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Maths 3 Marks Question

Model Paper 2 · Gujarat Board (GSEB) STD 9 (GSEB - English Medium). 8 questions.

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3 Marks Question

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Question 13 Marks
Draw the graphs of y = x and y = -x in the same graph. Also find the co-ordinates of the point where the two lines intersect.
Answer
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Question 23 Marks
In $\triangle A B C, D$ is the midpoint of BC . if $D L \perp A B$ and $D M \perp A C$ such that $DL = DM$. prove that $AB = AC$.
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Answer
Given that in a $\triangle ABC , D$ is the midpoint of $BC$ and $DL \perp A B$ and $D M \perp A C$ also, $DL = DM$
To prove $AB = AC$
Proof : IN right - angled $\Delta B L D$ and $\Delta C M D$
$\angle BLD=\angle CMD=90^{\circ}$
BD = CD
DL = DM 
Thus, by right angle hypotenuse side criterion of congruence, we have 
$\Delta BLD \cong \Delta CMD$
The corresponding parts of the congruent triangled are equal. 
$\angle ABD=\angle ACD$
In $\triangle ABC$, we have
$\Rightarrow AB=AC$
sides opposite to equal angles are equal
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Question 33 Marks
$\ce{\triangle ABC}$ is an isosceles triangle in which $\ce{AB = AC}$. Side $BA$ is produced to $D$ such that $\ce{AD = AB}$. Show that $\ce{\angle BCD}$ is a right angle.
Answer
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Given : $\ce{\triangle ABC}$ is an isosceles triangle in which $\ce{AB = AC}$.
Side $\ce{BA}$ is produced to $D$ such that $\ce{AD=AB}$.
To Prove : $\ce{\angle BCD}$ is a right angle.
Proof: As $\ce{ABC}$ is an isosceles triangle
$\ce{\angle ABC =\angle ACB}\ldots\ldots(1)$
$\ce{AC = AD}\ldots\ldots$[As given : $\ce{AB = AC}$ and $\ce{AD = AB}]$
In $\ce{\triangle ACD}$,
$\ce{\angle CDA =\angle ACD} \ldots \ldots[\angle s$  opposite to equal side of a  $\triangle]$
$\ce{\angle CBD =\angle ACD} \ldots(2)$
$\ce{\angle ABC + \angle CDB = \angle ACB  + \angle ACD} \ldots \ldots [$ Adding corresponding sides from $(1)$ and $(2)]$
$\ce{\angle ABC + \angle CDB =\angle BCD} \ldots\ldots (3)$
In $\ce{\triangle BCD}$
$\ce{\angle BCD +\angle DBC +\angle CDB}=180^{\circ} \ldots[$ Sum of three angles of a triangle $]$
$\ce{\therefore \angle BCD + \angle ABC + \angle CDB =180^{\circ}}$
$\ce{\angle BCD +\angle BCD} =180^{\circ} \ldots[$ From $(3)]$
$\ce{\therefore 2 \angle BCD} =180^{\circ}$
$\therefore \ce{\angle BCD} =90^{\circ}$
$\therefore \ce{\angle BCD}$ is a right angle proved.
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Question 43 Marks
A heap of wheat is in the form of a cone whose diameter is $10.5 m$ and height is $3 m.$ Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Answer
For heap of wheat
Diameter $= 10.5 m$
$\therefore$ Radius $( r )=\frac{10.5}{2} \ cm=5.25 m$
Height $( h )=3 m$
$\therefore$ Volume $=\frac{1}{3} \pi r^2 h$
$=\frac{1}{3} \times \frac{22}{7} \times(5.25)^2 \times 3$
$=86.625 m^3$
Slant height, $l=\sqrt{r^2+h^2}$
$=\sqrt{(5.25)^2+(3)^2}$
$=\sqrt{27.5625+9}$
$=\sqrt{36.5625}$
$=6.05 m$
$\therefore$ Curved surface area $=\pi rl$
$=\frac{22}{7} \times 5.25 \times 6.05$
$=99.825 m^2$
$\therefore$ The area of the canvas required is $99.825 m^2$
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Question 53 Marks
The sides of a triangular plot are in the ratio of $3 : 5 : 7$ and its perimeter is $300 m.$ Find its area.
Answer
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Suppose that the sides in metres are $3x, 5x$ and $7x.$
Then, we know that $3x + 5x + 7x = 300 ($Perimeter of the triangle$)$
Therefore, $15x = 300,$ which gives $x = 20.$
So the sides of the triangles are $3 \times 20 m, 5 \times 20 m$ and $7 \times 20 m$
i.e., $60m, 100m$ and $140m.$
We have $s=\frac{60+100+140}{2}=150 m$
and area will be $=\sqrt{150(150-60)(150-100)(150-140)}$
$=\sqrt{150 \times 90 \times 50 \times 10}$
$=1500 \sqrt{3} m^2$
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Question 63 Marks
The cost of leveling the ground in the form of a triangle having the sides $51 m, 37 m$ and $20 m$ at the rate of $Rs .3\ per\ m ^2$ is $Rs.918.$ State whether the statement is true or false and justify your answer.
Answer
True, Let $a = 51m, b = 37m, c = 20m$
$s=\frac{a+b+c}{2}=\frac{51+37+20}{2}=\frac{108}{2}=54 m$
$\therefore$ Area of triangular ground $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{54(54-51)(54-37)(54-20)}$
$=\sqrt{54 \times 3 \times 17 \times 34}$
$=\sqrt{9 \times 3 \times 2 \times 3 \times 17 \times 17 \times 2}$
$=3 \times 3 \times 2 \times 17$
$=306 m^2$
Cost of leveling the ground $=Rs.3 \times 306= Rs. 918$
Hence the cost of leveling the ground in the form of a triangle is $Rs. 918.$
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Question 73 Marks
If $x^2+\frac{1}{x^2}=34$, find $x^3+\frac{1}{x^3}-9$.
Answer
As we know
$\left(x+\frac{1}{x}\right)^2$
$=x^2+\frac{1}{x^2}+2$
$=34+2$
$=36$
$\left(x+\frac{1}{x}\right)=6$
On cubing both sides, we get
$\left(x+\frac{1}{x}\right)^3=6^3$
$\Rightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=216$
$\Rightarrow x^3+\frac{1}{x^3}+3 \times 6=216$
$\Rightarrow x^3+\frac{1}{x^3}=198$
$\Rightarrow x^3+\frac{1}{x^3}-9$
$=198-9$
$=189$
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Question 83 Marks
Locate $\sqrt{3}$ on the number line.
Answer
Let point $A$ represents $1$ as shown in Figure.
Clearly, $\text{OA}=1$ unit.
Now, draw a right triangle $\text{OAB}$ in which $\text{AB=OA}=1$ unit.
By Using Pythagoras theorem, we have
$\ce{OB^2 = OA^2 + AB^2}$
$=1^2+1^2$
$=2$
$\Rightarrow \text{OB}=\sqrt{2}$
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Taking $O$ as centre and $O B$ as a radius draw an arc intersecting the number line at point $P$.
Then $p$ corresponds to $\sqrt{2}$ on the number line.
Now draw $DB$ of unit length perpendicular to $OB .$
By using Pythagoras theorem, we have
$\ce{OD^2 = OB^2 + DB^2}$
$OD^2=(\sqrt{2})^2+12$
$=2+1=3$
$OD=\sqrt{3}$
Taking $O$ as centre and $OD$ as a radius draw an arc which intersects the number line at the point $Q.$
Clearly, $Q$ corresponds to $\sqrt{3}$.
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