Note:
- For drawing a tangent at a point of a circle, simply draw the radius through this point and draw a line perpendicular to this radius through this point.
- The two tangents to a circle from an external point are equal.
Steps of construction:
Thus, PA and PB are the required two tangents.
Note:
In case, the centre of the circle is not known, then to locate its centre, we take any two non-parallel chords and then find the point of intersection of their perpendicular bisectors.
In each of the following, give the justification of the construction also:
Q1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8.
Measure the two parts.
Ans: Steps of construction:
Thus, C divides AB in the ratio 5:8.
On measuring the two parts, we get: AC = 4.7 cm and BC = 2.9 cm.
Justification:
In Δ ABX13 and Δ ACX5, we have
∴
⇒ AC : CB = 5 : 8.
Q2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.
Ans: Steps of construction:
Thus, A′BC′ is the required triangle.
Justification:
By construction, we have:
⇒
But
⇒
⇒
Adding, 1 to both sides, we get
⇒
⇒
Now, in ΔBC′A′ and ΔBCA
we have CA ║ C′A′
∴ Using AA similarity, we have:
⇒ [each equal to 2/3]
Q3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
Ans: Steps of construction:
Q4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.
Ans: Steps of construction:
Q5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Ans: Steps of construction:
Thus, ΔA′BC′ is the required triangle.
Justification:
By construction, we have:
X4C ║ X3C′
[By BPT]
But [By construction]
⇒ ...(1)
Now, we also have
[By construction]
[using AA similarity]
⇒ [From (1)]
Q6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.
Ans: Steps of construction:
Thus, ΔA′BC′ is the required triangle.
Justification:
By construction, we have:
C′A′ ║ CA
∴ ΔABC ~ ΔA′BC′ [AA similarity]
⇒ ...(1)
Also, by construction,
X4C′ ║ X3C
Δ BX4C′ ~ Δ BX3C
⇒
But
⇒ ..(2)
From (1) and (2), we have:
Q7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Ans: Steps of construction:
Thus, ΔA′BC′ is the required triangle.
Justification:
By construction, we have:
C′A′ ║ CA
∴ ΔABC ~ Δ A′BC′ [By AA similarity ]
⇒ ...(1)
Also, X5C′ ║ X3C [By construction]
∴ Δ BX5C′ ~ Δ BX3C
⇒
But ...(2)
From (1) and (2) we get
276 docs|149 tests
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1. What is a tangent to a circle? |
2. How many tangents can a circle have? |
3. What is the relationship between the radius of a circle and the tangent drawn to it? |
4. Can a tangent be drawn to the inside of a circle? |
5. How can we construct a tangent to a circle at a given point? |
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