5 Marks Questions

Question 15 Marks

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

Answer

Given, tha dimension of rectangular lile is 50cm × 70cm
Area of rectangular tile = 50 × 70 = 3500cm²
The sides of a design of one triangle be,
a = 25cm, b = 17cm and c = 26cm
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 × 8 = 1632cm
Now, area of the desion = Total area of eight triangles
= 1632cm²
Also, remaining area of the tile = Area of the rectangle - Area of the design
= 3500 - 1632
= 1868cm
Hence, the total area of the design is 1632cm² and the remaining area of the tile is 1868cm²

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Question 25 Marks

A field is in the shape of a trapezium having parallel sides 90m and 30m. These sides meet the third side at right angles. The length of the fourth side is 100m. If it costs Rs. 4 to plough 1m² 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]

⇒ (100)² = (60)² + (EC)²

⇒ (EC)² = 10000 - 3600

⇒ (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) × 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 1m² = Rs. 4

$\therefore$ Cost of ploughing the field of 4800m² = 4800 × 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 44cm.

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 = 373.04cm²
Hence, the paper required for each shade to make a kite is red paper 242cm², yellow paper 484cm² and green paper 373.04cm²

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Maths 5 Marks Questions

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