question_answer

 

Directions: Q. 6 to 10

Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam , your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition, suggests two designs given below.

Observe these carefully.

Design I This design is made with a circle of radius 32 cm leaving equilateral triangle ABC in the middle as shown in the given figure.

Design II This Pookalam is made with 9 circular design each of radius 7 cm.

Refer Design I

The side of equilateral triangle is

A) \[12\sqrt{3}\,cm\]

B) \[32\sqrt{3}\,cm\]

C) 48cm

D) 64cm

Correct Answer: B

Solution :

Directions (i-ii) Given, radius of the circle = 32 cm Let the side of the equilateral \[\Delta ABC\]be a cm. Let h be the height of the triangle. We know that in an equilateral triangle, centroid and circumcentre coincide. \[\therefore \,\,\,AO=\frac{2}{3}\,h\,cm\] [\[\because \]centroid divides the median in the ratio 2 : 1] which is equal to the radius of circle. \[\therefore \,\,\,\,\frac{2}{3}h=32\,\,\Rightarrow h=48\,cm\]                      … (i) Now, we draw a perpendicular from vertex A to side BC which bisects BC at D. In right angled \[\Delta ADB\], \[A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}\] [by Pythagoras theorem] \[\Rightarrow \,\,{{a}^{2}}={{\left( \frac{a}{2} \right)}^{2}}+{{h}^{2}}\Rightarrow {{h}^{2}}={{a}^{2}}-\frac{{{a}^{2}}}{4}=\frac{3{{a}^{2}}}{4}\] \[\Rightarrow \,\,{{\left( 48 \right)}^{2}}=\frac{3{{a}^{2}}}{4}\]                      [from Eq.(i)] \[\Rightarrow \,\,\,{{a}^{2}}=3072\] \[\Rightarrow \,\,\,a=\sqrt{3072}\] [taking positive square root] \[=32\sqrt{3}\,\,cm\] (i) \[32\sqrt{3}\,cm\] (ii) 48cm