Compute the adjoint of each of the following matrices:
Question: Compute the adjoint of each of the following matrices: (i) $\left[\begin{array}{lll}1 2 2 \\ 2 1 2 \\ 2 2 1\end{array}\right]$ (ii) $\left[\begin{array}{ccc}1 2 5 \\ 2 3 1 \\ -1 1 1\end{array}\right]$ (iii) $\left[\begin{array}{ccc}2 -1 3 \\ 4 2 5 \\ 0 4 -1\end{array}\right]$ (iv) $\left[\begin{array}{ccc}2 0 -1 \\ 5 1 0 \\ 1 1 3\end{array}\right]$ Verify that $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$ for the above matrices. Solution: $(\mathrm{i}) A=\left[\begin{array}{...
Read More →In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm
Question: In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm with centre O. If POQ = 30, find the area of the shaded region. Solution: Area of the shaded portion = Area of sector OPQ Area of sector OAB $=\frac{30^{\circ}}{360^{\circ}} \times \pi(7)^{2}-\frac{30^{\circ}}{360^{\circ}} \times \pi(3.5)^{2}$ $=\frac{22}{7} \times \frac{1}{12}\left[(7)^{2}-(3.5)^{2}\right]$ $=\frac{22}{7} \times \frac{1}{12}\left[(7)^{2}-\left(\frac{7}{2}\right)...
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: $\frac{2}{3 x}-\frac{3}{2 x}=\frac{1}{12}$ Solution: $\frac{2}{3 x}-\frac{3}{2 x}=\frac{1}{12}$ or $\frac{4-9}{6 \mathrm{x}}=\frac{1}{12}$ or $\frac{-5}{6 \mathrm{x}}=\frac{1}{12}$ or $6 \mathrm{x}=-60$ or $\mathrm{x}=\frac{-60}{6}$ or $\mathrm{x}=-10$ Thus, $x=-10$ is the solution of the given equation. Check : Substituting $\mathrm{x}=-10$ in the given equation, we get: L. H. S. $=\frac{2}{3 \times(-10)}-\f...
Read More →In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm
Question: In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm with centre O. If POQ = 30, find the area of the shaded region. Solution: Area of the shaded portion = Area of sector OPQ Area of sector OAB $=\frac{30^{\circ}}{360^{\circ}} \times \pi(7)^{2}-\frac{30^{\circ}}{360^{\circ}} \times \pi(3.5)^{2}$ $=\frac{22}{7} \times \frac{1}{12}\left[(7)^{2}-(3.5)^{2}\right]$ $=\frac{22}{7} \times \frac{1}{12}\left[(7)^{2}-\left(\frac{7}{2}\right)...
Read More →n figure, ABCDE is any pentagon.
Question: n figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (ΔAPQ). Solution: Given ABCDE is a pentagon. BP || AC and EQ|| AD. To prove ar (ABCDE) = ar (APQ) Proof We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔADQ and ΔADE lie on the same base AD and between the same parallels AD and EQ. So, ar (ΔADQ) = ar (ΔADE) ...(i) Similarly, ΔACP a...
Read More →In the given figure, three sectors of a circle of radius 7 cm,
Question: In the given figure, three sectors of a circle of radius 7 cm, making angles of 60, 80 and 40 at the centre are shaded. Find the area of the shaded region. Solution: Area of the shaded region = Area of sector having central angle 60∘+ Area of sector having central angle 80∘+ Area of sector having central angle 40∘ $=\left[\frac{60^{\circ}}{360^{\circ}} \times \pi(7)^{2}\right]+\left[\frac{80^{\circ}}{360^{\circ}} \times \pi(7)^{2}\right]+\left[\frac{40^{\circ}}{360^{\circ}} \times \pi(...
Read More →In ΔABC, if L and M are the points on AB and AC,
Question: In ΔABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (ΔLOB) = ar (ΔMOC). Solution: Given In ΔABC, L and M are points on AB and AC respectively such that LM || BC. To prove $\operatorname{ar}(\Delta \mathrm{LOB})=\operatorname{ar}(\triangle \mathrm{MOC})$ Proof We know that, triangles on the same base and between the same parallels are equal in area. Hence, $\triangle B C$ and $\triangle M B C$ lie on the same base $B C$ and between the same pa...
Read More →In the given figure, OABC is a square of side 7 cm.
Question: In the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a circle with centre C find the area of the shaded region. Solution: Area of shaded region = Area of square OABC Area of quadrant COPB having radius OC $=(\text { Side })^{2}-\frac{1}{4}\left(\pi \times r^{2}\right)$ $=(7)^{2}-\frac{1}{4}\left[\frac{22}{7} \times 7^{2}\right]$ $=49-38.5$ $=10.5 \mathrm{~cm}^{2}$ Hence, the area of the shaded region is 10.5 cm2...
Read More →In the given figure, OABC is a square of side 7 cm.
Question: In the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a circle with centre C find the area of the shaded region. Solution: Area of shaded region = Area of square OABC Area of quadrant COPB having radius OC $=(\text { Side })^{2}-\frac{1}{4}\left(\pi \times r^{2}\right)$ $=(7)^{2}-\frac{1}{4}\left[\frac{22}{7} \times 7^{2}\right]$ $=49-38.5$ $=10.5 \mathrm{~cm}^{2}$ Hence, the area of the shaded region is 10.5 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:0.18(5x 4) = 0.5x+ 0.8 Solution: $0.18(5 x-4)=0.5 x+0.8$ or $0.9 \mathrm{x}-0.72=0.5 \mathrm{x}+0.8$ or $0.9 \mathrm{x}-0.5 \mathrm{x}=0.8+0.72$ or $0.4 \mathrm{x}=1.52$ or $\mathrm{x}=\frac{1.52}{0.4}$ or $\mathrm{x}=3.8$ Thus, $\mathrm{x}=3.8$ is the solution of the given equation. Check : Substituting $\mathrm{x}=3.8$ in the given equation, we get: L.H.S. $=0.18(5 \times 3.8-4)=0.18 \times 15=2.7$ R.H.S. $=...
Read More →ABCD is trapezium in which AB || DC,
Question: ABCD is trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA). Solution: Given in a trapezium $A B C D, A B \| D C, D C=30 \mathrm{~cm}$ and $A B=50 \mathrm{~cm}$. Also, $X$ and $Y$ are respectively the mid-points of $A D$ and $B C$. To prove ar $(D C Y X)=\frac{7}{9} \operatorname{ar}(X Y B A)$ Construction Join $D Y$ and extend it to meet produced $A B$ at $P$. Proof $\ln \triangle D C Y$...
Read More →From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm,
Question: From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semicircular portion with BC as diameter is cut off. Findthe area of the remaining paper. Solution: We know that the opposite sides of rectangle are equal AD = BC = 28 cm Now, Radius of semicircular portion $=\frac{1}{2} \mathrm{BC}=14 \mathrm{~cm}$ Area of remaining paper = Area of rectangular sheet Area of semicircular portion $=40 \times 28-\frac{1}{2}\left(\frac{22}{7} \times 14 \times 14\right)$ $=1120-308$ $...
Read More →From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm,
Question: From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semicircular portion with BC as diameter is cut off. Findthe area of the remaining paper. Solution: We know that the opposite sides of rectangle are equal AD = BC = 28 cm Now, Radius of semicircular portion $=\frac{1}{2} \mathrm{BC}=14 \mathrm{~cm}$ Area of remaining paper = Area of rectangular sheet Area of semicircular portion $=40 \times 28-\frac{1}{2}\left(\frac{22}{7} \times 14 \times 14\right)$ $=1120-308$ $...
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: $\frac{9 x+7}{2}-\left(x-\frac{x-2}{7}\right)=36$ Solution: $\frac{9 x+7}{2}-\left(x-\frac{x-2}{7}\right)=36$ or $\frac{63 \mathrm{x}+49-14 \mathrm{x}+2 \mathrm{x}-4}{14}=36$ or $\frac{51 \mathrm{x}+45}{14}=36$ or $51 \mathrm{x}+45=504$ or $51 \mathrm{x}=504-45$ or $\mathrm{x}=\frac{459}{51}=9$ Thus, $x=9$ is the solution of the given equation. Check : Substituting $\mathrm{x}=9$ in the given equation, we get...
Read More →In the given figure, ABCD is a square of side 4 cm.
Question: In the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region. [Use $\pi=3.14 .$ ] Solution: Area of the square $A B C D=(\text { Side })^{2}$ $=4^{2}$ $=16 \mathrm{~cm}^{2}$ Area of the circle $=\pi r^{2}$ Radius = 1 cm Area $=3.14 \times(1)^{2}$ $=3.14 \mathrm{~cm}^{2}$ Area of the quadrant of one circle $=\frac{1}{4} \pi r^{2}$ $=\frac{1}{...
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: $\frac{(1-2 x)}{7}-\frac{(2-3 x)}{8}=\frac{3}{2}+\frac{x}{4}$ Solution: $\frac{1-2 \mathrm{x}}{7}-\frac{2-3 \mathrm{x}}{8}=\frac{3}{2}+\frac{\mathrm{x}}{4}$ or $\frac{1-2 \mathrm{x}}{7}=\frac{3}{2}+\frac{\mathrm{x}}{4}+\frac{2-3 \mathrm{x}}{8}$ or $\frac{1-2 \mathrm{x}}{7}=\frac{12+2 \mathrm{x}+2-3 \mathrm{x}}{8}$ or $\frac{1-2 \mathrm{x}}{7}=\frac{14-x}{8}$ or $8-16 \mathrm{x}=98-7 x$ or $-16 \mathrm{x}+7 \m...
Read More →In the given figure, ABCD is a square of side 4 cm.
Question: In the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region. [Use $\pi=3.14 .$ ] Solution: Area of the square $A B C D=(\text { Side })^{2}$ $=4^{2}$ $=16 \mathrm{~cm}^{2}$ Area of the circle $=\pi r^{2}$ Radius = 1 cm Area $=3.14 \times(1)^{2}$ $=3.14 \mathrm{~cm}^{2}$ Area of the quadrant of one circle $=\frac{1}{4} \pi r^{2}$ $=\frac{1}{...
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: $\frac{(3 x+1)}{16}+\frac{(2 x-3)}{7}=\frac{(x+3)}{8}+\frac{(3 x-1)}{14}$ Solution: $\frac{3 \mathrm{x}+1}{16}+\frac{2 \mathrm{x}-3}{7}=\frac{\mathrm{x}+3}{8}+\frac{3 \mathrm{x}-1}{14}$ or $\frac{3 \mathrm{x}+1}{16}-\frac{\mathrm{x}+3}{8}=\frac{3 \mathrm{x}-1}{14}-\frac{2 \mathrm{x}-3}{7}$ or $\frac{3 \mathrm{x}+1-2 \mathrm{x}-6}{16}=\frac{3 \mathrm{x}-1-4 \mathrm{x}+6}{14}$ or $\frac{\mathrm{x}-5}{8}=\frac{-...
Read More →In figure, CD||AE and CY||BA.
Question: In figure, CD||AE and CY||BA. Prove that ar (ΔCBX) = ar (ΔAXY). Solution: Given In figure, CD||AE and CY || BA To prove ar (ΔCBX) = ar (ΔAXY) . Proof We know that, triangles on the same base and between the same parallels are equal . in areas. Here, ΔABY and ΔABC both lie on the same base AB and between the same parallels CY and BA. ar (ΔABY) = ar (ΔABC) = ar (ABX) + ar (AXY) = ar (ABX) + ar (CBX) = ar (AXY) = ar (CBX) [eliminating ar (ABX) from both sides]...
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: $\frac{5 x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$ Solution: $\frac{5 x}{3}-\frac{x-1}{4}=\frac{x-3}{5}$ or $\frac{20 x-3 x+3}{12}=\frac{x-3}{5}$ or $\frac{17 x+3}{12}=\frac{x-3}{5}$ or $85 x+15=12 x-36$ or $73 x=-51$ or $x=\frac{-51}{73}$ Check : L. H.S. $=\frac{5 \times \frac{-51}{73}}{3}-\frac{\frac{-51}{73}-1}{4}=\frac{-255}{219}-\frac{-124}{292}=\frac{-54}{73}$ R.H.S. $=\frac{\frac{-51}{73}-3}{5}=\frac{-54}...
Read More →Find the adjoint of each of the following matrices:
Question: Find the adjoint of each of the following matrices: (i) $\left[\begin{array}{cc}-3 5 \\ 2 4\end{array}\right]$ (ii) $\left[\begin{array}{ll}a b \\ c d\end{array}\right]$ (iii) $\left[\begin{array}{ll}\cos \alpha \sin \alpha \\ \sin \alpha \cos \alpha\end{array}\right]$ (iv) $\left[\begin{array}{cc}1 \tan \alpha / 2 \\ -\tan \alpha / 2 1\end{array}\right]$ Verify that $(\operatorname{adj} A) A=|A| I=A(\operatorname{adj} A)$ for the above matrices. Solution: Given below are the squares m...
Read More →The median BE and CF of a triangle ABC
Question: The median BE and CF of a triangle ABCintersect at G. Prove that the area of ΔGBC = area of the quadrilateral AFGE. Solution: Given In $\triangle A B C$, medians $B E$ and $C F$ intersect each other at $G$. To prove To prove $\operatorname{ar}(\Delta G B C)=\operatorname{ar}(A F G E)$ Proof Since, $B E$ is the median of $\triangle A B C$ and we know that a median of a triangle divides it into two parts of equal area. So, $\quad$ ar $(\Delta A B E)=\operatorname{ar}(\Delta C B E)$ $\Rig...
Read More →Two circular pieces of equal radii and maximum area,
Question: Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular cardboard of dimensions 14 cm ⨯ 7 cm. Find the area of the remaining cardboard. Solution: We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard. Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cmNow,Area of remaining cardboard = Are...
Read More →Two circular pieces of equal radii and maximum area,
Question: Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular cardboard of dimensions 14 cm ⨯ 7 cm. Find the area of the remaining cardboard. Solution: We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard. Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cmNow,Area of remaining cardboard = Are...
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: $\frac{3 x}{4}-\frac{(x-1)}{2}=\frac{(x-2)}{3}$ Solution: $\frac{3 x}{4}-\frac{(x-1)}{2}=\frac{(x-2)}{3}$ or $\frac{3 x-2 x+2}{4}=\frac{x-2}{3}$ or $4 x-8=3 x+6$ or $x=14$ Check: L. H.S. $=\frac{3 \times 14}{4}-\frac{14-1}{2}=\frac{21}{2}-\frac{13}{2}=\frac{8}{2}=4$ R.H.S. $=\frac{14-2}{3}=\frac{12}{3}=4$ Check: L. H.S. $=\frac{3 \times 14}{4}-\frac{14-1}{2}=\frac{21}{2}-\frac{13}{2}=\frac{8}{2}=4$ R.H.S. $=\...
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