Find the area of the shaded region in the given figure,
Question: Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of the circle. Solution: In right triangle ABCAC2= AB2+ BC2= 82+62= 64 + 36= 100 AC2= 100⇒ AC = 10 cm Now, Radius of circle $(\mathrm{OA})=\frac{1}{2} \mathrm{AC}=5 \mathrm{~cm}$ Area of the shaded region = Area of circle Area of rectangle OABC $=\pi(\mathrm{OA})^{2}-\mathrm{AB} \times \mathrm{BC}$ $=\frac{22}{7} \times(5)^{2}-8 \times 6$ $=78.57-48$ $=30.57 \math...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{3 x+5}{2 x+7}=4$ Solution: $\frac{3 x+5}{2 x+7}=4$ or $3 \mathrm{x}+5=8 \mathrm{x}+28$ or $8 \mathrm{x}+28=3 \mathrm{x}+5$ (After cross multipl ication) or $8 \mathrm{x}-3 \mathrm{x}=5-28$ or $5 \mathrm{x}=-23$ or $\mathrm{x}=\frac{-23}{5}$ $\therefore \mathrm{x}=\frac{-23}{5}$ is the solution of given equation. Check: Substituting $\mathrm{x}=\frac{-23}{5}$ in the given equation, we get: L.H.S. $=\frac{3 \times \frac{-23}{5}+...
Read More →In figure, if ∠OAB = 40°,
Question: In figure, if OAB = 40, then ACB is equal to From Eq. (i) ACB = ADB = 70 (a)50 (b)40 (c)60 (d)70 Solution: (a) In ΔQAB, OA = OB [both are the radius of a circle] OAB = OBA = OBA = 40 [angles opposite to equal sides are equal] Also, AOB + OBA + BAO = 180 [by angle sum property of a triangle] AOB + 40 + 40 = 180 = AOB = 180 80 = 100 We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. AOB = 2 ACB = ...
Read More →Find the inverse of each of the following matrices.
Question: Find the inverse of each of the following matrices. (i) $\left[\begin{array}{lll}1 2 3 \\ 2 3 1 \\ 3 1 2\end{array}\right]$ (ii) $\left[\begin{array}{ccc}1 2 5 \\ 1 -1 -1 \\ 2 3 -1\end{array}\right]$ (iii) $\left[\begin{array}{ccc}2 -1 1 \\ -1 2 -1 \\ 1 -1 2\end{array}\right]$ (iv) $\left[\begin{array}{ccc}2 0 -1 \\ 5 1 0 \\ 0 1 3\end{array}\right]$ (v) $\left[\begin{array}{ccc}0 1 -1 \\ 4 -3 4 \\ 3 -3 4\end{array}\right]$ (vi) $\left[\begin{array}{ccc}0 0 -1 \\ 3 4 5 \\ -2 -4 -7\end{a...
Read More →In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm
Question: In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm and all circles are of the same diameter. Solution: Area of the square $=\operatorname{Side}^{2}=14^{2}=196$ sq. $\mathrm{cm}$ Area of the circles $=4 \times \pi \times 3.5 \times 3.5=154$ sq. $\mathrm{cm}$ Area of the shaded region = Area of the squareArea of four circles $=196-154$ $=42 \mathrm{~cm}^{2}$...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{5 x-7}{3 x}=2$ Solution: $\frac{5 x-7}{3 x}=2$ or $6 \mathrm{x}=5 \mathrm{x}-7$ (After c ross multiplication) or $6 \mathrm{x}-5 \mathrm{x}=-7$ or $\mathrm{x}=-7$ $\therefore \mathrm{x}=-7$ is the solution of given equation. Check: Substituting $x=-7$ in the given equation, we get: L. H.S $=\frac{5 \times(-7)-7}{.3(-7)}=\frac{-35-7}{-21}=\frac{-42}{-21}=2$ R. H.S. $=2$ $\therefore$ L. H.S. $=$ R. H. S. for $x=-7$....
Read More →In figure, if AOB is a diameter
Question: In figure, if AOB is a diameter of the circle and AC = BC, then CAB is equal to (a)30 (b)60 (c)90 (d)45 Solution: (d) We know that, diameter subtends a right angle to the circle. $\therefore \quad \angle B C A=90^{\circ}$ ....(i) Given, $A C=B C$ $\Rightarrow$ $\angle A B C=\angle C A B$ ...(ii) [angles opposite to equal sides are equal] In $\triangle A B C$, $\quad \angle C A B+\angle A B C+\angle B C A=180^{\circ}$ [by angle sum property of a triangle] $\Rightarrow \quad \angle C A B...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{2-y}{y+7}=\frac{3}{5}$ Solution: $\frac{2-y}{y+7}=\frac{3}{5}$ or $10-5 y=3 y+21($ After c ross multipl ication $)$ or $3 y+5 y=10-21$ or $8 y=-11$ or $y=\frac{-11}{8}$ $\therefore y=\frac{-11}{8}$ is the solution of the given equation. Check : Substituting $\mathrm{y}=\frac{-11}{8}$ in the given equation, we get: L. H.S. $=\frac{2-\frac{-11}{8}}{\frac{-11}{8}+7}=\frac{16+11}{-11+56}=\frac{27}{45}=\frac{3}{5}$ R.H.S. $=\frac{3...
Read More →Find the area of a quadrant of a circle whose circumference is 44 cm.
Question: Find the area of a quadrant of a circle whose circumference is 44 cm. Solution: Let the radius of the circle ber.Now, Circumference $=44$ $\Rightarrow 2 \pi r=44$ $\Rightarrow r=7 \mathrm{~cm}$ Now, Area of quadrant $=\frac{1}{4} \pi r^{2}=\frac{1}{4} \times \frac{22}{7} \times(7)^{2}=38.5 \mathrm{~cm}^{2}$ Hence, the area of the quadrant of the circle is 38.5 cm2....
Read More →In figure, if ∠ABC = 20°,
Question: In figure, if $\angle A B C=20^{\circ}$, then $\angle A O C$ is equal to (a) $20^{\circ}$ (b) $40^{\circ}$ (c) $60^{\circ}$ (d) $10^{\circ}$ Thinking Process Use the theorem, that in a circale the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle and further simplify it. Solution: (b)Given, ABC = 20 We know that, angle subtended at the centre by an arc is twice the angle subtended by it at the remaining part of circle. AOC = ...
Read More →In the given figure, APB and AQO are semicircles and AO = OB.
Question: In the given figure, APB and AQO are semicircles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. Solution: Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB $\Rightarrow 40=\frac{1}{2} \times 2 \pi\left(\frac{\mathrm{AO}}{2}\right)+\frac{1}{2} \times 2 \pi(\mathrm{OB})+\mathrm{OB}$ $\Rightarrow 40=\frac{11}{7} \mathrm{AO}+\frac{22}{7} \mathrm{OB}+\mathrm{OB}$ $\Rightarrow 40=\frac{11}{7} \mathrm{OB}+\frac...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{2 x-3}{3 x+2}=-\frac{2}{3}$ Solution: $\frac{2 x-3}{3 x+2}=-\frac{2}{3}$ or $6 \mathrm{x}-9=-6 \mathrm{x}-4$ (After c ross multipl ication) or $6 x+6 x=-4+9$ or $\mathrm{x}=\frac{5}{12}$ $\therefore \mathrm{x}=\frac{5}{12}$ is the solution of the given equation. Check: L.H.S. $=\frac{2 \times \frac{5}{12}-3}{3 \times \frac{5}{12}+2}=\frac{\frac{5}{6}-3}{\frac{5}{4}+2}=\frac{\frac{-13}{6}}{\frac{13}{4}}=\frac{-4}{6}=\frac{-2}{3...
Read More →In the given figure, APB and AQO are semicircles and AO = OB.
Question: In the given figure, APB and AQO are semicircles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. Solution: Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB $\Rightarrow 40=\frac{1}{2} \times 2 \pi\left(\frac{\mathrm{AO}}{2}\right)+\frac{1}{2} \times 2 \pi(\mathrm{OB})+\mathrm{OB}$ $\Rightarrow 40=\frac{11}{7} \mathrm{AO}+\frac{22}{7} \mathrm{OB}+\mathrm{OB}$ $\Rightarrow 40=\frac{11}{7} \mathrm{OB}+\frac...
Read More →If AB = 12 cm, BC = 16 cm and AB
Question: If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is (a)6 cm (b)8 cm (c) 10 cm (d)12 cm Solution: (c) Given, $A B=12 \mathrm{~cm}$ and $B C=16 \mathrm{~cm}$ In a circle, $B C \perp A B$, it means that $A C$ will be a diameter of circle. [diameter of a circle subtends a right angle to the circle] Use Pythagoras theorem in right angled $\triangle A B C$, $A C^{2}=A B^{2}+B C^{2}$ $\Rightarrow$ $A C^{2}=(12)^{2}+(1...
Read More →In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle.
Question: In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. [Use = 3.14.] Solution: Construction: Join OB In right triangle AOBOB2= OA2+ AB2= 202+ 202= 400 + 400= 800 OB2= 800Area of the shaded region = Area of quadrant OPBQ Area of Square OABC $=\frac{1}{4} \pi(\mathrm{OB})^{2}-(\mathrm{OA})^{2}$ $=\frac{1}{4} \times 3.14 \times 800-400$ $=628-400$ $=228 \mathrm{~cm}^{2}$ Hence, the area of the shaded region is 228...
Read More →In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle.
Question: In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. [Use = 3.14.] Solution: Construction: Join OB In right triangle AOBOB2= OA2+ AB2= 202+ 202= 400 + 400= 800 OB2= 800Area of the shaded region = Area of quadrant OPBQ Area of Square OABC $=\frac{1}{4} \pi(\mathrm{OB})^{2}-(\mathrm{OA})^{2}$ $=\frac{1}{4} \times 3.14 \times 800-400$ $=628-400$ $=228 \mathrm{~cm}^{2}$ Hence, the area of the shaded region is 228...
Read More →In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively.
Question: In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region. [Use = 3.14] Solution: Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ $=\frac{1}{2} \times 2 \pi r_{1}+\frac{1}{2} \times 2 \pi r_{2}+\frac{1}{2} \times 2 \pi r_{3}$ $=\frac{1}{2} \times 2 \pi\left(\frac{7}{2}\right)+\frac{1}{2} \times 2 \pi\left(\frac{10}{2}\right)+\frac{1}{2} \times 2 \pi\le...
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: [(2x+ 3) + (x+ 5)]2+ [(2x+ 3) (x+ 5)]2= 10x2+ 92 Solution: $[(2 \mathrm{x}+3)+(\mathrm{x}+5)]^{2}+[(2 \mathrm{x}+3)-(\mathrm{x}+5)]^{2}=10 \mathrm{x}^{2}+92$ or $(3 \mathrm{x}+8)^{2}+(\mathrm{x}-2)^{2}=10 \mathrm{x}^{2}+92$ or $9 \mathrm{x}^{2}+48 \mathrm{x}+64+\mathrm{x}^{2}-4 \mathrm{x}+4=10 \mathrm{x}^{2}+92 \quad\left[(\mathrm{a}+\mathrm{b})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}+2 \mathrm{ab} \quad \mathrm{an...
Read More →In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively.
Question: In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region. [Use = 3.14] Solution: Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ $=\frac{1}{2} \times 2 \pi r_{1}+\frac{1}{2} \times 2 \pi r_{2}+\frac{1}{2} \times 2 \pi r_{3}$ $=\frac{1}{2} \times 2 \pi\left(\frac{7}{2}\right)+\frac{1}{2} \times 2 \pi\left(\frac{10}{2}\right)+\frac{1}{2} \times 2 \pi\le...
Read More →Find the inverse of each of the following matrices:
Question: Find the inverse of each of the following matrices: (i) $\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$ (ii) $\left[\begin{array}{ll}0 1 \\ 1 0\end{array}\right]$ (iii) $\left[\begin{array}{cc}a b \\ c \frac{1+b c}{a}\end{array}\right]$ (iv) $\left[\begin{array}{cc}2 5 \\ -3 1\end{array}\right]$ Solution: (i) $A=\left[\begin{array}{cc}\cos \theta \sin \theta \\ -\sin \theta \cos \theta\end{array}\right]$ $|A|=\cos ^{2} \theta+\sin ^{2} \th...
Read More →In the given figure, APB and CQD are semicircles of diameter 7 cm each,
Question: In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC an BSD are semicircles of diameter 14 cm each. Find the (i) perimeter,] (ii) area of the shaded region. Solution: (i) Perimeter of shaded region = Length of the arc APB + Length of the arc ARC + Length of the arc BSD + Length of the arc CQD $=\frac{1}{2} \times 2 \pi r_{1}+\frac{1}{2} \times 2 \pi r_{2}+\frac{1}{2} \times 2 \pi r_{3}+\frac{1}{2} \times 2 \pi r_{4}$ $=\frac{1}{2} \times 2 \pi\left(\frac{7}...
Read More →In the given figure, APB and CQD are semicircles of diameter 7 cm each,
Question: In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC an BSD are semicircles of diameter 14 cm each. Find the (i) perimeter,] (ii) area of the shaded region. Solution: (i) Perimeter of shaded region = Length of the arc APB + Length of the arc ARC + Length of the arc BSD + Length of the arc CQD $=\frac{1}{2} \times 2 \pi r_{1}+\frac{1}{2} \times 2 \pi r_{2}+\frac{1}{2} \times 2 \pi r_{3}+\frac{1}{2} \times 2 \pi r_{4}$ $=\frac{1}{2} \times 2 \pi\left(\frac{7}...
Read More →In figure, if OA = 5 cm,
Question: In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to (a) $2 \mathrm{~cm}$ (b) $3 \mathrm{~cm}$ (c) $4 \mathrm{~cm}$ (d) $5 \mathrm{~cm}$ Solution: (a)We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = AB = x 8 = 4 cm given OA = 5 cm AO2= AC2+ OC2 (5)2= (4)2+ OC2 25 = 16 + OC2 OC2= 25-16 = 9 OC = 3 cm [taking positive square root, because length is always positive] OA = OD [same radius of a circle] OD = 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:(3x 8)(3x+ 2) (4x 11)(2x+ 1) = (x 3)(x+ 7) Solution: $(3 \mathrm{x}-8)(3 \mathrm{x}+2)-(4 \mathrm{x}-11)(2 \mathrm{x}+1)=(\mathrm{x}-3)(\mathrm{x}+7)$ or $9 \mathrm{x}^{2}+6 \mathrm{x}-24 \mathrm{x}-16-8 \mathrm{x}^{2}-4 \mathrm{x}+22 \mathrm{x}+11=\mathrm{x}^{2}+7 \mathrm{x}-3 \mathrm{x}-21$ or $\mathrm{x}^{2}-5=\mathrm{x}^{2}+4 \mathrm{x}-21$ or $4 \mathrm{x}=-5+21$ or $\mathrm{x}=\frac{16}{4}=4$ Thus, $x=4$...
Read More →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 →