5 Marks Questions

Question 15 Marks

A design is made on a rectangular tile of dimensions $50cm \times 70cm$ as shown in the design shows $8$ triangles, each of sides $26\ cm, 17\ cm$ and $25\ cm$. Find the total area of the design and the remaining area of the tile.

Answer

Given, tha dimension of rectangular lile is $50 cm \times 70 cm$
Area of rectangular tile $=50 \times 70=3500 cm^2$
The sides of a design of one triangle be, $a=25 cm, b =17 cm$ and $c =26 cm$
Now, semi-perimeter, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{25+17+26}{2}=\frac{68}{2}=34$
$\therefore$ Area of one triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula]
$=\sqrt{34\times9\times17\times8}$
$=\sqrt{17\times2\times3\times3\times17\times2\times2\times2}$
$=17\times3\times2\times2=204\text{cm}^2$
$\therefore$ Total area of eight triangles $=204 \times 8=1632 cm$
Now, area of the desion $=$ Total area of eight triangles $=1632 cm^2$
Also, remaining area of the tile
$=$ Area of the rectangle - Area of the design
$=3500-1632=1868 cm$
Hence, the total area of the design is $1632 cm^2$
and the remaining area of the tile is $1868 cm^2$

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Question 25 Marks

A field is in the shape of a trapezium having parallel sides $90m$ and $30\ m$. These sides meet the third side at right angles. The length of the fourth side is $100\ m$. If it costs Rs. $4$ to plough $1m^2 $of the field, find the total cost of ploughing the field.

Answer

In trapezium $ABCD,$ we draw a perpendicular line $CE$ to the line $AB.$

We have, $DC = AE = 30m$
Now, $BE = AB - AE = 90 - 30 = 60m$ In right angled $\triangle\text{BEC},$
$ (BC)^2 = (BE)^2 + (EC)^2$ [Using pythagoras theorem]
$\Rightarrow (100)^2 = (60)^2 + (EC)^2$
$\Rightarrow (EC)^2 = 10000 - 3600$
$\Rightarrow (EC)^2 = 6400$
$\therefore\ \text{EC}=\sqrt{6400}=80\text{m}$ [taking positive square root because length is always positive]
$\therefore$ Area of trapezium
$ABCD =\frac{1}{2}$ (Sum of parallel sides) $\times$ Distance between parallel sides
$=\frac{1}{2}(\text{AB}\times\text{CD})\times\text{EC}$
$=\frac{1}{2}(90+30)\times80$
$=\frac{1}{2}\times(120)\times80$
$=4800\text{m}^2$
$\because$ Cost of ploughing the field of $1m^2 = Rs. 4$
$\therefore$ Cost of ploughing the field of $4800m^2 = 4800 \times 4 = Rs. 19200$
Hence, the total cost of ploughing the field is $Rs. 19200$

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Question 35 Marks

How much paper of each shade is needed to make a kite given in which $ABCD$ is a square with diagonal $44\ cm.$

Answer

We know that, all the sides of a square are always equal.
i.e., $\text{AB}=\text{BC}=\text{CD}=\text{DA}$ In
$\triangle\text{ABCD},\text{ AC}=44\text{cm},\angle\text{D}=90^\circ$
Using pythagoras theorem in $\triangle\text{ACD},$
$\text{AC}^2=\text{AD}^2+\text{DC}^2$
$\Rightarrow44^2=\text{AD}^2+\text{AD}^2$
$\Rightarrow2\text{AD}^2=44\times44$
$\Rightarrow\text{AD}^2=22\times44$
$\Rightarrow\text{AD}=\sqrt{22\times44}$
[taking positive square because length is always positive]
$\Rightarrow\text{AD}=\sqrt{2\times11\times4\times11}$
$\Rightarrow\text{AD}=22\sqrt{2}\text{cm}$
So, $\text{AB}=\text{BC}=\text{CD}=\text{DA}=22\sqrt{2}\text{cm}$
$\therefore$ Area od squre $\text{ABCD}=\text{Side}\times\text{Side}=22\sqrt{2}\times22\sqrt{2}=968\text{cm}^2$
$\therefore$ Area of the red portion $=\frac{968}{4}=242\text{cm}^2$ [Since, area of square is divided into four parts]
area of the yellow portion $=\frac{968}{2}=484\text{cm}^2$ In
$\triangle\text{PCQ},$ Side $PC = a = 20cm,
CQ = b = 20cm$ and $PQ = c = 14cm$
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{20+20+4}{2}=\frac{54}{2}=27\text{cm}$
$\therefore\ \text{Area of }\triangle\text{PCQ}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula]
$=\sqrt{27(27-20)(27-20)(27-14)}$
$=\sqrt{27\times7\times7\times13}$
$=\sqrt{3\times3\times3\times7\times7\times13}$
$=21\sqrt{39}=21\times6.24=131.04\text{cm}^2$
$\therefore$ Total area of the green portion $= 242 + 131.04 = 373.04cm^2$
Hence, the paper required for each shade to make a kite is red paper $242cm^2,$
yellow paper $484cm^2$
and green paper $373.04cm^2$

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