Write True or False and give reasons for your answer in each of the following:
A pair of tangents can be constructed to a circle inclined at an angle of .
Answer [True]
Solution
Tangent : - it is a straight line that touches the curve but not cross it. A pair of tangents can be constructed to a circle inclines at an angle greater than but less than .Here, the inclination angle is Hence it is possible.
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Write True or False and give reasons for your answer in each of the following:
A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the center.
Answer [False]
Solution
Radius, r = 3.5 cm
Point distance from center = 3 cm
But,
So, the point P lies inside the circle.
So, pair of tangents cannot be drawn to point P to a circle.
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Write True or False and give reasons for your answer in each of the following:
To construct a triangle similar to a given with its sides of the corresponding sides of , draw a ray BX making acute angle with BCand X lies on the opposite side of A with respect to BC. The points B_{1}, B_{2}, ....,B_{7} are located at equal distances on BX, B3 is joined to C and then a linesegment is drawn parallel to B_{3}C where lies on BC produced. Finally,line segment is drawn parallel to AC.
Answer: False
Solution
According to question:-
To construct a triangle similar to a given with its sides of the corresponding sides of
1. Draw a line segment BC
2. Taking B and C as centres draw two arcs of suitable radii intersecting each other at A.
3. Join BA and CA. ?ABC is the required triangle.
4. From B draw any ray BX downwards making an acute angle CBX.
5. Locate seven points B_{1}, B_{2}, b_{3}, …. B_{7} on BX such that BB_{1} = B_{1}B_{2} = B_{1}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}.
6. Join B_{3}C and from B_{7} draw a line B_{7}C’ ? B_{3}C intersecting the extended line segment BC at C’.
7. From point C’ draw C’A’ ? CA intersecting the extended line segment BA at A
But as given if we join B_{3}C and from B_{6} draw a line B_{6}C’ ? B_{3}C intersecting the extended line segment BC at C’.
Hence the sides are not in the ratio of 7:3
So, the required triangle can not be constructed in this way.
Hence the given statement is false.
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Write True or False and give reasons for your answer in each of the following:
By geometrical construction, it is possible to divide a line segment in the ratio
Answer [True]
Solution
To divide a line segment in the ratio, we need both positive integer.
So, we can be simplified it by multiply both the terms by .
We obtain
So, the required ratio is
Geometrical construction, is possible to divide a line segment in the ratio
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Write True or False and justify your answer in each of the following: In Fig. 10.10, if AOB is a diameter and ADC = 120°, then CAB = 30°.
True
Solution:
Given, AOB is a diameter and ADC = 120°
Join CA & CB
Since ADCB is a cyclic quadrilateral
ADC + CBA = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°)
120° + CBA = 180°
CBA = 60°
In ACB,
CAB + CBA +ACB = 180° (angle sum property of a triangle)
ACB = 90° (Since, ACB is an angle in a semi-circle)
CAB + 60° + 90° = 180°
CAB = 180° – 150° = 30°
Therefore the given statement is true.
View Full Answer(1)Write True or False and justify your answer in each of the following: If A, B, C and D are four points such that and , then A, B, C, D are concyclic.
True
Solution:
Definition of concyclic points: The points that lie on the same circle.
We know that angles in the same segment of a circle are equal
Since
We know that these angles are in the same segment of a circle.
This can be represented as shown below:
BC is the chord of the circle and hence A, B, C & D are concyclic.
Therefore the given statement is true.
View Full Answer(1)Write True or False and justify your answer in each of the following: If A, B, C, D are four points such that , then D is the centre of the circle through A, B and C.
True
Solution:
Given, ∠BAC = 30° and ∠BDC = 60°
We know that,
The angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.
Considering BC,
At centre,
At any other point on the circle,
Hence the above rule is justified.
Therefore the given statement is true.
View Full Answer(1)Write True or False and justify your answer in each of the following: ABCD is a cyclic quadrilateral such that .
False.
Solution:
In a cyclic Quadrilateral, the sum of opposite pairs of the angles in the given quadrilateral must be equal to 180°.
Here
Hence, it is not a cyclic Quadrilateral because the sum of the measures of opposite angles is not equal to 180°
Therefore the given statement is false.
View Full Answer(1)Write True or False and justify your answer in each of the following: If AOB is a diameter of a circle and C is a point on the circle, then AC^{2} + BC^{2 }= AB^{2.}
True
Solution:
Given: AOB is a diameter of a circle
AOB is a straight line
AOB = 180°
Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle
ACB is a Right angle
So follows Pythagoras theorem,
i.e., AB^{2 }= AC^{2} + BC^{2}
Therefore the given statement is true.
View Full Answer(1)Write True or False and justify your answer in each of the following: A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.
True
Solution:
Yes, we can draw a circle of radius 3 cm.
The points A & B are such that AB = 6 cm.
So AB will become the diameter of the required circle as AB = 2 (Radius)
Hence, 2 × 3 = 6 cm
Therefore the given statement is true.
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