 3 \\ 0 2 -1 \\ -4 5 2\end{array}\right]$ Now, $C_{11}=\left|\begin{array}{cc}2 -1 \\ 5 2\end{array}\right|=9, C_{12}=-\left|\begin{array}{cc}0 -1 \\ -4 2\end{array}\right|=4$ and $C_{13}=\left|\begin{array}{cc}0 2 \\ -4 5\end{array}\right|=8$ $C_{21}=-\left|\begin{array}{cc}-2 3 \\ 5 2\end{array}\right|=19, C_{22}=\left|\begin{...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{0.5(x-0.4)}{0.35}-\frac{0.6(x-2.71)}{0.42}=x+6.1$ Solution: $\frac{0.5(x-0.4)}{0.35}-\frac{0.6(x-2.71)}{0.42}=x+6.1$ or $\frac{(\mathrm{x}-0.4)}{0.7}-\frac{(\mathrm{x}-2.71)}{0.7}=\mathrm{x}+6.1$ or $\frac{\mathrm{x}-0.4-\mathrm{x}+2.71}{0.7}=\mathrm{x}+6.1$ or $-0.4+2.71=0.7 x+4.27$ or $0.7 \mathrm{x}=2.71-0.4-4.27$ or $\mathrm{x}=\frac{-1.96}{07}=-2.8$ Thus, $x=-2.8$ is the solution of the given equa...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{7 x-1}{4}-\frac{1}{3}\left(2 x-\frac{1-x}{2}\right)=\frac{10}{3}$ Solution: $\frac{7 \mathrm{x}-1}{4}-\frac{1}{3}\left(2 \mathrm{x}-\frac{1-\mathrm{x}}{2}\right)=\frac{10}{3}$ or $\frac{7 \mathrm{x}-1}{4}-\frac{2 \mathrm{x}}{3}+\frac{1-\mathrm{x}}{6}=\frac{10}{3}$ or $\frac{21 \mathrm{x}-3-8 \mathrm{x}+2-2 \mathrm{x}}{12}=\frac{10}{3}$ or $11 \mathrm{x}-1=40[$ Multiply ing both sides by 12$]$ or $11 \m...
Read More →Solve this
Question: If $A=\left[\begin{array}{ccc}-1 -2 -2 \\ 2 1 -2 \\ 2 -2 1\end{array}\right]$, show that adj $A=3 A^{\top}$. Solution: $A=\left[\begin{array}{ccc}-1 -2 -2 \\ 2 1 -2 \\ 2 -2 1\end{array}\right]$ Now, $C_{11}=\left|\begin{array}{cc}1 -2 \\ -2 1\end{array}\right|=-3, C_{12}=-\left|\begin{array}{cc}2 -2 \\ 2 1\end{array}\right|=-6$ and $C_{13}=\left|\begin{array}{cc}2 1 \\ 2 -2\end{array}\right|=-6$ $C_{21}=-\left|\begin{array}{cc}-2 -2 \\ -2 1\end{array}\right|=6, C_{22}=\left|\begin{arra...
Read More →In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm.
Question: In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm. Find the area of shaded region. Solution: Area of the shaded portion = (Area of quadrant DPBA + Area of quadrant DQBC) Area of Square ABCD $=\left[\frac{1}{4} \pi(7)^{2}+\frac{1}{4} \pi(7)^{2}\right]-(7)^{2}$ $=\frac{1}{2} \times \frac{22}{7}(7)^{2}-49$ $=28 \mathrm{~cm}^{2}$ Hence, the area of the shaded portion is 28 cm2....
Read More →In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm.
Question: In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm. Find the area of shaded region. Solution: Area of the shaded portion = (Area of quadrant DPBA + Area of quadrant DQBC) Area of Square ABCD $=\left[\frac{1}{4} \pi(7)^{2}+\frac{1}{4} \pi(7)^{2}\right]-(7)^{2}$ $=\frac{1}{2} \times \frac{22}{7}(7)^{2}-49$ $=28 \mathrm{~cm}^{2}$ Hence, the area of the shaded portion is 28 cm2....
Read More →In figure, ABCD and AEFD are two
Question: In figure, ABCD and AEFD are two parallelograms. Prove that ar (APEA) = ar(AQFO). Solution: Given, ABCD and AEFD are two parallelograms. To prove ar (APEA) = ar (AQFD) Proof In quadrilateral PQDA, AP || DQ [since, in parallelogram ABCD, AB || CD ] and PQ || AD [since, in parallelogram AEFD, FE || AD] Then, quadrilateral PQDA is a parallelogram. Also, parallelogram PQDA and AEFD are on the same base AD and between the same parallels AD and EQ. ar (parallelogram PQDA) = ar (parallelogram...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $5\left(\frac{7 x+5}{3}\right)-\frac{23}{3}=13-\frac{4 x-2}{3}$ Solution: $5\left(\frac{7 x+5}{3}\right)-\frac{23}{3}=13-\frac{4 x-2}{3}$ or $\frac{35 x+25}{3}+\frac{4 x-2}{3}=13+\frac{23}{3}$ or $\frac{35 x+25+4 x-2}{3}=\frac{39+23}{3}$ or $39 \mathrm{x}+23=62[$ Multiply ing both sides by 3$]$ or $39 \mathrm{x}=62-23$ or $\mathrm{x}=\frac{39}{39}$ or $\mathrm{x}=1$ Thus, $x=1$ is the solution of the given eq...
Read More →Solve this
Question: If $A=\left[\begin{array}{ccc}-4 -3 -3 \\ 1 0 1 \\ 4 4 3\end{array}\right]$, show that adj $A=A$ Solution: $A=\left[\begin{array}{ccc}-4 -3 -3 \\ 1 0 1 \\ 4 4 3\end{array}\right]$ Now, $C_{11}=\left|\begin{array}{ll}0 1 \\ 4 3\end{array}\right|=-4, C_{12}=-\left|\begin{array}{ll}1 1 \\ 4 3\end{array}\right|=1$ and $C_{13}=\left|\begin{array}{ll}1 0 \\ 4 4\end{array}\right|=4$ $C_{21}=-\left|\begin{array}{cc}-3 -3 \\ 4 3\end{array}\right|=-3, C_{22}=\left|\begin{array}{cc}-4 -3 \\ 4 3\e...
Read More →In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and
Question: In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and AOB = 90. If AO = OB = 42 cm, then find the perimeter of the top of the table. Solution: We have $r=42 \mathrm{~cm}$ and $\theta=360^{\circ}-90^{\circ}=270^{\circ}$ Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB $=\frac{\theta}{360^{\circ}} \times 2 \pi r+42+42$ $=\frac{270^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 42+84$ $...
Read More →In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and
Question: In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and AOB = 90. If AO = OB = 42 cm, then find the perimeter of the top of the table. Solution: We have $r=42 \mathrm{~cm}$ and $\theta=360^{\circ}-90^{\circ}=270^{\circ}$ Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB $=\frac{\theta}{360^{\circ}} \times 2 \pi r+42+42$ $=\frac{270^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 42+84$ $...
Read More →In figure X and Y are the mid-points of AC
Question: In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ΔABP) = ar (ΔACQ). Thinking Process Firstly, use the theorem that joining the mid-points of the two sides of a triangle is parallel to the third side. Further, use the theorem that triangles on the same base and between the sameparallels are equal in area. Use this theorem by considering different triangles and prove the required result Solution: Given X and Y are...
Read More →In the given figure, find the area of the shaded region,
Question: In the given figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC aresemicircles. Solution: Area of the shaded region = Area of Square ABCD (Area of semicircle APD + Area of semicircle BPC) $=(14)^{2}-\left[\frac{1}{2} \pi\left(\frac{14}{2}\right)^{2}+\frac{1}{2} \pi\left(\frac{14}{2}\right)^{2}\right]$ $=(14)^{2}-\frac{22}{7}(7)^{2}$ $=42 \mathrm{~cm}^{2}$ Hence, the area of the shaded region is 42 cm2....
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{(45-2 x)}{15}-\frac{(4 x+10)}{5}=\frac{(15-14 x)}{9}$ Solution: $\frac{45-2 \mathrm{x}}{15}-\frac{4 \mathrm{x}+10}{5}=\frac{15-14 \mathrm{x}}{9}$ or $\frac{45-2 \mathrm{x}-12 \mathrm{x}-30}{15}=\frac{15-14 \mathrm{x}}{9}$ or $\frac{15-14 \mathrm{x}}{5}=\frac{15-14 \mathrm{x}}{3}[$ Multiply ing both sides by 3$]$ or $45-42 \mathrm{x}=75-70 \mathrm{x}[$ After c ross multiplication $]$ or $70 \mathrm{x}-4...
Read More →If the medians of a AABC intersect at G,
Question: If the medians of a AABC intersect at G, then show that ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = 1/3ar(ΔABC). Thinking Process Use the property that median of a triangle divides it into two triangles of equal area. Further, apply above property by considering different triangles and prove the required result. Solution: Given In ΔABC, AD, BE and CF are medians and intersect at G. To prove ar $(\Delta A G B)=\operatorname{ar}(\Delta A G C)=\operatorname{ar}(\Delta B G C)=\frac{1}{3} \operator...
Read More →For the matrix
Question: For the matrix $A=\left[\begin{array}{ccc}1 -1 1 \\ 2 3 0 \\ 18 2 10\end{array}\right]$, show that $A(\operatorname{adj} A)=0$. Solution: $A=\left[\begin{array}{ccc}1 -1 1 \\ 2 3 0 \\ 18 2 10\end{array}\right]$ Now, $C_{11}=\left|\begin{array}{cc}3 0 \\ 2 10\end{array}\right|=30 \quad C_{12}=-\left|\begin{array}{cc}2 0 \\ 18 10\end{array}\right|=-20 \quad C_{13}=\left|\begin{array}{cc}2 3 \\ 18 2\end{array}\right|=-50$ $C_{21}=-\left|\begin{array}{cc}-1 1 \\ 2 10\end{array}\right|=12 \...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{4 x}{9}+\frac{1}{3}+\frac{13}{108} x=\frac{8 x+19}{18}$ Solution: $\frac{4 \mathrm{x}}{9}+\frac{1}{3}+\frac{13}{108} \mathrm{x}=\frac{8 \mathrm{x}+19}{18}$ or $\frac{48 \mathrm{x}+36+13 \mathrm{x}}{108}=\frac{8 \mathrm{x}+19}{18}$ or $\frac{61 \mathrm{x}+36}{108}=\frac{8 \mathrm{x}+19}{18}$ or $61 \mathrm{x}+36=6(8 \mathrm{x}+19)$ [Multiply ing both $s$ ides by 108$]$ or $61 \mathrm{x}+36=48 \mathrm{x}...
Read More →In the given figure, find the area of the shaded region,
Question: In the given figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC aresemicircles. Solution: Area of the shaded region = Area of Square ABCD (Area of semicircle APD + Area of semicircle BPC) $=(14)^{2}-\left[\frac{1}{2} \pi\left(\frac{14}{2}\right)^{2}+\frac{1}{2} \pi\left(\frac{14}{2}\right)^{2}\right]$ $=(14)^{2}-\frac{22}{7}(7)^{2}$ $=42 \mathrm{~cm}^{2}$ Hence, the area of the shaded region is 42 cm2....
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