## NCERT Exemplar Class 10 Maths Chapter 6 Triangles

### NCERT Exemplar Class 10 Maths Chapter 6 Exercise 6.1

Choose the correct answer from the given four options:

Question 1.

In the figure, if âˆ BAC = 90Â° and AD âŠ¥ BC. Then,

(A) BD . CD = BC^{2}

(B) AB . AC = BC^{2}

(C) BD . CD = AD^{2}

(D) AB . AC = AD^{2}

Solution:

(C)

In âˆ†ABC,

âˆ B + âˆ BAC + âˆ C = 180Â°

â‡’ âˆ B + 90Â° + âˆ C = 180Â°

â‡’ âˆ B = 90Â° – âˆ C

Similarly, In âˆ†ADC, âˆ D AC = 90Â° – âˆ C

In âˆ†ADB and âˆ†ADC,

âˆ D = âˆ D = 90Â°

âˆ DBA = âˆ D AC [each equal to (90Â° – âˆ C)

âˆ´ âˆ†ADB ~ âˆ†CDA

[by AA similarity criterion]

âˆ´ \(\frac{B D}{A D}=\frac{A D}{C D}\)

â‡’ BD . CD = AD^{2}

Question 2.

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is

(A) 9 cm

(B) 10 cm

(C) 8 cm

(D) 20 cm

Solution:

(B)

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

Given, AC = 16 cm and BD = 12 cm

âˆ´ AO = 8 cm, BO = 6 cm and âˆ AOB = 90Â°

In right angled âˆ†AOB,

AB^{2} = AO^{2} + OB^{2} [by Pythagoras theorem]

â‡’ AB^{2} = 8^{2} + 6^{2} = 64 + 36 = 100

âˆ´ AB = 10 cm

Question 3.

If âˆ†ABC ~ âˆ†EDFand âˆ†ABC is not similar to âˆ†DEF, then which of the following is not true?

(A) BC . EF = AC . FD

(B) AB . EF = AC . DE

(C) BC . DE = AB . EF

(D) BC . DE = AB . FD

Solution:

(C)

Given, âˆ†ABC ~ âˆ†EDF

Hence, option (B) is true.

Question 4.

If in two triangles ABC and PQR, \(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\) then

(A) âˆ†PQR ~ âˆ†CAB

(B) âˆ†PQR ~ âˆ†ABC

(C) âˆ†CBA ~ âˆ†PQR

(D) âˆ†BCA ~ âˆ†PQR

Solution:

(A)

Given, in triangles ABC and PQR,

\(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\)

which shows that sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are also equal, so by SSS similarity, triangles are similar i.e., âˆ†CAB ~ âˆ†PQR

Question 5.

In the figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, âˆ APB = 50Â° and âˆ CDP = 30Â°. Then, âˆ PBA is equal to

(A) 50Â°

(B) 30Â°

(C) 60Â°

(D) 100Â°

Solution:

(D): In âˆ†APB and âˆ†CPD, âˆ APB = âˆ CPD = 50Â°

[vertically opposite angles]

âˆ´ âˆ†APB ~ âˆ†DPC [by SAS similarity criterion]

âˆ´ âˆ A = âˆ D = 30Â° [corresponding angles of similar triangles]

In âˆ†APB, âˆ A + âˆ B + âˆ APB = 180Â° [sum of angles of a triangle = 180Â°]

â‡’ 30Â° + âˆ B + 50Â° = 180Â°

âˆ´ âˆ B = 180Â° – (50Â° + 30Â°) = 100Â°

i.e., âˆ PBA = 100Â°

Question 6.

If in two triangles DEF and PQR, âˆ D = âˆ Q and âˆ R = âˆ E, then which of the following is not true?

Solution:

(B)

Given, in âˆ†DEF and âˆ†PQR, âˆ D = âˆ Q, âˆ R = âˆ E

âˆ´ âˆ†DEF ~ âˆ†QRPÂ Â Â Â Â Â [by AAA similarity criterion]

â‡’ âˆ F = âˆ P

[corresponding angles of similar triangles]

âˆ´\(\frac{D F}{Q P}=\frac{E D}{R Q}=\frac{F E}{P R}\)

Question 7.

In âˆ†ABC and âˆ†DEF, âˆ B = âˆ E, âˆ F = âˆ C and AB = 3 DE. Then, the two triangles are

(A) congruent but not similar

(B) similar but not congruent

(C) neither congruent nor similar

(D) congruent as well as similar

Solution:

(B)

In âˆ†ABC and âˆ†DEF, âˆ B = âˆ E,

âˆ F = âˆ C and AB = 3DE

We know that, if in two triangles corresponding two angles are same, then they are similar by AA similarity criterion.

Since, AB â‰ DE

Therefore âˆ†ABC and âˆ†DEF are not congruent.

Question 8.

It is given that âˆ†ABC ~ âˆ†PQR with \(\frac{B C}{Q R}=\frac{1}{3}\) then \(\frac { { ar }(\Delta PRQ) }{ { ar }(\Delta BCA) } \) equal to

(A) 9

(B) 3

(C) \(\frac{1}{3}\)

(D) \(\frac{1}{9}\)

Solution:

(A)

Given, âˆ†ABC ~ âˆ†QR and \(\frac{B C}{Q R}=\frac{1}{3}\)

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

Question 9.

It is given that âˆ†ABC ~ âˆ†DFE, âˆ A =30Â°, âˆ C = 50Â°, AB = 5 cm, AC = 8 cm and DF= 7.5 cm. Then, the following is true:

(A) DE= 12 cm, âˆ F= 50Â°

(B) DE= 12 cm, âˆ F= 100Â°

(C) EF= 12 cm, âˆ D = 100Â°

(D) EF= 12 cm, âˆ D = 30Â°

Solution:

(B)

Given, âˆ†ABC ~ âˆ†DFE, then âˆ A = âˆ D = 30Â°, âˆ C = âˆ E = 50Â°

Hence, DE = 12 cm, âˆ F = 100Â°

Question 10.

If in âˆ†ABC and âˆ†DEF, \(\frac{A B}{D E}=\frac{B C}{F D}\), then they will be similar, when

(A) âˆ B = âˆ E

(B) âˆ A = âˆ D

(C) âˆ B = âˆ D

(D) âˆ A = âˆ F

Solution:

(C)

Given, in âˆ†ABC and âˆ†EDF,

So, âˆ†ABC ~ âˆ†EDF if âˆ B = âˆ D [By SAS similarity criterion]

Question 11.

If âˆ†ABC ~ âˆ†QRP, \(\frac { { ar }(\Delta ABC) }{ ar(\Delta PQR) } =\frac { 9 }{ 4 } \), AB= 18 cm and BC = 15 cm, then PR is equal to

(A) 10 cm

(B) 12 cm

(C) \(\frac{20}{3}\) cm

(D) 8 cm

Solution:

(A)

Given, âˆ†ABC ~ âˆ†QRP, AB = 18 cm and BC = 15 cm

Question 12.

If S is a point on side PQ of a âˆ†PQR such that PS = QS = RS, then

(A) PR – QR = RS^{2}

(B) QS^{2} + RS^{2} = QR^{2}

(C) PR^{2} + QR^{2} = PQ^{2}

(D) PS^{2} + RS^{2} = PR^{2}

Solution:

(C)

Given, in âˆ†PQR,

PS = QS = RS …………. (i)

In âˆ†PSR, PS = RS [from Eq(i)]

â‡’ âˆ 1 = âˆ 2 ………… (ii)

[Angles opposite to equal sides are equal]

Similarly, in âˆ†RSQ, RS = SQ

â‡’ âˆ 3 = âˆ 4 …………. (iii)

[angles opposite to equal sides are equal]

Now, in âˆ†PQR, sum of angles = 180Â°

â‡’ âˆ P + âˆ Q + âˆ P = 180Â°

â‡’ âˆ 2 + âˆ 4 + âˆ 1 + âˆ 3 = 180Â°

â‡’ âˆ 1 + âˆ 3 + âˆ 1 + âˆ 3 = 180Â°

â‡’ 2(âˆ 1 + âˆ 3) = 180Â°

â‡’ âˆ l + âˆ 3 = \(\frac{180^{\circ}}{2}\) = 90Â°

âˆ´ âˆ R = 90Â°

In âˆ†PQR, by Pythagoras theorem,

PR^{2} + QR^{2} = PQ^{2}

### NCERT Exemplar Class 10 Maths Chapter 6 Exercise – 6.2

Question 1.

Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.

Solution:

False

Let a = 25 cm, b = 5 cm and c = 24 cm

Now, b^{2} + c^{2} = (5)^{2} + (24)^{2}

= 25 + 576 = 601 â‰ (25)^{2}

Hence, given sides do not make a right triangle because it does not satisfy the property of Pythagoras theorem.

Question 2.

It is given that âˆ†DEF ~ âˆ†RPQ. Is it true to say that âˆ D = âˆ R and âˆ F = âˆ P? Why?

Solution:

False

We know that, if two triangles are similar, then their corresponding angles are equal.

âˆ´ âˆ D = âˆ R, âˆ E = âˆ P and âˆ F = Q

Question 3.

A and B are respectively the points on the sides PQ and PR of A PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is 4B||Q/?? Give reasons for your answer.

Solution:

True

Given, PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm

and \(\frac{P B}{B R}=\frac{4}{6}=\frac{2}{3}\)

From Eqs. (i) and (ii), \(\frac{P A}{A Q}=\frac{P B}{B R}\)

By converse of basic proportionality theorem, AB || QR

Question 4.

In the figure, BD and CE intersect each other at the point P. Is Aâˆ†PBC ~ âˆ†PDE?Why?

Solution:

True

In âˆ†PBC and âˆ†PDE,

âˆ BPC = âˆ EPD [vertically opposite angles]

Since, one angle of âˆ†PBC is equal to one angle of âˆ†PDE and the sides including these angles are proportional, so both triangles are similar.

Hence, âˆ†PBC ~ âˆ†PDE, by SAS similarity criterion.

Question 5.

In âˆ†PQR and âˆ†MST, âˆ P = 55Â°, âˆ Q = 25Â°, âˆ M = 100Â° and âˆ S = 25Â°. Is âˆ†QPR ~ âˆ†TSM? Why?

Solution:

False

We know that, the sum of three angles of a triangle is 180Â°.

In âˆ†PQR, âˆ P +âˆ Q +âˆ R = 180Â°

â‡’ 55Â° + 25 Â° + âˆ R = 180Â°

â‡’âˆ R = 180Â° – (55Â° + 25 Â°)

= 180Â° – 80Â° = 100Â°

In âˆ†TSM, âˆ T + âˆ S + âˆ M = 180Â°

â‡’ âˆ T + âˆ 25Â° + 100Â° = 180Â°

â‡’ âˆ T = 180Â° – (25Â° + 100Â°) = 180Â° – 125Â° = 55Â°

In âˆ†PQR and âˆ†TSM,

âˆ P = âˆ T, âˆ Q = âˆ S and âˆ R = âˆ M

âˆ´ âˆ PQR = âˆ TSM

[since, all corresponding angles are equal]

Hence, âˆ†QPR is not similar to âˆ†TSM, since correct correspondence is P ↔ T, Q ↔ S and R ↔ M.

Question 6.

Is the following statement true? Why? “Two quadrilaterals are similar, if their corresponding angles are equal”.

Solution:

False

Two quadrilaterals are similar if their corresponding angles are equal and corresponding sides must also be proportional.

Question 7.

Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?

Solution:

True

Here, the corresponding two sides and the perimeters of two triangles are proportional, then the third side of both triangles will also in proportion.

Question 8.

If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?

Solution:

True

Let two right angled triangles be âˆ†ABC and âˆ†PQR

In which, âˆ A = âˆ P = 90Â° and âˆ B = âˆ Q = acute angle (Given)

Then, by AA similarity criterion, âˆ†ABC ~ âˆ†PQR

Question 9.

The ratio of the corresponding altitudes of two similar triangles is \(\frac{3}{5}\).Is it correct to say that ratio of their areas is \(\frac{6}{5}\) ? Why?

Solution:

False

Ratio of corresponding altitudes of two triangles having areas A_{1} and A_{2} respectively is \(\frac{3}{5}\).

By the property of area of two similar triangles,

\(\Rightarrow\left(\frac{A_{1}}{A_{2}}\right)=\left(\frac{3}{5}\right)^{2} \Rightarrow \frac{9}{25} \neq \frac{6}{5}\)

So, the given statement is not correct.

Question 10.

D is a point on side QR of âˆ†PQR such that PD âŠ¥ QR. Will it be correct to say that âˆ†PQD ~ Aâˆ†RPD? Why?

Solution:

False

In âˆ†PQD and âˆ†RPD,

PD = PD [common side]

âˆ PDQ = âˆ PDR [each 90Â°]

Here, no other sides or angles are equal, so we can say that âˆ†PQD is not similar to âˆ†RPD. But if âˆ P = 90Â°, then âˆ DPQ = âˆ PRD

[each equal to 90Â° – âˆ Q and by ASA similarity criterion, âˆ†PQD ~ âˆ†RPD]

Question 11.

In the figure, if âˆ D = âˆ C, then is it true that âˆ†ADE ~ âˆ†ACB? Why?

Solution:

True

In âˆ†ADE and âˆ†ACB,

âˆ A = âˆ A [common angle]

âˆ D = âˆ C [given]

âˆ´ âˆ†ADE ~ âˆ†ACB [by AA similarity criterion]

Question 12.

Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.

Solution:

False

Because, according to SAS similarity criterion, if one angle of a triangle is equal to an angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Here, one angle and two sides of two triangles are equal but these sides not including equal angle, so given statement is not correct.

### NCERT Exemplar Class 10 Maths Chapter 6 Exercise – 6.3

Question 1.

In a âˆ†PQR, PR^{2} – PQ^{2} = QR^{2} and M is a point on side PR such that QMâŠ¥ PR. Prove that QM^{2} = PM Ã— MR.

Solution:

Given, In âˆ†PQR,

PR^{2} – PQ^{2} = QR^{2} and QM âŠ¥ PR

To prove : QM^{2} = PM Ã— MR

Proof : Since, PR^{2} – PQ^{2} = QR^{2}

â‡’ PR^{2} = PQ^{2} + QR^{2}

So, âˆ†PQR is right angled triangle right angle at Q.

In âˆ†QMR and âˆ†PMQ, âˆ M = âˆ M [each 90Â°]

âˆ MQR = âˆ QPM [each equal to 90Â° – âˆ R]

âˆ´ âˆ†QMR ~ âˆ†PMQ [by AA similarity criterion]

Now, using property of area of similar triangles, we get

Question 2.

Find the value of x for which DE || AB is given figure

Solution:

Given, DE || AB

âˆ´ \(\frac{C D}{A D}=\frac{C E}{B E}\) [by basic proportionality theorem]

\(\frac{x+3}{3 x+19}=\frac{x}{3 x+4}\)

â‡’ (x + 3)(3x + 4) = x (3x + 19)

â‡’ 3x^{2} + 4x + 9x + 12 = 3x^{2} + 19x

â‡’ 19x – 13x = 12

â‡’ 6x = 12

âˆ´ x = \(\frac{12}{6}\) = 2

Hence, the required value of x is 2.

Question 3.

In the figure, if âˆ 1 = âˆ 2and âˆ†NSQ â‰… âˆ†MTR, then prove that âˆ†PTS ~ âˆ†PRQ.

Solution:

Given âˆ†NSQ â‰… âˆ†MTR and âˆ 1 = âˆ 2

To prove : âˆ†PTS ~ âˆ†PRQ

Proof : Since, âˆ†NSQ â‰… âˆ†MTR

So, SQ = TR ………….. (i)

Also, âˆ 1 = âˆ 2 â‡’ PT = PS ………… (ii)

[since, sides opposite to equal angles are also equal]

From Eqs.(i) and (ii), \(\frac{P S}{S Q}=\frac{P T}{T R}\)

â‡’ ST || QRÂ Â Â Â Â Â [by converse of basic proportionality theorem]

âˆ´ âˆ 1 = âˆ PQRÂ Â Â Â Â [Corresponding angles]

and âˆ 2 = âˆ PRQ

In âˆ†PTS and âˆ†PRQ,

âˆ P = âˆ P [common angles]

âˆ 1 = âˆ PQR

âˆ 2 = âˆ PRQ

âˆ´ âˆ†PTS ~ âˆ†PRQÂ Â Â Â Â Â Â [by AAA similarity criterion]

Question 4.

Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS. Find the ratio of the areas of âˆ†POQ and âˆ†ROS.

Solution:

Given PQRS is a trapezium in which PQ || RS and PQ = 3RS

In âˆ†POQ and âˆ†ROS,

âˆ SOR = âˆ QOPÂ Â Â [vertically opposite angles]

âˆ SRP = âˆ RPQÂ Â Â [alternate angles]

âˆ´ âˆ†POQ ~ âˆ†ROSÂ Â Â Â [by AA similarity criterion]

By property of area of similar triangles,

Hence, the required ratio is 9 : 1

Question 5.

In the figure if AB || DC and AC and PQ intersect each other at the point O, prove that OA . CQ = OC . AP.

Solution:

Given AC an PQ intersect each other at point O and AB || DC

To prove : OA . CQ = OC . AP

Proof: In âˆ†AOP and âˆ†COQ,

âˆ AOP = âˆ COQ [vertically opposite angles]

âˆ APO = âˆ CQO

[since AB || DC and PQ is transversal, so alternate angles]

âˆ´ âˆ†AOP ~ âˆ†COQÂ Â Â Â Â [by AA similarity criterion]

Then, \(\frac{O A}{O C}=\frac{A P}{C Q}\)

[since, corresponding sides are proportional]

â‡’ OA . CQ = OC . AP Hence proved.

Question 6.

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

Solution:

Let ABC be an equilateral triangle of side 8 cm i.e., AB = BC = CA = 8 cm

Draw altitude AD which is perpendicular to BC. Then, D is the mid-point of BC.

âˆ´ BD = CD = \(\frac{1}{2}\) BC = \(\frac{8}{2}\) = 4 cm

Now, AB^{2} = AD^{2} + BD^{2} [by Pythagoras theorem]

â‡’ (8)^{2} = AD^{2} + (4)^{2}

â‡’ 64 = AD^{2} + 16

â‡’ AD^{2} = 64 – 16 = 48

â‡’ AD = \(\sqrt{48}\) = \(4 \sqrt{3}\) cm

Hence, altitude of an equilateral triangle is \(4 \sqrt{3}\) cm

Question 7.

If âˆ†ABC ~ âˆ†DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of âˆ†ABC.

Solution:

Given AB = 4 cm, DE = 6 cm and EF = 9 cm and FD = 12 cm

Also, âˆ†ABC ~ âˆ†DEF

Now, perimeter of âˆ†ABC = AB + BC + AC = 4 + 6 + 8 = 18 cm

Question 8.

In the figure, if DE || BC, find the ratio of ar(âˆ†ADE) and ar(âˆ†ECB).

Solution:

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

In âˆ†ABC and âˆ†ADE,

âˆ ABC = âˆ ADEÂ Â Â Â Â [corresponding angle]

and âˆ A = âˆ AÂ Â Â Â Â [common side]

âˆ´ âˆ†ABC ~ âˆ†ADEÂ Â Â Â Â [by AA similarity criterion]

Let ar(âˆ†ADE) = k, then ar(âˆ†ABC) = 4k

Now, ar(âˆ†ECB) = ar(ABC) – ar(ADE) = 4k – k = 3k

âˆ´ Required ratio = ar(ADE): ar(DECB)

= k : 3k = 1 : 3

Question 9.

ABCD is a trapezium in which AB || DC and P, 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.

Solution:

Given, a trapezium ABCD in which AB || DC. P and Q are points on AD and BC, respectively such that PQ || DC. Thus,

â‡’ AP = 42 cm.

Now; AD = AP + PD = 42 + 18 = 60

âˆ´ AD = 60 cm

Question 10.

Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm^{2}, find the area of the larger triangle.

Solution:

Given, ratio of corresponding sides of two similar triangles is 2 : 3 or \(\frac{2}{3}\)

Area of smaller triangle = 48 cm^{2}

By the property of area of two similar triangles,

Ratio of area of both triangles = (Ratio of their corresponding sides)^{2}

Question 11.

In a âˆ†PQR, N is a point on PR, such that QN âŠ¥ PR. If PN . NR = QN^{2}, prove that âˆ PQR = 90Â°

Solution:

Given, âˆ†PQR, N is a point on PR, such that QN âŠ¥ PR and PN . NR = QN^{2}

To prove : âˆ PQR = 90Â°

Proof: We have, PN . NR = QNc

â‡’ PN . NR = QN . QN

and âˆ PNQ = âˆ RNQ [each equal to 90Â° ]

âˆ´ âˆ†QNP ~ âˆ†RNQÂ Â Â Â [by SAS similarity criterion]

Then, âˆ†QNP and âˆ†RNQ are equiangulars.

i.e., âˆ PQN = âˆ QRN

â‡’ âˆ RQN-âˆ QPN

On adding both sides, we get

âˆ PQN + âˆ RQN = âˆ QRN + âˆ QPN

â‡’ âˆ PQR = âˆ QRN + âˆ QPN …………… (ii)

We know that, sum of angles of a triangle is 180Â°

In âˆ†PQR, âˆ PQR + âˆ QPR + âˆ QRP = 180Â°

â‡’ âˆ PQR + âˆ QPN + âˆ QRN = 180Â°

[ âˆµâˆ QPR = âˆ QPN and âˆ QRP = âˆ QRN]

â‡’ âˆ PQR + âˆ PQR = 180Â° [using Eq. (ii)]

â‡’ 2âˆ PQR = 180Â°

â‡’ âˆ PQR = \(\frac{180^{\circ}}{2}\) = 90Â°

âˆ´ âˆ PQR = 90Â° Hence proved.

Question 12.

Areas of two similar triangles are 36 cm^{2} and 100 cm^{2} . If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle.

Solution:

Given, area of smaller triangle = 36 cm^{2}

and area of larger triangle= 100 cm^{2}

Also, length of a side of the larger triangle = 20 cm

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

By property of area of similar triangles,

Question 13.

In the figure, if âˆ ACB = âˆ CDA, AC = 8 cm and AD = 3 cm, find BD.

Solution:

Given, AC = 8 cm, AD = 3 cm

and âˆ ACB = âˆ CDA

In âˆ†ACD and âˆ†ABC,

âˆ A = âˆ AÂ Â Â Â [Common angle]

âˆ ADC = âˆ ACBÂ Â Â [Given]

âˆ´ âˆ†ADC ~ âˆ†ACBÂ Â Â [By AA similarity criterion]

Question 14.

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.

Solution:

Let BC = 15 m be the tower and its shadow AB is 24 m. At that time âˆ CAB = Î¸. Again, let EF = h be a telephone pole and its shadow DE = 16 m. At the same time âˆ EDF = Î¸. Here, âˆ†ABC and âˆ†DEF both are right angled triangles.

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

Question 15.

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.

Solution:

Let AB be a vertical wall and AC = 10 m is a ladder. The top of the ladder reached to A and distance of ladder from the base of the wall BC is 6 m.

In right angled âˆ†ABC

AC^{2} = AB^{2} + BC^{2} [by Pythagoras theorem]

â‡’ (10)^{2} = AB^{2} + (6)^{2}

â‡’ 100 = AB^{2} + 36

â‡’ AB^{2} = 100 – 36 = 64

âˆ´ AB = \(\sqrt{64}\) = 8 m

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

### NCERT Exemplar Class 10 Maths Chapter 6 Exercise – 6.4

Question 1.

In the given figure, if âˆ A = âˆ C, AB = 6 cm, BP = 15 cm, AP = 12 cm and CP = 4 cm, then find the lengths of PD and CD.

Solution:

Given âˆ A = âˆ C, AB = 6 cm, BP = 15 cm,

AP = 12 cm and CP = 4 cm

In âˆ†APB and âˆ†CPD,

âˆ A = âˆ C [given]

âˆ APB = âˆ CPD [vertically opposite angles]

âˆ´ âˆ†APB ~ âˆ†CPD [by AA similarity criterion]

Hence, length of PD is 5 cm and length of CD is 2 cm.

Question 2.

It is given that âˆ†ABC ~ âˆ†EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.

Solution:

Given, âˆ†ABC ~ âˆ†EDF, so the corresponding sides of âˆ†ABC and âˆ†EDF are in the same ratio

i.e., \(\frac{A B}{E D}=\frac{A C}{E F}=\frac{B C}{D F}\) ………………. (i)

Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

Question 3.

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

Solution:

Let a âˆ†ABC in which a line DE parallel to BC intersects AB at D and AC at E.

To prove : DE divides the two sides in the same ratio.

Now, since, âˆ†BDE and âˆ†DEC lie between the same parallel lines DE and BC and on the same base DE

So, ar(âˆ†BDE) = ar(âˆ†DEC) ………….. (iii)

From Eqs. (i), (ii) and (iii),

\(\frac{A D}{D B}=\frac{A E}{E C}\)

Hence proved

Question 4.

In the figure, if PQRS is a parallelogram and AB || PS, then prove that OC || SR.

Solution:

Given PQRS is a parallelogram, so PQ || SR and PS || QR. Also AB || PS.

To prove : OC || SR

Proof : In âˆ†OPS and âˆ†OAB, PS || AB

âˆ POS = âˆ AOBÂ Â [common angle]

âˆ OSP = âˆ OBAÂ Â Â Â [corresponding angles]

âˆ´ âˆ†OPS ~ âˆ†OABÂ Â Â Â Â Â [by AA similarity criterion]

Then, \(\frac{P S}{A B}=\frac{O S}{O B}\) ………… (i)

In âˆ†CQE and âˆ†CAB, QR || PS || AB

âˆ QCR = âˆ ACB [common angle]

âˆ CRQ = âˆ CBA [corresponding angles]

âˆ´ âˆ†CQR ~ âˆ†CAB

On subtracting 1 from both sides, we get OB CB

By converse of basic proportionality theorem, SR || OC.

Hence proved

Question 5.

A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

Solution:

Let AC be the ladder of length 5 m and BC = 4 m be the height of the wall, which ladder is placed. If the foot of the ladder is moved 1.6 m towards the wall i.e, AD = 1.6 m, then the ladder is slide upward i.e., CE = x m.

In right angled âˆ†ABC,

AC^{2} = AB^{2} + BC^{2Â Â Â } [by Pythagoras theorem]

â‡’ (5)^{2} = (AB)^{2} + (4)^{2}

â‡’ AB^{2} = 25 – 16 = 9

â‡’ AB = 3m

Now, DB = AB – AD = 3 – 1.6 = 1.4 m

In right angled âˆ†EBD,

ED^{2} = EB^{2} + BD^{2Â Â Â Â } [by Pythagoras theorem]

â‡’ (5)^{2} = (EB)^{2} + (1.4)^{2} [ âˆµ BD = 1.4 m]

â‡’ 25 = (EB)^{2} + 1.96

â‡’ (EB)^{2} = 25 – 1.96 = 23.04

â‡’ EB = \(\sqrt{23.04}\) = 4.8

Now, EC = EB – BC = 4.8 – 4 = 0.8

Hence, the top of the ladder would slide upwards on the wall at distance is 0.8 m.

Question 6.

For going to a city B from city A, there is a route via city C such that AC âŠ¥ CB, AC = 2x km and CB = 2 (x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

Solution:

Given, AC âŠ¥ CB, AC = 2xkm,CB = 2(x + 7)km and AB = 26 km

On drawing the figure, we get the right angle âˆ†ACB right angled at C.

Now, In âˆ†ACB, by Pythagoras theorem,

AB^{2} = AC^{2} + BC^{2}

â‡’ (26)^{2} = (2x)^{2} + {2(x + 7)}^{2}

â‡’ 676 = 4x^{2} + 4(x^{2} + 49 + 11x)

â‡’ 676 = 4x^{2} + 4x^{2} + 196 + 56x

â‡’ 676 = 8x^{2} + 56x + 196

â‡’ 8x^{2} + 56x – 480 = 0

On dividing by 8, we get x^{2} + 7x – 60 = 0

â‡’ x^{2} + 12x-5x-60 = 0

â‡’ x(x + 12) – 5(x + 12) = 0

â‡’ (x + 12)(x – 5) = 0

âˆ´ x = -12, x = 5

Since, distance cannot be negative.

âˆ´ x = 5 [âˆµ x â‰ 12]

Now, AC = 2x = 10 km and BC = 2(x + 7) = 2(5 + 7) = 24 km

The distance covered to reach city B from city A via city C = AC + BC = 10 + 24 = 34 km

Distance covered to reach city B from city A after the construction of the highway is

BA = 26 km

Hence, the required saved distance is 34 – 26

i.e., 8 km

Question 7.

A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.

Solution:

Let BC = 18 m be the flag pole and its shadow be AB = 9.6 m. The distance of the top of the pole, C from the far end i.e., A of the shadow is AC

In right angled âˆ†ABC

AC^{2} = AB^{2} + BC^{2} [by Pythagoras theorem]

â‡’ AC^{2} = (9.6)^{2} + (18)^{2}

â‡’ AC^{2} = 92.16 + 324

â‡’ AC^{2} = 416.16

âˆ´ AC = \(\sqrt{416.16}\) = 20.4 m

Hence, the required distance is 20.4 m.

Question 8.

A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3m, find how far she is away from the base of the pole.

Solution:

Let A be the position of the street bulb fixed on a pole AB = 6 m and CD = 1.5 m be the height of a woman and her shadow be ED = 3 m. Let distance between pole and woman be x m.

Here, woman and pole both are standing vertically

So, CD || AB

In âˆ†CDE and âˆ†ABE,

âˆ E = âˆ E [common angle]

âˆ ABE = âˆ CDE [each equal to 90Â°]

âˆ´ âˆ†CDE ~ âˆ†ABE [by AA similarity criterion]

Then \(\frac{E D}{E B}=\frac{C D}{A B} \Rightarrow \frac{3}{3+x}=\frac{1.5}{6}\)

â‡’ 3 Ã— 6 = 1.5(3 + x)

â‡’ 18 = 1.5 Ã— 3 + 1.5x

â‡’ 1.5x = 18 – 4.5

âˆ´ x = \(\frac{13.5}{1.5}\) = 9 m

Hence, she is at the distance of 9 m from the base of the pole.

Question 9.

In the figure, ABC is a triangle right angled at B and BD âŠ¥ AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.

Solution:

Given, âˆ†ABC in which âˆ B = 90Â° and BD âŠ¥ AC

Also, AD = 4 cm and CD = 5 cm

In âˆ†DBA and âˆ†DCB,

âˆ ADB = âˆ CDB [each equal to 90Â°]

and âˆ BAD = âˆ DBC [each equal to 90Â° – âˆ C] ;.

âˆ´ âˆ†DBA ~ âˆ†DCB [by AA similarity criterion]

Question 10.

In the figure, PQR is a right triangle right angled at Q and QS âŠ¥ PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.

Solution:

Given, âˆ†PQR in which âˆ Q = 90Â°, QS âŠ¥ PR and PQ = 6 cm, PS = 4 cm

In âˆ†SQP and âˆ†SRQ,

âˆ PSQ = âˆ RSQ [each equal to 90Â°]

âˆ SPQ = âˆ SQR [each equal to 90Â° – âˆ R]

âˆ´ âˆ†SQP ~ âˆ†SRQ [By AA similarity criterion]

Then, \(\frac{S Q}{P S}=\frac{S R}{S Q}\)

â‡’ SQ^{2} = PS Ã— SR ………….. (i)

In right angled âˆ†PSQ,

PQ^{2} = PS^{2} + QS^{2} [by Pythagoras theorem]

â‡’ (6)^{2} = (4)^{2}.+ QS^{2}

â‡’ 36 = 16 + QS^{2}

â‡’ QS^{2} = 36 – 16 = 20

âˆ´ QS.= \(\sqrt{20}=2 \sqrt{5}\) cm

On putting the value of QS in Eq(i), we get

Question 11.

In âˆ†PQR, PD âŠ¥ QR such that D lies on QR . If PQ = a, PR = b, QD = c and DR = d, prove that [a + b)(a – b) = (c + d)(c – d).

Solution:

Given: In âˆ†PQR, PD âŠ¥ QR, PQ = a, PR = b,

QD = c and DR = d

To prove : (a + b)(a -b) = (c + d)(c – d)

Proof : In right angled âˆ†PDQ,

PQ^{2} = PD^{2} + QD^{2} [by Pythagoras theorem]

â‡’ a^{2} = PD^{2} + c^{2}

â‡’ PD^{2} = a^{2} – c^{2} …………. (i)

In right angled âˆ†PDR,

PR^{2} = PD^{2} + DR^{2} [by Pythagoras theorem]

â‡’ b^{2} = PD^{2} + d^{2}

â‡’ PD^{2} = b^{2} – d^{2} ………….. (ii)

From Eqs. (i) and (ii)

a^{2} – c^{2} = b^{2} – d^{2}

â‡’ a^{2} – b^{2} = c^{2} – d^{2}

â‡’ (a – b)(a + b) = (c – d)(c + d)

Hence proved.

Question 12.

In a quadrilateral ABCD, âˆ A + âˆ D = 90Â°. Prove that AC^{2} + BD^{2} = AD^{2} + BC^{2} [Hint: Produce AB and DC to meet at E]

Solution:

Given : Quadrilateral ABCD, in which âˆ A + âˆ D = 90Â°

To prove : AC^{2} + BD^{2} = AD^{2} + BC^{2}

Construct: Produce AB and CD to meet at E

Also join AC and BD

Proof: In âˆ†AED, âˆ A + âˆ D = 90Â° [given]

âˆ´ âˆ E = 180Â° – (âˆ A + âˆ D) = 90Â°

[ âˆµ sum of angles of a triangle = 180Â°]

Then, by Pythagoras theorem,

AD^{2} = AE^{2} + DE^{2}

In âˆ†BEC, by Pythagoras theorem,

BC^{2} = BE^{2} + EC^{2}

On adding both equations, we get

AD^{2} + BC^{2} = AE^{2} + DE^{2} + BE^{2} + CE^{2} ………… (i)

In âˆ†AEC, by Pythagoras theorem,

AC^{2} = AE^{2} + CE^{2}

and in âˆ†BED, by Pythagoras theorem,

BD^{2} = BE^{2} + DE^{2}

On adding both equations, we get

AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2} ………… (ii)

From Eqs. (i) and (ii)

AC^{2} + BD^{2} = AD^{2} + BC^{2}

Hence proved.

Question 13.

In the given figure, l || m and line segments AB, CD and EF are concurrent at point P. Prove that \(\frac{A E}{B F}=\frac{A C}{B D}=\frac{C E}{F D}\)

Solution:

Given l || m and line segments AB, CD and EF are concurrent at point P

Question 14.

In the figure, PA, QB, RC and SD are all perpendiculars to a line l,AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ,

Solution:

Given, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm

Also, PA, QB, RC and SD are all perpendiculars to line l

âˆ´ PA || QB || RC || SD By Basic proportionality theorem,

PQ : QR : RS = AB : BC : CD = 6 : 9 : 12

Let PQ = 6x, QR = 9x and RS = 12x

Since, length of PS = 36 cm

âˆ´ PQ + QR + RS = 36

â‡’ 6x + 9x + 12x = 36

â‡’ 27x = 36

âˆ´ x = \(\frac{36}{27}=\frac{4}{3}\)

Now, PQ = 6x = 6 Ã— \(\frac{4}{3}\) = 8 cm

QR = 9x = 9 Ã— \(\frac{4}{3}\)= 12 cm

and RS = 12x = 12 Ã— \(\frac{4}{3}\) = 16 cm

Question 15.

O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through 0, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

Solution:

Given ABCD is a trapezium. Diagonals AC and BD are intersect at O.

PQ || AB || DC

To prove : PO = QO

Proof : In âˆ†ABD and âˆ†POD, PO || AB [âˆµ PQ || AB]

âˆ D = âˆ D [common angle]

âˆ ABD = âˆ POD [corresponding angles]

âˆ´ âˆ†ABD ~ âˆ†POD[by AA similarity criterion]

Then, \(\frac{O P}{A B}=\frac{P D}{A D}\) …………… (i)

In âˆ†ABC and âˆ†OQC, OQ || AB

âˆ C = âˆ C [common angle]

âˆ B AC = âˆ QOC [corresponding angles]

âˆ´ âˆ†ABC ~ âˆ†OQC [by AA similarity criterion]

Question 16.

In the figure, line segment DF intersect the side AC of a âˆ†ABC at the point E such that E is the mid-point of CA and âˆ AEF = âˆ AFE. Prove that \(\frac{B D}{C D}=\frac{B F}{C E}\)

[Hint:Take point G on AB such that CG || DF]

Solution:

Given âˆ†ABC, E is the mid-point of CA and âˆ AEF = âˆ AFE

To prove : \(\frac{B D}{C D}=\frac{B F}{C E}\)

Construction : Take a point G on AB such that CG || DF

Proof : Since, E is the mid-point of CA

âˆ´ CE = AE

In âˆ†ACG, CG || EF and E is mid-point of CA

So, CE = GF …………… (ii) [by mid-point theorem]

Now, in âˆ†BCG and âˆ†BDF, CG || DF

Question 17.

Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

Solution:

Let ABC be a right triangle, right angled at B and AB = y, BC = x

Three semi-circles are drawn on the sides AB,

BC and AC, respectively with diameters AB,

BC and AC, respectively

Again, let area of circles with diameters AB,

BC and AC are respectively A_{1}, A_{2} and A_{3}

To prove : A_{3} = A_{1} + A_{2}

Proof : In âˆ†ABC, by Pythagoras theorem,

Question 18.

Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

Solution:

Lett a right triangle BAC in which âˆ A is right angle and AC = y, AB = x

Three equilateral triangles âˆ†AEC, âˆ†AFB and âˆ†CBD are drawn on the three sides of âˆ†ABC.

Again, let area of triangles made on AC, AB and BC are A_{1}, A_{2} and A_{3}, respectively.

To prove : A_{3} = A_{1} + A_{2}

Proof : In âˆ†CAB, by Pythagoras theorem,

BC^{2} = AC^{2} + AB^{2}

â‡’ BC^{2} = y^{2} + x^{2}

â‡’ BC = \(\sqrt{y^{2}+x^{2}}\)