Chapter 15 - Similarity (With Applications to Maps and Models) Exercise Ex. 15(E)

Question 1(i)

In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.

 

 

 

 

Solution 1(i)

 

Question 1(ii)

In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.

 

 

 

 

Solution 1(ii)

 

Question 2(iii)

In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: PE

 

Solution 2(iii)

Question 2(i)

In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: CE

 

 

 

 

Solution 2(i)

 

Question 2(ii)

In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5,cm, AP = 6 cm and BE = 15 cm, Calculate: AF

 

Solution 2(ii)

 

Question 3

In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

 

 

 

Solution 3

 

Question 4

Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: .

Solution 4

 

 

Question 5

 

Solution 5

 

 

Question 6

 

Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that:.

Solution 6

 

 

Question 7

In the following figure, ∠AXY = ∠AYX. If show that triangle ABC is isosceles.

 

 

 

Solution 7

 

Question 8

In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown. 

Prove that: 

Solution 8

Join AR.

In ACR, BX || CR. By Basic Proportionality theorem,

In APR, XQ || AP. By Basic Proportionality theorem,

From (1) and (2), we get,

Question 9

 

 

Solution 9

 

Question 10(i)

In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : EF

 

 

 

Solution 10(i)

 

Question 10(ii)

In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate : AC

 

 

Solution 10(ii)

 

Question 11(i)

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

Prove that ΔACD is similar to ΔBCA.

 

 

Solution 11(i)

 

Question 11(ii)

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

Find BC and CD.

 

Solution 11(ii)

 

Question 11(iii)

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.

Find area of ΔACD : area of ΔABC.

 

Solution 11(iii)

 

Question 12

In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle QRP is similar to triangle ABC.

Solution 12

In ABC, PR || BC. By Basic proportionality theorem,

Also, in PAR and ABC,

Similarly, 

Question 13

In the following figure, AD and CE are medians of ABC. DF is drawn parallel to CE. Prove that:

(i) EF = FB,

(ii) AG: GD = 2: 1

Solution 13

(i)

(ii) In AFD, EG || FD. Using Basic Proportionality theorem,

 … (1)

Now, AE = EB (as E is the mid-point of AB)

AE = 2EF (Since, EF = FB, by (i))

From (1),

Hence, AG: GD = 2: 1.

Question 14

Two similar triangles are equal in area. Prove that the triangles are congruent.

Solution 14

Let us assume two similar triangles as ABC  PQR

Question 15

The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their:

(i) medians. (ii) perimeters. (iii) areas.

Solution 15

The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Required ratio = 3: 5

(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Required ratio = 3: 5

(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

Required ratio = (3)2 : (5)2 = 9: 25

Question 16

The ratio between the areas of two similar triangles is 16: 25. Find the ratio between their:

(i) perimeters. (ii) altitudes. (iii) medians.

Solution 16

The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.

So, the ratio between the sides of the two triangles = 4: 5

(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides.

Required ratio = 4: 5

Question 17

The following figure shows a triangle PQR in which XY is parallel to QR. If PX: XQ = 1: 3 and QR = 9 cm, find the length of XY.

Further, if the area of PXY = x cm²; find, in terms of x, the area of:

(i) triangle PQR. (ii) trapezium XQRY.

Solution 17

In PXY and PQR, XY is parallel to QR, so corresponding angles are equal.

Hence,  (By AA similarity criterion)

 (i) We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

 (ii) Ar (trapezium XQRY) = Ar (PQR) - Ar (PXY)

= (16x - x) cm²

= 15x cm²

Question 18

On a map drawn to a scale of 1: 20000, a rectangular plot of land ABCD is measured as AB = 24 cm and BC = 32 cm. Calculate:

(i) the diagonal distance of the plot in kilometre.

(ii) the area of the plot in sq. km.

Solution 18

Scale :- 1 : 20000

1 cm represents 20000 cm=  = 0.2 km

(i)

= 

= 576 + 1024 = 1600

AC = 40 cm

 

Actual length of diagonal = 40  0.2 km = 8 km

(ii)

1 cm represents 0.2 km

1 cm2 represents 0.2  0.2 

The area of the rectangle ABCD = AB  BC

= 24  32 = 768

Actual area of the plot = 0.2  0.2  768  = 30.72 km²

Question 19

The dimensions of the model of a multistoreyed building are 1 m by 60 cm by 1.20 m. If the scale factor is 1: 50, find the actual dimensions of the building. Also, find:

(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq. cm.

(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90 m³.

Solution 19

The dimensions of the building are calculated as below.

Length = 1  50 m = 50 m

Breadth = 0.60  50 m = 30 m

Height = 1.20  50 m = 60 m

Thus, the actual dimensions of the building are 50 m  30 m  60 m.

(i)

Floor area of the room of the building 

(ii)

Volume of the model of the building

Question 20

In a triangle PQR, L and M are two points on the base QR, such that LPQ = QRP and RPM = RQP. Prove that:

(i) 

(ii) 

(iii) 

Solution 20

(i)

(ii)

(iii)

Question 21

A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to DEF such that the longest side of DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of DEF.

Solution 21

Triangle ABC is enlarged to DEF. So, the two triangles will be similar.

Longest side in ABC = BC = 6 cm

Corresponding longest side in DEF = EF = 9 cm

Scale factor = = 1.5

Question 22

Two isosceles triangles have equal vertical angles. Show that the triangles are similar.

If the ratio between the areas of these two triangles is 16: 25, find the ratio between their corresponding altitudes.

Solution 22

Let ABC and PQR be two isosceles triangles.

Then, 

Also, A = P (Given)

Let AD and PS be the altitude in the respective triangles.

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

Question 23

In ABC, AP: PB = 2: 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:

(i) area APO : area ABC.

(ii) area APO : area CQO.

Solution 23

In triangle ABC, PO || BC. Using Basic proportionality theorem,

(i)

(ii)

Question 24

The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively. Show that:

Solution 24

Question 25

In the given figure, ABC is a triangle with EDB = ACB. Prove that ABC  EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of BED = 9 cm2. Calculate the :

(i) length of AB.

(ii) area of ABC.

Solution 25

In ABC and EBD,

ACB = EDB (given)

ABC = EBD (common)

 (by AA- similarity)

(i) We have, 

(ii) 

Question 26

In the given figure, ABC is a right angled triangle with m∠BAC = 90°

1. Prove that ∆ADB ~ ∆CDA
2. If BD = 18 cm and CD = 8 cm, find AD.
3. Find the ratio of the area of ∆ADB to the area of ∆CDA.

Solution 26

Question 27(i)

In the given figure, AB and DE are perpendiculars to BC.

Prove that : ΔABC ~  ΔDEC 

 

 

Solution 27(i)

 

Question 27(ii)

In the given figure, AB and DE are perpendiculars to BC.

If AB = 6 cm, DE = 4 cm and AC = 15 cm. Calculate CD.

 

 

Solution 27(ii)

 

Question 27(iii)

In the given figure, AB and DE are perpendiculars to BC.

Find the ratio of the area of a ΔABC : area of ΔDEC.

 

 

Solution 27(iii)

 

 

 

 

 

 

Question 28(i)

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :

 

 

 

 

ΔADE ~ ΔACB.

Solution 28(i)

 

Question 28(ii)

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :

 

 

 

 

If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.

Solution 28(ii)

 

Question 28(iii)

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :

 

 

 

 

Find, area of ΔADE : area of quadrilateral BCED.

Solution 28(iii)

Question 29

Given: AB || DE and BC || EF. Prove that:

(i) 

(ii) 

 

Solution 29

(i) In AGB, DE || AB , by Basic proportionality theorem,

 .... (1)

 

In GBC, EF || BC, by Basic proportionality theorem,

 .... (2)

 

From (1) and (2), we get,

 

(ii)

From (i), we have:

Question 30

PQR is a triangle. S is a point on the side QR of ΔPQR such that ∠PSR = ∠QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.

i. Prove ΔPQR ∼ ΔSPR.

ii. Find the lengths of QR and PS.

iii. 

 

Solution 30

i.

In ∆PQR and ∆SPR,

∠PSR = ∠QPR … given

∠PRQ = ∠PRS … common angle

⇒ ∆PQR ∼ ∆SPR  (AA Test)

ii. Find the lengths of QR and PS.

Since ∆PQR ∼ ∆SPR … from (i)

iii.