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Mathematics

Use graph paper for this question.

The points A(2, 3), B(4, 5) and C(7, 2) are the vertices of ΔABC.

(i) Write down the co-ordinates of A', B', C' if ΔA'B'C' is the image of ΔABC when reflected in the origin.

(ii) Write down the co-ordinates of A", B", C" if ΔA"B"C" is the image of ΔABC when reflected in the x-axis.

(iii) Mention the special name of the quadrilateral BCC"B" and find its area.

Reflection

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Answer

The graph is shown below:

The points A(2, 3), B(4, 5) and C(7, 2) are the vertices of ΔABC. Reflection, RSA Mathematics Solutions ICSE Class 10.

(i) From graph we get,

The coordinates of A', B', C' are (-2, -3), (-4, -5) and (-7, -2) respectively.

(ii) From graph we get,

The coordinates of A", B", C" are (2, -3), (4, -5) and (7, -2) respectively.

(iii) From graph we get,

BB" // CC" and BC = B"C" (As on reflection the length between the points do not changes)

BCC"B" formed is an isosceles trapezium.

We know that,

Area of trapezium=12×sum of parallel sides×distance between them=12×(10+4)×3=12×14×3=12×42=21sq. units.\text{Area of trapezium} = \dfrac{1}{2} \times \text{sum of parallel sides} \times \text{distance between them} \\[1em] = \dfrac{1}{2} \times (10 + 4) \times 3 \\[1em] = \dfrac{1}{2} \times 14 \times 3 \\[1em] = \dfrac{1}{2} \times 42 \\[1em] = 21 \text{sq. units}.

Hence, BCC"B" formed is an isosceles trapezium and area of BCC"B" = 21 sq.units.

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