Geometry and Mensuration: Level 1 Test 4

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Geometry and Mensuration: Level 1 Test 4

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Question 1

In a triangle ΔABC, the side BC is extended up to D. Such that CD= AC, if ∠BAD= 109^o and ACB= 72^o then the value of ∠ABC is

A	35^o
B	60^o
C	40^o
D	45^o

Question 1 Explanation:

$\displaystyle \begin{array}{l}Since,\text{ }AC=CD~\\\angle CAD\text{ }=\angle ADC~=\frac{72}{2}=36\\BAC\text{ }=\text{ }109-36\text{ }=\text{ }73.\\ABC\text{ }=\text{ }180-BAC-ACB\text{ }=\text{ }35.\\Correct\text{ }option\text{ }is\text{ }\left( a \right)\end{array}$

Question 2

The area of a square is 2.25 cm². What is its perimeter?

A	9.0 cms
B	6.0 cms
C	1.5 cms
D	4.5 cms.

Question 2 Explanation:

$\displaystyle \begin{array}{l}Perimeter~=\text{ }4~=\sqrt{area}\\\text{=}\,4\times 2.25~=\text{ }4\times 1.5\text{ }=\text{ }6\text{ }cm\\Correct\text{ }option\text{ }is\text{ }\left( b \right)\end{array}$

Question 3

The length of a rectangle is 20% more than its breadth. What will be the ratio of the area of this rectangle to the area of a square whose side is equal to the breadth of the rectangle?

A	5: 6
B	6: 5
C	2: 1
D	Data inadequate

Question 3 Explanation:

$\begin{array}{l}The\,\,required\,\,ratio\,\,will\,\,be\,\,1.2\times breadt{{h}^{2}}:breadt{{h}^{2}}=6:5\\Correct\,\,option\,\,is\,\,(b)\end{array}$

Question 4

The sum of three altitudes of a triangle is

A	equal to the sum of three sides
B	less than the sum of sides
C	greater than the sum of sides
D	twice the sum at sides

Question 4 Explanation:

Since the question is based on a geometrical identity it must be valid for all triangles.
Thus we can assume an equilateral triangle of side a cm.
Altitude = √3a/2.
Sum of all altitudes = √3a/2 cm
1.5 X 1.73 a= 2.595a cm
Sum of the three sides = 3a cm.

Question 5

A rectangular carpet has an area of 120sq. metres and a perimeter of 46 metres. The length of its diagonal (in metres) is:

A	11
B	13
C	15
D	17

Question 5 Explanation:

$\begin{array}{l}Diagonal=\sqrt{lengt{{h}^{2}}+breadt{{h}^{2}}}\\=\sqrt{{{\left( length+breadth \right)}^{2}}-2\,length\times breadth}\\=\sqrt{{{\left( \frac{46}{2} \right)}^{2}}-2\times 120}=17m\end{array}$

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