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#6.3

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

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#6.3

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.

Answer : [108 cm²]

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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#6.3

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.

Answer : [AD=60cm]

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

To prove:- Find AD

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

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

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

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

$AP=42 \; cm$

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

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#6.3

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

Answer : 1: 3

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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#6.3

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

Answer :

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

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

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.

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

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

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

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

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

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

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

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#6.3

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.

Answer : [12 cm ]

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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#6.3

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.

Answer : [10 m]

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.

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