#### Exercise 4.5

**1.Â In fig. 4.136 given below \(\Delta ACB \sim \Delta APQ\). If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, and AP = 2.8 cm find CA and AQ.**

Fig: 4.136

**Solution:**

Given,

\(\Delta ACB \sim \Delta APQ\)

BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, and AP = 2.8 cm

To find= CA and AQ

We know that,

\(\Delta ACB \sim \Delta APQ\)

\(\frac{BA}{AQ} = \frac{CA}{AP} = \frac{BC}{PQ}\)

Then,Â \(\frac{6.5}{AQ} = \frac{8}{4}\)

AQ = \(\frac{6.5 x 4}{8}\)

AQ = 3.25 cm

Similarly,

\(\frac{CA}{AP} = \frac{BC}{PQ}\)

\(\frac{CA}{2.8} = \frac{8}{4}\)

CA = 2.8 x 2

CA = 5.6 cm

Therefore,, CA = 5.6 cm and AQ = 3.25 cm.

**2. In fig.4.137 given, \(AB \parallel QR\), find the length of PB.**

Fig: 4.137

**Solution:**

Given,

\(AB \parallel PB\)

AB = 3 cm, QR = 9 cm and PR = 6 cm

To find= PB

We know that,

\(\frac{AB}{QR} = \frac{PB}{PR}\)

i.e., \(\frac{3}{9} = \frac{PB}{6}\)

Therefore, PB = 2 cm

**3. In fig.Â 4.138 given, \(XY \parallel BC\). Find the length of XY.**

Fig: 4.138

**Solution:**

Given,

\(XY \parallel BC\)

AX = 1 cm, XB = 3 cm, and BC = 6 cm

To find= XY

We know that,

\(\Delta AXY \sim \Delta ABC\)

\(\frac{XY}{BC} = \frac{AX}{AB}\)Â (AB = AX + XBÂ = 4)

\(\frac{XY}{6} = \frac{1}{4}\)

\(\frac{XY}{1} = \frac{6}{4}\)

Therefore, XY = 1.5 cm

**4. In a right-angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.**

**Solution: **

ConsiderÂ \(\Delta ABC\)Â to be a right angle triangle having sides a and b and hypotenuse c. Let BD be the altitude drawn on the hypotenuse AC.

To prove: ab = cx

We know that,

Since the altitude is perpendicular to the hypotenuse, both the triangles are similar

\(\frac{AB}{BD} = \frac{AC}{BC}\)

\(\frac{a}{x} = \frac{c}{b}\)

xc = ab

Therefore, ab = cx

**5. In fig. 4.139 given, \(\angle ABC\) = 90 and \(BD \perp AC\). If BD = 8 cm, and AD = 4 cm, find CD.**

Fig: 4.139

**Solution:**

Given,

\(\angle ABC\) Â = 90 and \(BD \perp AC\)

BD = 8 cm

AD = 4 cm

To find= CD.

We know that,

ABC is a right angled triangle and \(BD \perp AC\).

Then, \(\Delta DBA \sim \Delta DCB\)Â (AA similarity)

\(\frac{BD}{CD} = \frac{AD}{BD}\)

BD^{2} = AD x DC

(8)^{2} = 4 x DC

DC = 64/4 = 16 cm

Therefore, CD = 16 cm

**6. In fig.4.140 given, \(\angle ABC\) = 90 and \(BD \perp AC\). If AC = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, Find BC.**

Fig: 4.140

**Solution:**

Given:

\(BD \perp AC\). AC = 5.7 cm

BD = 3.8 cm

CD = 5.4 cm

\(\angle ABC\) = 90.

To find= BC,

We know that,

\(\Delta ABC \sim \Delta BDC\)

\(\frac{AB}{BD} = \frac{BC}{CD}\)

\(\frac{5.7}{3.8} = \frac{BC}{5.4}\)

\(\frac{BC}{1} = \frac{5.7 x 5.4}{3.8}\)

Therefore, BC = 8.1 cm

**7. In the fig.4.141 given, \(DE \parallel BC\) such that AE = (1/4)AC. If AB = 6 cm, find AD.**

Fig: 4.141

**Solution:**

Given:

\(DE \parallel BC\)

AE = (1/4)AC

AB = 6 cm.

To find= AD.

We kknow that,

\(\Delta ADE \sim \Delta ABC\)

Then,

\(\frac{AD}{AB} = \frac{AE}{AC}\)

\(\frac{AD}{6} = \frac{1}{4}\)

4 x AD = 6

AD = 6/4

Therefore, AD = 1.5 cm

**8. In the fig.4.142 given, if Â \(AB \perp BC\), \(DC \perp BC\), and \(DE \perp AC\), prove that \(\Delta CED \sim \Delta ABC\)**

Fig: 4.142

**Solution:**

Given:

\(AB \perp BC\)

\(DC \perp BC\)

\(DE \perp AC\)

To prove:Â \(\Delta CED \sim \Delta ABC\)

We know that,

FromÂ \(\Delta\)ABC and \(\Delta\)CED

\(\angle B\) Â = \(\angle E\) Â = 90Â Â (given)

\(\angle A\) Â = \(\angle ECD\) Â Â Â (alternate angles)

Therefore,Â \(\Delta CED \sim \Delta ABC\) (A-A similarity)

**9. Diagonals AC and BD of a trapezium ABCD with \(AB \parallel DC\) intersect each other at the point O. Using similarity criterion for two triangles, show that \(\frac{OA}{OC} = \frac{OB}{OD}\) Â **

**Solution:**

Given: OC is the point of intersection of AC and BD in the trapezium ABCD, with \(AB \parallel DC\).

To prove:Â \(\frac{OA}{OC} = \frac{OB}{OD}\)

We know that,

FromÂ \(\Delta\)AOB and \(\Delta\)COD

\(\angle AOB\) Â = \(\angle COD\) Â Â Â (VOA)

\(\angle OAB\) Â Â = \(\angle OCD\) Â Â (alternate angles)

Then, \(\Delta AOB \sim \Delta COD\)

Therefore, \(\frac{OA}{OC} = \frac{OB}{OD}\)Â Â (corresponding sides are proportional)

**10. If \(\Delta\)ABC and \(\Delta\)AMP are two right angled triangles, at angle B and M, repec. Such that \(\angle MAP\) = \(\angle BAC\). Prove that :**

**(i) \(\Delta ABC \sim \Delta AMP\)**

**(ii)** **\(\frac{CA}{PA} = \frac{BC}{MP}\)**

**Solution:**

**(i)**Â Given:

\(\Delta\) ABC and \(\Delta\) AMP are the two right angled triangle.

\(\angle MAP\) = \(\angle BAC\)

We know that,

\(\angle AMP\) Â = \(\angle B\) Â = 90

\(\Delta ABC \sim \Delta AMP\)Â Â Â (A-A similarity)

**(ii)**Â Since,Â \(\Delta ABC \sim \Delta AMP\)

Therefore,Â \(\frac{CA}{PA} = \frac{BC}{MP}\)Â (corresponding sides are proportional)

**11. A vertical stick 10 cm long casts a shadow 8 cm long. At the same time, a tower casts a shadow 30 m long. Determine the height of the tower.**

**Solution.:Â **To find = the height of the tower = PQ.

According to the question,

We know that,

\(\Delta ABC \sim \Delta PQR\)Â Â (A-A similarity)

\(\frac{AB}{BC} = \frac{PQ}{QR}\)

\(\frac{10}{8} = \frac{PQ}{3000}\)

PQ = \(\frac{3000 x 10}{8}\)

PQ = \(\frac{30000}{8}\)

PQ = \(\frac{3750}{100}\)

Therefore, PQ = 37.5 m

**12. In fig.4.143 given, \(\angle A\) = \(\angle CED\), prove that \(\Delta CAB \sim \Delta CED\). Also find the value of x.**

Fig: 4.143

**Solution:**

According to the question,

When we compareÂ \(\DeltaÂ with CAB \Delta CED\),

We get,

\(\frac{CA}{CE} Â = \frac{AB}{ED}\)Â (corresponding sides are in same proportions for similar triangles)

\(\frac{15}{10} = \frac{9}{x}\)

\(\frac{x}{1} = \frac{9 x 10}{15}\)

Therefore, x = 6 cm

**13. The perimeters of two similar triangles are 25 cm and 15 cm, respect. If one side of the first triangle is 9 cm, what is the corresponding side of the other triangle?**

**Solution:**

Given:

perimeter of two similar triangles are 25 cm, 15 cm

one side of the triangle =9 cm

To find= other side.

Let the corresponding side of other triangle = x cm

We know that the ratio of perimeter = ratio of corresponding side.

Then,

\(\frac{25}{15} = \frac{9}{x}\)

25 x (x) = 9 x 15

x = 135/25

Therefore, x =Â 5.4 cm

**14. In \(\Delta ABC and \Delta DEF\), it is being given that AB = 5 cm, BC = 4 cm, CA = 4.2 cm, DE = 10 cm, EF = 8 cm, and FD = 8.4 cm. If** **\(AL \perp BC\), \(DM \perp EF\), find AL : Dm.**

**Solution:**

Given:

AB = 5 cm

BC = 4 cm

CA = 4.2 cm

DE = 10 cm

EF = 8 cm

FD = 8.4 cm

To find= AL : DM

According to the question,

both the triangles are similar,

i.e.,Â \(\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} Â = \frac{1}{2} \)

We know that in similar triangle the ratio of corresponding altitude is same as the ratio of the corresponding sides.

Therefore,Â Â AL : DM = 1 : 2

**15. D and E are the points on the sides AB and AC respectively, of a \(\Delta ABC \) such that AD = 8 cm, DB = 12 cm, AE = 6 cm, and CE = 9 cm. Prove that BC = 5/2 DE. **

**Solution: **

Given:

AD = 8 cm

AE = 6 cm

CE = 9 cm

To prove: BC = 5/2 DE.

According to the question and figure,

\(\frac{AD}{AB} = \frac{AE}{AC} = \frac{2}{5}\)

And also, \(\Delta ADE \sim \Delta ABC\)Â Â (SAS similarity)

\(\frac{BC}{DE} = \frac{AB}{AD}\)

\(\frac{BC}{DE} = \frac{1}{(\frac{AD}{AB})}\)Â Â Â (\(\frac{AD}{AB} = \frac{2}{5}\))

\(\frac{BC}{DE} = \frac{5}{2}\)

Therefore, BC = 5/2 DE

**16. D is the midpoint of side BC of a \(\Delta ABC \). AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE: EX = 3 : 1**

**Solution.:**

Given: ABC is a triangle in which D is the midpoint of BC, E is the midpoint of AD and the line segment BE produced meets AC at X.

To prove: BE: EX = 3 : 1

According to the question,

From \(\Delta\) BCXÂ andÂ \(\Delta\)DCY

We know that,

\(\angle\)CBXÂ = \(\Delta\)CBYÂ Â (corresponding angles)

\(\angle\)CXBÂ = \(\Delta\)CYDÂ Â (corresponding angles)

\(\Delta BCX \sim \Delta DCY\)Â Â (angle-angle similarity)

Since the corresponding sides of similar sides of similar triangles are proportional

We have, \(\frac{BC}{DC} = \frac{BX}{DY} = \frac{CX}{CY}\)

\(\frac{BX}{DY} = \frac{BC}{DC}\)

\(\frac{BX}{DY} = \frac{2 DC}{DC}\)Â (As D is the midpoint of BC)

\(\frac{BX}{DY} = \frac{2}{1}\)….(i)

Similarly,

From\(\Delta\)AEX and \(\Delta\)ADY,

\(\angle\)AEXÂ = \(\Delta\)ADYÂ Â (corresponding angles)

\(\angle\)AXE = \(\Delta\)AYDÂ Â (corresponding angles)

\(\Delta\)AEX â€“ \(\Delta\)ADYÂ Â (angle-angle similarity)

Since the corresponding sides of similar sides of similar triangles are proportional

We have, \(\frac{AE}{AD} = \frac{EX}{DY} = \frac{AX}{AY}\)

\(\frac{EX}{DY} = \frac{AE}{AD}\)

\(\frac{EX}{DY} = \frac{AE}{2 AE}\)Â (As D is the midpoint of BC)

\(\frac{EX}{DY} = \frac{1}{2}\)…(ii)

On dividing eqation. (i) by eqn. (ii)

\(\frac{BX}{EX} = \frac{4}{1}\)

BX = 4EX

BE + EX = 4EX

BE = 3EX

Therefore, BE : EX = 3:1

**17.Â ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.**

**Solution:**

Given: ABCD is a parallelogram and APQ is a straight line meeting both BC at P and DC produced at Q.

To prove: the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.

i.e., to prove: BP x DQ = AB x BC

According to the question,

Form \(\Delta\)ABP and \(\Delta\)QCP,

\(\angle\)ABPÂ = \(\Delta\)QCPÂ Â (alternate angles as ABÂ DC)

\(\angle\)BPA = \(\Delta\)QPCÂ Â (VOA)

\(\Delta ABP \sim \Delta QCP\) \(\Delta\)Â Â (AA similarity)

Since the corresponding sides of similar triangles are proportional

We have, \(\frac{AE}{AD} = \frac{EX}{DY} = \frac{AX}{AY}\)

Therefore, \(\frac{EX}{DY} = \frac{AE}{AD}\)

**18. In \(\Delta ABC \), AL and CM are the perpendiculars from the vertices A and C to BC and AB respect. If Al and CM intersec at O, prove that:**

**(i) Â \(\Delta OMA \sim \Delta OLC\)**

**(ii) \(\frac{OA}{OC} = \frac{OM}{OL}\) **

**Solution:Â **

**(i)** According to the question,

\(\Delta\)OMA and \(\Delta\)OLC,

\(\angle\)AOM = \(\angle\)COLÂ (VOA)

\(\angle\)OMA = \(\angle\)OLCÂ (90 each)

Therefore, \(\Delta OMA \sim \Delta OLC\)Â Â (A-A similarity)

**(ii)** We know that, \(\Delta OMA \sim \Delta OLC\) by A-A similarity, then

\(\frac{OM}{OL} = \frac{OA}{OC}Â = \frac{MA}{LC}\)Â (corresponding sides of similar triangles are proportional)

Therefore, \(\frac{OA}{OC} = \frac{OM}{OL}\)

**19. ABCD is a quadrilateral in which AD = BC. If P,Q,R, S be the midpoints of AB, AC, CD and BD respect. Show that PQRS is a rhombus. **

**Solution.:Â **

Given: ABCD is a quadrilateral in which AD = BC

PÂ is the mid point of AB

Q is the mid point of AC

R is the mid point of CD

S is the mid point of BD.

To prove: PQRS is a rhombus

Proof,

According to the question,

FromÂ \(\Delta\)ABC, P and Q are the mid points of the sides B and AC respectively

Using midpoint theorem, we get,

\(PQ \parallel BC\),Â PQ = 1/2 BC.

We know that,

In \(\Delta\)ADC, Q and R are the mid points of the sides AC and DC respectively

Using the mid point theorem, we get,

\(QR \parallel AD\) and QR = 1/2 AD = 1/2 BCÂ Â Â (AD = BC)

We know that,

In \(\Delta\)BCD,

Similarly,

Using the mid point theorem, we get,

\(RS \parallel BC\) and RS = 1/2 AD = 1/2 BCÂ Â Â (AD = BC)

From above derivations and equations, we can conclude that.

PQ = QR = RS

Therefore, PQRS is a rhombus.

**20. In an isosceles \(\Delta ABC\), the base AB is produced both ways to P and Q such that AP x BQ = AC ^{2} . Prove that \(\Delta APC \sim \Delta BCQ\).**

**Solution:**

Given:Â \(\Delta\)ABC is isosceles and AP x BQ = AC^{2}

To prove:Â \(\Delta APC \sim \Delta BCQ\).

Proof:

Given:Â \(\Delta\)ABC is an isosceles triangle AC = BC.

We know that,

AP x BQ = AC^{2}

AP x BQ = AC x AC

\(\frac{AP}{AC} = \frac{AC}{BQ}\)

\(\frac{AP}{AC} = \frac{BC}{BQ}\)

Also, \(\angle\)CAB = \(\angle\)CBAÂ Â (equal sides have angles opposite to them)

180 â€“ CAP = 180 â€“ CBQ

\(\angle\)CAP = \(\angle\)CBQ

Therefore, \(\Delta APC \sim \Delta BCQ\)**Â **(SAS similarity)

**21. A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/sec. If the lamp is 3.6m above the ground, find the length of her shadow after 4 seconds.**

**Solution:**

Given:

Girlâ€™s height = 90 cm

Speed = 1.2m/sec

Height of lamp = 3.6 m

To find: the length of her shadow after 4 sec.

From the figure,

Let us assume that AB be the lamp post and CD be the girl,

And let DE be the length of her shadow

i.e., DE = x

andÂ BD = 1.2 x 4

BD = 4.8 m

According to the question,

\(\Delta\)ABE and \(\Delta\)CDE

We have,

\(\angle\)B = \(\angle\)D

\(\angle\)E = \(\angle\)E

So, Using A-A similarity criterion,

\(\Delta ABE \sim \Delta CDE\)

\(\frac{BE}{DE} = \frac{AB}{CD}\)

\(\frac{4.8 + x}{x} = \frac{3.6}{0.9}\)Â = 4

3x = 4.8

Hence, x = 1.6

Therefore, the length of her shadow after 4 second = 1.6 m

**22. A vertical stick of length 6m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28m long. Find the height of the tower. **

**Solution:**

Given: length of vertical stick = 6m

To find: height of the tower

Let AB be the height of the tower and BC be its shadow.

\(\Delta ABC \sim \Delta PCR\)Â (B = Q and A = P)

\(\frac{AB}{BC} = \frac{PQ}{QR}\)

\(\frac{AB}{28} = \frac{6}{4}\)

AB = (28 x 6)/4

AB = 42m

Therefore, the height of tower is 42m.

**23. In the fig.4.144 given, \(\Delta\)ABC is a right angled triangle at C and \(DE \perp AB\). Prove thatÂ \(\Delta ABC \sim \Delta ADE\).**

Fig: 4.144

**Solution:**

Given:

\(\Delta\)ACB is right angled triangle

C = 90

To prove:\(\Delta ABC \sim \Delta ADE\)

To find: length of AE and DE.

\(\Delta ABC \sim \Delta ADE\)

\(\angle\)A = \(\angle\)AÂ (common angle)

\(\angle\)C = \(\angle\)EÂ Â (90)

So, by the criterion of A-A similarity, we get,

According to the question,

\(\Delta ABC \sim \Delta ADE\)

\(\frac{AB}{AD} = \frac{BC}{DE} = \frac{AC}{AE}\)

\(\frac{13}{3} = \frac{12}{DE} = \frac{5}{AE}\)

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

= 5^{2} + 12^{2}

=Â 13^{2}

Therefore, DE = 36/13 cm

andÂ Â AE = 15/13 cm

**25. In fig. 4.146, we have \(AB \parallel CD \parallel EF \). If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm, and DE = y cm. Calculate the values of x and y.**

Fig: 4.146

**Solution:**

Given:

AB = 6 cm

CD = x cm

EF = 10 cm.

To find: the values of x and y

According to the question,

\(\Delta\)ADBÂ andÂ \(\Delta\)DEF,

\(\angle\)ADB = \(\angle\)EDF (VOA)

\(\angle\)ABD = \(\angle\)DEFÂ (alt. Interior angles)

\(\frac{EF}{AB} = \frac{OE}{OB}\)

\(\frac{10}{6} = \frac{y}{4}\)

Y = 40/6

Y = 6.67 cm

Similarly,

From \(\Delta\)ABE , we get

\(\frac{OC}{AB} = \frac{OE}{OB}\)

\(\frac{4}{6.7} = \frac{x}{6}\)

6.7 x (x) = 6 x 4

x = 24/6.7

x = 3.75 cm

Therefore, x = 3.75 cm and y = 6.67 cm