Similar Figures Questions in English

(A) $(i)$ All circles have the same shape but not necessarily the same size,therefore they are similar.
$(ii)$ All squares have the same shape (all angles are $90^{\circ}$ and sides are in the same ratio),therefore they are similar.
$(iii)$ All equilateral triangles have angles of $60^{\circ}$ and equal side ratios,therefore they are similar.
$(iv)$ Two polygons are similar if $(a)$ their corresponding angles are equal and $(b)$ their corresponding sides are proportional.

(N/A) Two figures are said to be similar if they have the same shape but not necessarily the same size.
Example $1$: Two equilateral triangles with side lengths $1\, cm$ and $2\, cm$ respectively. Since all equilateral triangles have angles of $60^{\circ}$,they have the same shape.
Example $2$: Two squares with side lengths $1\, cm$ and $2\, cm$ respectively. Since all squares have interior angles of $90^{\circ}$ and proportional sides,they have the same shape.

(N/A) Two figures are said to be non-similar if they do not have the same shape,even if they have the same number of sides.
Example $1$: $A$ trapezium and a square are non-similar figures because their corresponding angles are not equal and their corresponding sides are not in the same ratio.
Example $2$: $A$ triangle and a parallelogram are non-similar figures because they have a different number of sides and different geometric properties.

(N/A) Two polygons are similar if:
$1$. Their corresponding angles are equal.
$2$. Their corresponding sides are in the same ratio (proportional).
In quadrilateral $PQRS$ and $ABCD$:
- The sides are in the ratio $PQ/AB = QR/BC = RS/CD = SP/DA = 1.5/3 = 1/2$. Thus,the sides are proportional.
- However,the angles of $PQRS$ are not $90^{\circ}$,while all angles of $ABCD$ are $90^{\circ}$.
Since the corresponding angles are not equal,the quadrilaterals $PQRS$ and $ABCD$ are not similar.

(B) The statement is $False$.
Two polygons (including quadrilaterals) are considered similar if and only if two conditions are met:
$1$. Their corresponding angles are equal.
$2$. Their corresponding sides are in the same ratio (proportional).
For example,a square and a rectangle have equal corresponding angles ($90^{\circ}$ each),but they are not similar because their sides are not proportional.

(N/A) Let $ABCD$ be a square of side $a$.
Therefore,its diagonal $= \sqrt{2} a$.
Two equilateral triangles are formed: $\Delta ABE$ (on side $AB$) and $\Delta DBF$ (on diagonal $DB$).
Side of equilateral triangle $\Delta ABE$ described on side $AB = a$.
Side of equilateral triangle $\Delta DBF$ described on diagonal $DB = \sqrt{2} a$.
Since all equilateral triangles are similar,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area of } \Delta ABE}{\text{Area of } \Delta DBF} = \left( \frac{a}{\sqrt{2} a} \right)^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$.
Thus,$\text{Area of } \Delta ABE = \frac{1}{2} \times \text{Area of } \Delta DBF$.

(B) Let $AB$ be the ladder and $CA$ be the wall with the window at $A$. The wall and the ground form a right angle at $C$. Thus,$\triangle ABC$ is a right-angled triangle with $\angle C = 90^\circ$.
Given: $BC = 2.5\, m$ (distance of the foot of the ladder from the wall) and $CA = 6\, m$ (height of the window from the ground).
According to the Pythagoras Theorem,in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides:
$AB^2 = BC^2 + CA^2$
$AB^2 = (2.5)^2 + (6)^2$
$AB^2 = 6.25 + 36$
$AB^2 = 42.25$
$AB = \sqrt{42.25}$
$AB = 6.5\, m$
Therefore,the length of the ladder is $6.5\, m$.

(B) We know that all equilateral triangles have all their angles equal to $60^{\circ}$ and all their sides of the same length. Therefore,all equilateral triangles are similar to each other.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Let the side length of $\triangle ABC$ be $x$.
Since $D$ is the mid-point of $BC$,the side length of $\triangle BDE$ is $\frac{x}{2}$.
Therefore,the ratio of the areas is:
$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle BDE)} = \left(\frac{x}{\frac{x}{2}}\right)^{2} = (2)^{2} = \frac{4}{1}$.
Hence,the ratio is $4:1$,and the correct answer is $(B)$.

(N/A) Distance travelled by the plane flying towards north in $1 \frac{1}{2}$ hours $= 1000 \times 1.5 = 1500\, km$.
Distance travelled by the plane flying towards west in $1 \frac{1}{2}$ hours $= 1200 \times 1.5 = 1800\, km$.
Let the airport be at the origin $O$. The plane flying north reaches point $A$ and the plane flying west reaches point $B$.
Since north and west directions are perpendicular,$\triangle AOB$ is a right-angled triangle with $\angle AOB = 90^\circ$.
By Pythagoras theorem,the distance between the two planes is the hypotenuse $AB = \sqrt{OA^2 + OB^2}$.
$AB = \sqrt{1500^2 + 1800^2} = \sqrt{2250000 + 3240000} = \sqrt{5490000}$.
$AB = \sqrt{90000 \times 61} = 300\sqrt{61}\, km$.
Thus,the distance between the two planes after $1 \frac{1}{2}$ hours is $300\sqrt{61}\, km$.

(13 M) Let $CD$ and $AB$ be the poles of height $11 \, m$ and $6 \, m$ respectively.
Draw a line segment $AP$ parallel to $BD$ such that $P$ lies on $CD$. Then $AP = BD = 12 \, m$ and $PD = AB = 6 \, m$.
Now,$CP = CD - PD = 11 \, m - 6 \, m = 5 \, m$.
In the right-angled triangle $\triangle APC$,by applying the Pythagoras theorem:
$AC^2 = AP^2 + CP^2$
$AC^2 = (12 \, m)^2 + (5 \, m)^2$
$AC^2 = 144 \, m^2 + 25 \, m^2 = 169 \, m^2$
$AC = \sqrt{169} \, m = 13 \, m$.
Therefore,the distance between their tops is $13 \, m$.

(A) In the given correspondence $ABC \leftrightarrow RPQ$,the vertices are mapped as follows:
$A$ corresponds to $R$
$B$ corresponds to $P$
$C$ corresponds to $Q$
Therefore,$\angle B$ corresponds to $\angle P$.

(C) In a square,the length of a diagonal is given by the formula: $\text{Diagonal} = \sqrt{2} \times \text{side}$.
Given the side length $AB = 4 \sqrt{2}$.
Substituting the value of the side into the formula:
$\text{Diagonal} = \sqrt{2} \times (4 \sqrt{2})$
$= 4 \times (\sqrt{2} \times \sqrt{2})$
$= 4 \times 2 = 8$.
Therefore,the length of the diagonal is $8$.