Geometry and Mensuration: Level 2 Test 7

∠ADC = 70 + (180 – 70)/2 = 55+70
Thus ∠ADC = (180-125)/2 = 55/2 = 27.5
Correct option is (b)

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)

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)

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

$ \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}$