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## Geometry and Mensuration: Level 2 Test 1

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 Question 1
The area of a square is 1444 square metre. The breadth of a rectangle is 1/4th of the side of the square and the length of the rectangle is thrice the breadth. What is the difference between the area of the square and the area of the rectangle?
 A 1152.38 sq. mtr. B 1169.33 sq. mtr. C 1181.21 sq. mtr. D 1173.25 sq. mtr.
Question 1 Explanation:
The side of the square = 38 m.
The breath of the rectangle = 38/4 m
Length = 3x38/4 m
Difference between area of square and rectangle = 1444 –(3x38/4 x 38/4) = 1173.25 sq.m
 Question 2
If area of an equilateral triangle is a and height b, then value of b2/a is:
 A 3 B 1/3 C √3 D 1/√3
Question 2 Explanation: $\begin{array}{l}Let\text{ }the\text{ }side\text{ }of\text{ }the\text{ }equilateral\text{ }triangle\text{ }be\text{ }x\text{ }units.\\Area\text{ }=\text{ }a\text{ }=\surd 3/4\text{ }{{x}^{2}}~\\and\text{ }Height=b=\text{ }\surd 3/2\\\frac{{{b}^{2}}}{a}=\frac{3}{4}{{x}^{2}}\times \frac{4}{\sqrt{3}}\times \frac{1}{{{x}^{{}}}}=\sqrt{3}\end{array}$
 Question 3
Circumference of a circle-A is 11/7 times perimeter of a square. Area of the square is 784 sq. cm. What is the area of another circle-B whose diameter is half the radius of the circle-A?
 A 38.5 sq. cm B 156 sq. cm C 35.8 sq. cm D None of these
Question 3 Explanation:
The side of the square = 28 cm.
Perimeter = 4 X 28 = 112 cm
The circumference of the circle = 11/7 x 112 = 22/7 x 2 x 28
Half the radius of Circle-A = 14 cm.
Radius of Circle-B = 14/2 = 7 cm
The area = 22/7 x 7 x 7 = 54 cm2
 Question 4
An isosceles triangle ABC is right-angled at B.D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the sides AB and AC respectively of ΔABC. If AP= a cm, AQ= b cm and ∠BAD= 15o, sin 75o=
 A $\displaystyle \frac{2b}{\sqrt{3}\,\,a}$ B $\displaystyle \frac{a}{2b}$ C $\displaystyle \frac{\sqrt{3}\,\,a}{2b}$ D $\displaystyle \frac{2\,\,a}{\sqrt{3}\,b}$
Question 4 Explanation: $\displaystyle \begin{array}{l}To\text{ }find\text{ }Sin\text{ }{{75}^{0}}we\text{ }need\text{ }AD\\So\text{ }\\From\,\,\,\,\Delta AQD\\\sin {{60}^{o}}\,\,\,=\frac{AQ}{AD}\\\frac{\sqrt{3}}{2}=\frac{b}{AD}\\\Rightarrow AD=\frac{2b}{\sqrt{3}}\\Now\,\,Sin\theta =\frac{perpendicular}{hypotenuse}\\so\,\,Sin{{75}^{0}}=\frac{a}{AD}\\OR\\\sin {{75}^{o}}=\frac{AP}{AD}=\frac{a}{\frac{2b}{\sqrt{3}}}=\frac{\sqrt{3}a}{2b}\end{array}$
 Question 5
A horse is tethered to a peg with a 14 metre long rope at the corner of a 40 metre long and 24 metre wide rectangular grass field. What area of the field will the horse graze?
 A 154 m2 B 308 m2 C 240 m2 D 480 m2
Question 5 Explanation: $\displaystyle \begin{array}{l}The\text{ }area\text{ }which\text{ }the\text{ }horse\text{ }can\text{ }graze\text{ }is\\=\frac{\frac{22}{7}\times 14\times 14}{4}=154{{m}^{2}}\end{array}$
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