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Geometry and Mensuration: Level 2 Test 7

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Question 1
In the following figure, find ∠ADC. 94
A
55o
B
27.5o
C
60o
D
30o
Question 1 Explanation: 
95
∠ADC = 70 + (180 – 70)/2 = 55+70
Thus ∠ADC = (180-125)/2 = 55/2 = 27.5
Correct option is (b)
Question 2
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 2 cm2, then the area of the hexagon is
A
2 cm2
B
3 cm2
C
4 cm2
D
6 cm2
Question 2 Explanation: 
Let the individual sides of the hexagon be 1 cm
Thus the perimeter =6 and the side of the equilateral triangle is 2cm.
Thus the area of the equilateral triangle = √3/4 x 4 = √3cm
Thus the area of the hexagon= {(6 x √3)/4} x 12 = (3/2)√3
Thus the ratio of the triangle to hexagon is 2:3.
Thus the area of the hexagon = 2 x 3/2 = 3 cm.
The correct option is (b)
Question 3
In a right angled ABC, ABC= 90o; BN is perpendicular to AC, AB= 6cm, AC= 10 cm. Then AN: NC is
A
3: 4
B
9: 16
C
3: 16
D
1: 4
Question 3 Explanation: 
96
Before starting calculation, one can eliminate two choices = a
and since 1 cannot be broken down in the required ratios.
By Pythagoras theorem the third side = 8 cm
The length of BN = 48/10 =4.8 cm
Thus the length of CN according to Pythagoras theorem is √40.96 =6.4 cm
The ratio = (10-6.4): 6.4 = 9:16
Correct option is (b)
Question 4
In a triangle PQR, R = 90o and Q is mid-point of RP. The value of PS2 – QS2 is equal to
A
PQ2
B
2PQ2
C
3PR2
D
4RQ2
Question 4 Explanation: 
By Pythagoras Theorem
PS2 = PR2 + RS2
QS2 = QR2 + RS2
PS2 – QS2
= PR2 + RS2 – QR2 –RS2
= PR2 + QR2
= (PR– QR) (PR + QR)
=(2QR – QR) (2QR + QR)
= QR × 3QR = 3QR2
Question 5
If the interior angle of a regular polygon is double the measure of exterior angle, then the number of sides of the polygon is
A
6
B
8
C
10
D
12
Question 5 Explanation: 
$ \displaystyle \begin{array}{l}\begin{array}{*{35}{l}} Let\text{ }the\text{ }number\text{ }of\text{ }sides\text{ }of\text{ }the\text{ }polygon\text{ }=\text{ }n \\ We\text{ }know\text{ }that\text{ }the\text{ }interior\text{ }angle\text{ }= \\ \end{array}\\=\left( \frac{2n-4}{n} \right)\times {{90}^{o}}\\And\,\,the\,\,exterior\,\,angle\,\,=\frac{{{360}^{o}}}{n}\\\therefore \,\,from\,\,question\,\,\\\frac{2n-4}{n}\times {{90}^{o}}=\frac{2\times 360}{n}\\\Rightarrow 2n-4=8\\\Rightarrow 2n=12\\\Rightarrow n=6=Number\,\,\,of\,\,\,sides\end{array}$
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