Notes
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Topics to be learn
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Construction of Similar Triangle :
If two triangles are similar then,
- Corresponding sides of similar triangles are in proportion.
- Corresponding angles are congruent.
Using this property, a triangle similar to the given triangle can be constructed.
Examples :
(i) Δ ABC ~ Δ PQR, in Δ ABC, AB = 5.4 cm, BC = 4.2 cm, AC = 6.0 cm. AB : PQ = 3 : 2. Construct Δ ABC and Δ PQR.
Answer :
For Δ ABC the length of the three sides are given.
∴ Δ ABC can be constructed.
Δ ABC and Δ PQR are similar.
∴ their corresponding sides are in proportion.
∴ \(\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{3}{2}\)
∴ \(\frac{5.4}{PQ}=\frac{4.2}{QR}=\frac{6.0}{PR}=\frac{3}{2}\)
PQ length :
\(\frac{5.4}{PQ}=\frac{3}{2}\)
∴ PQ = \(\frac{5.4×2}{3}=\frac{10.8}{3}\) = 3.6 cm
QR length :
\(\frac{4.2}{QR}=\frac{3}{2}\)
∴ QR = \(\frac{4.2×2}{3}=\frac{8.4}{3}\) = 2.8 cm
PR length :
\(\frac{6.0}{PR}=\frac{3}{2}\)
∴ PR = \(\frac{6.0×2}{3}=\frac{12}{3}\) = 4.0 cm
For Δ PQR, the length of three sides are now known
Δ PQR can be constructed.
| Remember : While drawing the triangle similar to the given triangle, sometimes the lengths of the sides we obtain by calculation are not easily measureable by a scale. In such a situation we can use the construction ‘To divide the given segment in the given number of eqaul parts’.
For example, if length of side AB is \(\frac{11.6}{3} cm, then by dividing the line segment of length 11.6 cm in three equal parts, we can draw segment AB. |
Note : In the above example (i) there was no common vertex in the given triangle and the triangle to be constructed. If there is a common vertex, it is convenient to follow the method in the following example.
(ii) Construct any Δ ABC. Construct Δ A’BC’ such that AB : A’B = 5 : 3 and
Δ ABC ~ Δ A’BC’
Answer :
The length of three sides of Δ ABC are known,
∴ Δ ABC can be constructed.
Δ ABC ~ Δ A’BC’ such that \(\frac{AB}{A'B}{BC}{C'B}=\frac{5}{3}\)
∴ sides of Δ A’BC’ are smaller than corresponding sides of Δ ABC and
∠ ABC ≅ ∠ A’BC’ …..(Corresponding angles of similar triangle)
Δ ABC and Δ A’BC’ can have common angle B.
Consider the given analytical figure.
If we divide BC into 5 equal parts, then BC’ would be equal to three equal parts. Thus point C’ can be located on seg BC.
Now A’ is the point of intersection of AB and a line through C’ parallel to CA.
As, ∠ BCA ≅ ∠ BC’A ….. (Corresponding angles of similar triangles)
∴ at point C’, we draw line C’A’ || side AC intersecting side AB at A’. Thus we obtain Δ A’BC’.
Steps of construction :
(1) Construct any Δ ABC.
(2) Divide segment BC in 5 equal parts.
(3) Name the end point of third part of seg BC as C’ ∴ BC’ = \(\frac{3}{5}\)BC
(4) Now draw a line parallel to AC through C’. Name the point where the parallel line intersects AB as A’.
(5) Δ A’BC’ is the required trinangle similar to Δ ABC
Note :
- To divide segment BC, in five equal parts, it is convenient to draw a ray from B, on the side of line BC in which point A does not lie.
- Take points T1, T2, T3, T4, T5 on the ray such that BT1 = T1T2 = T2T3 = T3T4 = T4T5
- Join T5C and draw lines parallel to T5C through T1, T2, T3, T4.
(iii) Δ PQR ~ Δ PMN. In Δ PQR, PQ = 4 cm, QR = 5 cm and PR = 6 cm. Construct Δ PQR and Δ PMN such that \(\frac{PR}{PN}=\frac{3}{5}\)
Answer :
The length of three sides of Δ PQR are known.
∴ Δ PQR can be constructed.
∴ Δ PQR ~ Δ PMN such that \(\frac{PR}{PN}=\frac{3}{5}\)
∴ sides of Δ PMN are larger than corresponding sides of Δ PQR and
∠ QPR ≅ ∠MPN …. (Corresponding angles of similar triangles)
∴ Δ PQR and Δ PMN have common angle P.
Consider the given analytical figure.
If we divide PR into 3 equal parts then PN will be 5 times each part of PR on the same line. We can thus locate N such that P—R—N
∠ PRQ ≅ ∠ PNM …. (Corresponding angles of similar triangles)
at point N, we draw line NM || side QR intersecting side PQ at M (P—Q—M). Thus we obtain Δ PMN.
Steps of construction :
(1) Construct Δ PQR such that PQ = 4 cm, PR = 6 cm and QR = 5 cm.
(2) Draw a ray PX making a suitable angle with side PR.
(3) Take equal parts PP1, P1P2, P2P3, P3P4 and P4P5 on ray PX.
(4) Join the points P3 and R.
(5) Draw P5N parallel to PQR and intersecting line PR in the point N.
(6) Draw NM parallel to side RQ and intersecting line PQ in the point M.
(7) Δ PMN is the required triangle similar to Δ PQR
Construction of a tangent to a circle at a point on the circle :
(i) Using the centre of the circle.
Let’s understand, how to construct a tangent at a given point on the circle using the centre of the circle. Consider the following example.
Example :
Suppose we want to construct a tangent l passing through a point P on the circle with centre C.
Analysis :
A circle of radius CP can be drawn.
Let the centre of the given circle be C and line l be the required tangent.
We know, converse of tangent theorem states that, ‘A line perpendicular to radius at its outer end is tangent’.
∴ we construct perpendicular to radius CP at point P then line l is the required tangent.
Steps of construction :
(1) Draw a circle with centre C.
(2) Take any point P on it.
(3) Draw ray CP.
(4) Draw line l ⊥ ray OP at point P.
Line l is the required tangent to the circle Construction :
Construction :
(ii) Without using the centre of the circle.
Let's understand how to construct the tangent at a given point on the circle without using the centre of the circle.
Example:
Construct a circle of any radius. Take any point C on it. Construct a tangent to the circle without using centre of the circle.
Analysis :
Through C, a chord can be drawn. Let it be CB.
Draw any ∠CAB in the alternate segment.
Now an ∠BCD can be constructed congruent to ∠CAB.
By converse of tangent- secant theorem, if we draw the line CD such that, ∠CAB ≅ ∠BCD, then it will be the required tangent.
Steps of Construction :
(1) Draw a circle of a suitable radius. Take any point C on it.
(2) Draw chord CB and an inscribed ∠CAB .
(3) With the centre A and any convenient radius draw an arc intersecting the sides of ∠BAC in points M and N.
(4) Using the same radius and centre C, draw an arc intersecting the
chord CB at point R.
(5) Taking the radius equal to d(MN) and centre R, draw an arc intersecting the
arc drawn in the previous step. Let D be the point of intersection of these arcs. Draw line CD. Line CD is the required tangent to the circle.
Note that ∠MAN and ∠BCD in the above figure are congruent. If we draw seg MN and seg RD, then Δ MAN and Δ RCD are congruent by SSS test. ∴ ∠MAN ≅ ∠BCD
To construct tangents to a circle from a point outside the circle :
Example :
Draw a circle of radius 4.5 cm and centre O. Mark a point P at a distance of 8 cm from the centre. Draw tangents to the circle from point P.
Analysis :
A circle of radius 4.5 cm can be drawn and point P at a distance of 8 cm can be located. Consider the given analytical figure.
Suppose tangents through P touch the circle at point A and B, then ∠OAP = ∠ OBP = 90° (Tangent theorem)
we know, ‘Angle inscribed in a semicircle is a right angle.’
∴ A and B lie on the semicircular arcs whose diameter is OP.
A and B, therefore, would be the points of intersection of those semicircular arcs with the circle.
∴ on drawing the perpendicular bisector of seg OP we can obtain the centre and the radius of the semicircular arcs.
Points of intersection of semicircular arcs and the circle are points A and B.
∴ tangents PA and PB can be drawn.
Steps of construction :
(1) Draw a circle of radius 3.5 cm.
(2) Take a point P in the exterior of the circle such that d(O, P) = 8 cm.
(3) Draw seg OP. Draw perpendicular bisector of seg OP to get its midpoint M.
(4) Draw an arc with radius OM and centre M.
(5) Mark the points of intersection of the arc with the circle as A and B.
(6) Draw line PA and PB.