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#### Question 1:

Choose the correct answer from the given four options:
In the figure, ∠BAC = 90° and AD ⊥ BC. Then, (A) BD · CD = BC2
(B) AB · AC = BC2
(C) BD · CD = AD2
(D) AB · AC = AD2 In $△$ABC,
∠B + ∠BAC + ∠C = 180°         (Angle sum property)
⇒ ∠B + 90° + ∠C = 180°
⇒ ∠B  = ∠DBA = 90° – ∠C

Similarly, using angle sum property of triangles in $△$ADC,
∠DAC = 90° – ∠C

In $△$ADB and $△$ADC,
∠D = ∠D = 90°                                                   (∵ AD ⊥ BC)
∠DBA = ∠DAC                                                  [each angle equals to 90° $-$ ∠C ]
∴ $△$ADB ∼ $△$ADC                                           [Now by AA similarity criteria]

$⇒$$\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{AD}}{\mathrm{CD}}$
$⇒$ BD · CD = AD2

Hence, the correct answer is option C.

#### Question 2:

Choose the correct answer from the given four options:
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) 8cm
(D) 20 cm

Consider the rhombus ABCD such that AC = 16 cm and BD = 12 cm. Also, ∠AOB = 90°. Now, a rhombus is a quadrilateral whose four sides are of same length and diagonals bisect each other at right angles.
∴ AC and BD bisects each other
$\mathrm{AO}=\frac{1}{2}\mathrm{AC}$ and $\mathrm{BO}=\frac{1}{2}\mathrm{BD}$
⇒ AO = 8 cm and BO = 6 cm

Now, in right angled $△$AOB,
Using Pythagoras theorem,
$\begin{array}{rcl}& ⇒& {\mathrm{AB}}^{2}={\mathrm{AO}}^{2}+{\mathrm{OB}}^{2}\\ & =& {8}^{2}+{6}^{2}\\ & =& 100\end{array}$

Thus, the side of rhombus is 10 cm.
Hence, the correct answer is option B.

#### Question 3:

Choose the correct answer from the given four options:
If $△$ABC ~ $△$EDF and $△$ABC is not similar to $△$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

Given: $△$ABC ∼ $△$EDF
$⇒\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{AC}}{\mathrm{EF}}$                                                (By similarity property) Taking first two terms, we get
$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DF}}$
⇒ AB · DF = ED · BC
So, option (D) is true

Taking last two terms, we get
$\frac{\mathrm{BC}}{\mathrm{DF}}=\frac{\mathrm{AC}}{\mathrm{EF}}$
⇒ BC · EF = AC · DF
So, option (A) is true.

Taking first and last terms, we get
$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{AC}}{\mathrm{EF}}$
⇒ AB · EF = ED · AC
Hence, option (B) is true.

Thus, all the options are true except C.
Hence, the correct answer is option (C).

#### Question 4:

Choose the correct answer from the given four options:

If in two triangles ABC and PQR, $\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}},$ then
(A) $△$PQR ~ $△$CAB
(B) $△$PQR ~ $△$ABC
(C) $△$CAB ~ $△$PQR
(D) $△$BCA ~ $△$PQR

In $△$ABC and $△$PQR,
$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}},$

If the sides of one triangle are proportional to the sides of the other triangle and their corresponding angles are also equal, then by SSS similarity criterion, both the triangles are similar.
Therefore, by SSS Similarity,
$△$CAB $~$ $△$PQR Hence, the correct answer is option (C).

#### Question 5:

Choose the correct answer from the given four options:

In Fig. 6.3, 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, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to (A) 50°
(B) 30°
(C) 60°
(D) 100°

In $△$APB and $△$CPD,
∠APB = ∠CPD = 50°                       [vertically opposite angles]

Now,
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{6}{5}$                                             .....(1)

$\frac{\mathrm{BP}}{\mathrm{CP}}=\frac{3}{2.5}=\frac{6}{5}$                                   .....(2)

From (1) and (2),
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{BP}}{\mathrm{CP}}$

$△$APB ∼ $△$DPC                           [by SAS similarity criterion]
⇒ ∠A = ∠D = 30°                             [corresponding angles of similar triangles]

In $△$APB,
∠A + ∠B + ∠APB = 180°                 [Sum of angles of a triangle = 180°]
⇒ 30° + ∠B + 50° = 180°
∴ ∠B = 180° $-$ (50° + 30°)
⇒ ∠B = 180 – 80° = 100°
⇒ ∠PBA = 100°
Hence, the correct answer is option D.

#### Question 6:

Choose the correct answer from the given four options:
If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?

(A) $\frac{\mathrm{EF}}{\mathrm{PR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$

(B) $\frac{\mathrm{DE}}{\mathrm{PQ}}=\frac{\mathrm{EF}}{\mathrm{RP}}$

(C) $\frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$

(D) $\frac{\mathrm{EF}}{\mathrm{RP}}=\frac{\mathrm{DE}}{\mathrm{QR}}$

Given that, $△$DEF and $△$PQR such that ∠D = ∠Q and ∠R = ∠E. $△$DEF ∼ $△$QRP                                         [ By AA similarity criteria]

$⇒\frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}=\frac{\mathrm{FE}}{\mathrm{PR}}$                                        [ Corresponding sides of similar triangles are in same ratio]
From the above ratio, we can see that all options are true except B.
Hence, the correct answer is option B.

#### Question 7:

Choose the correct answer from the given four options:

In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. 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

In $△$ABC and $△$DEF,
∠B = ∠E,
∠C = ∠F,
∴ $△$ABC $~$$△$DEF                                             [by AA similarity criteria]

As AB = 3DE,
$\therefore$ AB $\ne$ DE
For the triangles to be congruent, AB has to be equal to DE.
Thus, the triangles are similar but not congruent.

Hence, the correct answer is option B.

#### Question 8:

Choose the correct answer from the given four options:

It is given that $△$ABC ~ $△$PQR with $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}.$ Then, $\frac{\mathrm{ar}\left(\mathrm{PRQ}\right)}{\mathrm{ar}\left(\mathrm{BCA}\right)}$ is equal to
(A) 9

(B) 3

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

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

Given that $△$ABC ~ $△$PQR with $\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}.$

By property of similar triangles, the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.
$\begin{array}{rcl}& ⇒& \frac{\mathrm{ar}\left(△\mathrm{PQR}\right)}{\mathrm{ar}\left(△\mathrm{ABC}\right)}=\frac{{\left(\mathrm{QR}\right)}^{2}}{{\left(\mathrm{BC}\right)}^{2}}\\ & =& {\left(\frac{3}{1}\right)}^{2}\\ & =& \frac{9}{1}=9\end{array}$

Hence, the correct answer is option A.

#### Question 9:

Choose the correct answer from the given four options:

It is given that $△$ABC ~ $△$DFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true:
(A) DE = 12 cm, ∠F = 50°
(B) DE = 12 cm, ∠F = 100°
(C) EF = 12 cm, ∠D = 100°
(D) EF = 12 cm, ∠D = 30°

Given that, $△$ABC ∼ $△$DFE, ∠A = 30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm.

Now, corresponding angles of similar triangles are equal.
∴ ∠A = ∠D = 30°
and ∠C = ∠E = 50° By angle sum property of triangles, the sum of angles of a triangle is 180°.
∴ ∠B = ∠F = 180° $-$ (30° + 50°) = 100°

Also, the ratio of the sides of two similar triangles are equal.

∴ DE = 12 cm and ∠F = 100°.
Hence, the correct answer is option B.

#### Question 10:

Choose the correct answer from the given four options:

If in triangles ABC and DEF, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}},$ then they will be similar, when
(A) ∠B = ∠E
(B) ∠A = ∠D
(C) ∠B = ∠D
(D) ∠A = ∠F

Given that, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}$.
∴ $△$ABC ~ $△$EDF Now, for the triangles to be similar by SAS criteria, the angle between the proportional sides must be equal.
∴ ∠B = ∠D
Hence, the correct answer is option C.

#### Question 11:

Choose the correct answer from the given four options:

If $△$ABC ~ $△$QRP, $\frac{\mathrm{ar}\left(\mathrm{ABC}\right)}{\mathrm{ar}\left(\mathrm{PQR}\right)}=\frac{9}{4}$, AB = 18 cm and BC = 15 cm, then PR is equal to
(A) 10 cm

(B) 12 cm

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

(D) 8 cm

Given that, $△$ABC ~ $△$QRP, $\frac{\mathrm{ar}\left(\mathrm{ABC}\right)}{\mathrm{ar}\left(\mathrm{PQR}\right)}=\frac{9}{4}$, AB = 18 cm and BC = 15 cm. The ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.
$\therefore \frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{QRP}\right)}=\frac{{\left(\mathrm{BC}\right)}^{2}}{{\left(\mathrm{RP}\right)}^{2}}$

Now, $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{QRP}\right)}=\frac{9}{4}$

Hence, the correct answer is option A.

#### Question 12:

Choose the correct answer from the given four options:

If S is a point on side PQ of a $△$PQR such that PS = QS = RS, then
(A) PR · QR = RS2
(B) QS2 + RS2 = QR2
(C) PR2 + QR2 = PQ2
(D) PS2 + RS2 = PR2

In $△$PQR
PS = QS = RS                                   .....(1)
In $△$PSR,
PS = RS                                             .....[from (1)]
⇒ ∠1 = ∠2                                        .....(2)

Similarly,
In $△$RSQ, Angles opposite to equal sides are equal
⇒ ∠3 = ∠4                                        .....(3) Now in $△$PQR, the sum of angles is 180°.
⇒ ∠P + ∠Q + ∠R = 180°
⇒ ∠2 + ∠4 + ∠1 + ∠3 = 180°
⇒ ∠1 + ∠3 + ∠1 + ∠3 = 180°
⇒ 2(∠1 + ∠3) = 180°
$\begin{array}{rcl}& ⇒& \angle 1+\angle 3=\frac{180°}{2}\\ & =& 90°\end{array}$
∴ ∠R = 90°

In $△$PQR, by Pythagoras theorem,
PR2 + QR2 = PQ2

Hence, the correct answer is option B.

#### Question 1:

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

Let us suppose that a = 25 cm, b = 5 cm and c = 24 cm.
Now, if the triangle is right-angle triangle, then a should be hypotenuse.

According to the Pythagoras theorem, the sum of square of two sides must be equal to square of the hypotenuse.
Here,
$\begin{array}{rcl}{b}^{2}+{c}^{2}& =& {\left(5\right)}^{2}+{\left(24\right)}^{2}\\ & =& 25+576=601\end{array}$

But ${a}^{2}={\left(25\right)}^{2}=625\ne 601$
Thus, the given sides do not make a right triangle as they do not satisfy the property of Pythagoras theorem.

Hence, the triangle with sides 25cm, 5cm and 24cm is not a right triangle.

#### Question 2:

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

In similar triangles, the corresponding angles are also equal.

Considering $△$DEF ~ $△$RPQ,
∠D = ∠R,
∠E = ∠P and
∠F = ∠Q
Hence, the given statement is false.

#### Question 3:

A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR?

Given that, PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Then,
QA = QP – PA
= 12.5 – 5
= 7.5 cm

Considering the ratio for the converse of basic proportionality theorem to hold true.
.....(1)
And,                 .....(2)
$⇒\frac{\mathrm{PA}}{\mathrm{AQ}}=\frac{\mathrm{PB}}{\mathrm{BR}}$

According to the converse of basic proportionality theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Hence, AB || QR.

#### Question 4:

In Fig 6.4, BD and CE intersect each other at the point P. Is $△$PBC ~ $△$PDE? Why? In $△$PBC and $△$PDE,
∠BPC = ∠EPD                       [vertically opposite angles]
Now,          .....(1)
And, $\frac{\mathrm{PC}}{\mathrm{EP}}=\frac{6}{12}=\frac{1}{2}$                 .....(2)

From (1) and (2), we get
$\frac{\mathrm{PB}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{EP}}$

Since, ∠BPC of $△$PBC is equal to ∠EPD of $△$PDE and the sides including these angles are proportional.
Hence, $△$PBC ∼ $△$PDE by SAS similarity criteria.

#### Question 5:

In triangles PQR and MST, ∠P = 55°, ∠Q = 25°, ∠M = 100° and ∠S = 25°. Is $△$QPR ~ $△$TSM? Why?

The sum of three angles of a triangle is 180°.
In $△$PQR,
∠P + ∠Q + ∠R = 180°
⇒ 55° + 25° + ∠R = 180°
⇒ ∠R = 180° $-$ (55° + 25°)
= 180° $-$ 80°
= 100°

Similarly, in $△$TSM,
∠T + ∠S + ∠M = 180°
⇒ ∠T + 25° + 100° = 180°
⇒ ∠T = 180° $-$ (25° + 100°)
⇒ ∠T = 180° $-$ 125°
= 55°  In $△$PQR and $△$TSM,
∠P = ∠T,
∠Q = ∠S and
∠R = ∠M
So, $△$PQR ∼ $△$TSM                         ( by  AAA criteria of similarity )
Hence, ∆QPR is not similar to $△$TSM as $△$PQR is similar to $△$TSM.

#### Question 6:

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

Two quadrilaterals are similar if their corresponding angles are equal and the ratio of corresponding sides are proportional.

Having corresponding angles equal isn't enough to say that the quadrilaterals would be similar.
Hence, the given statement is false.

#### 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?

Let the sides of first triangle be x, y and z.
As per given information, sides of second triangle will be 3x, 3y and z'.

Since, the perimeter of other triangle is three times the perimeter of first triangle.
$⇒$3x + 3y + z' = 3(x + y + z)
$⇒$3x + 3y + z' = 3x + 3y + 3z
$⇒$z' = 3z
Here, the corresponding three sides of triangle are in proportion.
Hence, the two triangles are similar by SSS criteria of similarity.

#### 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?

Let two right angled triangles be $△$LMO and $△$RST. Given that,
∠L = ∠R = 90°
∠M = ∠S       [Acute angles equal]

Hence, by AA similarity criteria, we have $△$LMO ∼ $△$RST.
Hence, we can say that the two triangles will be similar.

#### 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?

Given that, the ratio of altitudes of similar triangles is $\frac{3}{5}$.
By the property of area of two similar triangles, the ratio of the area of two similar triangles equal to the square of the ratio of their altitudes.
$\left(\frac{{\mathrm{Area}}_{1}}{{\mathrm{Area}}_{2}}\right)={\left(\frac{{\mathrm{Altitude}}_{1}}{{\mathrm{Altitude}}_{2}}\right)}^{2}$

$⇒\frac{9}{25}\ne \frac{6}{5}\phantom{\rule{0ex}{0ex}}$
So, given statement is not correct because the ratio of their areas is equal to $\frac{9}{25}$.

#### Question 10:

D is a point on side QR of $△$PQR such that PD ⊥ QR. Will it be correct to say that $△$PQD ~ $△$RPD? Why?

In $△$PQD and $△$RPD,
PD = PD                          [common side]
∠PDQ = ∠PDR = 90$°$ Here, neither the other sides are proportional nor the angles are equal, so we can say that $△$PQD is not similar to $△$RPD.

#### Question 11:

In Fig. 6.5, if ∠D = ∠C, then is it true that $△$ADE ~ $△$ACB? Why? In $△$ADE and $△$ACB,
∠D = ∠C                                [given]
∠A = ∠A                                [common angle]

The sum of all the angles of a triangle is 180$°$.
So, by angle sum property of triangle, the third angle of both triangles must be equal.
∠E = ∠B
$△$ADE ∼ $△$ACB                    [by AAA similarity criterion]
Hence, the given statement is true.

#### 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.

We know that by SAS similarity criteria, if one angle of a triangle is equal to one angle of the other triangle and the sides including these are proportional, then the two triangles are similar.

Here, one angle of a triangle is equal to one angle of the other triangle but the sides which are proportional are not the ones necessarily including equal angle.
Hence, the given statement is not correct.

#### Question 1:

In a $△$PQR, PR2 – PQ2 = QR2 and M is a point on side PR such that QM ⊥ PR. Prove that
QM2 = PM × MR.

Given that, in a $△$PQR,
PR2 – PQ2 = QR2
⇒ PR= PQ2 + QR2
So, $△$PQR is right angled triangle at Q.       (By converse of Pythagoras Theorem) In $△$QMR and $△$PMQ,
∠QMR = ∠PMQ = 90°
∠MQR = ∠QPM = 90°$-$∠R
$△$QMR ∼ $△$PMQ                                (by AA similarity criteria)

Now, using property of area of similar triangles, we get,
$⇒\frac{\mathrm{ar}\left(△\mathrm{QMR}\right)}{\mathrm{ar}\left(△\mathrm{PMQ}\right)}=\frac{{\left(\mathrm{QM}\right)}^{2}}{{\left(\mathrm{PM}\right)}^{2}}$                                                                      .....(1)

Also,
.....(2)

From (1) and (2),
$⇒\frac{{\left(\mathrm{QM}\right)}^{2}}{{\left(\mathrm{PM}\right)}^{2}}=\frac{\frac{1}{2}×\mathrm{RM}×\mathrm{QM}}{\frac{1}{2}×\mathrm{PM}×\mathrm{QM}}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(\mathrm{QM}\right)}^{2}}{{\left(\mathrm{PM}\right)}^{2}}=\frac{\mathrm{RM}}{\mathrm{PM}}\phantom{\rule{0ex}{0ex}}⇒{\left(\mathrm{QM}\right)}^{2}=\mathrm{RM}×\mathrm{PM}$
Hence, proved.

#### Question 2:

Find the value of x for which DE ∥ AB in Fig. 6.8. If DE ∥ AB, then by Basic Proportionality Theorem,
$⇒\frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{CE}}{\mathrm{BE}}$
$⇒\frac{x+3}{3x+19}=\frac{x}{3x+4}$
⇒ (x + 3) (3x + 4) = x (3x + 19)
⇒ 3x2 + 4x + 9x + 12 = 3x2 + 19x
⇒ 19x − 13x = 12
⇒ 6x = 12
x = 2
Hence, the required value of x is 2.

#### Question 3:

In Fig. 6.9, if ∠1 = ∠2 and $△$NSQ ≅ $△$MTR, then prove that $△$PTS ~ $△$PRQ. Given that, $△$NSQ ≅ $△$MTR and ∠1 = ∠2.

Since,
$△$NSQ = $△$MTR
Now, by CPCT,
SQ = TR                            .....(1)

Also, sides opposite to equal angles are also equal and ∠1 = ∠2.
PT = PS                             .....(2)

From (1) and (2),
$\frac{\mathrm{PS}}{\mathrm{SQ}}=\frac{\mathrm{PT}}{\mathrm{TR}}$

Now, by converse of basic proportionality theorem,
ST ∥ QR
∴ ∠1 = ∠PQR
∠2 = ∠PRQ

In $△$PTS and $△$PRQ,
∠P = ∠P                                 (Common)
∠1 = ∠PQR                            (proved)
∠2 = ∠PRQ                            (proved)
$△$PTS ~ $△$PRQ                 (By AAA similarity criteria)
Hence, proved.

#### Question 4:

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

Given that, PQRS is a trapezium in which PQ ∥ RS and PQ = 3RS. In $△$POQ and $△$ROS,
∠SOR = ∠QOP                                         (vertically opposite angles)
∠SRP = ∠RPQ                                          (alternate angles)
$△$POQ ∼ $△$ROS                                      (by AA similarity criterion)

By property of areas of similar triangles, the ratio of areas of similar triangles is equal to the square of the ratio of their corresponding sides.
$\begin{array}{rcl}\frac{\mathrm{ar}\left(△\mathrm{POQ}\right)}{\mathrm{ar}\left(△\mathrm{SOR}\right)}& =& \frac{{\left(\mathrm{PQ}\right)}^{2}}{{\left(\mathrm{RS}\right)}^{2}}\\ & =& \frac{{3}^{2}}{{1}^{2}}\\ & =& \frac{9}{1}\end{array}$

Hence, the required ratio is 9 : 1.

#### Question 5:

In Fig. 6.10, if AB ∥ DC and AC and PQ intersect each other at the point O, prove that OA ∙ CQ = OC ∙ AP. Given that, AC and PQ intersect each other at the point O and AB DC.

In $△$AOP and $△$COQ,
∠AOP = ∠COQ                                              (vertically opposite angles)
∠APO = ∠CQO                                              (since, AB DC and PQ is transversal, so alternate angles)
$△$AOP ∼ $△$COQ                                       (by AA similarity criterion)
Then,
$\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{AP}}{\mathrm{CQ}}$                                                      (since, corresponding sides are proportional of similar triangle)
⇒ OA × CQ = OC × AP
Hence, proved.

#### Question 6:

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

Let $△$LMN be an equilateral triangle of side 8 cm.
∴ LM = MN = NL = 8 cm Now, LD is drawn perpendicular to MN.
Since, in equilateral triangle altitude and median are same,
∴ D is the mid-point of MN.
⇒ MD = ND = $\frac{1}{2}$MN
⇒ MD = $\frac{8}{2}$ = 4 cm

Now, by Pythagoras theorem,
LM2 = LD2 + MD2
⇒ (8)2 = LD+ (4)2
⇒ 64 = LD2 + 16
⇒ LD =
Hence, altitude of an equilateral triangle is .

#### Question 7:

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

Given that, $△$ABC ~ $△$DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm.

The sides of the two similar triangles are in same proportion,
$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$
$⇒\frac{4}{6}=\frac{\mathrm{BC}}{9}=\frac{\mathrm{AC}}{12}$

By taking first two terms, we have
$\frac{4}{6}=\frac{\mathrm{BC}}{9}$

And by taking last two terms, we have
$\frac{\mathrm{BC}}{9}=\frac{\mathrm{AC}}{12}$

Now,
Perimeter of $△$ABC = AB + BC + AC
= 4 + 6 + 8
= 18 cm
Thus, the perimeter of $△$ABC is 18 cm.

#### Question 8:

In Fig. 6.11, if DE ∥ BC, find the ratio of ar(ADE) and ar(DECB). Given that, DE  BC, DE = 6 cm and BC = 12 cm.
In $△$ABC and $△$ADE,
∠ABC = ∠ADE                            (corresponding angles)
∠ACB = ∠AED                            (corresponding angles)
∠A = ∠A                                       (common)
$△$ABC ∼ $△$AED                      (by AAA similarity criterion)

By property of area of similar triangle, ratio of area of two triangles is equal to the square of the ratio of their sides.
$\frac{\mathrm{ar}\left(△\mathrm{ADE}\right)}{\mathrm{ar}\left(△\mathrm{ABC}\right)}=\frac{{\left(1\right)}^{2}}{{\left(2\right)}^{2}}=\frac{1}{4}$

So, let ar($△$ADE) = and ar($△$ABC) = 4k

Now,
Ar(DECB) = ar(ABC) – ar(ADE) = 4kk = 3k
∴ Required ratio = ar(ADE) : ar(DECB) = k : 3k = 1 : 3
Therefore, ar(ADE) : ar(DECB) = 1 : 3

#### Question 9:

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.

ABCD is a trapezium in which AB  DC, P and Q are points on AD and BC respectively such that PQ  DC.
Thus,
AB  PQ  DC. In $△$ABD,
PO  AB            (∵ PQ  AB)
(By basic proportionality theorem)          .....(1)

In $△$BDC,
OQ  DC             (∵ PQ  DC)
(By basic proportionality theorem)
$⇒\frac{\mathrm{QC}}{\mathrm{BQ}}=\frac{\mathrm{DO}}{\mathrm{OB}}$                                                      .....(2)

From (1) and (2),
$\frac{\mathrm{DP}}{\mathrm{AP}}=\frac{\mathrm{QC}}{\mathrm{BQ}}$
$⇒\frac{18}{\mathrm{AP}}=\frac{15}{35}$
$⇒\mathrm{AP}=\frac{\left(18×35\right)}{15}=42$
∴ AD = AP + DP
⇒ AD = 42 + 18 = 60 cm
Hence, AD is 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 cm2, find the area of the larger triangle.

Ratio of corresponding sides of two similar triangles = 2 : 3
Area of smaller triangle = 48 cm2

By the property of areas of two similar triangles, the ratio of the areas of two triangles is equal to the square of the ratio of their corresponding sides.

Hence, the area of the larger triangle is 108 cm.

#### Question 11:

In a triangle PQR, N is a point on PR such that QN ⊥ PR. If PN NR = QN2, prove that ∠PQR = 90°.

In $△$PQR, N is a point on PR, such that QN ⊥ PR and PN NR = QN2. ⇒ PN NR = QN QN
.....(1)

In $△$QNP and $△$RNQ,
$⇒\frac{\mathrm{PN}}{\mathrm{QN}}=\frac{\mathrm{QN}}{\mathrm{NR}}$                       [from (1)]
∠PNQ = ∠RNQ = 90$°$
$△$QNP ∼ $△$RNQ            (by SAS criteria of similarity)

Then, corresponding angles $△$QNP and $△$RNQ are equal.
∠PQN = ∠QRN                     .....(2)
∠RQN = ∠QPN                     .....(3)

Adding (2) and (3),
∠PQN + ∠RQN = ∠QRN + ∠QPN
⇒ ∠PQR = ∠QRN + ∠QPN  .....(4)

The sum of all angles of a triangle is 180°.

In $△$PQR,
∠PQR + ∠QPR + ∠QRP = 180°
⇒ ∠PQR + ∠QPN + ∠QRN = 180°               (∵ ∠QPR = ∠QPN and ∠QRP = ∠QRN)
⇒ ∠PQR + ∠PQR = 180°                               [using (4)]
⇒ 2∠PQR = 180°
∴ ∠PQR = 90°
Hence, proved.

#### Question 12:

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.

Area of smaller triangle = 36 cm2
Area of larger triangle = 100 cm2
Also, length of a side of the larger triangle = 20 cm
Let length of the corresponding side of the smaller triangle be x cm.

By property of area of similar triangle,

$\begin{array}{rcl}& ⇒& {x}^{2}=\frac{{\left(20\right)}^{2}×36}{100}\\ & =& \frac{400×36}{100}\end{array}$

Hence, the length of corresponding side of the smaller triangle is 12 cm.

#### Question 13:

In Fig. 6.12, if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, find BD. Given that, AC = 8 cm, AD = 3 cm and ∠ACB = ∠CDA.

In $△$ACD and $△$ABC,
∠A = ∠A                 (Common)
∠ADC = ∠ACB      (Given)
$△$ADC ~ $△$ACB      [By AA similarity criterion]

Now,

Hence, the value of BD is $\frac{55}{3}$ cm.

#### 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.

Given that, QR = 15 m (height of tower)
PQ = 24 m (shadow of tower)
XY = 16 m (shadow of the pole) At that time, ∠RPQ = ∠YXZ = θ
Again, let YZ = h be a telephone pole.

Here, $△$PQR and $△$XYZ both are right angles triangles.

In $△$PQR and $△$XYZ,
∠RPQ = ∠YXZ = θ
∠Q = ∠Y = 90°
$△$PQR ∼ $△$XYZ                      (by AA similarity criterion)

Then, corresponding sides of the two triangles are proportionate.
$\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{\mathrm{YZ}}$
$⇒\frac{24}{16}=\frac{15}{h}$

Hence, the height of the telephone pole is 10 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.

Let PQ be a vertical wall and PR = 10 m is a ladder.
The top of the ladder reaches to P and distance of ladder from the base of the wall QR is 6 m. In right angled $△$PQR, by Pythagoras theorem,
PR2 = PQ2 + QR2
⇒ (10)2 = PQ2 + (6)2
⇒ 100 = PQ2 + 36
⇒ PQ2 = 100 – 36 = 64

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

#### Question 1:

In Fig. 6.16, if ∠A = ∠C, AB = 6 cm, BP = 15 cm, AP = 12 cm and CP = 4 cm, then find the lengths of PD and CD. In $△$APB and $△$CPD,
∠A = ∠C                                   (given)
∠APB = ∠CPD                         (vertically opposite angles)
∴ $△$APD ∼ $△$CPD                  (by AA similarity criteria)

The ratio of their corresponding sides will be same.
$⇒\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PB}}{\mathrm{PD}}=\frac{\mathrm{AB}}{\mathrm{CD}}$
$⇒\frac{12}{4}=\frac{15}{\mathrm{PD}}=\frac{6}{\mathrm{CD}}$

By taking first two terms,
$\frac{12}{4}=\frac{15}{\mathrm{PD}}$

By taking first and last terms,

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

#### Question 2:

It is given that $△$ABC ~ $△$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.

Given that, $△$ABC ∼ $△$EDF.

So, the corresponding sides of $△$ABC and $△$EDF are in the same ratio.
$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{AC}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{DF}}$                                           .....(1) Also, AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm.

Substituting these values in (1),
$\frac{5}{12}=\frac{7}{\mathrm{EF}}=\frac{\mathrm{BC}}{15}$

On taking first and second terms,
$⇒\frac{5}{12}=\frac{7}{\mathrm{EF}}$

On taking first and third terms, we get
$⇒\frac{5}{12}=\frac{\mathrm{BC}}{15}$

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.

Let $△$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.
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ Construction join BE, CD and draw EF ⊥ AB and DG ⊥ AC.
$\frac{\mathrm{ar}\left(△\mathrm{ADE}\right)}{\mathrm{ar}\left(△\mathrm{BDE}\right)}=\frac{\frac{1}{2}×\mathrm{AD}×\mathrm{EF}}{\frac{1}{2}×\mathrm{DB}×\mathrm{EF}}$            (area of triangle = $\frac{1}{2}$× base × height)
$⇒\frac{\mathrm{ar}\left(△\mathrm{ADE}\right)}{\mathrm{ar}\left(△\mathrm{BDE}\right)}=\frac{\mathrm{AD}}{\mathrm{DB}}$                  .....(1)

Similarly,
$\frac{\mathrm{ar}\left(△\mathrm{ADE}\right)}{\mathrm{ar}\left(△\mathrm{DEC}\right)}=\frac{\frac{1}{2}×\mathrm{AE}×\mathrm{GD}}{\frac{1}{2}×\mathrm{EC}×\mathrm{GD}}$          (area of triangle = $\frac{1}{2}$× base × height)

$⇒\frac{\mathrm{ar}\left(△\mathrm{ADE}\right)}{\mathrm{ar}\left(△\mathrm{DEC}\right)}=\frac{\mathrm{AE}}{\mathrm{EC}}$                 .....(2)

Since, $△$BDE and $△$DEC lie between the same parallel DE and BC and on the same base DE.
So, ar($△$BDE) = ar($△$DEC)    .....(3)

From (1), (2) and (3),
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
Hence, proved.

#### Question 4:

In Fig 6.17, if PQRS is a parallelogram and AB  PS, then prove that OC  SR. Given, PQRS is a parallelogram,
So, PQ  SR and PS  QR.
Also, AB  PS. To prove: OC  SR
Proof:
In $△$OPS and $△$OAB,
PS AB
∠POS = ∠AOB                                 (common angle)
∠OSP = ∠OBA                                 (corresponding angles)
$△$OPS ∼ $△$OAB                          (by AA similarity criteria)

Then,
$\frac{\mathrm{PS}}{\mathrm{AB}}=\frac{\mathrm{OS}}{\mathrm{OB}}$                           (corresponding sides of similar triangle have same ratio)           .....(1)

In $△$CQR and $△$CAB,
QR  PS AB
∠QCR = ∠ACB                               (common angle)
∠CRQ = ∠CBA                               (corresponding angles)
$△$CQR ∼ $△$CAB                         (by AA similarity criteria)

$⇒\frac{\mathrm{QR}}{\mathrm{AB}}=\frac{\mathrm{CR}}{\mathrm{CB}}\phantom{\rule{0ex}{0ex}}$

Now, PS = QR as PQRS is a parallelogram.

From (1) and (2),
$\frac{\mathrm{OS}}{\mathrm{OB}}=\frac{\mathrm{CR}}{\mathrm{CB}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{OB}}{\mathrm{OS}}=\frac{\mathrm{CB}}{\mathrm{CR}}$

Subtracting 1 from both sides, we get
$\frac{\mathrm{OB}}{\mathrm{OS}}-1=\frac{\mathrm{CB}}{\mathrm{CR}}-1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{OB}-\mathrm{OS}}{\mathrm{OS}}=\frac{\mathrm{CB}-\mathrm{CR}}{\mathrm{CR}}$
$⇒\frac{\mathrm{BS}}{\mathrm{OS}}=\frac{\mathrm{BR}}{\mathrm{CR}}$
By converse of basic proportionality theorem in $△$OBC, 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.

Let AC be the ladder of length 5 m, BC = 4 m be the height of the wall, against which ladder is placed. In right angled $△$ABC, by Pythagoras theorem,
AC2 = AB2 + BC2
⇒ (5)= (AB)2 + (4)2
⇒ AB = 3 m
⇒ BD = AB $-$ AD
= 3 $-$ 1.6
= 1.4 m

In right angled $△$EBD, by Pythagoras theorem
ED2 = EB2 + BD2
⇒ (5)= (EB)2 + (1.4)2                             (∵ BD = 1.4)
⇒ 25 = (EB)2 + 1.96
⇒ (EB)2 = 25 – 1.96
= 23.04

Now, EC = EB – BC
= 4.8 – 4
= 0.8 m
Hence, the top of the ladder would slide upwards on the wall by distance 0.8 m.

#### Question 6:

For going to a city B from city A, there is a route via city C such that AC 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.

Given that, AC CB, AC = 2x km, CB = 2 (x + 7) km and AB = 26 km.
On drawing the figure, we get the right angled $△$ACB right angled at C. Now, in $△$ACB, by Pythagoras theorem,
AB2 = AC2 + BC2
⇒ (26)2 = (2x)2 + [2(x + 7)]2
⇒ 676 = 4x2 + 4(x2 + 49 + 14x)
⇒ 676 = 4x2 + 4x2 + 196 + 56x
⇒ 676 = 8x2 + 56x + 196
⇒ 8x2 + 56x – 480 = 0
⇒ x2 + 7x – 60 = 0
x2 + 12x $-$ 5x $-$ 60 = 0
x(x + 12) $-$ 5(x + 12) = 0
⇒ (x + 12)(x $-$ 5) = 0
x$-$12, x = 5
As the distance can’t be negative.
x = 5                                                                   (∵ x$-$12)

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

The distance covered to each 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 = BA = 26 km

Hence, the required saved distance is 34 – 26 = 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.

Let MN = 18 m be the flag pole and its shadow be LM = 9.6 m.
The distance of the top of the pole N from the far end L of the shadow is LN. In right angled $△$LMN, by Pythagoras theorem,
LN2 = LM2 + MN2
⇒ LN2 = (9.6)+ (18)2
⇒ LN2 = 92.16 + 324
⇒ LN2 = 416.16

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 3 m, find how far she is away from the base of the pole.

Let P be the position of the street bulb fixed on a pole PQ = 6 m.
CD = 1.5 m be the height of a woman
ED = 3 m be the shadow of woman
Let distance between pole and woman be x meters. Here, woman and pole both are standing vertically.
So, CD  PQ

In $△$CDE and $△$PQE,
∠E = ∠E                                                      (common angle)
∠PQE = ∠CDE = 90°
$△$CDE ∼ $△$PQE                                         (by AA similarity criterion)
Then,
(corresponding sides of similar triangles are in same ratio)

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

#### Question 9:

In Fig. 6.18, ABC is a triangle right angled at B and BD AC. If AD = 4 cm, and CD = 5 cm, find BD and AB. Given that, $△$ABC in which ∠B = 90° , BD AC, AD = 4 cm and CD = 5 cm.

In $△$ADB and $△$CDB,
∠ADB = ∠CDB = 90°
∠BAD = ∠DBC = 90° $-$ ∠C
$△$DBA ∼ $△$DCB                           (by AA similarity criteria)

Then,
$\frac{\mathrm{DB}}{\mathrm{DA}}=\frac{\mathrm{DC}}{\mathrm{DB}}$                                       (corresponding sides of similar triangle are in proportion)

In right angled $△$BDC, by Pythagoras theorem,
BC2 = BD2 + CD2
$⇒$BC2 = (2√5)2 + (5)2
$⇒$BC2 = 20 + 25 = 45
$⇒$ BC = √45
= 3√5 cm

Again,

Hence, BD = 2√5 cm and AB = 6 cm.

#### Question 10:

In Fig. 6.19, PQR is a right triangle right angled at Q and QS ⏊ PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR. Given, $△$PQR in which ∠Q = 90°, QS  PR, PQ = 6 cm, PS = 4 cm.

In $△$SQP and $△$SRQ,
∠PSQ = ∠RSQ = 90°
∠SPQ = ∠SQR = 90° $-$ ∠R
$△$SQP ∼ $△$SRQ

Then,
$\frac{\mathrm{SQ}}{\mathrm{PS}}=\frac{\mathrm{SR}}{\mathrm{SQ}}$
$⇒{\mathrm{SQ}}^{2}=\mathrm{PS}×\mathrm{SR}$                        .....(1)

In right angled $△$PSQ, by Pythagoras theorem
PQ2 = PS2 + QS2
$⇒$ (6)2 = (4)2 + QS2
$⇒$ 36 = 16 + QS2
$⇒$ QS2 = 36 $-$ 16 = 20
∴ QS = √20 = 2√5 cm

On putting the value of QS in (1), we get

In right angled $△$QSR,
QR2 = QS2 + SR2
$⇒$QR2 = (2√5)2 + (5)2
$⇒$ QR2 = 20 + 25
∴ QR = √45 = 3√5cm

Hence, QS = 2√5 cm, RS = 5 cm and QR = 3√5 cm.

#### Question 11:

In $△$PQR, PD ⏊ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (ab) = (c + d) (cd).

Given that, in $△$PQR, PD ⏊ QR, PQ = a, PR = b, QD = c and DR = d.

In right angled $△$PDQ, by Pythagoras theorem,
PQ2 = PD2 + QD2
$⇒$ a2 = PD2 + c2
$⇒$ PD2 = a– c2                     .....(1) In right angled $△$PDR, by Pythagoras theorem,
PR2 = PD2 + DR2
$⇒$b2 = PD2 + d2
$⇒$PD2 = b2 $-$ d2                     .....(2)

From (1) and (2),
a2 $-$ c2 = b2 $-$ d2
a2 $-$ b2 = c2 $-$ d2
$⇒$(ab) (a + b) = (cd) (c + d)
Hence, proved.

#### Question 12:

In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2.

A quadrilateral ABCD is such that ∠A + ∠D = 90°.

Construct AB and CD to meet at E.
Also, join AC and BD. In $△$AED,
∠A + ∠D = 90°                                             (given)
∴ ∠E = 180° $-$ (∠A + ∠D) = 90°                  (∵ sum of angles of a triangle is 180°)

Then, by Pythagoras theorem,
AD2 = AE2 + DE2

In $△$BEC, by Pythagoras theorem,
BC2 = BE2 + EC2

On adding both equations, we get
AD2 + BC2 = AE2 + DE2 + BE2 + CE2                      .....(1)

In $△$AEC, by Pythagoras theorem,
AC2 = AE2 + CE2
And in $△$BED,
BD2 = BE2 + DE2
On adding both equations, we get
AC2 + BD2 = AE2 + CE2 + BE2 + DE2               .....(2)

From (1) and (2),
AC2 + BD2 = AD2 + BC2
Hence, proved.

#### Question 13:

In fig. 6.20, l || m and line segments AB, CD and EF are concurrent at point P.

Prove that $\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}.$ Given l || m and line segments AB, CD and EF are concurrent at point P.

In $△$APC and $△$BPD,
∠APC = ∠BPD                                                 [vertically opposite angles]
∠PAC = ∠PBD                                                  [alternate angles]
$△$APC ∼ $△$BPD                                           [by AA similarity criterion]

Then,
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{PC}}{\mathrm{PD}}$                                                .....(1)

In $△$APE and $△$BPF,
∠APE = ∠BPF                                                [vertically opposite angles]
∠PAE = ∠PBF                                                 [alternate angles]
$△$APE ∼ $△$BPF                                          [by AA similarity criterion]
Then,
.....(2)

In $△$PEC and $△$PED,
∠EPC = ∠FPD                                                  [vertically opposite angles]
∠PCE = ∠PDF                                                  [alternate angles]
$△$PEC ∼ $△$PFD                                           [by AA similarity criterion]

Then,
.....(3)

From (1), (2) and (3),
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{PE}}{\mathrm{PF}}=\frac{\mathrm{EC}}{\mathrm{FD}}$
$⇒\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}$
Hence, proved.

#### Question 14:

In Fig. 6.21, 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, QR and RS. 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
∴ $\mathrm{PA}\parallel \mathrm{QB}\parallel \mathrm{RC}\parallel \mathrm{SD}$

By basic proportionality theorem,
PA : QB : RC = AB : BC : CD = 6 : 9 : 12

Let PQ = 6x, QR = 9x and RS = 12x.
Since, length of PS = 36 km
∴ PQ + QR + RS = 36
⇒ 6x + 9x + 12x = 36
⇒ 27x = 36
$⇒x=\frac{36}{27}=\frac{4}{3}$
Now,

#### Question 15:

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

Given: ABCD is a trapezium, diagonals AC and BD intersect at O.
PQ  AB  DC In $△$ABD and $△$POD,
PO  AB                                                                [∵ PQ  AB]
∠D = ∠D                                                               [common angle]
∠ABD = ∠POD                                                    [corresponding angles]
$△$ABD ∼ $△$POD                                             [by AA similarity criterion]
Then,
$\frac{\mathrm{OP}}{\mathrm{AB}}=\frac{\mathrm{PD}}{\mathrm{AD}}$                                                                         .....(1)

In $△$ABC and $△$OQC,
OQ  AB                                                               [∵ OQ  AB]
∠C = ∠C                                                             [common angle]
∠BAC = ∠QOC                                                  [corresponding angle]
$△$ABC ∼ $△$OQC                                            [by AA similarity criterion]

Then,
$\frac{\mathrm{OQ}}{\mathrm{AB}}=\frac{\mathrm{QC}}{\mathrm{BC}}$                                                                         .....(2)

Now, in $△$ADC, OP  DC
$\therefore \frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{OA}}{\mathrm{OC}}$          [by basic proportionality theorem]    .....(3)

In $△$ABC, OQ  AB
$\therefore \frac{\mathrm{BQ}}{\mathrm{QC}}=\frac{\mathrm{OA}}{\mathrm{OC}}$        [by basic proportionality theorem]    .....(4)

From (3) and (4),
$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{BQ}}{\mathrm{QC}}$
Adding 1 on both sides, we get,
$\frac{\mathrm{AP}}{\mathrm{PD}}+1=\frac{\mathrm{BQ}}{\mathrm{QC}}+1$

Hence, proved.

#### Question 16:

In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE. Prove that $\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}}.$ Given: In $△$ABC, E is the mid-point of CA and ∠AEF = ∠AFE. Construction: Take a point G on AB such that CG  EF.
Now, CE = AE                                     [E is the mid-point of CA]            .....(1)

In $△$ACG, CG  EF and E is mid-point of CA.
So, AF = GF                               [by mid-point theorem]                           .....(2)
Also, AE = AF                      [Sides opposite to equal angles are equal]   .....(3)

From (1), (2) and (3),
CE = GF                                                                                                      .....(4)

Now, in $△$BDF, CG  DF.
$\therefore \frac{\mathrm{BC}}{\mathrm{CD}}=\frac{\mathrm{BG}}{\mathrm{GF}}$                            [by Basic Proportionality Theorem]
$⇒\frac{\mathrm{BC}}{\mathrm{CD}}+1=\frac{\mathrm{BG}}{\mathrm{GF}}+1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BC}+\mathrm{CD}}{\mathrm{CD}}=\frac{\mathrm{BG}+\mathrm{GF}}{\mathrm{GF}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{GF}}$

$⇒\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}}$                           [from (4)]

Hence, proved.

#### 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.

Let $△$RST be a right triangle at S and RS = y and ST = x. Three semi-circles are drawn on the sides RS, ST and RT respectively.
To prove: A3 = A1 + A2

In $△$RST, by Pythagoras theorem,
RT2 = RS2 + ST2
⇒RT2 = y2 + x2
$⇒\mathrm{RT}=\sqrt{\left({y}^{2}+{x}^{2}\right)}$

Now,
Area of a semi-circle with radius r = $\frac{\mathrm{\pi }{r}^{2}}{2}$

∴ Area of semi-circle drawn on RT,
$\begin{array}{rcl}{\mathrm{A}}_{3}& =& \left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\mathrm{RT}}{2}\right)}^{2}\\ & =& \left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\sqrt{\left({x}^{2}+{y}^{2}\right)}}{2}\right)}^{2}\end{array}$
$⇒{\mathrm{A}}_{3}=\frac{\mathrm{\pi }\left({x}^{2}+{y}^{2}\right)}{8}$                                            .....(1)

Now, the area of semi-circle drawn on RS,
${A}_{1}=\left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\mathrm{RS}}{2}\right)}^{2}$
$⇒{A}_{1}=\frac{\mathrm{\pi }\left({y}^{2}\right)}{8}$                                                   .....(2)

And area of semi-circle drawn on ST,
${A}_{2}=\left(\frac{\mathrm{\pi }}{2}\right)×{\left(\frac{\mathrm{ST}}{2}\right)}^{2}$
$⇒{A}_{2}=\frac{\mathrm{\pi }\left({x}^{2}\right)}{8}$                                                   .....(3)

Adding (2) and (3),
${A}_{{}^{2}}+{A}_{1}=\frac{\mathrm{\pi }\left({x}^{2}+{y}^{2}\right)}{8}={A}_{3}$                             [from(1)]

Hence, proved.

#### 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.

Consider a right triangle $△$QPR in which $\angle \mathrm{RPQ}$ is right angle and PR = y and PQ = x. Three equilateral triangles $△$PER, $△$PFQ and $△$RQD are drawn on the three sides of $△$PQR.
Again, let area of triangles made on PR, PQ and RQ be A1, A2 and A3 respectively.
To prove:  A3 = A1 + A2

In $△$RPQ, by Pythagoras theorem,
QR2 = PR2 + PQ2
⇒ QR2 = y2 + x2

Now, area of an equilateral triangle = $\frac{\sqrt{3}}{4}×{\left(\mathrm{side}\right)}^{2}$

∴ Area of equilateral $△$PER,
${\mathrm{A}}_{1}=\frac{\sqrt{3}}{4}×{y}^{2}$                                                             .....(1)
And area of equilateral $△$PFQ,
${\mathrm{A}}_{2}=\frac{\sqrt{3}}{4}×{x}^{2}$                                                             .....(2)

Now, area of equilateral $△$RDQ,
${\mathrm{A}}_{3}=\frac{\sqrt{3}}{4}×{\mathrm{QR}}^{2}$
$⇒{\mathrm{A}}_{3}=\frac{\sqrt{3}}{4}×\left({x}^{2}+{y}^{2}\right)={\mathrm{A}}_{2}+{\mathrm{A}}_{1}$

Hence, proved.

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