3 Marks Question

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

y=x
We have, $y=x$
Let $x=1: y=1$
Let $x =2: y =2$
Let $x=3: y=3$
Thus, we have the following table :

x	1	2	2
y	1	2	3

By plotting the points (1, 1), (2, 2) and (3, 3) on the graph paper and joining them by a line, we obtain the graph of y = x.

y = -x
We have, $y=-x$
Let $x=1: y=-1$
Let $x=2: y=-2$
Let $x=-2 : y=-(-2)=2$
Thus, we have the following table exhibiting the abscissa and ordinates of the points of the line represented by the equation y = -x.

x	1	2	-2
y	-1	-2	2

Now, plot the points $(1,-1),(2,-2)$ and $(-2,2)$ and join them by a line to obtain the line represented by the equation $y=-x$.
The graphs of the lines $y=x$ and $y=-x$ are shown in figure.
Two lines intersect at $O (0,0)$.

View full question & answer→

Question 23 Marks

$\text{BE}$ and $\text{CF}$ are two equal altitudes of a triangle $\text{ABC.}$ Using $\text{RHS}$ congruence rule, prove that the triangle $\text{ABC}$ is isosceles.

Answer

Given: $\text{BE}$ and $\text{CF}$ are two equal altitudes of a triangle $\text{ABC .}$
To Prove: $\triangle ABC$ is isosceles.
Proof : In right $\triangle BEC$ and right $\triangle CFB$
side $\text{BE} =$ side $\text{CF} ...[$Given$]$
$\text{BC = CB} \ldots[$ Common$]$
$\triangle BEC \cong \triangle CFB \ldots[$ By $\text{RHS}$ rule $]$
$\therefore \angle BCE =\angle CBF \ldots[\text { c.p.c.t.] }$
$\therefore \text{AB = AC} \ldots[$ Sides opposite to equal angles of a triangle are equal $]$
$\therefore \triangle ABC$ is isosceles.

View full question & answer→

Question 33 Marks

$\text{ABCD}$ is a square and $\text{DEC}$ is an equilateral triangle. Prove that $\text{AE = BE.}$

Answer

In $\text{DEDA}$ and $\text{DECB,}$
$\text{DE = CE} \ldots \ldots [$ Sides of an equilateral triangle $]$
$\text{AD = BC} \ldots \ldots [$ Sides of a square $]$
$\angle EDA =\angle ECB \ldots[$ As $\angle EDC =\angle ECD$ and $\angle ADC =\angle BCD ]$
$\angle EDC +\angle ADC =\angle ECD +\angle BCD \ldots[$ By addition $]$
$\Rightarrow \angle EDA =\angle ECB$
$\therefore DEDA \cong DECB \ldots[$ By $\text{SAS}$ property $]$
$\therefore AE = BE \ldots \text { [c.p.c.t. }]$

View full question & answer→

Question 43 Marks

The internal and external diameters of a hollow hemispherical vessel are $20 \ cm$ and $28 \ cm$ respectively. Find the cost of painting the vessel all over at $35$ paise per $cm ^2$.

Answer

We are given that ,
Outer radius of the vessel, $R = 14 \ cm.$
Inner radius of the vessel, $r = 10 \ cm.$
Area of the outer surface $=\left(2 \pi R^2\right)$ sq units
$=(2 \pi \times 14 \times 14) \ cm^2=(392 \pi) \ cm^2 .$
Area of the inner surface $=\left(2 \pi r^2\right)$ sq units
$=(2 \pi \times 10 \times 10) \ cm^2=(200 \pi) \ cm^2 .$
Area of the ring at the top $=\pi\left( R ^2- r ^2\right)$ sq units
$=\pi\left[(14)^2-(10)^2\right] \ cm ^2$
$=\pi(14+10)(14-10) \ cm ^2=(96 \pi) \ cm ^2$
Total area to be painted $=(392 \pi+200 \pi+96 \pi) \ cm ^2=(688 \pi) \ cm ^2$
Cost of painting $= ₹\left(688 \pi \times \frac{35}{100}\right)$
$= ₹\left(688 \times \frac{22}{7} \times \frac{35}{100}\right)$
$= ₹ \frac{3784}{5}= ₹756.80$

View full question & answer→

Question 53 Marks

Find the area of the shaded region in figure.

Answer

In right triangle $\text{PSQ,}$
$\ce{PQ^2 = PS^2 + QS^2} \ldots[$ By Pythagoras theorem $]$
$=(12)^2+(16)^2$
$=144+256=400$
$\Rightarrow PQ =\sqrt{400}=20 \ cm$
Now, for $\Delta PQR$
$a = 20 \ cm, b = 48 \ cm, c = 52 \ cm$
$\therefore s =\frac{a+b+c}{2}$
$=\frac{20+48+52}{2}$
$=60 \ cm$
$\therefore$ Area of $\Delta P Q R$
$=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{60(60-20)(60-48)(60-52)}$
$=\sqrt{60(40)(12)(8)}$
$=\sqrt{(6 \times 10)(4 \times 10)(6 \times 2)(8)}$
$=6 \times 10 \times 8=480 \ cm^2$
Area of $\Delta PSQ =\frac{1}{2} \times$ Base $\times$ Altitude
$=\frac{1}{2} \times 16 \times 12=96 \ cm^2$
$\therefore$ Area of the shaded portion
$=$ Area of $\Delta PQR -$ Area of $\Delta PSQ$
$=480-96$
$=384 \ cm^2$

View full question & answer→

Question 63 Marks

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122 m, 22 m$ and $120 m ($see Fig.$).$ The advertisements yield an earning of $₹ 5000\ per\ m^2$ per year. A company hired one of its walls for $3$ months. How much rent did it pay?

Answer

Given: $= 122 m, = 22 m$ and $= 120 m$
Semi-perimeter of triangle $(s)=\frac{122+22+120}{2}=\frac{264}{2}=132 m$
Using Heron's Formula,
Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{132(132-122)(132-22)(132-120)}$
$=\sqrt{132 \times 10 \times 110 \times 12}$
$=\sqrt{11 \times 12 \times 10 \times 10 \times 11 \times 12}$
$=10 \times 11 \times 12$
$=1320 m^2$
$\because$ Rent for advertisement on wall for $1$ year $= Rs. 5000\ per\ m^2$
$\therefore$ Rent for advertisement on wall for $3$ months for $1320 m^2$ ;
$\frac{5000}{12} \times 3 \times 1320$
$= Rs.1650000$
Hence rent paid by company $= Rs. 16,50,000$

View full question & answer→

Question 73 Marks

Find the value of the polynomial $3 x^3-4 x^2+7 x+5$, when $x =3$ and also when $x =-3$.

Answer

$\text { Let } p(x)=3 x^3-4 x^2+7 x-5$
$\therefore p(3)=3(3)^3-4(3)^2+7(3)-5$
$=3(27)-4(9)+21-5$
$=81-36+21-5$
$=61$
$\text { Now, } p(x)=3 x^3-4 x^2+7 x-5$
$p(-3)=3(-3)^3-4(-3)^2+7(-3)-5$
$=3(-27)-4(9)-21-5$
$=-81-36-21-5$
$=-143$

View full question & answer→

Question 83 Marks

You know that $\frac{1}{7}=0 . \overline{142857}$. Can you predict what the decimal expansions of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$ are, without actually doing the long division? If so, how?

Answer

Yes, We can predict the decimal expansions of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$, without actually doing the long division as follows:
$\frac{2}{7}=2 \times \frac{1}{7}=2 \times 0 . \overline{142857}=0 . \overline{285714}$
$\frac{3}{7}=3 \times \frac{1}{7}=3 \times 0 . \overline{142857}=0 . \overline{428571}$
$\frac{4}{7}=4 \times \frac{1}{7}=4 \times 0 . \overline{142857}=0 . \overline{571428}$
$\frac{5}{7}=5 \times \frac{1}{7}=5 \times 0 . \overline{142857}=\overline{0.714285}$
$\frac{6}{7}=6 \times \frac{1}{7}=6 \times 0 . \overline{142857}=0 . \overline{857142}$

View full question & answer→

Maths 3 Marks Question

Test yourself on this topic