Free Triangles 03 Practice Test - 10th Grade 

The ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding heights
So,
$\frac{(Heightofsmaller△{)}^2}{(Heightofbigger△{)}^2}=\frac{12}{48}=\frac{1}{4}$

Let the height of bigger triangle be  x

$\therefore$$\frac{(2.1{)}^2}{x^2}=\frac{1}{4}\iff x=\sqrt{4\times (2.1{)}^2}$ 
                           $=(2\times 2.1)=4.2cm$

All congruent figures are similar but the similar figures need not be congruent as in case of similar figures only shape is considered whereas, in the case of congruent figures, both shape and sizes are considered. Hence, the statement is correct.

Both the figures are similar as all the corresponding angles are equal and all the corresponding sides are in the same ratio as all the sides of both the figures are equal.
Ratio of corresponding sides = $\frac{4.2}{2.1}=2$

$\frac{PS}{SQ}=\frac{PT}{TR}$ ...... (given)

$ST\parallel QR$ ...... (converse of BPT)

Therefore, $\angle PST=\angle RQRand\angle PTQ=\angle PTR$

Therefore $△PST\sim △PQR$ ...... (AA similarity)

Therefore, the information provided in the question is sufficient to prove that the triangles are similar.

We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

So $\frac{9}{16}$ =$\frac{54}{areaof△DEF}$

area of $△$ DEF = 96$c{m}^2$

We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given, the sides of two triangles are in the ratio of 4 : 9.
Therefore, the ratio of areas of these triangles is equal to $(\frac{4}{9}{)}^2$, which is, $\frac{16}{81}$.

 Pythagoras theorem states that $a^2+b^2=c^2$

So he did correct i.e. he divided the land equally among his two sons.

In triangle ABC, by applying pythagoras rule,

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=8^2+6^2=64+36$

$AC=\sqrt{64+36}=\sqrt{100}=10cm$

Since, M is the mid-point of AC,

$AM=MC=\frac{AC}{2}=\frac{10}{2}=5cm$

We know that in a right angled triangle, mid-point of the hypotenuse is always at equidistance from all the three vertices of the triangle.

 $\Rightarrow$ AM = MC = MB = 5 cm

We are given two triangles ABC and PQR such that $△ABC\sim △PQR$.

For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below.

Now,
$\text{area of}△ABC=\frac{1}{2}\times BC\times AM\text{and}\text{area of}△PQR=\frac{1}{2}\times QR\times PN$

So,
$\frac{areaof△ABC}{areaof△PQR}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}=\frac{BC\times AM}{QR\times PN}\cdots \left(1\right)$

Now, in $△ABM$ and $△PQN$,
$\angle B=\angle Q$    (As $△ABC\sim △$PQR)
and $\angle AMB=\angle PNQ={90}^{\circ }$.

So, $△ABM\sim △PQN$
( By AA similarity criterion)
$\therefore \frac{AM}{PN}=\frac{AB}{PQ}\cdots \left(2\right)$
Also, $△ABC\sim △PQR$   (Given)
So, $\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\cdots \left(3\right)$

From (1) and (3), we get,
$\frac{areaof\left(ABC\right)}{areaof\left(PQR\right)}=\frac{AB\times AM}{PQ\times PN}=\frac{AB\times AB}{PQ\times PQ}=\frac{A{B}^2}{P{Q}^2}$

Now using (3), we get,
 $\frac{areaof△ABC}{areaof△PQR}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$