Question 13 Marks
In fig find the vertices' co-ordinates of $\triangle ABC$
Answer
View full question & answer→(A) (0, 0), (B) (3, 4), (c) (-4, 4)
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Question 23 Marks
Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Answer
View full question & answer→Given $\triangle A B C$ in which $E$ and $F$ are mid points of side $A B$ and $A C$ respectively.
To prove: $EF \| BC$
Construction: Produce EF to D such that EF = FD. Join CD
Proof: In $\triangle A E F$ and $\triangle C D F$
$AF = FC [\because F$ is mid - point of $A C]$
$\angle 1=\angle 2$ [vertically opposite angles]
EF=FD [By construction]
$\therefore \triangle A E F \cong \triangle C D F[B y S A S]$
And $\therefore A E=C D[B y C P C T]$
$AE = BE [\because E$ is the mid-point $]$
And $\therefore B E=c d$
$A B \| C D[\therefore \angle B A C=\angle A C D]$
$\therefore B C D E$ is a paralle $\log$ ram
$E F \| B C$ Henceproved
To prove: $EF \| BC$
Construction: Produce EF to D such that EF = FD. Join CD
Proof: In $\triangle A E F$ and $\triangle C D F$
$AF = FC [\because F$ is mid - point of $A C]$
$\angle 1=\angle 2$ [vertically opposite angles]
EF=FD [By construction]
$\therefore \triangle A E F \cong \triangle C D F[B y S A S]$
And $\therefore A E=C D[B y C P C T]$
$AE = BE [\because E$ is the mid-point $]$
And $\therefore B E=c d$
$A B \| C D[\therefore \angle B A C=\angle A C D]$
$\therefore B C D E$ is a paralle $\log$ ram
$E F \| B C$ Henceproved
Question 33 Marks
Through A, B and C, lines RQ, PR and QP have been drawn, respectively parallel to sides BC, CA and AB of a $\triangle ABC$ as shown in Fig., Show that $BC =\frac{1}{2} QR$
Answer
Given,
$PQ \| AB , PR \| AC$ and $RQ \| BC$.
In quadrilateral BCAR,
$BR \| CA$ and $BC \| RA$
$\therefore$ BCAR is a parallelogram
$\therefore B C=A R \ldots( i )$
Now, in quadrilateral BCQA,;
$BC \| AQ$ and $AB \| QC$
$\therefore B C Q A$ is a parallelogram
$\therefore B C=A Q \ldots$ (ii)
Adding Eqn. (i) and (ii), we get
2BC = AR + AQ
2BC = RQ
$BC =\frac{ QR }{2}$
Hence proved.
View full question & answer→Given,
$PQ \| AB , PR \| AC$ and $RQ \| BC$.
In quadrilateral BCAR,
$BR \| CA$ and $BC \| RA$
$\therefore$ BCAR is a parallelogram
$\therefore B C=A R \ldots( i )$
Now, in quadrilateral BCQA,;
$BC \| AQ$ and $AB \| QC$
$\therefore B C Q A$ is a parallelogram
$\therefore B C=A Q \ldots$ (ii)
Adding Eqn. (i) and (ii), we get
2BC = AR + AQ
2BC = RQ
$BC =\frac{ QR }{2}$
Hence proved.
Question 43 Marks
Write linear equation 3x + 2y =18 in the form of ax + by + c = 0. Also write the values of a, b and c. Are (4, 3) and (1, 2) solution of this equation?
Answer
View full question & answer→We have the equation as 3x + 2y = 18
In standard form
3x + 2y - 18 = 0
or 3x + 2y + (-18) =0
But standard linear equation is
ax + by + c = 0
on comparison we get, a = 3, b = 2, c = -18
If (4, 3) lie on the line, i.e., solution of the equation LHS=RHS
$\therefore 3(4)+2(3)=18$
12 + 6 = 18
18 = 18
As LHS = RHS, Hence (4, 3) is the solution of given equation.
Again for (1,2)
3x + 2y = 18
$\therefore 3(1)+2(2)=18$
3 + 4 = 18
7 = 18
LHS $\neq$ RHS
Hence (1, 2) is not the solution of given equation.
Therefore (4,3) is the point where the equation of the line 3x + 2y = 18 passes through where as the line for the equation 3x + 2y=18 does not pass through the point (1,2).
In standard form
3x + 2y - 18 = 0
or 3x + 2y + (-18) =0
But standard linear equation is
ax + by + c = 0
on comparison we get, a = 3, b = 2, c = -18
If (4, 3) lie on the line, i.e., solution of the equation LHS=RHS
$\therefore 3(4)+2(3)=18$
12 + 6 = 18
18 = 18
As LHS = RHS, Hence (4, 3) is the solution of given equation.
Again for (1,2)
3x + 2y = 18
$\therefore 3(1)+2(2)=18$
3 + 4 = 18
7 = 18
LHS $\neq$ RHS
Hence (1, 2) is not the solution of given equation.
Therefore (4,3) is the point where the equation of the line 3x + 2y = 18 passes through where as the line for the equation 3x + 2y=18 does not pass through the point (1,2).
Question 53 Marks
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $13 m, 14 m$ and $15 m$. The advertisements yield an earning of $Rs\ 2000$ per $m ^2$ a year. A company hired one of its walls for $6$ months. How much rent did it pay?
Answer
View full question & answer→The sides of triangular side walls of flyover which have been used for advertisements are $13 m, 14 m, 15 m.$
$s=\frac{13+14+15}{2}=\frac{42}{2}=21 m$
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21 \times 8 \times 7 \times 6}$
$=\sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 3 \times 2}$
$=7 \times 3 \times 2 \times 2=84 m^2$
It is given that the advertisement yield an earning of $₹ 2,000$ per $m^2$ a year.
$\therefore$ Rent for $1 m^2$ for 1 year $= ₹\ 2000$
So, rent for $1 m^2$ for $6$ months or $\frac{1}{2}$ year $=R s\left(\frac{1}{2} \times 2000\right)= ₹\ 1,000 $.
$\therefore$ Rent for $84 m^2$ for $6$ months $=$ ₹ $(1000 \times 84)=$ ₹ $84,000$.
$s=\frac{13+14+15}{2}=\frac{42}{2}=21 m$
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21 \times 8 \times 7 \times 6}$
$=\sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 3 \times 2}$
$=7 \times 3 \times 2 \times 2=84 m^2$
It is given that the advertisement yield an earning of $₹ 2,000$ per $m^2$ a year.
$\therefore$ Rent for $1 m^2$ for 1 year $= ₹\ 2000$
So, rent for $1 m^2$ for $6$ months or $\frac{1}{2}$ year $=R s\left(\frac{1}{2} \times 2000\right)= ₹\ 1,000 $.
$\therefore$ Rent for $84 m^2$ for $6$ months $=$ ₹ $(1000 \times 84)=$ ₹ $84,000$.
Question 63 Marks
One side of an equilateral triangle is $8 \ cm.$ Find its area by using Heron's Formula. Find its altitude also.
Answer
View full question & answer→$a = 8 \ cm, b = 8 \ cm, c = 8 \ cm.$
$s=\frac{a+b+c}{2}$
$\therefore \frac{8+8+8}{2} m=12 \ cm$
$\therefore$ Area of the equilateral triangle
$=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{12(12-8)(12-8)(12-8)}$
$=\sqrt{12(4)(4)(4)}$
$=\sqrt{(4)(3)(4)(4)(4)}$
$=16 \sqrt{3} \ cm^2$
$\text { Area }=\frac{1}{2} \times \text { Base } \times \text { Altitude }$
$=16 \sqrt{3}=\frac{1}{2} \times 8 \times \text { Altitude }$
$=16 \sqrt{3}=4 \text { Altitude }$
$\text { Altitude }=\frac{16 \sqrt{3}}{4}$
$=4 \sqrt{3} \ cm$
$s=\frac{a+b+c}{2}$
$\therefore \frac{8+8+8}{2} m=12 \ cm$
$\therefore$ Area of the equilateral triangle
$=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{12(12-8)(12-8)(12-8)}$
$=\sqrt{12(4)(4)(4)}$
$=\sqrt{(4)(3)(4)(4)(4)}$
$=16 \sqrt{3} \ cm^2$
$\text { Area }=\frac{1}{2} \times \text { Base } \times \text { Altitude }$
$=16 \sqrt{3}=\frac{1}{2} \times 8 \times \text { Altitude }$
$=16 \sqrt{3}=4 \text { Altitude }$
$\text { Altitude }=\frac{16 \sqrt{3}}{4}$
$=4 \sqrt{3} \ cm$
Question 73 Marks
If $x+y+z=0$, show that $x^3+y^3+z^3=3 x y z$.
Answer
View full question & answer→We know that
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$\left(\right.$ Using Identity $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right)$
$=(0)\left(x^2+y^2+z^2-x y-y z-z x\right)(\because x+y+z=0)$
$=0$
$\Rightarrow x^3+y^3+z^3=3 x y z$
$x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
$\left(\right.$ Using Identity $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right)$
$=(0)\left(x^2+y^2+z^2-x y-y z-z x\right)(\because x+y+z=0)$
$=0$
$\Rightarrow x^3+y^3+z^3=3 x y z$
Question 83 Marks
Find the values of $a$ and $b$ in each of $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a-b \sqrt{3}$
Answer
View full question & answer→$\text{LHS} =\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}$
$=\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$
$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{(7)^2-(4 \sqrt{3})^2}$
$=\frac{35-20 \sqrt{3}+14 \sqrt{3}-24}{49-48}$
$=\frac{11-6 \sqrt{3}}{1}=11-6 \sqrt{3}$
Now, $11-6 \sqrt{3}=a-b \sqrt{3}$
$\Rightarrow a=11 , b=6$
$=\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$
$=\frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{(7)^2-(4 \sqrt{3})^2}$
$=\frac{35-20 \sqrt{3}+14 \sqrt{3}-24}{49-48}$
$=\frac{11-6 \sqrt{3}}{1}=11-6 \sqrt{3}$
Now, $11-6 \sqrt{3}=a-b \sqrt{3}$
$\Rightarrow a=11 , b=6$