Factorize each of the following expressions:
Question: Factorize each of the following expressions:6xy+ 6 9y 4x Solution: $6 x y+6-9 y-4 x=(6 x y-4 x)+(6-9 y) \quad$ [Regrouping the expressions] $=2 x(3 y-2)+3(2-3 y)$ $=2 x(3 y-2)-3(3 y-2)$ $[\because(2-3 y)=-(3 y-2)]$ $=(2 x-3)(3 y-2)$ $[$ Taking $(3 y-2)$ as the common factor $]$...
Read More →The length of a rectangular field is 12 m and the length of its diagonal is 15 m.
Question: The length of a rectangular field is 12 m and the length of its diagonal is 15 m. The area of the field is(a) 108 m2(b)180 m2 (c) $30 \sqrt{3} \mathrm{~m}^{2}$ (d) $12 \sqrt{15} \mathrm{~m}^{2}$ Solution: (a) 108 m2Length of the rectangular field = 12 mDiagonal = 15 m Diagonal $^{2}=$ Length $^{2}+$ Breadth $^{2}$ Breadth $=\sqrt{\text { Diagonal }^{2}-\text { Length }^{2}}$ $=\sqrt{15^{2}-12^{2}}$ $=\sqrt{225-144}$ $=9 \mathrm{~m}$ $\therefore$ Area of the field $=$ Length $\times$ Br...
Read More →If Q is a point oh the side SR of a ΔPSR
Question: If Q is a point oh the side SR of a ΔPSRsuch that PQ = PR, then prove that PS PQ. Thinking Process Use the property of a triangle that if two sides are equal then their opposite angles are also equal. Also use the property that side opposite to greater angle is longer. Solution: Given in $\triangle P S R, Q$ is a point on the side $S R$ such that $P Q=P R$. To prove $P SP Q$ Proof $\ln \Delta P R Q$, $P Q=P R$ [given] $\Rightarrow$ $\angle R=\angle P Q R$ ...(i) [angles opposite to equ...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:x3y2+xx2y2 Solution: $x^{3}-y^{2}+x-x^{2} y^{2}$ $=\left(x^{3}+x\right)-\left(x^{2} y^{2}+y^{2}\right)$ [Regrouping the expressions] $=x\left(x^{2}+1\right)-y^{2}\left(x^{2}+1\right)$ $=\left(x-y^{2}\right)\left(x^{2}+1\right)$ $\left[\right.$ Taking $\left(x^{2}+1\right)$ as the common factor $]$...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:lm2mn2lm +n2 Solution: l m^{2}-m n^{2}-l m+n^{2}=\left(l m^{2}-l m\right)+\left(n^{2}-m n^{2}\right) \quad[\text { Regrouping the expressions }] $=\operatorname{lm}(m-1)+n^{2}(1-m)$ $=\operatorname{lm}(m-1)-n^{2}(m-1)$ $[\because(1-\mathrm{m})=-(\mathrm{m}-1)]$ $=\left(\operatorname{lm}-n^{2}\right)(m-1)$ $[$ Taking $(\mathrm{m}-1)$ as the common factor $]$...
Read More →If ω is a non-real cube root of unity and n is not a multiple of 3, then
Question: If $\omega$ is a non-real cube root of unity and $n$ is not a multiple of 3, then $\Delta=\left|\begin{array}{ccc}1 \omega^{n} \omega^{2 n} \\ \omega^{2 n} 1 \omega^{n} \\ \omega^{n} \omega^{2 n} 1\end{array}\right|$ is equal to (a) 0 (b) $\omega$ (c) $\omega^{2}$ (d) 1 Solution: (a) 0 $\Delta=\mid 1 \quad w^{n} \quad w^{2 n}$ $w^{2 n} \quad 1 \quad w^{n}$ $w^{n} \quad w^{2 n} \quad 1 \mid$ $=\mid 1+w^{n}+w^{2 n} \quad w^{n} \quad w^{2 n}$ $w^{2 n}+1+w^{n} \quad 1 \quad w^{n}$ $w^{n}+w...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:axy + bcxy az bcz Solution: $a x y+b c x y-a z-b c z$ $=(a x y+b c x y)-(a z+b c z)$ [Grouping the expressions] $=x y(a+b c)-z(a+b c)$ $=(x y-z)(a+b c)$ $[$ Taking $(a+b c)$ as the common factor $]$...
Read More →In figure, BA ⊥ AC, DE ⊥ DF
Question: In figure, BA AC, DE DF such that BA = DE and BF = EC. Show that ΔABC ΔDEF. Thinking Process Use the RHS congruence rule to show the given result Solution: Given in figure, $B A \perp A C, D E \perp D F$ such that $B A=D E$ and $B F=E C$. To show $\triangle A B C \cong \triangle D E F .$ Proof Since, $B F=E C$ On adding CF both sides, we get $B F+C F=E C+C F$ $\Rightarrow$ $B C=E F$ $\ldots(1)$ In $\triangle A B C$ and $\triangle D E F$, $\angle A=\angle D=90^{\circ}$ $[\because B A \p...
Read More →The length of a rectangular field is 23 m more than its breadth.
Question: The length of a rectangular field is 23 m more than its breadth. If the perimeter of the field is 206 m, then its area is(a) 2420 m2(b) 2520 m2(c) 2480 m2(d) 2620 m2 Solution: (b) 2520 m2Let the breadth of the field bexm. Length = (x+ 23) mNow, Perimeter $=2($ Length $+$ Breadth $)=2(x+x+23)=(4 x+46) \mathrm{m}$ Thus, we have: $4 x+46=206$ $\Rightarrow 4 x=206-46=160$ $\Rightarrow x=\frac{160}{4}=40$ Breadth =x = 40 m Length $=x+23=40+23=63 \mathrm{~m}$ Area $=$ Length $\times$ Breadth...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:ab by ay +y2 Solution: $a b-b y-a y+y^{2}$ $=(a b-a y)+\left(y^{2}-b y\right)$ [Grouping the expressions] $=a(b-y)+y(y-b)$ $=a(b-y)-y(b-y)$ $[\because(\mathrm{y}-\mathrm{b})=-(\mathrm{b}-\mathrm{y})]$ $=(a-y)(b-y)$ $[$ Taking $(b-y)$ as the common factor $]$...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:2ax+bx+ 2ay+by Solution: $2 a x+b x+2 a y+b y$ $=(2 a x+b x)+(2 a y+b y)$ [Grouping the expressions] $=x(2 a+b)+y(2 a+b)$ $=(x+y)(2 a+b)$ [Taking $(2 a+b)$ as the common factor]...
Read More →The length of a rectangular hall is 5 m more than its breadth.
Question: The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 750 m2, then its length is(a) 15 m(b) 20 m(c) 25 m(d) 30 m Solution: (d) 30 mLet the length of the rectangle bex m. $\therefore$ Breadth of the rectangle $=(x-5) \mathrm{m}$ Area $=x(x-5)=x^{2}-5 x$ $\Rightarrow x^{2}-5 x=750$ $\Rightarrow x^{2}-5 x-750=0$ $\Rightarrow x^{2}-30 x+25 x-750=0$ $\Rightarrow x(x-30)+25(x-30)=0$ $\Rightarrow(x+25)(x-30)=0$ $\Rightarrow x+25=0$ and $x-30=0$ $\Rightarrow...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:x2+xy + xz + yz Solution: $x^{2}+x y+x z+y z$ $=\left(x^{2}+x y\right)+(x z+y z)$ [Grouping the expressions] $=x(x+y)+z(x+y)$ $=(x+z)(x+y)$ $[$ Taking $(x+y)$ as the common factor $]$ $=(x+y)(x+z)$...
Read More →The length of a rectangular hall is 5 m more than its breadth.
Question: The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 750 m2, then its length is(a) 15 m(b) 20 m(c) 25 m(d) 30 m Solution: (d) 30 mLet the length of the rectangle bex m. $\therefore$ Breadth of the rectangle $=(x-5) \mathrm{m}$ Area $=x(x-5)=x^{2}-5 x$ $\Rightarrow x^{2}-5 x=750$ $\Rightarrow x^{2}-5 x-750=0$ $\Rightarrow x^{2}-30 x+25 x-750=0$ $\Rightarrow x(x-30)+25(x-30)=0$ $\Rightarrow(x+25)(x-30)=0$ $\Rightarrow x+25=0$ and $x-30=0$ $\Rightarrow...
Read More →In the given figure,
Question: In the given figure, $\triangle C D E$ is an equilateral triangle formed on a side $C D$ of a square $A B C D$. Show that $\triangle A D E$ $\cong \triangle \mathrm{BCE}$. Solution: Given in figure $\triangle C D E$ is an equilateral triangle formed on a side $C D$ of a square $A B C D$. To show $\triangle \mathrm{ADE} \cong \triangle B C E$ Proof $\ln \triangle A D E$ and $\triangle B C E$, $D E=C E$ [sides of an equilateral triangle] $\angle A D E=\angle B C E$ $\left[\begin{array}{l...
Read More →If the determinant
Question: If the determinant $\left|\begin{array}{ccc}a b 2 a \alpha+3 b \\ b c 2 b \alpha+3 c \\ 2 a \alpha+3 b 2 b \alpha+3 c 0\end{array}\right|=0$, then (a) $a, b, c$ are in H.P. (b) $\alpha$ is a root of $4 a x^{2}+12 b x+9 c=0$ or $a, b, c$ are in G.P. (c) $a, b, c$ are in G.P. only (d) $a, b, c$ are in A.P. Solution: (b) $\alpha$ is a root of $4 a x^{2}+12 b x+9 c=0$ or $a, b, c$ are in G.P. Let $\Delta=\left|\begin{array}{lllllll}a b 2 a \alpha+3 b b c 2 b \alpha+3 c 2 a \alpha+3 b 2 b \...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:xa2+xb2ya2yb2 Solution: $x a^{2}+x b^{2}-y a^{2}-y b^{2}$ $=\left(x a^{2}+x b^{2}\right)-\left(y a^{2}+y b^{2}\right)$ [Grouping the expressions] $=x\left(a^{2}+b^{2}\right)-y\left(a^{2}+b^{2}\right)$ $=(x-y)\left(a^{2}+b^{2}\right)$ $\left[\right.$ Taking $\left(a^{2}+b^{2}\right)$ as the common factor $]$...
Read More →Find the area of a trapezium whose parallel sides are 11 ma and 25 m long,
Question: Find the area of a trapezium whose parallel sides are 11 ma and 25 m long, and the nonparallel sides are 15 m and 13 m long. Solution: Draw $D E \| B C$ and $D L$ perpendicular to $A B$. The opposite sides of quadrilateralDEBCare parallel. Hence,DEBCis a parallelogram.DE=BC= 13 mAlso $A E=(A B-E B)=(A B-D C)=(25-11)=14 \mathrm{~m}$ For $\Delta D A E$ Let:AE=a=14 mDE=b= 13 mDA=c=15 m Thus, we have: $s=\frac{a+b+c}{2}$ $s=\frac{14+13+15}{2}=21 \mathrm{~m}$ Area of $\Delta D A E=\sqrt{s(s...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions:ax + ay bx by Solution: $a x+a y-b x-b y$ $=(a x+a y)-(b x+b y)$ [Grouping the expressions] $=a(x+y)-b(x+y)$ $=(a-b)(x+y)$ $[$ Taking $(x+y)$ as the common factor $]$...
Read More →Find the area of a trapezium whose parallel sides are 11 ma and 25 m long,
Question: Find the area of a trapezium whose parallel sides are 11 ma and 25 m long, and the nonparallel sides are 15 m and 13 m long. Solution: Draw $D E \| B C$ and $D L$ perpendicular to $A B$. The opposite sides of quadrilateralDEBCare parallel. Hence,DEBCis a parallelogram.DE=BC= 13 mAlso $A E=(A B-E B)=(A B-D C)=(25-11)=14 \mathrm{~m}$ For $\Delta D A E$ Let:AE=a=14 mDE=b= 13 mDA=c=15 m Thus, we have: $s=\frac{a+b+c}{2}$ $s=\frac{14+13+15}{2}=21 \mathrm{~m}$ Area of $\Delta D A E=\sqrt{s(s...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions: 1 +x+xy+x2y Solution: $1+x+x y+x^{2} y$ $=(1+x)+\left(x y+x^{2} y\right)$ [Grouping the expressions] $=(1+x)+x y(1+x) $=(1+x y)(1+x)$ $[$ Taking $(1+x)$ as the common factor $]$...
Read More →Factorize each of the following expressions:
Question: Factorize each of the following expressions: qr pr + qs ps Solution: $p^{2} q-p r^{2}-p q+r^{2}$ $=\left(p^{2} q-p q\right)+\left(r^{2}-p r^{2}\right) \quad[$ Grouping the expressions $]$ $=p q(p-1)+r^{2}(1-p)$ $=p q(p-1)-r^{2}(p-1)$ $[\because(1-p)=-(p-1)]$ $=\left(p q-r^{2}\right)(p-1)$ $[$ Taking $(p-1)$ as the common factor $]$...
Read More →If a, b, c are distinct,
Question: If $a, b, c$ are distinct, then the value of $x$ satisfying $\left|\begin{array}{ccc}0 x^{2}-a x^{3}-b \\ x^{2}+a 0 x^{2}+c \\ x^{4}+b x-c 0\end{array}\right|=0$ is (a) $c$ (b) $a$ (c) $b$ (d) 0 Solution: (d) 0 When we put $x=0$ in the given matrix, then it turns out to be the skew symmetric matrix of order 3 and the determinant of the skew symmetric matrix of odd order is always 0 ....
Read More →The shape of the cross section of a canal is a trapezium.
Question: The shape of the cross section of a canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the bottom and the area of its cross section is 640 m2, find the depth of the canal. Solution: Area of the canal $=640 \mathrm{~m}^{2}$ Area of trapezium $=\frac{1}{2} \times($ Sum of parallel sides $) \times($ Distance between them $)$ $\Rightarrow 640=\frac{1}{2} \times(10+6) \times h$ $\Rightarrow \frac{1280}{16}=h$ $\Rightarrow h=80 \mathrm{~m}$ Therefore, the depth of the ca...
Read More →In figure, D and E are points on side
Question: In figure, D and E are points on side BC of a ΔABC such that BD = CE and AD = AE. Show that ΔABD ΔACE. Solution: Given $D$ and $E$ are the points on side $B C$ of a $\triangle A B C$ such that $B D=C E$ and $A D=A E$. To show $\triangle A B D \equiv \triangle A C E$ Proof We have, $A D=A E$ [given] $\Rightarrow$ $\angle A D E=\angle A E D$ $\ldots(1)$ [since, angles opposite to equal sides are equal] We have, $\angle A D B+\angle A D E=180^{\circ}$ [linear pair axiom] $\Rightarrow$ $\a...
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