Geometry and Mensuration: Level 2 Test 4

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Geometry and Mensuration: Level 2 Test 4

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Question 1

$ \displaystyle \begin{array}{l}Suppose\,\vartriangle ABC\text{ }be\text{ }a\text{ }right-angled\text{ }triangle\text{ }where\,\angle A=\text{ }{{90}^{o}}and\,AD\bot BC.\text{ }\\If\,\vartriangle ABC=\text{ }40\text{ }c{{m}^{2}},\vartriangle ACD=\text{ }10\text{ }c{{m}^{2}}and\,\,\overline{AC}=\text{ }9\text{ }cm,\text{ }\\then\text{ }the\text{ }length\text{ }of\text{ }BC\text{ }is\end{array}$

A	12 cm
B	18 cm
C	4 cm
D	6 cm

Question 1 Explanation:

The area of ABD = 40-10 = 30 .
Therefore,AB²= (9²x30)/10 = 243
Thus
BC²=243 +81 =325
Therefore BC = 18 approx

Question 2

If length of the rectangle is increased by 50% and breadth is decreased by 20%. Then what is the percentage change in the area?

A	By decrease 20%
B	20% increase
C	80% increase
D	30% decrease

Question 2 Explanation:

The new rectangle is 1.5 x 0.8 = 1.2 of the previous rectangle.
Thus the increase in area = 20 %
Correct option is (b)

Question 3

The length and breadth of the floor of a room are 20 feet and 10 feet respectively. Square tiles of 2 feet length of three different colours are to be laid on the floor. Black tiles are laid in the first row on all sides. If white titles are laid in the one-third of the remaining and blue tiles in the rest, how many blue tiles will be there?

A	16
B	32
C	48
D	24

Question 3 Explanation:

Area covered by blue and white tiles = (20-4)(10-4) = 16 x 6 sq. ft. The area covered by the blue tiles = (16 x 6 x 2)/3 = 64 ft²
The total number of blue tiles= 64/4 = 16
Correct option is (a)

Question 4

The sum of the interior angles of a polygon is 1620^o. The numbers of sides of the polygon are

A	9
B	11
C	15
D	12

Question 4 Explanation:

180 (no. of sides -2 ) = 1620
=> No of sides = (162 /18)+2
=> No of sides = 11.
Correct option is (b)

Question 5

In a triangle ABC, BAC= 90^o and AD is perpendicular to BC. If AD= 6 cm and BD = 4cm, then the length of BC is

A	8 cm
B	10 cm
C	9 cm
D	13 cm

Question 5 Explanation:

$ \displaystyle \begin{array}{l}Let\,\,DC=x\,\,\,cm\\AB=\sqrt{52}\,\,by\,\,pythagoras\,\,Theorem\,\,\\{{(4+x)}^{2}}={{(4+x)}^{2}}=\sqrt{{{6}^{2}}+{{x}^{2}}}+52\\x=9\\Thus\,\,BC=9+4=13cm\end{array}$

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