Study the following statements
Question: Study the following statements Two intersecting lines cannot be perpendicular to the same line. Check whether it is an equivalent version to the Euclids fifth postulate. Solution: Two equivalent versions of Euclids fifth postulate are (i)For every line l and for every point P not lying on Z, there exists a unique line m passing through P and parallel to Z. (ii)Two distinct intersecting lines cannot be parallel to the same line. From above two statements it is clear that given statement...
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Question: Read the following statements: An equilateral triangle is a polygon made up of three line segments out of which two line segments are equal to the third-one and all its angles are 60 each. Define the terms used in this definition which you feel necessary. Are there any undefined terms in this? Can you justify that all sides and all angles are equal in a equilateral triangle. Solution: The terms need to be defined are (i)Polygon A closed figure bounded by three or more line segments. (i...
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Question: $x+y+z+1=0$ $a x+b y+c z+d=0$ $a^{2} x+b^{2} y+x^{2} z+d^{2}=0$ Solution: These equations can be written as $x+y+z=-1$ $a x+b y+c z=-d$ $a^{2} x+b^{2} y+x^{2} z=-d^{2}$ $D=\left|\begin{array}{ccc}1 1 1 \\ a b c \\ a^{2} b^{2} c^{2}\end{array}\right|$ $=\left|\begin{array}{ccc}1 0 0 \\ a a-b b-c \\ a^{2} a^{2}-b^{2} b^{2}-c^{2}\end{array}\right|$ [Applying $C_{2} \rightarrow C_{1}-C_{2}, C_{3} \rightarrow C_{2}-C_{3}$ ] Taking $(b-a)$ and $(c-a)$ common from $C_{1}$ and $C_{2}$, respect...
Read More →Simplify: (3x − 2)(2x − 3) + (5x − 3)(x + 1)
Question: Simplify:(3x 2)(2x 3) + (5x 3)(x+ 1) Solution: To simplify, we will proceed as follows: (3x 2)(2x 3) + (5x 3)(x + 1) $=[(3 x-2)(2 x-3)]+[(5 x-3)(x+1)]$ $=[3 x(2 x-3)-2(2 x-3)]+[5 x(x+1)-3(x+1)] \quad$ (Distributive law) $=6 x^{2}-9 x-4 x+6+5 x^{2}+5 x-3 x-3$ $=6 x^{2}+5 x^{2}-9 x-4 x+5 x-3 x-3+6 \quad$ (Rearranging) $=11 x^{2}-11 x+3 \quad$ (Combining like terms) Thus, the answer is $11 x^{2}-11 x+3$....
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Question: In the adjoining figure (i) $A B=B C, M$ is the mid-point of $A B$ and $N$ is the mid-point of $B C$. Show that $A M=N C$. (ii) $B M=B N, M$ is the mid-point of $A B$ and $N$ is the mid-point of $B C$. Show that $A B=B C$. Solution: (i) Given,$A B=B C$ $\ldots$ (i) $M$ is the mid-point of $A B$. $\therefore$ $A M=M B=\frac{1}{2} A B$ ...(ii) and $N$ is the mid-point of $B C$. $\therefore$ $B N=N C=\frac{1}{2} B C$ ....(iii) According to Euclid's axiom, things which are halves of the sa...
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Question: Simplify:(2x2+ 3x 5)(3x2 5x+ 4) Solution: To simplify, we will proceed as follows: (2x2+ 3x 5)(3x2 5x + 4) $=2 x^{2}\left(3 x^{2}-5 x+4\right)+3 x\left(3 x^{2}-5 x+4\right)-5\left(3 x^{2}-5 x+4\right) \quad$ (Distributive law) $=6 x^{4}-10 x^{3}+8 x^{2}+9 x^{3}-15 x^{2}+12 x-15 x^{2}+25 x-20$ $=6 x^{4}-10 x^{3}+9 x^{3}+8 x^{2}-15 x^{2}-15 x^{2}+12 x+25 x-20 \quad$ (Rearranging) $=6 x^{4}-x^{3}-22 x^{2}+36 x-20$ (Combining like terms) Thus, the answer is $6 x^{4}-x^{3}-22 x^{2}+36 x-20$...
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Question: $x+y=1$ $x+z=-6$ $x-y-2 z=3$ Solution: These equations can be written as $x+y+0 z=1$ $x+0 y+z=-6$ $x-y-2 z=3$ $D=\left|\begin{array}{ccc}1 1 0 \\ 1 0 1 \\ 1 -1 -2\end{array}\right|$ $=1(0+1)-1(-2-1)+0(-1-0)$ $=4$ $D_{1}=\left|\begin{array}{ccc}1 1 0 \\ -6 0 1 \\ 3 -1 -2\end{array}\right|$ $=1(0+1)-1(12-3)+0(6-0)$ $=-8$ $D_{2}=\left|\begin{array}{ccc}1 1 0 \\ 1 -6 1 \\ 1 3 -2\end{array}\right|$ $=1(12-3)-1(-2-1)+0(3+6)$ $=12$ $D_{3}=\left|\begin{array}{ccc}1 1 1 \\ 1 0 -6 \\ 1 -1 3\end{...
Read More →Simplify: (5 − x)(6 − 5x)( 2 − x)
Question: Simplify:(5 x)(6 5x)( 2 x) Solution: To simplify, we will proceed as follows: (5 x)(6 5x)( 2 x) $=[(5-x)(6-5 x)](2-x)$ $=[5(6-5 x)-x(6-5 x)](2-x)$ (Distributive law) $=\left(30-25 x-6 x+5 x^{2}\right)(2-x)$ $=\left(30-31 x+5 x^{2}\right)(2-x)$ $=2\left(30-31 x+5 x^{2}\right)-x\left(30-31 x+5 x^{2}\right)$ $=60-62 x+10 x^{2}-30 x+31 x^{2}-5 x^{3}$ $=60-62 x-30 x+10 x^{2}+31 x^{2}-5 x^{3} \quad$ (Rearranging) $=60-92 x+41 x^{2}-5 x^{3} \quad$ (Combining like terms) Thus, the answer is $6...
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Question: $x+y=1$ $x+z=-6$ $x-y-2 z=3$ Solution: These equations can be written as $x+y+0 z=1$ $x+0 y+z=-6$ $x-y-2 z=3$ $D=\left|\begin{array}{ccc}1 1 0 \\ 1 0 1 \\ 1 -1 -2\end{array}\right|$ $=1(0+1)-1(-2-1)+0(-1-0)$ $=4$ $D_{1}=\left|\begin{array}{ccc}1 1 0 \\ -6 0 1 \\ 3 -1 -2\end{array}\right|$ $=1(0+1)-1(12-3)+0(6-0)$ $=-8$ $D_{2}=\left|\begin{array}{ccc}1 1 0 \\ 1 -6 1 \\ 1 3 -2\end{array}\right|$ $=1(12-3)-1(-2-1)+0(3+6)$ $=12$ $D_{3}=\left|\begin{array}{ccc}1 1 1 \\ 1 0 -6 \\ 1 -1 3\end{...
Read More →Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm.
Question: Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle. Solution: Let the height of the triangle behcm. Each of the equal sides measures $a=(h+2) \mathrm{cm}$ and $b=12 \mathrm{~cm}$ (base). Now,Area of the triangle = Area of the isosceles triangle $\frac{1}{2} \times$ base $\times$ height $=\frac{1}{4} \times b \sqrt{4 a^{2}-b^{2}}$ $\Rightarrow \frac{1}{2} \times 12 \times h=\frac...
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Question: In the adjoining figure, if OX = XY, PX = XZ and OX = PX, show that XY = XZ. Solution: Given OX= XY = 2 OX = XY .(i) PX= XZ = 2 PX = XZ .(ii) and OX = PX (iii) According to Euclids axiom, things which are double of the same things are equal to one another. On multiplying Eq. (iii) by 2, we get 2 OX = 2 PX XY=XZ. [from Eqs. (i) and (ii)]...
Read More →Simplify: (5x + 3)(x − 1)(3x − 2)
Question: Simplify:(5x+ 3)(x 1)(3x 2) Solution: To simplify, we will proceed as follows: (5x+ 3)(x 1)(3x 2) $=[(5 x+3)(x-1)](3 x-2)$ $=[5 x(x-1)+3(x-1)](3 x-2)$ (Distributive law) $=\left[5 x^{2}-5 x+3 x-3\right](3 x-2)$ $=\left[5 x^{2}-2 x-3\right](3 x-2)$ $=3 x\left(5 x^{2}-2 x-3\right)-2\left(5 x^{2}-2 x-3\right)$ $=15 x^{3}-6 x^{2}-9 x-\left[10 x^{2}-4 x-6\right]$ $=15 x^{3}-6 x^{2}-9 x-10 x^{2}+4 x+6$ $=15 x^{3}-6 x^{2}-10 x^{2}-9 x+4 x+6 \quad$ (Rearranging) $=15 x^{3}-16 x^{2}-5 x+6 \quad...
Read More →In the adjoining figure,
Question: In the adjoining figure, we have AC = DC and CB = CE. Show that AB = DE. Solution: Given, AC = DC (i) and C6 = CE (ii) According to Euclids axiom, if equals are added to equals, then wholes are also equal. So, on adding Eqs.(i) and (ii), we get AC + CB = DC + CE = AB = DE...
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Question: In the adjoining figure, we have ABC = ACB and 3 = 4. Show that BD = DC. Solution: Given, ABC = ACB (i) and 4 = 3 (ii) According to Eulids axiom, if equals are subtracted from equals, then remainders are also equal. On subtracting Eq. (ii) from Eq. (i), we get ABC 4 = ACB 3 =1 = 2 Now, in ABDC, 1=2 = DC =BD [sides opposite to equal angles are equal] BD = DC....
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Question: $2 x-3 y-4 z=29$ $-2 x+5 y-z=-15$ $3 x-y+5 z=-11$ Solution: Given: $2 x-3 y-4 z=29$ $-2 x+5 y-z=-15$ $3 x-y+5 z=-11$ $D=\left|\begin{array}{ccc}2 -3 -4 \\ -2 5 -1 \\ 3 -1 5\end{array}\right|$ $=2(25-1)+3(-10+3)-4(2-15)$ $=2(24)+3(-7)-4(-13)$ $=79$ $D_{1}=\left|\begin{array}{ccc}29 -3 -4 \\ -15 5 -1 \\ -11 -1 5\end{array}\right|$ $=29(25-1)+3(-75-11)-4(15+55)$ $=29(24)+3(-86)-4(70)$ $=158$ $D_{2}=\left|\begin{array}{ccc}2 29 -4 \\ -2 -15 -1 \\ 3 -11 5\end{array}\right|$ $=2(-75-11)-29(-...
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Question: In the adjoining figure, we have 1 =3 and 2 = 4. Show that A = C. Solution: Given,1 = 3 (i) and 2 = 4 (ii) According to Euclids axiom, if equals are added to equals, then wholes are also equal. On adding Eqs. (i) and (ii), we get 1 + 2 = 3 +4 = A = C...
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Question: Simplify:(x3 2x2+ 5x 7)(2x 3) Solution: To simplify, we will proceed as follows: (x3 2x2+ 5x 7)(2x 3) $=2 x\left(x^{3}-2 x^{2}+5 x-7\right)-3\left(x^{3}-2 x^{2}+5 x-7\right)$ $=2 x^{4}-4 x^{3}+10 x^{2}-14 x-3 x^{3}+6 x^{2}-15 x+21$ $=2 x^{4}-4 x^{3}-3 x^{3}+10 x^{2}+6 x^{2}-14 x-15 x+21 \quad$ (Rearranging) $=2 x^{4}-7 x^{3}+16 x^{2}-29 x+21$ (Combining like terms) Thus, the answer is $2 x^{4}-7 x^{3}+16 x^{2}-29 x+21$....
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Question: Simplify:(x3 2x2+ 5x 7)(2x 3) Solution: To simplify, we will proceed as follows: (x3 2x2+ 5x 7)(2x 3) $=2 x\left(x^{3}-2 x^{2}+5 x-7\right)-3\left(x^{3}-2 x^{2}+5 x-7\right)$ $=2 x^{4}-4 x^{3}+10 x^{2}-14 x-3 x^{3}+6 x^{2}-15 x+21$ $=2 x^{4}-4 x^{3}-3 x^{3}+10 x^{2}+6 x^{2}-14 x-15 x+21 \quad$ (Rearranging) $=2 x^{4}-7 x^{3}+16 x^{2}-29 x+21$ (Combining like terms) Thus, the answer is $2 x^{4}-7 x^{3}+16 x^{2}-29 x+21$....
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Question: In the adjoining figure, we have 1 = 2 and 2= 3. Show that 1 = 3. Solution: Given, 1 =2 (i) and 2 = 3 (ii) According to Euclids axiom, things which are equal to the same thing are equal to one another. From Eqs. (i) and (ii), 1 = 3...
Read More →The base of an isosceles triangle measures 80 cm and its area is 360 cm2.
Question: The base of an isosceles triangle measures 80 cm and its area is 360 cm2. Find the perimeter of the triangle. Solution: Given :Base = 80 cm Area $=360 \mathrm{~cm}^{2}$ Area of an isosceles triangle $=\left(\frac{1}{4} b \sqrt{4 a^{2}-b^{2}}\right)$ $\Rightarrow 360=\frac{1}{4} \times 80 \sqrt{4 a^{2}-80^{2}}$ $\Rightarrow 360=20 \sqrt{4 a^{2}-6400}$ $\Rightarrow 18=2 \sqrt{a^{2}-1600}$ $\Rightarrow 9=\sqrt{a^{2}-1600}$ Squaring both the sides, we get: $\Rightarrow 81=a^{2}-1600$ $\Rig...
Read More →In the adjoining figure,
Question: In the adjoining figure, if BX = AB, BY = BC and AB = BC, then show that BX = BY. Thinking Process Apply the Euclids axiom, things which are double of the same things are equal to one another. Solution: Given, BX = AB =2 BX = AB ,..(i) BY = BC = 2 BY = BC (ii) and AB = BC (iii) On putting the values from Eqs. (i) and (ii) in Eq. (iii), we get 2BX = 2BY According to Euclids axiom, things which are double of the same things are equal to one another. BX = BY...
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Question: Simplify:x2(xy)y2(x+ 2y) Solution: To simplify, we will proceed as follows: x2(xy)y2(x+ 2y) $=\left[x^{2}(x-y)\right]\left[y^{2}(x+2 y)\right]$ $=\left(x^{3}-x^{2} y\right)\left(x y^{2}+2 y^{3}\right)$ $=x^{3}\left(x y^{2}+2 y^{3}\right)-x^{2} y\left(x y^{2}+2 y^{3}\right)$ $=x^{4} y^{2}+2 x^{3} y^{3}-\left[x^{3} y^{3}+2 x^{2} y^{4}\right]$ $=x^{4} y^{2}+2 x^{3} y^{3}-x^{3} y^{3}-2 x^{2} y^{4}$ $=x^{4} y^{2}+x^{3} y^{3}-2 x^{2} y^{4}$ Thus, the answer is $x^{4} y^{2}+x^{3} y^{3}-2 x^{2...
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Question: In the adjoining figure, we have X and Y are the mid-points of AC and BC and AX = CY. Show that AC = BC. Solution: Given, X is the mid-point of AC AX = CX = AC = 2AX =2CX = AC (i) and Y is the mid-point of BC. BY =CY = BC = 2BY=2CY = BC (ii) Also, given AX=CY (iii) According to Euclids axiom, things which are double of the same things are equal to one another. From Eq. (iii), 2AX = 2CY = AC = BC [from Eqs. (i) and (ii)]...
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Question: Simplify:a2b2(a+ 2b)(3a+b) Solution: To simplify, we will proceed as follows: a2b2(a+ 2b)(3a+b) $=\left[a^{2} b^{2}(a+2 b)\right](3 a+b)$ $=\left(a^{3} b^{2}+2 a^{2} b^{3}\right)(3 a+b)$ $=3 a\left(a^{3} b^{2}+2 a^{2} b^{3}\right)+b\left(a^{3} b^{2}+2 a^{2} b^{3}\right)$ $=3 a^{4} b^{2}+6 a^{3} b^{3}+a^{3} b^{3}+2 a^{2} b^{4}$ $=3 a^{4} b^{2}+7 a^{3} b^{3}+2 a^{2} b^{4}$ Thus, the answer is $3 a^{4} b^{2}+7 a^{3} b^{3}+2 a^{2} b^{4}$....
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Question: $5 x-7 y+z=11$ $6 x-8 y-z=15$ $3 x+2 y-6 z=7$ Solution: Given $5 x-7 x+7=11$ $6 x-8 y-z=15$ $3 x+2 y-6 z=7$ $D=\left|\begin{array}{ccc}5 -7 1 \\ 6 -8 -1 \\ 3 2 -6\end{array}\right|$ $=5(48+2)+7(-36+3)+1(12+24)$ $=5(50)+7(-33)+1(36)$ $=55$ $D_{1}=\left|\begin{array}{ccc}11 -7 1 \\ 15 -8 -1 \\ 7 2 -6\end{array}\right|$ $=11(48+2)+7(-90+7)+1(30+56)$ $=11(50)+7(-83)+1(86)$ $=55$ $D_{2}=\left|\begin{array}{ccc}5 11 1 \\ 6 15 -1 \\ 3 7 -6\end{array}\right|$ $=5(-90+7)-11(-36+3)+1(42-45)$ $=5...
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