Answer :

(i) T

Two similar figures have the same shape but not necessarily the same size. Therefore, all circles are similar.

(ii) F

Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).

Consider an example,

Let a rectangle have sides 2cm and 3cm and another rectangle have sides 2cm and 5cm.

Here, the corresponding angles are equal but the corresponding sides are not in the same ratio.

(iii) F

Two triangles are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio (or proportion).

(iv) T

Midpoint Theorem states that the line joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

(v) F

Two triangles are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio (or proportion).

But here, the corresponding sides are

AB/DE = 6/12 = 1/2 and AC/DF = 8/9

AB/DE ≠ AC/DF

(vi) F

The polygon formed by joining the midpoints of sides of any quadrilateral is a parallelogram.

(vii) T

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

(viii) T

The perimeters of the two triangles are in the same ratio as the sides. The corresponding medians also will be in this same ratio.

(ix) T

Let us construct perpendiculars OP, OQ, OR and OS from point O.

Let us take LHS = OA^{2} + OC^{2}

From Pythagoras theorem,

= (AS^{2} + OS^{2}) + (OQ^{2} + QC^{2})

As also AS = BQ, QC = DS and OQ = BP = OS,

= (BQ^{2} + OQ^{2}) + (OS^{2} + DC^{2})

Again by Pythagoras theorem,

= OB^{2} + OD^{2} = RHS

As LHS = RHS, hence proved.

(x) T

In rhombus ABCD, AB = BC = CD = DA.

We know that diagonals of a rhombus bisect each other perpendicularly.

i.e. AC ⊥ BD, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° and

OA = OC = AC/2, OB = OD = BD/2

Let us consider right angled triangle AOB.

By Pythagoras theorem,

AB^{2} = OA^{2} + OB^{2}

⇒ AB^{2} = (AC/2)^{2} + (BD/2)^{2}

⇒ AB^{2} = AC^{2}/4 + BD^{2}/4

⇒ 4AB^{2} = AC^{2}+ BD^{2}

⇒ AB^{2} + AB^{2} + AB^{2} + AB^{2} = AC^{2}+ BD^{2}

∴ AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2}+ BD^{2}

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