5 Marks Questions

Question 15 Marks

Find the integral roots of the polynomial $f(x)=x^3+6 x^2+11 x+6$.

Answer

Given, that $f(x)=x^3+6 x^2+11 x+6$
Clearly we can say that, the polynomial $f(x)$ with an integer coefficient and the highest degree term coefficient which is known as leading factor is $1.$
So, the roots of $f(x)$ are limited to integer factor of $6 ,$
they are $\pm 1, \pm 2, \pm 3, \pm 6$
Let $x = -1$
$f(-1)=(-1)^3+6(-1)^2+11(-1)+6$
$=-1+6-11+6$
$=0$
Let $x = -2$
$f(-2)=(-2)^3+6(-2)^2+11(-2)+6$
$=-8-(6 * 4)-22+6$
$=-8+24-22+6$
$=0$
Let $x = – 3$
$f(-3)=(-3)^3+6(-3)^2+11(-3)+6$
$=-27-(6 \times 9)-33+6$
$=-27+54-33+6$
$=0$
But from all the given factors only $-1,-2,-3$ gives the result as zero.
Furher, since $f(x)$ is a polynomial of degree $3 ,$
therefore, it has almost $3$ roots.
Therefore, the integral roots of $x^3+6 x^2+11 x+6$ are $-1,-2,-3$.

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Question 25 Marks

The sides of a triangle are in the ratio $5 : 12 : 13$ and its perimeter is $150 m.$ Find the area of the triangle.

Answer

Given that the sides of a triangle are in the ratio $5: 12: 13$ and its perimeter is $150 m$
Let the sides of the triangle be $5x m, 12x m$ and $13x m.$
We know:
Perimeter $=$ Sum of all sides
or, $150=5 x+12 x+13 x$
or, $30 x =150$
or, $x=5$
Thus, we obtain the sides of the triangle.
$5 \times 5=25 m$
$12 \times 5=60 m$
$13 \times 5=65 m$
Now,
Let: $a = 25 m, b = 60 m$ and $c = 65 m$
$\therefore s=\frac{150}{2}=75 m$
$\Rightarrow s =75 m$
By Heron's formula, we have
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{75(75-25)(75-60)(75-65)}$
$=\sqrt{75 \times 50 \times 15 \times 10}$
$=\sqrt{15 \times 5 \times 5 \times 10 \times 15 \times 10}$
$=15 \times 5 \times 10$

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Question 35 Marks

Find the percentage increase in the area of a triangle if its each side is doubled.

Answer

Let $a, b, c$ be the sides of the old triangle and s be its semi$-$perimeter. Then,
$s=\frac{1}{2}(a+b+c)$
The sides of the new triangle are $2a, 2b$ and $2c.$
Let $s$ be its semi$-$perimeter. Then,
$s^{\prime}=\frac{1}{2}(2 a+2 b+2 c)=a+b+c=2 s$
$\Rightarrow s^{\prime}=2 s$
Let $\Delta$ and $\Delta^{\prime}$ be the areas of the old and new triangles respectively. Then
$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
and,
$\Delta^{\prime}=\sqrt{s^{\prime}\left(s^{\prime}-2 a\right)\left(s^{\prime}-2 b\right)\left(s^{\prime}-2 c\right)}$
$\Rightarrow \Delta^{\prime}=\sqrt{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)} \quad\left[\because s^{\prime}=2 s\right]$
$\Rightarrow \Delta^{\prime}=\sqrt{16 s(s-a)(s-b)(s-c)}$
$\Rightarrow \Delta^{\prime}=4 \sqrt{s(s-a)(s-b)(s-c)}=4 \Delta[$ from $(1)]$
$\therefore$ Increase in the area of the triangle $=\Delta^{\prime}-\Delta=4 \Delta-\Delta=3 \Delta$
Hence, percentage increase in area $=\left(\frac{3 \Delta}{\Delta} \times 100\right)=300 \%$

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Question 45 Marks

An iron pillar consists of a cylindrical portion $2.8\ m$ high and $20 \ cm$ in diameter and a cone $42 \ cm$ high is surmounting it. Find the weight of the pillar, given that $1 \ cm^3$ of iron weighs $7.5\ g.$

Answer

We are Given that,
An iron pillar consists of a cylindrical portion and a cone mounted on it.
The height of the cylindrical portion of the pillar, $H = 2.8\ m = 280 \ cm$.
The height of the conical portion of the pillar, $h = 42 \ cm.$
The diameter of the cylindrical portion of the pillar = diameter of the circular base of cone $= D = 20 \ cm.$
The radius of the circular base of cylinder/ cone $r=\frac{D}{2}=10 \ cm$.
Now, we have,
Volume of the pillar, $(V) =$ Volume of the cylindrical portion of pillar $+$ volume of the conical portion of the pillar.
$\Rightarrow V =\pi r^2 H+\frac{1}{3} \pi r^2 h$
$\Rightarrow V=\left(\frac{22}{7} \times 10^2 \times 280+\frac{1}{3} \times \frac{22}{7} \times 10^2 \times 42\right) \ cm ^3$
$\Rightarrow V=(22 \times 100 \times 40+22 \times 100 \times 2) \ cm ^3$
$\Rightarrow V=(88000+4400) \ cm ^3$
$\Rightarrow V=92400 \ cm^3$
Hence, volume of iron pillar is $92400 \ cm^3$
Given,
Weight of $1 \ cm^3$ iron $=7.5\ gm$.
Hence, weight of $92400 \ cm^3$ iron $=7.5 \times 92400\ gm$.
$= 693000\ gm.$
$= 693 \ Kg.$
Since, $1 \ Kg =1000 gm$.
Hence, the weight of iron piller is $693 \ Kg .$

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Question 55 Marks

In the given figure, $AB \| CD$ and $\angle A O C=x^{\circ}$. If $\angle O A B=104^{\circ}$ and $\angle O C D=116^{\circ}$, find the value of $x.$

Answer

Through $O$ draw $OE \| AB \| CD$
Then, $\angle A O E+\angle C O E=x^{\circ}$
Now, $AB \| OE$ and $AO$ is the transversal
$\therefore \angle O A B+\angle A O E=180^{\circ}$
$\Rightarrow 104^{\circ}+\angle A O E=180^{\circ}$
$\Rightarrow \angle A O E=(180-104)^{\circ}=76^{\circ}\ldots\ldots (1)$
Again, $C D \| O E$ and $O C$ is the transversal
$\therefore \angle C O E+\angle O C D=180^{\circ}$
$\Rightarrow \angle C O E+116^{\circ}=180^{\circ}$
$\Rightarrow \angle C O E=\left(180^{\circ}-116^{\circ}\right)=64^{\circ}\ldots\ldots (2)$
$\therefore \angle A O C=\angle A O E+\angle C O E=\left(76^{\circ}+64^{\circ}\right)=140^{\circ}[$ from $(1)$ and $(2)]$
Hence, $x^{\circ}=140^{\circ}$

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Question 65 Marks

In each of the figures given below, $AB \| CD$. Find the value of $x ^{\circ}$ in each case.

Answer

Draw $E F\|A B\| C D$
Now, $AB \| EF$ and $BE$ is the transversal.
Then,
$\angle A B E=\angle B E F [$ Alternate Interior Angles $]$
$\Rightarrow \angle B E F=35^{\circ}$
Again, $EF \| CD$ and $DE$ is the transversal
Then,
$\angle D E F=\angle F E D$
$\Rightarrow \angle F E D=65^{\circ}$
$\therefore x^{\circ}=\angle B E F+\angle F E D$
$x ^{\circ}=35^{\circ}+65^{\circ}$
$x ^{\circ}=100^{\circ}$

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