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प्रश्न
Construct a Δ PQR in which ∠ R = 90 ° , PQ = 5.2 cm and QR= 2.6 cm. Complete the figure taking PR as the line of symmetry and name the figure.
उत्तर
Steps of oonstruction:
(i) Draw a line segment QR = 2.6 cm
(ii) At R draw a perpendicular to QR.
(iii) With Q as centre and radius 5.2 cm cut an arc on perpendicular to R at P.
(iv) Join PQ . Δ PQR is the required triangle.
(v) Produce QR to S such that RS = 2.6 cm
(vi) With S as centre and radius 5.2 cm cut an arc on perpendicular to Rat P.
(vii) Join PS. Δ PSR is the triangle which is the reflection of &PQR.
Δ PQS is the required triangle and is an equilateral triangle.
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संबंधित प्रश्न
Use a graph paper for this question taking 1 cm = 1 unit along both the x and y-axis :
1) Plot the points A(0, 5), B(2, 5), C(5, 2), D(5, -2), E(2, -5) and F(0, -5).
2) Reflect the points B, C, D and E on the y-axis and name them respectively as B’, C’, D’ and E’.
3) Write the coordinates of B’, C’, D’ and E’.
4) Name the figure formed by B C D E E’ D’ C’ B’.
5) Name a line of symmetry for the figure formed.
Use graph paper for this question) (4)
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(1) Plot the triangle on a graph sheet taking 2 cm = 1 unit on both the axes.1
(2) Plot D the reflection of B in the Y-axis, and write its coordinates.
(3) Give the geometrical name of the figure ABOD.
(4) Write the equation of the line of symmetry of the figure ABOD.
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