Geometry  1
Quizsummary
0 of 5 questions completed
Questions:
 1
 2
 3
 4
 5
Information
There are 5 Questions
Time Limit : 10 min
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 5 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Geometry 0%
 Geometry 0%
 1
 2
 3
 4
 5
 Answered
 Review

Question 1 of 5
1. Question
In a triangle with one angle equal to 120 degrees, the lengths of the sides form an Arithmetic Progression. If the length of the greatest side is 7 cm, the radius of the circumcircle of the triangle (in cm) is
Correct
The side opposite to greatest angle will be longest. So, side opposite to 120° is 7.
Therefore, a/Sin A=2R⟹R=1/2 x 7/Sin 120= 7/√3 = 7√3/3Incorrect
The side opposite to greatest angle will be longest. So, side opposite to 120° is 7.
Therefore, a/Sin A=2R⟹R=1/2 x 7/Sin 120= 7/√3 = 7√3/3 
Question 2 of 5
2. Question
In a triangle ABC if 1/(a+c) + 1/(b+c) = 3/(a+b+c), then C (in degrees) is equal to
Correct
The given relation can be written as
(a+b+2c)/{(a+c)(b+c)} = 3/(a+b+c)
⇒ (a + b + 2c) (a + b + c) = 3(a + c) (b + c)
⇒ (a + b)^{2} + 3c (a + b) + 2c^{2} = 3(ab + ac + bc + c^{2})
⇒ a^{2} + b^{2} – c^{2} = ab
∴ cos C = (a^{2}+b^{2}c^{2})/2ab = ab/2ab = 1/2
⇒ C = 60°Incorrect
The given relation can be written as
(a+b+2c)/{(a+c)(b+c)} = 3/(a+b+c)
⇒ (a + b + 2c) (a + b + c) = 3(a + c) (b + c)
⇒ (a + b)^{2} + 3c (a + b) + 2c^{2} = 3(ab + ac + bc + c^{2})
⇒ a^{2} + b^{2} – c^{2} = ab
∴ cos C = (a^{2}+b^{2}c^{2})/2ab = ab/2ab = 1/2
⇒ C = 60° 
Question 3 of 5
3. Question
In a parallelogram ABCD, E is a point on AB such that EC is the bisector of angle BCD. The length of ED is 3 cm, and angle EAD is equal to angle ADE. BC is 4 cm. What is the length of BE?
Correct
Angle DCE = Angle ECB (given)
But Angle DCE = Angle BEC (AB and CD are parallel sides)
Therefore, Angle BEC = Angle ECB
So, BE = BC = 4 cmIncorrect
Angle DCE = Angle ECB (given)
But Angle DCE = Angle BEC (AB and CD are parallel sides)
Therefore, Angle BEC = Angle ECB
So, BE = BC = 4 cm 
Question 4 of 5
4. Question
A rectangle of the maximum possible area is cut from a semicircle of perimeter 72 cm. What is the area of the rectangle cut out (in sq. cm.)?
Correct
Incorrect

Question 5 of 5
5. Question
Find the sum of all values of ‘k’ for which the graphs x^{2}+y^{2}8x10y+16=0 and x^{2}+y^{2}12x10y+k=0 have only one common tangent.
Correct
x^{2}+y^{2}8x10y+16=0⟹(x4)^{2}+(y5)^{2}=5^{2}
And, x^{2}+y^{2}12x10y+k=0⟹(x6)^{2}+(y5)^{2}=√(61−k)^{ 2}
So, the circles will have only one tangent when they are touching each other internally.
⟹r5=2⟹r=7 or 3
∴√(61−k)=7⟹k=6149=12
And, √(61−k) =9⟹k=619=52
Sum of values of k= 12 + 52 = 64Incorrect
x^{2}+y^{2}8x10y+16=0⟹(x4)^{2}+(y5)^{2}=5^{2}
And, x^{2}+y^{2}12x10y+k=0⟹(x6)^{2}+(y5)^{2}=√(61−k)^{ 2}
So, the circles will have only one tangent when they are touching each other internally.
⟹r5=2⟹r=7 or 3
∴√(61−k)=7⟹k=6149=12
And, √(61−k) =9⟹k=619=52
Sum of values of k= 12 + 52 = 64