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Maharashtra State BoardSSC (English Medium) 10th Standard

Construct ∆Pyq Such That, Py = 6.3 Cm, Yq = 7.2 Cm, Pq = 5.8 Cm. If Y Z Y Q = 6 5 , Then Construct ∆Xyz Similar to ∆Pyq. - Geometry Mathematics 2

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Question

Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If \[\frac{YZ}{YQ} = \frac{6}{5},\] then construct ∆XYZ similar to ∆PYQ.

Sum
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Solution


Rough Figure

Analysis:

As shown in the figure,

Let Y – Q – Z and Y – P – X.

∆XYZ ~ ∆PYQ          …[Given]

∴ ∠XYZ ≅ ∠PYQ          …[Corresponding angles of similar triangles]

\[\frac{XY}{PY} = \frac{YZ}{YQ} = \frac{XZ}{PQ}\]          …(i)   [Corresponding sides of similar triangles]

But, \[\frac{YZ}{YQ} = \frac{6}{5}\]        …[From (i) and (ii)]

∴ Sides of∆XYZ are longer than corresponding sides of ∆PYQ.

∴ If seg YQ is divided into 5 equal parts, then seg YZ will be 6 times each part of seg YQ.

So, if we construct ∆PYQ, point Z will be on side YQ, at a distance equal to 6 parts from Y.

Now, point X is the point of intersection of ray YP and a line through Z, parallel to PQ.

∆XYZ is the required triangle similar to ∆PYQ.

Steps of construction:

  1. Draw ∆ PYQ of given measure. Draw ray YT making an acute angle with side YQ.
  2. Taking convenient distance on compass, mark 6 points Y1, Y2, Y3, Y4, Y5 and Y6 such that YY1 = Y1Y2 = Y2Y3 = Y3Y4 = Y4Y5 = Y5Y6.
  3. Join Y5Q. Draw line parallel to Y5Q through Y6 to intersects ray YQ at Z.
  4. Draw a line parallel to side PQ through Z. Name the point of intersection of this line and ray YP as X.

∆XYZ is the required triangle similar to ∆PYQ.

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Chapter 4: Geometric Constructions - Problem Set 4 [Page 99]

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Balbharati Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 4 Geometric Constructions
Problem Set 4 | Q 8 | Page 99

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