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#### In a triangle PQR, N is a point on PR such that QN  $\perp$ PR . If PN. NR = QN2, prove that $\angle$ PQR = 90° .

Given:- In a triangle PQR, N is a point on PR such that QN $\perp$ PR and PN.NR = QN2

To prove:-$\angle$ PQR = 90°

Proof:

We have PN.NR = QN2

$\\\Rightarrow PN.NR=QN.QN\\\frac{PN}{QN}=\frac{QN}{NR}\; \; \; \; \; \; \; ....(1)$

$In \Delta QNP \; and\; \Delta RNQ$

$\frac{PN}{QN}=\frac{QN}{NR}$

and $\angle$PNQ = $\angle$RNQ          (each equal to 90°)

we know that if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar by SAS similarity criterion.

$\therefore \Delta QNP \sim \Delta RNQ$

Then $\Delta$QNP and $\Delta$RNQ are equiangular.

i.e.   $\angle PQN = \angle QRN \; \; \; \; \; ....(2)$

$\angle RQN = \angle QPN \; \; \; \; \; ....(3)$

adding equation (2) and (3) we get

$\angle PQN + \angle RQN = \angle QRN + \angle QPN$

$\Rightarrow \angle PQR = \angle QRN + \angle QPN \; \; \; \; \; ...(4)$

In triangle PQR

$\angle PQR + \angle QPR + \angle QRP = 180^{o}$

$\angle PQR + \angle QPN + \angle QRN = 180^{o}$

$\begin{Bmatrix} \because \angle QPR=\angle QPN \\\angle QRP=\angle QRN \end{Bmatrix}$

$\angle PQR + \angle PQR = 180^{o}$

[Using equation (4)]

$2\angle PQR=180^{o}$

$\angle PQR=\frac{180^{o}}{2}$

$\angle PQR=90^{o}$

Hence proved

#### Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle.

Given:- Corresponding sides of two similar triangles are in the ratio of 2 : 3.

Area of smaller triangle = 48 cm2

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

i.e $\frac{ar\left ( \text {smaller triangle} \right )}{ar\left ( \text {larger triangle} \right )}=\left ( \frac{2}{3} \right )^{2}$

$\frac{48}{ar\left ( \text {larger triangle} \right )}=\frac{4}{9}$

$ar\left ( \text {larger triangle} \right )=\frac{48 \times 9}{4}=12 \times 9=108\; cm^{2}$

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#### ABCD is a trapezium in which AB ||DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.

Given:- ABCD is a trapezium in which AB || DC, P and Q are points on AD and BC. Such that PQ || DC.

PD = 18 cm, BQ = 35, QC = 15 cm

Proof:-

Construction:- Join BD

In $\Delta$ ABD, PO || AB

And we know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then according to basic proportionality theorem the other two sides are divided in the same ratio.

$\therefore \frac{DP}{AP}=\frac{DO}{OB}$

$\Rightarrow \frac{DP}{AP}=\frac{OD}{OB}\; \; \; \; \; \; \; \; \; .....(1)$

In $\Delta$ BDC, OQ || DC

Similarly by using basic proportionality theorem.

$\frac{BQ}{QC}=\frac{OB}{OD}$

$\Rightarrow \frac{QC}{BQ}=\frac{OD}{OB}\; \; \; \; \; \; \; \; \; .....(2)$

from equation (1) and (2) we get

$\frac{DP}{AP}=\frac{QC}{BQ}$

$\Rightarrow \frac{18}{AP}=\frac{15}{35}$

$AP=\frac{18 \times 35}{15}$

$AP=42 \; cm$

$\\AD = AP + PD\\ 42 + 18 = 60 cm\\ \therefore AD = 60 cm$

#### In Fig., if DE || BC, find the ratio of ar (ADE) and ar (DECB).

Given:- DE || BC and DE = 6 cm, BC = 12 cm

$In$ $\Delta ABC \; and\; \Delta ADE$$\angle ABC = \angle ADE$  $\text{ (corresponding angle) }$
$\angle A = \angle A$  $\text{ (common angle) }$

As we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.

$\therefore \; \; \; \; \; \Delta ABC \sim \Delta ADE$

Then

$\frac{ar\left ( \Delta ADE \right )}{ar\left ( \Delta ABC \right )}=\left ( \frac{DE}{BC} \right )^{2}=\left ( \frac{6}{12} \right )^{2}=\left ( \frac{1}{2} \right )^{2}=\frac{1}{4}$

$Let \; ar\left ( \Delta ADE \right )=\text {k then ar}\left ( \Delta ABC \right )=4k$

$Now\; ar(\Delta ECB) = ar(\Delta ABC) = ar(\Delta ADE) = 4k - k = 3k$

$\therefore Required \; ratio = ar(ADE) : ar (DECB)$

$\\k:3k\\1:3$

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#### In Fig., if $\angle ACB = \angle CDA$, AC = 8 cm and AD = 3 cm, find BD.

$\left [ \frac{55}{3}\; cm\right ]$

$\text{Given : }$ $\angle ACB=\angle CDA$

AC = 8 cm and AD = 3 cm

$\text{In}$ $\Delta ACD\; and\; \Delta ABC$

$\angle A = \angle A$

$\angle ADC = \angle ACB$

We know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.

$\therefore \; \; \; \; \Delta ADC\sim \Delta ACB$

$\Rightarrow \frac{AC}{AD}=\frac{AB}{AC}$  $\text{ [corresponding sides are proportional]}$

$=\frac{8}{3}=\frac{AB}{8}$

$AB=\frac{8 \times 8}{3}=\frac{64}{3}cm$

$BD+AD=\frac{64}{3}\; \; \; \; \; \; \; \; \; \; \; \left ( Q\; AB=BD+AD \right )$

$BD=\frac{64}{3}-AD$

$BD=\frac{64}{3}-3=\frac{64-9}{3}=\frac{55}{3}\; cm$

#### Areas of two similar triangles are 36 cm2 and 100 cm2. If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle.

Given:- area of smaller triangle = 36 cm2

area of larger triangle = 100 cm2

length of the side of larger triangle = 20 cm

let the length of the corresponding side of the smaller triangle = x cm

According to the property of area of similar triangles,

$\frac{ar\left ( \text {larger triangle} \right )}{ar\left ( \text {smaller triangle} \right )}=\frac{\left ( \text {side of larger triangle} \right )^{2}}{\left ( \text {Side of smaller triangle} \right )^{2}}$

$\frac{100}{36}=\frac{\left ( 20 \right )^{2}}{x^{2}}$

$x^{2}=\frac{20 \times 20 \times 36}{100}$

$x=\sqrt{144}$

$x=12\; cm$

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#### A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

Let BC = 15 m, AB = 24 m in $\Delta$ABC and $\angle$A = Q

Again let DE = 16m and $\angle$EDF = Q in $\Delta$DEF

In $\Delta$ABC and $\Delta$DEF

$\angle A=\angle D=Q$

$\angle B=\angle E=90^{o}$

We know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.

$\therefore \; \; \; \; \; \Delta ABC\sim \Delta DEF$

$\text{Then }\frac{AB}{DE}=\frac{BC}{EF}$   $\text{(corresponding sides are proportional)}$

$\frac{24}{16}=\frac{15}{h}$

$h=\frac{15 \times 16}{24}=10$

$h=10$

Hence the height of the point on the wall where the top of the laden reaches is 10 m.

#### Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Here    AC = 10 m is a ladder

BC = 6 m distance from the base of the wall.

In right-angle triangle ABC use Pythagoras theorem

$\left ( AC \right )^{2}=\left ( AB \right )^{2}+\left ( BC \right )^{2}$

$\left ( 10 \right )^{2}=\left ( AB \right )^{2}+\left ( 6 \right )^{2}$

$100= AB^{2}+36$

$100-36= AB^{2}$

$64= AB^{2}$

$AB=\sqrt{64}$

$AB=8\; m$

Hence, the height of the point on the wall where the top of the ladder reaches is 8 m.

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#### If $\\ \Delta ABC \sim \Delta DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm \; and \; FD = 12 cm,$ find the perimeter of $\Delta ABC.$

Given : $\Delta ABC \sim \Delta DEF$

and      AB = 4cm , DE = 6 cm

EF = 9 cm, FD = 12 cm

Here $\Delta ABC \sim \Delta DEF$   (given)

$\therefore \frac{AB}{ED}=\frac{BC}{EF}=\frac{AC}{DF}$

$\frac{4}{6}=\frac{BC}{9}=\frac{AC}{12}$

Taking first two terms we get

$\frac{4}{6}=\frac{BC}{9}$

$\frac{9 \times 4}{6}=BC$

$BC=6 \; cm$

Taking first and last terms we get

$\frac{4}{6}=\frac{AC}{12}$

$AC=\frac{4 \times 12}{6}=8$

$Perimeter \; of\; \Delta ABC = AB + BC + CA$

$= 4 + 6 + 8 = 18 \; cm.$

#### Find the altitude of an equilateral triangle of side 8 cm.

Let ABC be an equilateral triangle of side 8 cm

i.e. AB = BC = AC = 8 cm and AD $\perp$ BC

Then D is the mid-point of BC.

$\therefore BD=DC=\frac{1}{2}BC=\frac{8}{2}=4cm$

Apply Pythagoras theorem in triangle ABD we get

$\left ( AB \right )^{2}=\left ( BD \right )^{2}+\left ( AD \right )^{2}$

$\left ( 8 \right )^{2}=\left ( 4 \right )^{2}+\left ( AD \right )^{2}$

$64=16+\left ( AD \right )^{2}$

$64-16=\left ( AD \right )^{2}$

$48=\left ( AD \right )^{2}$

$\sqrt{48}=AD$

$4\sqrt{3}=AD$

$\text{Hence the altitude of the equilateral triangle is}$ $4\sqrt{3}cm$