In the given figure, the side of square is 28 cm
Question: In the given figure, the side of square is 28 cm and radius of each circle is half of the length of the side of the square, whereOandO' are centres of the circles. Find the area of shaded region. Solution: We have, Side of square $=28 \mathrm{~cm}$ and radius of each circle $=\frac{28}{2} \mathrm{~cm}$ Area of the shaded region= Area of the square + Area of the two circles Area of the two quadrants $=(28)^{2}+2 \times \pi \times\left(\frac{28}{2}\right)^{2}-2 \times \frac{1}{4} \times ...
Read More →Solve this
Question: If $\left|\begin{array}{ll}2 x+5 3 \\ 5 x+2 9\end{array}\right|=0$, find $x$ Solution: $\left|\begin{array}{ll}2 x+5 3 \\ 5 x+2 9\end{array}\right|=0$ $\Rightarrow 9(2 x+5)-3(5 x+2)=0$ $\Rightarrow 18 x+45-15 x-6=0$ $\Rightarrow 3 x+39=0$ $\Rightarrow 3 x=-39$ $\Rightarrow x=\frac{-39}{3}=-13$...
Read More →Write the cofactor
Question: Write the cofactor of $a_{12}$ in the following matrix $\left[\begin{array}{ccc}2 -3 5 \\ 6 0 4 \\ 1 5 -7\end{array}\right]$. Solution: Given: $\left[\begin{array}{ccc}2 -3 5 \\ 6 0 4 \\ 1 5 -7\end{array}\right]$ Here, $a_{12}=-3$ Cofactor of $a_{12}=(-1)^{1+2}\left|\begin{array}{rr}6 4 \\ 1 -7\end{array}\right|=-(-42-4)=46$...
Read More →The area of the parallelogram ABCD is
Question: The area of the parallelogram ABCD is 90 cm2. Find ar (ABEF) ar (ΔABD) ar (ΔBEF) Solution: Given, area of parallelogram, ABCD = 90 cm2. We know that, parallelograms on the same base and between the same parallel are equal in areas.Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB and CF.So, ar (ΔBEF) = ar (ABCD) = 90 cm2 We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then area of triangle is eq...
Read More →Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm
Question: Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region. Solution: Area of the shaded region= Area of the semi-circle with diameter of 9 cm Areas of two semi-circles with diameter 3 cm Area of the circle with diameter 4.5 cm + Area of semi-circle with diameter 3 cm= Area of the semi-circle with radius of 4.5 cm 2 Area of semi-circle with radius 1.5 cm Area of the circle with...
Read More →Divide the sum of
Question: Divide: $\frac{1}{4} x^{2}-\frac{1}{2} x-12$ by $\frac{1}{2} x-4$ Solution: $\frac{\frac{1}{4} x^{2}-\frac{1}{2} x-12}{\frac{1}{2} x-4}$ $=\frac{\frac{1}{2} x\left(\frac{1}{2} x-4\right)+3\left(\frac{1}{2} x-4\right)}{\frac{1}{2} x-4}$ $=\frac{\left(\frac{1}{2} x-4\right)\left(\frac{1}{2} x+3\right)}{\left(\frac{1}{2} x-4\right)}$ $=\frac{1}{2} x+3$...
Read More →X and Y are points on the side L N of
Question: $X$ and $Y$ are points on the side $L N$ of the triangle $L M N$ such that $L X=X Y=Y N$. Through $X$, a line is drawn parallel to $L M$ to meet $M N$ at $Z$ (see figure). Prove that ar $(\Delta L Z Y)=\operatorname{ar}(M Z Y X)$. Thinking Process Use the property that the triangles on the same base and between the same two parallel lines are equal in area. Further prove the required result. Solution: Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ...
Read More →Write the value
Question: Write the value of $\left|\begin{array}{cc}a+i b c+i d \\ -c+i d a-i b\end{array}\right|$. Solution: $\left|\begin{array}{cc}a+i b c+i d \\ -c+i d a-i b\end{array}\right|$ Here, $i^{2}=-1$ $\Rightarrow\left|\begin{array}{cc}a+i b c+i d \\ -c+i d a-i b\end{array}\right|=a^{2}+b^{2}+c^{2}+d^{2}$...
Read More →Divide the sum of
Question: Divide:(a2+ 2ab+b2) (a2+ 2ac+c2) by 2a+b+c Solution: $\frac{\left(a^{2}+2 a b+b^{2}\right)-\left(a^{2}+2 a c+c^{2}\right)}{(2 a+b+c)}$ $=\frac{(a+b)^{2}-(a+c)^{2}}{(2 a+b+c)}$ $=\frac{(a+b+a+c)(a+b-a-c)}{(2 a+b+c)}$ $=\frac{(2 a+b+c)(b-c)}{(2 a+b+c)}$ $=b-c$...
Read More →In the figure, PSDA is a parallelogram.
Question: In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. Prove that ar (PQE) =ar (CFD). Thinking Process Firstly, use the formula, area of parallelogram = Base x Altitude Further, prove that A PQE = AD CF, by ASA congruent rule. At the end use the property that congruent figures have same area. Solution: Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) ...
Read More →Divide the sum of
Question: Divide:acx2+ (bc + ad)x + bdby (ax + b) Solution: $\frac{a c x^{2}+(b c+a d) x+b d}{(a x+b)}$ $=\frac{a c x^{2}+b c x+a d x+b d}{(a x+b)}$ $=\frac{c x(a x+b)+d(a x+b)}{(a x+b)}$ $=\frac{(a x+b)(c x+d)}{(a x+b)}$ $=c x+d$...
Read More →If A and B are square matrices of order 3 such that |A| = − 1, |B| = 3,
Question: If $A$ and $B$ are square matrices of order 3 such that $|A|=-1,|B|=3$, then find the value of $|3 A B|$. Solution: $|K A|=K^{n}|A|$$[n$ is the order of $A]$ $\Rightarrow|3 A B|=3^{3}|A B| \quad \ldots(1)$ If $A$ and $B$ are square matrices of the same order, then $|A B|=|A||B| .$ So, $|3 A B|=3^{3}|A||B| \quad$ [From eq. (1)] $=27 \times-1 \times 3$ $=-81$...
Read More →Divide the sum of
Question: Divide:x4y4byx2y2 Solution: $\frac{x^{4}-y^{4}}{x^{2}-y^{2}}$ $=\frac{\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}}{\left(x^{2}-y^{2}\right)}$ $=\frac{\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)}{\left(x^{2}-y^{2}\right)}$ $=x^{2}+y^{2}$...
Read More →In the figure, ABCD and EFGD are two parallelograms
Question: In the figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then, ar (ΔDPC) = ar (EFGD). Solution: False In the given figure, join PG. Since, G is the mid-point of CD. Thus, PG is a median of ΔDPC and it divides the triangle into parts of equal areas. Then,$\operatorname{ar}(\Delta D P G)=\operatorname{ar}(\Delta G P C)=\frac{1}{2} \operatorname{ar}(\Delta D P C)$...(i) Also, we know that, if a parallelogram and a triangle lie on the same base and between the same...
Read More →In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm.
Question: In the given figure, two concentric circles with centreOhave radii 21 cm and 42 cm. IfAOB= 60, find the area of the shaded region.$\left[\right.$ Use $\left.\pi=\frac{22}{7}\right]$ Solution: Given: Radius of the inner circle with radius OC,r= 21 cmRadius of the inner circle with radius OA,R= 42 cmAOB = 60Area of the circular ring $=\pi R^{2}-\pi r^{2}$ $=\pi\left[R^{2}-r^{2}\right]$ $=\pi\left[42^{2}-21^{2}\right] \mathrm{cm}^{2}$ Area of ACDB = area of sector AOB area of COD $=\frac{...
Read More →Divide the sum of
Question: Divide:ax2ay2byax + ay Solution: $\frac{a x^{2}-a y^{2}}{a x+a y}$ $=\frac{a\left(x^{2}-y^{2}\right)}{a(x+y)}$ $=\frac{a(x+y)(x-y)}{a(x+y)}$ $=x-y$...
Read More →A matrix of order 3 × 3 has determinant 2.
Question: A matrix of order $3 \times 3$ has determinant 2 . What is the value of $|A(3 D)|$, where $/$ is the identity matrix of order $3 \times 3$. Solution: Let $A$ be the given matrix. Then, $|A|=2 \quad[$ Order $=n=3]$ $|I|=1 \quad[I$ is an identity matrix $]$ $3(\mathrm{I})=3$ $|A 3(\mathrm{I})|=|3 A|=3^{3}|A|$ $[A$ being of order 3$]$ $=27 \times 2=54$ $\Rightarrow|A 3(\mathrm{I})|=54$...
Read More →Divide the sum of
Question: Divide:x2 5x+ 6 byx 3 Solution: $\frac{x^{2}-5 x+6}{x-3}$ $=\frac{x^{2}-3 x-2 x+6}{x-3}$ $=\frac{x(x-3)-2(x-3)}{(x-3)}$ $=\frac{(x-3)(x-2)}{(x-3)}$ $=x-2$...
Read More →ABC and BDE are two equilateral triangles
Question: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then, ar (ΔBDE) = ar (ΔABC). Solution: True Given, $\triangle A B C$ and $\triangle B D E$ are two equilateral triangles. $\therefore$ Area of an equilateral $\triangle A B C=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}=\frac{\sqrt{3}}{4}(B C)^{2}$ $\ldots(1)$ $[\because$ in equilateral $\triangle A B C, A B=B C=A C]$ Also, given $D$ is the mid-point of $B C$. $\therefore$ $B D=D C=\frac{1}{2} B C$ ...(ii) N...
Read More →Find whether the first polynomial is a factor of the second.
Question: Find whether the first polynomial is a factor of the second. (i)x+ 1, 2x2+ 5x+ 4 (ii)y 2, 3y3+ 5y2+ 5y+ 2 (iii) 4x2 5, 4x4+ 7x2+ 15 (iv) 4 z, 3z2 13z+ 4 (v) 2a 3, 10a2 9a 5 (vi) 4y+ 1, 8y2 2y+ 1 Solution: (i) $\frac{2 x^{2}+5 x+4}{x+1}$ $=\frac{2 x(x+1)+3(x+1)+1}{x+1}$ $=\frac{(x+1)(2 x+3)+1}{(x+1)}$ $=(2 x+3)+\frac{1}{x+1}$ $\because$ Remainder $=1$ Therefore, $(\mathrm{x}+1)$ is not a factor of $2 \mathrm{x}^{2}+5 \mathrm{x}+4$ (ii) $\frac{3 y^{3}+5 y^{2}+5 y+2}{y-2}$ $=\frac{3 \math...
Read More →If A and B are non-singular matrices of the same order, write whether AB is singular or non-singular.
Question: IfAandBare non-singular matrices of the same order, write whetherABis singular or non-singular. Solution: LetABbe non-singular matrices of ordern. $|A| \neq 0$ and $|B| \neq 0 \quad$ [By definition] Since they are of same order, $\begin{aligned}|A B| =|A||B| \\|A B| =0 \text { iff either }|A|=0 \text { or }|B|=\end{aligned}$ $=0$ But it is not the case here. Thus, $|A B|$ is non-zero and $A B$ is non-singular matrix....
Read More →In the given figure, ABCD is a rectangle of dimensions 21 cm × 14 cm.
Question: In the given figure,ABCDis a rectangle of dimensions 21 cm 14 cm. A semicircle is drawn withBCas diameter. Find the area and the perimeter of the shaded region in the figure. Solution: Area of the shaded region= Area of the rectangle Area of the semicircle $=21 \times 14-\left\{\frac{1}{2} \times \pi \times\left(\frac{14}{2}\right)^{2}\right\}$ $=294-\left\{\frac{1}{2} \times \frac{22}{7} \times 7 \times 7\right\}$ $=294-77$ $=217 \mathrm{~cm}^{2}$ Therefore, area of shaded region is 2...
Read More →PQRS is a parallelogram whose area
Question: PQRS is a parallelogram whose area is 180 cm2and A is any point on the diagonal QS. The area of ΔASR = 90 cm2. Solution: False Given, area of parallelogram PQRS = 180 cm2and QS is its diagonal which divides it into two triangles of equal area....
Read More →Find the value of the determinant
Question: Find the value of the determinant $\left|\begin{array}{lll}2^{2} 2^{3} 2^{4} \\ 2^{3} 2^{4} 2^{5} \\ 2^{4} 2^{5} 2^{6}\end{array}\right|$. Solution: $\begin{array}{lll}\mid 2^{2} 2^{3} 2^{4} \\ 2^{3} 2^{4} 2^{5} \\ 2^{4} 2^{5} 2^{6} \mid\end{array}$ $=2^{2} \times 2^{3} \times 2^{4} \mid 1 \quad 2 \quad 2^{2}$ $\begin{array}{lll}1 2 2^{2}\end{array}$ $\begin{array}{lll}1 2 2^{2}\end{array}$ [Taking out common factors from $R_{1}, R_{2}$ and $R_{3}$ ] $=2^{2} \times 2^{3} \times 2^{4} \...
Read More →PQRS is a rectangle inscribed in a quadrant
Question: PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm and A isany point on PQ. If PS = 5 cm, then ar (ΔPAS) = 30 cm2. Solution: True Given,PS = 5 cm radius of circle = SQ = 13 cm In right angled $\triangle S P Q, \quad S Q^{2}=P Q^{2}+P S^{2} \quad$ [by Pythagoras theorem] $(13)^{2}=P Q^{2}+(5)^{2}$ $\Rightarrow$$P Q^{2}=169-25=144$ $\Rightarrow$ $P Q=12 \mathrm{~cm}$ [taking positive square root, because length is always positive] Now, area of $\triangle A P S=\frac{...
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