# ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity

ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity are given here. The solutions are comprehensive and prepared by our expert faculty team to help students clear all their doubts and help in their exam preparation to obtain good marks in Maths. The ML Aggarwal Solutions for Class 10 Maths Chapter 13 given here are available for free. Students can download these PDFs and start practising offline.

Chapter 13 – Similarity. Similarity is an idea in geometry. It means that two polygons, line segments, or other figures have the same shape. Similar objects do not need to have the same size. Two shapes are similar if their angles have the same measure and their sides are proportional. Some of the important topics discussed in this ML Aggarwal Solutions for Class 10 Maths Chapter 13 are,

• Similarity of Triangles
• Axioms of Similarity of Triangles
• Basic Theorem of Proportionality
• Relation Between Areas of Similar Triangles

## ML Aggarwal Solutions for Class 10 Maths Chapter 13 :-

### Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 13 Similarity

Exercise 13.1

1. State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):

Solution:-

(i) From the Î”ABC and Î”PQR

AB/PQ = 3.2/4

= 32/40

Divide both numerator and denominator by 8 we get,

= 4/5

AC/PR = 3.6/4.5

= 36/45

Divide both numerator and denominator by 9 we get,

= 4/5

BC/QR = 3/5.4

= 30/54

Divide both numerator and denominator by 6 we get,

= 5/9

By comparing all the results, the side are not equal.

Therefore, the triangles are not equal.

(ii) From the Î”DEF and Î”LMN

âˆ E = âˆ N = 40o

Then, DE/LN = 4/2

Divide both numerator and denominator by 2 we get,

= 2

EF/MN = 4.8/2.4

= 48/24

Divide both numerator and denominator by 24 we get,

= 2

Therefore, Î”DEF ~ Î”LMN

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:-

From the question is given that, âˆ†DEF ~ âˆ†RPQ

âˆ D = âˆ R and âˆ F = âˆ Q not âˆ P

No, âˆ F â‰  âˆ P

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

Solution:-

From the figure, two line segments are intersecting each other at P.

In Î”BCP and Î”DPE

5/10 = 6/12

Dividing LHS and RHS by 2 we get,

Â½ = Â½

Therefore, Î”BCD ~ Î”DEP

5. 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:-

As per the dimensions give in the questions,

From the question it is given that,

Î”DEF ~ Î”LMN

So, AB/ED = AC/EF = BC/DF

Consider AB/ED = AC/EF

5/12 = 7/EF

By cross multiplication,

EF = (7 Ã— 12)/5

EF = 16.8 cm

Now, consider AB/ED = BC/DF

5/12 = BC/15

BC = (5 Ã— 15)/12

BC = 75/12

BC = 6.25

6.

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

Solution:-

As per the dimensions give in the questions,

Now, we have to find out the perimeter of Î”ABC

Let Î”ABC ~ Î”DEF

So, AB/DE = AC/DF = BC/EF

Consider, AB/DE = AC/DE

4/6 = AC/12

By cross multiplication we get,

AC = (4 Ã— 12)/6

AC = 48/6

AC = 8 cm

Then, consider AB/DE = BC/EF

4/6 = BC/9

BC = (4 Ã— 9)/6

BC = 36/6

BC = 6 cm

Therefore, the perimeter of Î”ABC = AB + BC + AC

= 4 + 6 + 8

= 18 cm

(b) If âˆ†ABC ~ âˆ†PQR, Perimeter of âˆ†ABC = 32 cm, perimeter of âˆ†PQR = 48 cm and PR = 6 cm, then find the length of AC.

Solution:-

From the question it is given that,

âˆ†ABC ~ âˆ†PQR

Perimeter of âˆ†ABC = 32 cm

Perimeter of âˆ†PQR = 48 cm

So, AB/PQ = AC/PR = BC/QR

Then, perimeter of âˆ†ABC/perimeter of âˆ†PQR = AC/PR

32/48 = AC/6

AC = (32 Ã— 6)/48

AC = 4

Therefore, the length of AC = 4 cm.

7. Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.

Solution:-

Let us assume that, âˆ†ABC ~ âˆ†DEF

âˆ†ABC is BC = 6cm

âˆ†ABC ~ âˆ†DEF

So, AB/DE = BC/EF = AC/DF

Consider AB/DE = BC/EF

AB/8 = 6/4

AB = (6 Ã— 8)/4

AB = 48/4

AB = 12

Now, consider BC/EF = AC/DF

6/4 = AC/7

AC = (6 Ã— 7)/4

AC = 42/4

AC = 21/2

AC = 10.5 cm

8.

(a) In the figure given below, AB || DE, AC = 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.

Solution:-

From the question it is given that,

AB||DE

AC = 3 cm

CE = 7.5 cm

BD = 14 cm

From the figure,

âˆ ACB = âˆ DCE [because vertically opposite angles]

âˆ BAC = âˆ CED [alternate angles]

Then, âˆ†ABC ~ âˆ†CDE

So, AC/CE = BC/CD

3/7.5 = BC/CD

By cross multiplication we get,

7.5BC = 3CD

Let us assume BC = x and CD = 14 â€“ x

7.5 Ã— x = 3 Ã— (14 – x)

7.5x = 42 â€“ 3x

7.5x + 3x = 42

10.5x = 42

x = 42/10.5

x = 4

Therefore, BC = x = 4 cm

CD = 14 â€“ x

= 14 â€“ 4

= 10 cm

(b) In the figure (2) given below, CA || BD, the lines AB and CD meet at G.

(i) Prove that âˆ†ACO ~ âˆ†BDO.

(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.

Solution:-

(i) We have to prove that, âˆ†ACO ~ âˆ†BDO.

So, from the figure

Consider âˆ†ACO and âˆ†BDO

Then,

âˆ AOC = âˆ BOD [from vertically opposite angles]

âˆ A = âˆ B

Therefore, âˆ†ACO = âˆ†BDO

Given, BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm, AC = 3.6 cm,

âˆ†ACO ~ âˆ†BOD

So, AO/OB = CO/OD = AC/BD

Consider AC/BD = AO/OB

3.6/2.4 = AO/3.2

AO = (3.6 Ã— 3.2)/2.4

AO = 4.8 cm

Now, consider AC/BD = CO/OD

3.6/2.4 = CO/4

CO = (3.6 Ã— 4)/2.4

CO = 6 cm

9. (a) In the figure
(i) given below, âˆ P = âˆ RTS.
Prove that âˆ†RPQ ~ âˆ†RTS.

Solution:-

From the given figure, âˆ P = âˆ RTS

So we have to prove that âˆ†RPQ ~ âˆ†RTS

In âˆ†RPQ and âˆ†RTS

âˆ R = âˆ R (common angle for both triangle)

âˆ P = âˆ RTS (from the question)

âˆ†RPQ ~ âˆ†RTS

(b) In the figure (ii) given below, âˆ ADC = âˆ BAC. Prove that CAÂ² = DC x BC

Solution:-

From the figure, âˆ ADC = âˆ BAC

So, we have to prove that, CAÂ² = DC x BC

âˆ C = âˆ C (common angle for both triangle)

âˆ BAC = âˆ ADC (from the question)

Therefore, CA/DC = BC/CA

We know that, corresponding sides are proportional,

Therefore, CA2 = DC Ã— BC

10. (a) In the figure (1) given below, AP = 2PB and CP = 2PD.
(i) Prove that âˆ†ACP is similar to âˆ†BDP and AC || BD.
(ii) If AC = 4.5 cm, calculate the length of BD.

Solution:-

From the question it is give that,

AP = 2PB, CP = 2PD

(i) We have to prove that, âˆ†ACP is similar to âˆ†BDP and AC || BD

AP = 2PB

AP/PB = 2/1

Then, CP = 2PD

CP/PD = 2/1

âˆ APC = âˆ BPD [from vertically opposite angles]

So, âˆ†ACP ~ âˆ†BDP

Therefore, âˆ CAP = âˆ PBD [alternate angles]

Hence, AC || BD

(ii) AP/PB = AC/BD = 2/1

AC = 2BD

2BD = 4.5 cm

BD = 4.5/2

BD = 2.25 cm

(b) In the figure (2) given below,

(i) Prove that âˆ†s ABC and AED are similar.

(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.

Solution:-

From the given figure,

(i) âˆ A = âˆ A (common angle for both triangles)

âˆ ACB = âˆ ADE [given]

Therefore, âˆ†ABC ~ âˆ†AED

(ii) from (i) proved that, âˆ†ABC ~ âˆ†AED

So, BC/DE = AB/AE = AC/AD

= 6 â€“ 1 = 5

6/3 = AC/5

AC = (6 Ã— 5)/3

AC = 30/3

AC = 10 cm

(c) In the figure (3) given below, âˆ PQR = âˆ PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.

Solution:-

From the figure,

âˆ P = âˆ P (common angle for both triangles)

âˆ PQR = âˆ PRS [from the question]

So, âˆ†PQR ~ âˆ†PRS

Then, PQ/PR = PR/PS = QR/SR

Consider PQ/PR = PR/PS

PQ/8 = 8/4

PQ = (8 Ã— 8)/4

PQ = 64/4

PQ = 16 cm

11. In the given figure, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM âŠ¥ AB and PN âŠ¥ AC. Prove that BM x NP = CN x MP.

Solution:-

From the question it is given that, ABC is a triangle in which AB = AC.

P is a point on the side BC such that PM âŠ¥ AB and PN âŠ¥ AC.

We have to prove that, BM x NP = CN x MP

Consider the âˆ†ABC

AB = AC â€¦ [from the question]

âˆ B = âˆ C â€¦ [angles opposite to equal sides]

Then, consider âˆ†BMP and âˆ†CNP

âˆ M = âˆ N

Therefore, âˆ†BMP ~ âˆ†CNP

So, BM/CN = MP/NP

By cross multiplication we get,

BM x NP = CN x MP

Hence it is proved.

12. Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

Solution:-

Consider the two triangles, âˆ†MNO and âˆ†XYZ

From the question it is given that, two triangles are similar triangles

So, âˆ†MNO ~ âˆ†XYZ

If two triangles are similar, the corresponding angles are equal and their corresponding sides are proportional.

MN/XY = NO/YZ = MO/XZ

Perimeter of âˆ†MNO = MN + NO + MO

Perimeter of âˆ†XYZ = XY + YZ + XZ

Therefore, (MN/XY = NO/YZ = MO/XZ) = (MN/XY + NO/YZ + MO/XZ)

= Perimeter of âˆ†MNO/perimeter of âˆ†XYZ

13. In the adjoining figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/OD

Using the above result, find the values of x if OA = 3x â€“ 19, OB = x â€“ 4, OC = x â€“ 3 and OD = 4.

Solution:-

From the given figure, ABCD is a trapezium in which AB || DC,

The diagonals AC and BD intersect at O.

So we have to prove that, AO/OC = BO/OD

Consider the âˆ†AOB and âˆ†COD,

âˆ AOB = âˆ COD â€¦ [vertically opposite angles]

âˆ OAB = âˆ OCD

Therefore, âˆ†AOB ~ âˆ†COD

So, OA/OC = OB/OD

Now by using above result we have to find the value of x if OA = 3x â€“ 19, OB = x â€“ 4, OC = x â€“ 3 and OD = 4.

OA/OC = OB/OD

(3x â€“ 19)/(x â€“ 3) = (x – 4)/4

By cross multiplication we get,

(x – 3) (x – 4) = 4(3x – 19)

X2 â€“ 4x â€“ 3x + 12 = 12x â€“ 76

X2 â€“ 7x + 12 â€“ 12x + 76 = 0

X2 â€“ 19x + 88 = 0

X2 â€“ 8x â€“ 11x + 88 = 0

X(x – 8) â€“ 11(x – 8) = 0

(x – 8) (x – 11) = 0

Take x â€“ 8 = 0

X = 8

Then, x â€“ 11= 0

X = 11

Therefore, the value of x is 8 and 11.

14. In âˆ†ABC, âˆ A is acute. BD and CE are perpendicular on AC and AB respectively. Prove that AB x AE = AC x AD.

Solution:-

Consider the âˆ†ABC,

So, we have to prove that, AB Ã— AE = AC Ã— AD

Now, consider the âˆ†ADB and âˆ†AEC,

âˆ A = âˆ A [common angle for both triangles]

âˆ ADB = âˆ AEC [both angles are equal to 90o]

By cross multiplication we get,

AB Ã— AE = AC Ã— AD

15. In the given figure, DB âŠ¥ BC, DE âŠ¥ AB and AC âŠ¥ BC. Prove that BE/DE = AC/BC

Solution:-

From the figure, DB âŠ¥ BC, DE âŠ¥ AB and AC âŠ¥ BC

We have to prove that, BE/DE = AC/BC

Consider the âˆ†ABC and âˆ†DEB,

âˆ C = 90o

âˆ A + âˆ ABC = 90o [from the figure equation (i)]

Now in âˆ†DEB

âˆ DBE + âˆ ABC = 90o [from the figure equation (ii)]

From equation (i), we get

âˆ A = âˆ DBE

Then, in âˆ†ABC and âˆ†DBE

âˆ C = âˆ E [both angles are equal to 90o]

So, âˆ†ABC ~ âˆ†DBE

Therefore, AC/BE = BC/DE

By cross multiplication, we get

AC/BC = BE/DE

16.

(a) In the figure (1) given below, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that âˆ†ABE ~ âˆ†CFB.

Solution:-

From the figure, ABCD is a parallelogram,

Then, E is a point on AD and produced and BE intersects CD at F.

We have to prove that âˆ†ABE ~ âˆ†CFB

Consider âˆ†ABE and âˆ†CFB

âˆ A = âˆ C [opposite angles of a parallelogram]
âˆ ABE = âˆ BFC [alternate angles are equal]
âˆ†ABE ~ âˆ†CFB

(b) In the figure (2) given below, PQRS is a parallelogram; PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.

(i) Prove that triangle RLQ is similar to triangle PLN. Hence, find PN.

Solution:-

From the question it is give that,

Consider the âˆ†RLQ and âˆ†PLN,

âˆ RLQ = âˆ NLP [vertically opposite angles are equal]

âˆ RQL = âˆ LNP [alternate angle are equal]

Therefore, âˆ†RLQ ~ âˆ†PLN

So, QR/PN = RL/LP = 2/3

QR/PN = 2/3

10/PN = 2/3

PN = (10 Ã— 3)/2

PN = 30/2

PN = 15 cm

Therefore, PN = 15 cm

(ii) Name a triangle similar to triangle RLM. Evaluate RM.

Solution:-

From the figure,

Consider âˆ†RLM and âˆ†QLP

Then, âˆ RLM = âˆ QLP [vertically opposite angles are equal]

âˆ LRM = âˆ LPQ [alternate angles are equal]

Therefore, âˆ†RLM ~ âˆ†QLP

Then, RM/PQ = RL/LP = 2/3

So, RM/16 = 2/3

RM = (16 Ã— 2)/3

RM = 32/3

RM =

17. The altitude BN and CM of âˆ†ABC meet at H. Prove that

(i) CN Ã— HM = BM Ã— HN

(ii) HC/HB = âˆš[(CN Ã— HN)/(BM Ã— HM)]

(iii) âˆ†MHN ~ âˆ†BHC

Solution:-

Consider the âˆ†ABC,

Where, the altitude BN and CM of âˆ†ABC meet at H.

and construction: join MN

(i) We have to prove that, CN Ã— HM = BM Ã— HN

In âˆ†BHM and âˆ†CHN

âˆ BHM = âˆ CHN [because vertically opposite angles are equal]

âˆ M = âˆ N [both angles are equal to 90o]

Therefore, âˆ†BHM ~ âˆ†CHN

So, HM/HN = BM/CN = HB/HC

Then, by cross multiplication we get

CN Ã— HM = BM Ã— HN

(ii) Now, HC/HB = âˆš(HN Ã— CN)/(HM Ã— BM)

= âˆš(CN Ã— HN)/(BM Ã— HM)

Because, M and N divide AB and AC in the same ratio.

(iii) Now consider âˆ†MHN and âˆ†BHC

âˆ MHN = âˆ BHC [because vertically opposite angles are equal]

âˆ MNH = âˆ HBC [because alternate angles are equal]

Therefore, âˆ†MHN ~ âˆ†BHC

18. In the given figure, CM and RN are respectively the medians of âˆ†ABC and âˆ†PQR. If âˆ†ABC ~ âˆ†PQR, prove that:

(i) âˆ†AMC ~ âˆ†PQR

(ii) CM/RN = AB/PQ

(iii) âˆ†CMB ~ âˆ†RNQ

Solution:-

From the given figure it is given that, CM and RN are respectively the medians of âˆ†ABC and âˆ†PQR.

(i) We have to prove that, âˆ†AMC ~ âˆ†PQR

Consider the âˆ†ABC and âˆ†PQR

As âˆ†ABC ~ âˆ†PQR

âˆ A = âˆ P, âˆ B = âˆ Q and âˆ C = âˆ R

And also corresponding sides are proportional

AB/PQ = BC/QR = CA/RP

Then, consider the âˆ†AMC and âˆ†PNR,

âˆ A = âˆ P

AC/PR = AM/PN

Because, AB/PQ = Â½ AB/Â½PQ

AB/PQ = AM/PN

Therefore, âˆ†AMC ~ âˆ†PNR

(ii) From solution(i) CM/RN = AM/PN

CM/RN = 2AM/2PN

CM/RN = AB/PQ

(iii)Now consider the âˆ†CMB and âˆ†RNQ

âˆ B = âˆ Q

BC/QP = BM/QN

Therefore, âˆ†CMB ~ âˆ†RNQ

19. In the adjoining figure, medians AD and BE of âˆ†ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1

Solution:-

From the figure it is given that, medians AD and BE of âˆ†ABC meet at the point G, and DF is drawn parallel to BE.

(i) We have to prove that, EF = FC

From the figure, D is the midpoint of BC and also DF parallel to BE.

So, F is the midpoint of EC

Therefore, EF = FC

= Â½ EC

EF = Â½ AE

(ii) Now consider the âˆ†AGE and âˆ†ADF

Then, (BG or GE) ||DF

So, AG/GD = AE/EF

AG/GD = 1/Â½

AG/GD = 1 Ã— (2/1)

Therefore, AG: GD = 2: 1

20.

(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate
(i) EF
(ii) AC.

Solution:-

From the figure it is given that, AB, EF and CD are parallel lines.

(i) Consider the âˆ†EFG and âˆ†CGD

âˆ EGF = âˆ CGD [Because vertically opposite angles are equal]

âˆ FEG = âˆ GCD [alternate angles are equal]

Therefore, âˆ†EFG ~ âˆ†CGD

Then, EG/GC = EF/CD

5/10 = EF/18

EF = (5 Ã— 18)/10

Therefore, EF = 9 cm

(ii) Now, consider the âˆ†ABC and âˆ†EFC

EF ||AB

So, âˆ†ABC ~ âˆ†EFC

Then, AC/EC = AB/EF

AC/(5 + 10) = 15/9

AC/15 = 15/9

AC = (15 Ã— 15)/9

Therefore, AC = 25 cm

(b) In the figure given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm, BE = x and AE = y. Find the values of x and y.

Solution:-

From the figure, AF, BE and CD are parallel lines.

Consider the âˆ†AEF and âˆ†CED

âˆ AEF and âˆ CED [because vertically opposite angles are equal]

âˆ F = âˆ C [alternate angles are equal]

Therefore, âˆ†AEF ~ âˆ†CED

So, AF/CD = AE/ED

7.5/4.5 = y/3

By cross multiplication,

y = (7.5 Ã— 3)/4.5

y = 5 cm

So, similarly in âˆ†ACD, BE ||CD

Therefore, âˆ†ABE ~ âˆ†ACD

x/CD = y/y + 3

x/4.5 = 5/(5 + 3)

x/4.5 = 5/8

x = (4.5 Ã— 5)/8

x = 22.5/8

x = 225/80

x = 45/16

x =

21. In the given figure, âˆ A = 90Â° and AD âŠ¥ BC If BD = 2 cm and CD = 8 cm, find AD.

Solution:-

From the figure, consider âˆ†ABC,

So, âˆ A = 90o

âˆ BAC = 90o

Then, âˆ BAD + âˆ DAC = 90o â€¦ [equation (i)]

So, âˆ DCA + âˆ DAC = 90o â€¦ [equation (ii)]

From equation (i) and equation (ii)

We have,

âˆ BAD + âˆ DAC = âˆ DCA + âˆ DAC

âˆ BAD = âˆ DCA â€¦ [equation (iii)]

âˆ BDA = âˆ ADC â€¦ [both the angles are equal to 90o]

âˆ BAD = âˆ DCA â€¦ [from equation (iii)]

Because, corresponding sides of similar triangles are proportional

By cross multiplication we get,

AD2 = 2 Ã— 8 = 16

22. A 15 metres high tower casts a shadow of 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:-

From the question it is given that,

Height of a tower PQ = 15m

Itâ€™s shadow QR = 24 m

Let us assume the height of a telephone pole MN = x

Itâ€™s shadow NO = 16 m

Given, at the same time,

âˆ†PQR ~ âˆ†MNO

Therefore, PQ/MN = ON/RQ

15/x = 24/16

By cross multiplication we get,

x = (15 Ã— 16)/24

x = 240/24

x = 10

Therefore, height of pole = 10 m.

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

Solution:-

From the question it is given that,

Height of pole (PQ) = 6m

Height of a woman (MN) = 1.5m

Therefore, pole and woman are standing in the same line

PM ||MR

âˆ†PRQ ~ âˆ†MNR

So, RQ/RN = PQ/MN

(3 + x)/3 = 6/1.5

(3 + x)/3 = 60/15

(3 + x)/3 = 4/1

(3 + x) = 12

X = 12 â€“ 3

X = 9m

Therefore, women is 9m away from the pole.

Exercise 13.2

1. (a) In the figure (i) given below if DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm. Find (i) AE : EC (ii) DE.

Solution:-

From the figure,

DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm

(i) AE: EC

AE/EC = Â¾

AE: EC = 3: 4

âˆ D = âˆ B

âˆ E = âˆ C

DE/5 = 3/(3 + 4)

DE/5 = 3/7

DE = (3 Ã— 5)/7

DE = 15/7

DE =

(b) In the figure (ii) given below, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm. Find CQ and BQ.

Solution:-

From the figure,

PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm

âˆ BQP = âˆ BCA â€¦ [because alternate angles are equal]

Also, âˆ B = âˆ B â€¦ [common for both the triangles]

Therefore, âˆ†ABC ~ âˆ†BPQ

Then, BQ/BC = BP/AB = PQ/AC

BQ/BC = 6/(6 + 4) = PQ/AC

BQ/BC = 6/10 = PQ/AC

BQ/8 = 6/10 = PQ/AC â€¦ [because BC = 8 cm given]

Now, BQ/8 = 6/10

BQ = (6/10) Ã—8

BQ = 48/10

BQ = 4.8 cm

Also, CQ = BC â€“ BQ

CQ = (8 â€“ 4.8) cm

CQ = 3.2cm

Therefore, CQ = 3.2 cm and BQ = 4.8 cm

(c) In the figure (iii) given below, if XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.

Solution:-

From the figure,

XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm,

So, PX/QX = PY/YR

1/3 = PY/4.5

By cross multiplication we get,

(4.5 Ã— 1)/3 = PY

PY = 45/30

PY = 1.5

Then, âˆ X = âˆ Q

âˆ Y = âˆ R

So, âˆ†PXY ~ âˆ†PQR

Therefore, XY/QR = PX/PQ

XY/9 = 1/(1 + 3)

XY/9 = Â¼

XY = 9/4

XY = 2.25

2. In the given figure, DE || BC.

(i) If AD = x, DB = x â€“ 2, AE = x + 2 and EC = x â€“ 1, find the value of x.
(ii) If DB = x â€“ 3, AB = 2x, EC = x â€“ 2 and AC = 2x + 3, find the value of x.

Solution:-

(i) From the figure, it is given that,

Consider the âˆ†ABC,

x/(x – 2) = (x + 2)/(x – 1)

By cross multiplication we get,

X(x – 1) = (x – 2) (x + 2)

x2 â€“ x = x2 â€“ 4

-x = -4

x = 4

(ii) From the question it is given that,

DB = x â€“ 3, AB = 2x, EC = x â€“ 2 and AC = 2x + 3

Consider the âˆ†ABC,

2x/(x – 2) = (2x + 3)/(x – 3)

By cross multiplication we get,

2x(x – 2) = (2x + 3) (x – 3)

2x2 â€“ 4x = 2x2 â€“ 6x + 3x â€“ 9

2x2 â€“ 4x â€“ 2x2 + 6x â€“ 3x = -9

-7x + 6x = -9

-x = – 9

x = 9

3. E and F are points on the sides PQ and PR respectively of a âˆ†PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.

Solution:-

From the given dimensions,

Consider the âˆ†PQR

So, PE/EQ = 3.9/3

= 39/30

= 13/10

Then, PF/FR = 8/9

By comparing both the results,

13/10 â‰  8/9

Therefore, PE/EQ â‰  PF/FR

So, EF is not parallel to QR

(ii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

Solution:-

From the dimensions given in the question,

Consider the âˆ†PQR

So, PQ/PE = 1.28/0.18

= 128/18

= 64/9

Then, PR/PF = 2.56/0.36

= 256/36

= 64/9

By comparing both the results,

64/9 = 64/9

Therefore, PQ/PE = PR/PF

So, EF is parallel to QR.

4. 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? Give reasons for your answer.

Solution:-

From the dimensions given in the question,

Consider the âˆ†PQR

So, PQ/PA = 12.5/5

= 2.5/1

PR/PB = (PB + BR)/PB

= (4 + 6)/4

= 10/4

= 2.5

By comparing both the results,

2.5 = 2.5

Therefore, PQ/PA = PR/PB

So, AB is parallel to QR.

5.

(a) In figure (i) given below, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.

Solution:-

From the question it is given that,

DE || BC and BD = CE

So, we have to prove that ABC is an isosceles triangle.

Consider the triangle ABC,

Given, DB = EC â€¦ [equation (i)]

Then, AD = AE â€¦ [equation (ii)]

By adding equation (i) and equation (ii) we get,

AD + DB = AE + EC

So, AB = AC

Therefore, âˆ†ABC is an isosceles triangle.

(b) In figure (ii) given below, AB || DE and BD || EF. Prove that DCÂ² = CF x AC.

Solution:-

From the figure it is given that, AB || DE and BD || EF.

We have to prove that, DCÂ² = CF x AC

Consider the âˆ†ABC,

DC/CA = CE/CB â€¦ [equation (i)]

Now, consider âˆ†CDE

CF/CD = CE/CB â€¦ [equation (ii)]

From equation (i) and equation (ii),

DC/CA = CF/CD

DC/AC = CF/DC

By cross multiplication we get,

DC2 = CF x AC

6.

(a) In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.

Solution:-

From the given figure, CD || LA and DE || AC,

Consider the âˆ†BCA,

BE/BC = BD/BA

By using the corollary of basic proportionality theorem,

BE/(BE + EC) = BD/AB

4/(4 + 2) = BD/AB â€¦ [equation (i)]

Then, consider the âˆ†BLA

BC/BL = BD/AB

By using the corollary of basic proportionality theorem,

6/(6 + CL) = BD/AB â€¦ [equation (ii)]

Now, combining the equation (i) and equation (ii), we get

6/(6 + CL) = 4/6

By cross multiplication we get,

6 x 6 = 4 x (6 + CL)

24 + 4CL = 36

4CL = 36 â€“ 24

CL = 12/4

CL = 3 cm

Therefore, the length of CL is 3 cm.

(b) In the give figure, âˆ D = âˆ E and AD/BD = AE/EC. Prove that BAC is an isosceles triangle.

Solution:-

From the given figure, âˆ D = âˆ E and AD/BD = AE/EC,

We have to prove that, BAC is an isosceles triangle

âˆ D = âˆ E â€¦ [from the question]

AD = AE â€¦ [sides opposite to equal angles]

Consider the âˆ†ABC,

Then, AD/DB = AE/EC â€¦ [equation (i)]

Therefore, DE parallel to BC

DB = EC â€¦ [equation (ii)]

By adding equation (i) and equation (ii) we get,

AD + DB = AE + EC

AB = AC

Therefore, âˆ†ABC is an isosceles triangle.

7. In the adjoining given below, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. show that BC || QR.

Solution:-

Consider the âˆ†POQ

AB || PQ â€¦ [given]

So, OA/AP = OB/BQ â€¦ [equation (i)]

Then, consider the âˆ†OPR

AC || PR

OA/AP = OC/CR â€¦ [equation (ii)]

Now by comparing both equation (i) and equation (ii),

OB/BQ = OC/CR

Then, in âˆ†OQR

OB/BQ = OC/CR

Therefore, BC || QR

8. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove thatÂ AO/BO = CO/DO

Solution:-

From the question it is given that,

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O

Now consider the âˆ†OAB and âˆ†OCD,

âˆ AOB = âˆ COD [because vertically opposite angles are equal]

âˆ OBA = âˆ ODC [because alternate angles are equal]

âˆ OAB = âˆ OCD [because alternate angles are equal]

Therefore, âˆ†OAB ~ âˆ†OCD

Then, OA/OC = OB/OD

AO/OB = CO/DO â€¦ [by alternate angles]

9.

(a) In the figure (1) given below, AB || CR and LM || QR.

(i) Prove that BM/MC = AL/LQ

(ii) Calculate LM : QR, given that BM : MC = 1 : 2.

Solution:-

From the question it is given that, AB || CR and LM || QR

(i) We have to prove that, BM/MC = AL/LQ

Consider the âˆ†ARQ

LM || QR â€¦ [from the question]

So, AM/MR = AL/LQ â€¦ [equation (i)]

Now, consider the âˆ†AMB and âˆ†MCR

âˆ AMB = âˆ CMR â€¦ [because vertically opposite angles are equal]

âˆ MBA = âˆ MCR â€¦ [because alternate angles are equal]

Therefore, AM/MR = BM/MC â€¦ [equation (ii)]

From equation (i) and equation (ii) we get,

BM/MR = AL/LQ

(ii) Given, BM : MC = 1 : 2

AM/MR = BM/MC

AM/MR = Â½ â€¦ [equation (iii)]

LM || QR â€¦ [given from equation]

AM/MR = LM/QR â€¦ [equation (iv)]

AR/AM = QR/LM

(AM + MR)/AM = QR/LM

1 + MR/AM = QR/LM

1 + (2/1) = QR/LM

3/1 = QR/LM

LM/QR = 1/3

Therefore, the ratio of LM: QR is 1: 3.

(b) In the figure (2) given below AD is bisector of âˆ BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC

Solution:-

From the question it is given that,

AD is bisector of âˆ BAC

AB = 6 cm, AC = 4 cm and BD = 3cm

Construction, from C draw a straight line CE parallel to DA and join AE

âˆ 1 = âˆ 2 â€¦ [equation (i)]

By construction CE || DE

So, âˆ 2 = âˆ 4 â€¦ [because alternate angles are equal] [equation (ii)]

Again by construction CE || DE

âˆ 1 = âˆ 3 â€¦ [because corresponding angles are equal] [equation (iii)]

By comparing equation (i), equation (ii) and equation(iii) we get,

âˆ 3 = âˆ 4

So, AC = AE â€¦ [equation (iv)]

Now, consider the âˆ†BCE,

CE || DE

BD/DC = AB/AE

BD/DC = AB/AC

3/DC = 6/4

By cross multiplication we get,

3 Ã— 4 = 6 Ã— DC

DC = (3 Ã— 4)/6

DC = 12/6

DC = 2

Therefore, BC = BD + DC

= 3 + 2

= 5 cm

Exercise 13.3

1. Given that âˆ†s ABC and PQR are similar.
Find:
(i) The ratio of the area of âˆ†ABC to the area of âˆ†PQR if their corresponding sides are in the ratio 1 : 3.
(ii) the ratio of their corresponding sides if area of âˆ†ABC : area of âˆ†PQR = 25 : 36.

Solution:-

From the question it is given that,

(i) The area of âˆ†ABC to the area of âˆ†PQR if their corresponding sides are in the ratio 1 : 3

Then, âˆ†ABC ~ âˆ†PQR

area of âˆ†ABC/area of âˆ†PQR = BC2/QR2

So, BC : QR = 1 : 3

Therefore, âˆ†ABC/area of âˆ†PQR = 12/32

= 1/9

Hence the ratio of the area of âˆ†ABC to the area of âˆ†PQR is 1: 9

(ii) The area of âˆ†ABC to the area of âˆ†PQR if their corresponding sides are in the ratio 25 : 36

Then, âˆ†ABC ~ âˆ†PQR

area of âˆ†ABC/area of âˆ†PQR = BC2/QR2

area of âˆ†ABC/area of âˆ†PQR = BC2/QR2 = 25/36

= (BC/QR)2 = (5/6)2

BC/QR = 5/6

Hence the ratio of their corresponding sides is 5 : 6

2. âˆ†ABC ~ DEF. If area of âˆ†ABC = 9 sq. cm., area of âˆ†DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.

Solution:-

From the question it is given that,

âˆ†ABC ~ DEF

Area of âˆ†ABC = 9 sq. cm

Area of âˆ†DEF =16 sq. cm

We know that,

area of âˆ†ABC/area of âˆ†DEF = BC2/EF2

area of âˆ†ABC/area of âˆ†DEF = BC2/EF2

9/16 = BC2/EF2

9/16 = (2.1)2/x2

2.1/x = âˆš9/âˆš16

2.1/x = Â¾

By cross multiplication we get,

2.1 Ã— 4 = 3 Ã— x

8.4 = 3x

x = 8.4/3

x = 2.8

Therefore, EF = 2.8 cm

3. âˆ†ABC ~ âˆ†DEF. If BC = 3 cm, EF = 4 cm and area of âˆ†ABC = 54 sq. cm. Determine the area of âˆ†DEF.

Solution:-

From the question it is given that,

âˆ†ABC ~ âˆ†DEF

BC = 3 cm, EF = 4 cm

Area of âˆ†ABC = 54 sq. cm.

We know that,

Area of âˆ†ABC/ area of âˆ†DEF = BC2/EF2

54/area of âˆ†DEF = 32/42

54/area of âˆ†DEF = 9/16

By cross multiplication we get,

Area of âˆ†DEF = (54 Ã— 16)/9

= 6 Ã— 16

= 96 cm

4. The area of two similar triangles are 36 cmÂ² and 25 cmÂ². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

Solution:-

From the question it is given that,

The area of two similar triangles are 36 cmÂ² and 25 cmÂ².

Let us assume âˆ†PQR ~ âˆ†XYZ, PM and XN are their altitudes.

So, area of âˆ†PQR = 36 cm2

Area of âˆ†XYZ = 25 cm2

PM = 2.4 cm

Assume XN = a

We know that,

area of âˆ†PQR/area of âˆ†XYZ = PM2/XN2

36/25 = (2.4)2/a2

By cross multiplication we get,

36a2 = 25 (2.4)2

a2 = 5.76 Ã— 25/36

a2 = 144/36

a2 = 4

a = âˆš4

a = 2 cm

Therefore, altitude of the other triangle XN = 2 cm.

5.

(a) In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of âˆ†POB = 120 cmÂ², find the area of âˆ†QOA.

Solution:-

From the question it is given that, PO = 6 cm, QO = 9 cm and the area of âˆ†POB = 120 cmÂ²

From the figure,

Consider the âˆ†AOQ and âˆ†BOP,

âˆ OAQ = âˆ OBP â€¦ [both angles are equal to 90o]

âˆ AOQ = âˆ BOP â€¦ [because vertically opposite angles are equal]

Therefore, âˆ†AOQ ~ âˆ†BOP

Then, area of âˆ†AOQ/area of âˆ†BOP = OQ2/PO2

Area of âˆ†AOQ/120 = 92/62

Area of âˆ†AOQ/120 = 81/36

Area of âˆ†AOQ = (81 Ã— 120)/36

Area of âˆ†AOQ = 270 cm2

b) In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm.

(i) Prove that âˆ†OAB ~ âˆ†OCD.

(ii) Find CD and OB.

(iii) Find the ratio of areas of âˆ†OAB and âˆ†OCD.

Solution:-

From the question it is given that,

AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm

(i) We have to prove that, âˆ†OAB ~ âˆ†OCD

So, consider the âˆ†OAB and âˆ†OCD

âˆ AOB = âˆ COD â€¦ [because vertically opposite angles are equal]

âˆ OBA = âˆ OCD â€¦ [because alternate angles are equal]

Therefore, âˆ†OAB ~ âˆ†OCD â€¦ [from AAA axiom]

(ii) Consider the âˆ†OAB and âˆ†OCD

OA/OC = OB/OD = AB/CD

Now consider OA/OC = OB/OD

10/5 = OB/2.8

OB = (10 Ã— 2.8)/5

OB = 2 Ã— 2.8

OB = 5.6 cm

Then, consider OA/OC = AB/CD

10/5 = 6.5/CD

CD = (6.5 Ã— 5)/10

CD = 32.5/10

CD = 3.25 cm

(iii) We have to find the ratio of areas of âˆ†OAB and âˆ†OCD.

From (i) we proved that, âˆ†OAB ~ âˆ†OCD

Then, area of (âˆ†OAB)/area of âˆ†OCD

AB2/CD2 = (6.5)2/(3.25)2

= (6.5 Ã— 6.5)/(3.25 Ã— 3.25)

= 2 Ã— 2/1

= 4/1

Therefore, the ratio of areas of âˆ†OAB and âˆ†OCD = 4: 1.

6.

(a) In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of âˆ†ADE = 28 sq. cm, find the area of âˆ†ABC.

Solution:-

From the question it is given that,

DE || BC, DE = 6 cm, BC = 9 cm and area of âˆ†ADE = 28 sq. cm

From the fig, âˆ D = âˆ B and âˆ E = âˆ C â€¦ [corresponding angles are equal]

Now consider the âˆ†ADE and âˆ†ABC,

âˆ A = âˆ A â€¦ [common angles for both triangles]

Then, area of âˆ†ADE/area of âˆ†ABC = (DE)2/(BC)2

28/area of âˆ†ABC = (6)2/(9)2

28/area of âˆ†ABC = 36/81

area of âˆ†ABC = (28 Ã— 81)/36

area of âˆ†ABC = 2268/36

area of âˆ†ABC = 63 cm2

(b) In the figure (ii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of âˆ†ADE and trapezium DBCE.

Solution:-

From the question it is given that, DE || BC and AD : DB = 1 : 2,

âˆ D = âˆ B, âˆ E = âˆ C â€¦ [corresponding angles are equal]

âˆ A = âˆ A â€¦ [common angles for both triangles]

Now, adding 1 for both side LHS and RHS,

(DB/AD) + 1 = (2/1) + 1

Area of âˆ†ADE/area of âˆ†ABC = (1/3)2

Area of âˆ†ADE/area of âˆ†ABC = 1/9

Area of âˆ†ABC = 9 area of âˆ†ADE

Area of trapezium DBCE

Area of âˆ†ABC â€“ area of âˆ†ADE

Therefore, area of âˆ†ADE/area of trapezium = 1/8

Then area of âˆ†ADE : area of trapezium DBCE = 1: 8

7.

In the given figure, DE || BC.
(i) Prove that âˆ†ADE and âˆ†ABC are similar.

(ii) Given that AD = Â½ BD, calculate DE if BC = 4.5 cm.

(iii) If area of âˆ†ABC = 18cm2, find the area of trapezium DBCE

Solution:-

(i) From the question it is given that, DE || BC

We have to prove that, âˆ†ADE and âˆ†ABC are similar

âˆ A = âˆ A â€¦ [common angle for both triangles]

âˆ ADE = âˆ ABC â€¦ [because corresponding angles are equal]

Therefore, âˆ†ADE ~ âˆ†ABC â€¦ [AA axiom]

(ii) From (i) we proved that, âˆ†ADE ~ âˆ†ABC

Then, AD/AB = AB/AC = DE/BC

(Â½ BD)/ ((Â½BD) + BD) = DE/4.5

(Â½ BD)/ ((3/2)BD) = DE/4.5

Â½ Ã— (2/3) = DE/4.5

1/3 = DE/4.5

Therefore, DE = 4.5/3

DE = 1.5 cm

(iii) From the question it is given that, area of âˆ†ABC = 18 cm2

Then, area of âˆ†ADE/area of âˆ†ABC = DE2/BC2

area of âˆ†ADE/18 = (1/3)2 = 1/9

area of âˆ†ADE = 18 Ã— 1/9

So, area of trapezium DBCE = area of âˆ†ABC â€“ area of âˆ†ADE

= 18 â€“ 2

= 16 cm2

8. In the given figure, AB and DE are perpendicular to BC.
(i) Prove that âˆ†ABC ~ âˆ†DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of âˆ†ABC : area of âˆ†DEC.

Solution:-

(i) Consider the âˆ†ABC and âˆ†DEC,

âˆ ABC = âˆ DEC â€¦ [both angles are equal to 90o]

âˆ C = âˆ C â€¦ [common angle for both triangles]

Therefore, âˆ†ABC ~ âˆ†DEC â€¦ [by AA axiom]

(ii) AC/CD = AB/DE

Corresponding sides of similar triangles are proportional

15/CD = 6/4

CD = (15 Ã— 4)/6

CD = 60/6

CD = 10 cm

(iii) we know that, area of âˆ†ABC/area of âˆ†DEC = AB2/DE2

area of âˆ†ABC/area of âˆ†DEC = 62/42

area of âˆ†ABC/area of âˆ†DEC = 36/16

area of âˆ†ABC/area of âˆ†DEC = 9/4

Therefore, the ratio of the area of âˆ†ABC : area of âˆ†DEC is 9 : 4.

9. In the adjoining figure, ABC is a triangle. DE is parallel to BC and AD/DB = 3/2,

(i) Determine the ratios AD/AB, DE/BC

(ii) Prove that âˆ†DEF is similar to âˆ†CBF. Hence, find EF/FB.

(iii) What is the ratio of the areas of âˆ†DEF and âˆ†CBF?

Solution:-

(i) We have to find the ratios AD/AB, DE/BC,

From the question it is given that, AD/DB = 3/2

Now add 1 for both LHS and RHS we get,

(DB/AD) + 1 = (2/3) + 1

From the figure (DB + AD) = AB

Now, consider the âˆ†ADE and âˆ†ABC,

âˆ ADE = âˆ B â€¦ [corresponding angles are equal]

âˆ AED = âˆ C â€¦ [corresponding angles are equal]

Therefore, âˆ†ADE ~ âˆ†ABC â€¦ [by AA similarity]

Then, AD/AB = DE/BC = 3/5

(ii) Now consider the âˆ†DEF and âˆ†CBF

âˆ EDF = âˆ BCF â€¦ [because alternate angles are equal]

âˆ DEF = âˆ FBC â€¦ [because alternate angles are equal]

âˆ DFE = âˆ ABFC â€¦ [because vertically opposite angles are equal]

Therefore, âˆ†DEF ~ âˆ†CBF

So, EF/FB = DE/BC = 3/5

(iii) we have to find the ratio of the areas of âˆ†DEF and âˆ†CBF,

We know that, Area of âˆ†DFE/Area of âˆ†BFC = DE2/BC2

Area of âˆ†DFE/Area of âˆ†BFC = (DE/BC)2

Area of âˆ†DFE/Area of âˆ†BFC = (3/5)2

Area of âˆ†DFE/Area of âˆ†BFC = 9/25

Therefore, the ratio of the areas of âˆ†DEF and âˆ†CBF is 9: 25.

10. In âˆ†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 âˆ†APO : area âˆ†ABC.
(ii) area âˆ†APO : area âˆ†CQO.

Solution:-

From the question it is given that,

PB = 2: 3

PO is parallel to BC and is extended to Q so that CQ is parallel to BA.

(i) we have to find the area âˆ†APO: area âˆ†ABC,

Then,

âˆ A = âˆ A â€¦ [common angles for both triangles]

âˆ APO = âˆ ABC â€¦ [because corresponding angles are equal]

Then, âˆ†APO ~ âˆ†ABC â€¦ [AA axiom]

We know that, area of âˆ†APO/area of âˆ†ABC = AP2/AB2

= AP2/(AP + PB)2

= 22/(2 + 3)2

= 4/52

= 4/25

Therefore, area âˆ†APO: area âˆ†ABC is 4: 25

(ii) we have to find the area âˆ†APO: area âˆ†CQO

Then, âˆ AOP = âˆ COQ â€¦ [because vertically opposite angles are equal]

âˆ APQ = âˆ OQC â€¦ [because alternate angles are equal]

Therefore, area of âˆ†APO/area of âˆ†CQO = AP2/CQ2

area of âˆ†APO/area of âˆ†CQO = AP2/PB2

area of âˆ†APO/area of âˆ†CQO = 22/32

area of âˆ†APO/area of âˆ†CQO = 4/9

Therefore, area âˆ†APO: area âˆ†CQO is 4: 9

11.

(a) In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of âˆ†AOB and âˆ†COD.

Solution:-

From the question it is given that,

ABCD is a trapezium in which AB || DC and AB = 2 CD,

Then, âˆ OAB = âˆ OCD â€¦ [because alternate angles are equal]

âˆ OBA = âˆ ODC

Then, âˆ†AOB ~ âˆ†COD

So, area of âˆ†AOB/area of âˆ†COD = AB2/CD2

= (2CD)2/CD2 â€¦ [because AB = 2 CD]

= 4CD2/CD2

= 4/1

Therefore, the ratio of the areas of âˆ†AOB and âˆ†COD is 4: 1.

(b) In the figure (ii) given below, ABCD is a parallelogram. AM âŠ¥ DC and AN âŠ¥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cmÂ², find
(i) AB
(ii) BC
(iii) area of âˆ†ADM : area of âˆ†ANB.

Solution:-

From the question it is given that,

ABCD is a parallelogram, AM âŠ¥ DC and AN âŠ¥ CB

AM = 6 cm

AN = 10 cm

The area of parallelogram ABCD is 45 cmÂ²

Then, area of parallelogram ABCD = DC Ã— AM = BC Ã— AN

45 = DC Ã— 6 = BC Ã— 10

(i) DC = 45/6

Divide both numerator and denominator by 3 we get,

= 15/2

= 7.5 cm

Therefore, AB = DC = 7.5 cm

(ii) BC Ã— 10 = 45

BC = 45/10

BC = 4.5 cm

(iii) Now, consider âˆ†ADM and âˆ†ABN

âˆ D = âˆ B â€¦ [because opposite angles of a parallelogram]

âˆ M = âˆ N â€¦ [both angles are equal to 90o]

= BC2/AB2

= 4.52/7.52

= 20.25/56.25

= 2025/5625

= 81/225

= 9/25

Therefore, area of âˆ†ADM : area of âˆ†ANB is 9: 25

(c) In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(ii) area of âˆ†BEF : area of âˆ†ABD
(iii) area of âˆ†ABD : area of trapezium AFED
(iv) area of âˆ†FEO : area of âˆ†OBC.

Solution:-

From the question it is given that, ABCD is a parallelogram.

E is a point on AB, CE intersects the diagonal BD at O.

AE : EB = 2 : 3

(i) We have to find EF : AD

AE/EB = 2/3 â€¦ [given]

Now add 1 to both LHS and RHS we get,

(AE/EB) + 1 = (2/3) + 1

(AE + EB)/EB = (2 + 3)/3

AB/EB = 5/3

EB/AB = 3/5

Therefore, EF : AD is 3: 5

(ii) we have to find area of âˆ†BEF: area of âˆ†ABD,

Then, area of âˆ†BEF/area of âˆ†ABD = (EF)2/(AD)2

area of âˆ†BEF/area of âˆ†ABD = 32/52

= 9/25

Therefore, area of âˆ†BEF: area of âˆ†ABD is 9: 25

(iii) From (ii) area of âˆ†ABD/area of âˆ†BEF = 25/9

25 area of âˆ†BEF = 9 area of âˆ†ABD

25(area of âˆ†ABD â€“ area of trapezium AEFD) = 9 area of âˆ†ABD

25 area of âˆ†ABD â€“ 25 area of trapezium AEFD = 9 area of âˆ†ABD

25 area of trapezium AEFD = 25 area of âˆ†ABD â€“ 9 area of âˆ†ABD

25 area of trapezium AEFD = 16 area of âˆ†ABD

area of âˆ†ABD/area of trapezium AEFD = 25/16

Therefore, area of âˆ†ABD : area of trapezium AFED = 25: 16

(iv) Now we have to find area of âˆ†FEO : area of âˆ†OBC

So, consider âˆ†FEO and âˆ†OBC,

âˆ EOF = âˆ BOC â€¦ [because vertically opposite angles are equal]

âˆ F = âˆ OBC â€¦ [because alternate angles are equal]

âˆ†FEO ~ âˆ†OBC

Then, area of FEO/area of âˆ†OBC = EF2/BC2

Therefore, area of âˆ†FEO: area of âˆ†OBC = 9: 25.

12. In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q.Â If area of âˆ†CPQ = 20 cmÂ², find

(i) area of âˆ†BPQ.
(ii) area âˆ†CDP.
(iii) area of parallelogram ABCD.

Solution:-

From the question it is given that, ABCD is a parallelogram.

BP: PC = 1: 2

area of âˆ†CPQ = 20 cmÂ²

Construction: draw QN perpendicular CB and Join BN.

Then, area of âˆ†BPQ/area of âˆ†CPQ = ((Â½BP) Ã— QN)/((Â½PC) Ã— QN)

= BP/PC = Â½

(i) So, area âˆ†BPQ = Â½ area of âˆ†CPQ

= Â½ Ã— 20

Therefore, area of âˆ†BPQ = 10 cm2

(ii) Now we have to find area of âˆ†CDP,

Consider the âˆ†CDP and âˆ†BQP,

Then, âˆ CPD = âˆ QPD â€¦ [because vertically opposite angles are equal]

âˆ PDC = âˆ PQB â€¦ [because alternate angles are equal]

Therefore, âˆ†CDP ~ âˆ†BQP â€¦ [AA axiom]

area of âˆ†CDP/area of âˆ†BQP = PC2/BP2

area of âˆ†CDP/area of âˆ†BQP = 22/12

area of âˆ†CDP/area of âˆ†BQP = 4/1

area of âˆ†CDP = 4 Ã— area âˆ†BQP

Therefore, area of âˆ†CDP = 4 Ã— 10

= 40 cm2

(iii) We have to find the area of parallelogram ABCD,

Area of parallelogram ABCD = 2 area of âˆ†DCQ

= 2 area (âˆ†DCP + âˆ†CPQ)

= 2 (40 + 20) cm2

= 2 Ã— 60 cm2

= 120 cm2

Therefore, the area of parallelogram ABCD is 120 cm2.

13. (a) In the figure (i) given below, DE || BC and the ratio of the areas of âˆ†ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

Solution:-

From the question it is given that,

DE || BC

The ratio of the areas of âˆ†ADE and trapezium DBCE is 4 : 5

Now, consider the âˆ†ABC and âˆ†ADE

âˆ A = âˆ A â€¦ [common angle for both triangles]Â

âˆ D = âˆ B and

âˆ E = âˆ CÂ  â€¦ [because corresponding angles are equal]Â

So, area of âˆ†ADE/area of âˆ†ABC = (DE)2/(BC)2 â€¦ [equation (i)]

Then, area of âˆ†ADE/area of trapezium DBCE = 4/5

area of trapezium DBCE/area of âˆ†ADE = 5/4

Add 1 for both LHS and RHS we get,

(area of trapezium DBCE/area of âˆ†ADE) + 1 = (5/4) + 1

(area of trapezium DBCE + area of âˆ†ADE)/area of âˆ†ADE = (5 + 4)/4

area of âˆ†ABC/area of âˆ†ADE = 9/4

area of âˆ†ADE/area of âˆ†ABC = 4/9

From equation (i),

area of âˆ†ADE/area of âˆ†ABC = (DE)2/(BC)2

area of âˆ†ADE/area of âˆ†ABC = (DE)2/(BC)2 = 42/92

area of âˆ†ADE/area of âˆ†ABC = (DE)2/(BC)2 = 2/3

Therefore, DE: BC = 2: 3

(b) In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find (i) ED (ii) BE (iii) area of âˆ†EDC : area of trapezium ABCD.

Solution:-

From the question it is given that,

AB || DC

AB = 2 DC, AD = 3 cm, BC = 4 cm

Now consider âˆ†EAB,

EA/DA = EB/CB = AB/DC = 2DC/DC = 2/1

(i) EA = 2, DA = 2 Ã— 3 = 6 cm

Then, ED = EA â€“ DA

= 6 â€“ 3

= 3 cm

(ii) EB/CB = 2/1

EB = 2 CB

EB = 2 Ã— 4

EB = 8 cm

(iii) Now, consider the âˆ†EAB, DC || AB

So, âˆ†EDC ~ âˆ†EAB

Therefore, area of âˆ†EDC/area of âˆ†ABE = DC2/AB2

area of âˆ†EDC/area of âˆ†ABE = DC2/(2DC)2

area of âˆ†EDC/area of âˆ†ABE = DC2/4DC2

area of âˆ†EDC/area of âˆ†ABE = Â¼

Therefore, area of ABE = 4 area of âˆ†EDC

Then, area of âˆ†EDC + area of trapezium ABCD = 4 area of âˆ†EDC

Area of trapezium ABCD = 3 area of âˆ†EDC

So, area of âˆ†EDC/area of trapezium ABCD = 1/3

Therefore, area of âˆ†EDC: area of trapezium ABCD = 1: 3

14. (a) In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., findÂ (i) BPÂ (ii) the ratio of areas of âˆ†APB and âˆ†DPC.Â

Solution:-

From the question it is given that,

DC is parallel to AB

AB = 9 cm, DC = 6 cm and BB = 12 cm

(i) Consider the âˆ†APB and âˆ†CPD

âˆ APB = âˆ CPD â€¦ [because vertically opposite angles are equal]

âˆ PAB = âˆ PCD â€¦ [because alternate angles are equal]

So, âˆ†APB ~ âˆ†CPD

Then, BP/PD = AB/CD

BP/(12 – BP) = 9/6

6BP = 108 â€“ 9BP

6BP + 9BP = 108

15BP = 108

BP = 108/15

Therefore, BP = 7.2 cm

(ii) We know that, area of âˆ†APB/area of âˆ†CPD = AB2/CD2

area of âˆ†APB/area of âˆ†CPD = 92/62

area of âˆ†APB/area of âˆ†CPD = 81/36

By dividing both numerator and denominator by 9, we get,

area of âˆ†APB/area of âˆ†CPD = 9/4

Therefore, the ratio of areas of âˆ†APB and âˆ†DPC is 9: 4

(b) In the figure given below, âˆ ABC = âˆ DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.Â (i) Prove that âˆ†ACD is similar to âˆ†BCAÂ (ii) Find BC and CDÂ (iii) Find the area of âˆ†ACD : area of âˆ†ABC.

Solution:-

From the question it is given that,

âˆ ABC = âˆ DAC

AB = 8 cm, AC = 4 cm, AD = 5 cm

(i) Now, consider âˆ†ACD and âˆ†BCA

âˆ C = âˆ C â€¦ [common angle for both triangles]

âˆ ABC = âˆ CAD â€¦ [from the question]

So, âˆ†ACD ~ âˆ†BCA â€¦ [by AA axiom]

(ii) AC/BC = CD/CA = AD/AB

4/BC = 5/8

BC = (4 Ã— 8)/5

BC = 32/5

BC = 6.4 cm

CD/4 = 5/8

CD = (4 Ã— 5)/8

CD = 20/8

CD = 2.5 cm

(iii) from (i) we proved that, âˆ†ACD ~ âˆ†BCA

area of âˆ†ACB/area of âˆ†BCA = AC2/AB2

= 42/82

= 16/64

By dividing both numerator and denominator by 16, we get,

= Â¼

Therefore, the area of âˆ†ACD : area of âˆ†ABC is 1: 4.

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

(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.Â

Solution:-

From the question it is given that,

âˆ ABC = 90Â°

AB and DE is perpendicular to AC

(i) Consider the âˆ†ADE and âˆ†ACB,

âˆ A = âˆ A â€¦ [common angle for both triangle]

âˆ B = âˆ E â€¦ [both angles are equal to 90o]

(ii) from (i) we proved that, âˆ†ADE ~ âˆ†ACB

So, AE/AB = AD/AC = DE/BC â€¦ [equation (i)]

Consider the âˆ†ABC, is a right angle triangle

From Pythagoras theorem, we have

AC2 = AB2 + BC2

132 = AB2 + 52

169 = AB2 + 25

AB2 = 169 â€“ 25

AB2 = 144

AB = âˆš144

AB = 12 cm

Consider the equation (i),

Now, take AE/AB = DE/BC

4/12 = DE/5

1/3 = DE/5

DE = (5 Ã— 1)/3

DE = 5/3

DE = 1.67 cm

(iii) Now, we have to find area of âˆ†ADE : area of quadrilateral BCED,

We know that, Area of âˆ†ADE = Â½ Ã— AE Ã— DE

= Â½ Ã— 4 Ã— (5/3)

= 10/3 cm2

Then, area of quadrilateral BCED = area of âˆ†ABC â€“ area of âˆ†ADE

= Â½ Ã— BC Ã— AB â€“ 10/3

= Â½ Ã— 5 Ã— 12 â€“ 10/3

= 1 Ã— 5 Ã— 6 â€“ 10/3

= 30 â€“ 10/3

= (90 – 10)/3

= 80/3 cm2

So, the ratio of area of âˆ†ADE : area of quadrilateral BCED = (10/3)/(80/3)

= (10/3) Ã— (3/80)

= (10 Ã— 3)/(3 Ã— 80)

= (1 Ã— 1)/(1 Ã— 8)

= 1/8

Therefore, area of âˆ†ADE : area of quadrilateral BCED is 1: 8.

16. Two isosceles triangles have equal vertical angles and their areas are in the ratio

7: 16. Find the ratio of their corresponding height.

Solution:-

Consider the two isosceles triangle PQR and XYZ,

âˆ P = âˆ X â€¦ [from the question]

So, âˆ Q + âˆ R = âˆ Y + âˆ Z

âˆ Q = âˆ R and âˆ Y = âˆ Z [because opposite angles of equal sides]

Therefore, âˆ Q = âˆ Y and âˆ R = âˆ Z

âˆ†PQR ~ âˆ†XYZ

Then, area of âˆ†PQR/area of âˆ†XYZ = PM2/XN2 â€¦ [from corollary of theorem]

PM2/XN2 = 7/16

PM/XN = âˆš7/âˆš16

PM/XN = âˆš7/4

Therefore, ratio of PM: DM = âˆš7: 4

17. On a map drawn to a scale of 1 : 250000, a triangular plot of land has the following measurements :Â AB = 3 cm, BC = 4 cm and âˆ ABC = 90Â°. CalculateÂ

(i) the actual length of AB in km.Â

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

Solution:-

From the question it is given that,

Map drawn to a scale of 1: 250000

AB = 3 cm, BC = 4 cm and âˆ ABC = 90o

(i) We have to find the actual length of AB in km.Â

Let us assume scale factor K = 1: 250000

K = 1/250000

Then, length of AB of actual plot = 1/k Ã— length of AB on the map

= (1/(1/250000)) Ã— 3

= 250000 Ã— 3

To covert cm into km divide by 100000

= (250000 Ã— 3)/(100 Ã— 1000)

= 15/2

length of AB of actual plot = 7.5 km

(ii) We have to find the area of the plot in sq. km

Area of plot on the map = Â½ Ã— AB Ã— BC

= Â½ Ã— 3 Ã— 4

= Â½ Ã— 12

= 1 Ã— 6

= 6 cm2

Then, area of actual plot = 1/k2 Ã— 6 cm2

= 2500002 Ã— 6

To covert cm into km divide by (100000)2

= (250000 Ã— 250000 Ã— 6)/(100000 Ã— 100000)

= (25/4) Ã— 6

= 75/2

= 37.5 km2

18. On a map drawn to a scale of 1 : 25000, a rectangular plot of land, ABCD has the following measurements AB = 12 cm and BG = 16 cm.Â Calculate:Â

(i) the distance of a diagonal of the plot in km.Â

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

Solution:-

From the question it is given that,

Map drawn to a scale of 1: 25000

AB = 12 cm, BG = 16 cm

Consider the âˆ†ABC,

From the Pythagoras theorem,

AC2 = AB2 + BC2

AC = âˆš(AB2 + BC2)

= âˆš((12)2 + (16)2)

= âˆš144 + 256

= âˆš400

= 20 cm

Then, area of rectangular plot ABCD = AB Ã— BC

= 12 Ã— 16

= 192 cm2

(i) We have to find the distance of a diagonal of the plot in km.Â .Â

Let us assume scale factor K = 1: 25000

K = 1/25000

Then, length of AB of actual plot = 1/k Ã— length of diagonal of rectangular plot

= (1/(1/25000)) Ã— 3

= 25000 Ã— 20

To covert cm into km divide by 100000

= (25000 Ã— 20)/(100 Ã— 1000)

= 5 km

(ii) We have to find the area of the plot in sq. km.

Then, area of actual plot = 1/k2 Ã— 192 cm2

= 250002 Ã— 192

To covert cm into km divide by (100000)2

= (25000 Ã— 25000 Ã— 192)/(100000 Ã— 100000)

= 12 km2

19. The model of a building is constructed with the scale factor 1 : 30.Â

(i) If the height of the model is 80 cm, find the actual height of the building in metres.Â

(ii) If the actual volume of a tank at the top of the building is 27 mÂ³, find the volume of the tank on the top of the model.

Solution:-

From the question it is given that,

The model of a building is constructed with the scale factor 1 : 30

So, Height of the model/Height of actual building = 1/30

(i) Given, the height of the model is 80 cm

Then, 80/H = 1/30

H = (80 Ã— 30)

H = 2400 cm

H = 2400/100

H = 24 m

(ii) Given, the actual volume of a tank at the top of the building is 27 mÂ³

Volume of model/Volume of tank = (1/30)3

V/27 = 1/27000

V = 27/27000

V = 1/1000 m3

Therefore, Volume of model = 1000 cm3

20. A model of a ship is made to a scale of 1 : 200.Â

(i) If the length of the model is 4 m, find the length of the ship.Â

(ii) If the area of the deck of the ship is 160000 mÂ², find the area of the deck of the model.Â

(iii) If the volume of the model is 200 liters, find the volume of the ship in mÂ³.Â (100 liters = 1 mÂ³)

Solution:-

From the question it is given that, a model of a ship is made to a scale of 1 : 200

(i) Given, the length of the model is 4 m

Then, length of the ship = (4 Ã— 200)/1

= 800 m

(ii) Given, the area of the deck of the ship is 160000 mÂ²

Then, area of deck of the model = 160000 Ã— (1/200)2

= 160000 Ã— (1/40000)

= 4 m2

(iii) Given, the volume of the model is 200 liters

Then, Volume of ship = 200 Ã— (200/1)3

= 200 Ã— 8000000

= (200 Ã— 8000000)/100

= 1600000 m3

Chapter test

1. In the adjoining figure, âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4. Show that PT x QR = PR x ST.

Solution:-

From the question it is given that,

âˆ 1 = âˆ 2 and âˆ 3 = âˆ 4

We have to prove that, PT x QR = PR x ST

Given, âˆ 1 = âˆ 2

Adding âˆ 6 to both LHS and RHS we get,

âˆ 1 + âˆ 6 = âˆ 2 + âˆ 6

âˆ SPT = âˆ QPR

Consider the âˆ†PQR and âˆ†PST,

From above âˆ SPT = âˆ QPR

âˆ 3 = âˆ 4

Therefore, âˆ†PQR ~ âˆ†PST

So, PT/PR = ST/QR

By cross multiplication we get,

PT x QR = PR x ST

Hence, it is proved that PT x QR = PR x ST

2. In the adjoining figure, AB = AC. If PM âŠ¥ AB and PN âŠ¥ AC, show that PM x PC = PN x PB.

Solution:-

From the given figure,

AB = AC. If PM âŠ¥ AB and PN âŠ¥ AC

We have to show that, PM x PC = PN x PB

Consider the âˆ†ABC,

AB = AC â€¦ [given]

âˆ B = âˆ C

Then, consider âˆ†CPN and âˆ†BPM

âˆ N = âˆ M â€¦ [both angles are equal to 90o]

âˆ C = âˆ B â€¦ [from above]

Therefore, âˆ†CPN ~ âˆ†BPM â€¦ [from AA axiom]

So, PC/PB = PN/PM

By cross multiplication we get,

PC x PM = PN x PB

Therefore, it is proved that, PM x PC = PN x PB

3.

(a) In the figure given below. âˆ AED = âˆ ABC. Find the values of x and y.

Solution:-

From the figure it is given that,

âˆ AED = âˆ ABC

âˆ AED = âˆ ABC â€¦ [from the figure]

âˆ A = âˆ A â€¦ [common angle for both triangles]

Therefore, âˆ†ABC ~ âˆ†ADE â€¦ [by AA axiom]

3/(4 + 2) = y/10

3/6 = y/10

By cross multiplication we get,

y = (3 x 10)/6

y = 30/6

y = 5

Now, consider AB/AE = BC/DE

(3 + x)/4 = 10/y

Substitute the value of y,

(3 + x)/4 = 10/5

By cross multiplication,

5(3 + x) = 10 x 4

15 + 5x = 40

5x = 40 â€“ 15

5x = 25

X = 25/5

x = 5

Therefore, the value of x = 5 cm and y = 5 cm

(b) In the figure given below, CD = Â½ AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :Â

(i) CE || AGÂ

(ii) 3 ED = GD

Solution:-

From the question it is given that,

CD = Â½ AC

BF || AG

(i) We have to prove that, CE || AGÂ

Consider, CD = Â½ AC

AC = 2BC â€¦ [because from the figure B is mid-point of AC]

So, CD = Â½ (2BC)

CD = BC

Hence, CE || BF â€¦ [equation (i)]

Given, BF || AG â€¦ [equation (ii)]

By comparing the results of equation (i) and equation (ii) we get,

CE || AG

(ii) We have to prove that, 3 ED = GD

Consider the âˆ†AGD,

CE || AG â€¦ [above it is proved]

AD = AB + BC + DC

= DC + DC + DC

= 3DC

So, ED/GD = DC/(3DC)

ED/GD = 1/(3(1))

ED/GD = 1/3

3ED = GD

Hence it is proved that, 3ED = GD

4. In the adjoining figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:Â

(i) DF || BHÂ

(ii) AH = 3 AF.Â

Solution:-

From the question it is given that, 2 AD = BD, EC || BH

(i) Given, E is mid-point of BD

2DE = BD â€¦ [equation (i)]

2AD = BD â€¦ [equation (ii)]

From equation (i) and equation (ii) we get,

Also given that, F is mid-point of AC

DF || EC â€¦ [equation (iii)]

Given, EC || BH â€¦ [equation (iv)]

By comparing equation (iii) and equation (iv) we get,

DF || BH

(ii) We have to prove that, AH = 3 AF,

Given, E is mid-point of BD and EC || BH

And c is midpoint of AH,

Then, FC = CH â€¦ [equation (v)]

Also given F is mid-point of AC

AF = FC â€¦ [equation (vi)]

By comparing both equation (v) and equation (vi) we get,

FC = AF = CH

AF = (1/3)AH

By cross multiplication we get,

3AF = AH

Therefore, it is proved that 3AF = AH

5. In a âˆ†ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.

Solution:-

From the question it is given that, In a âˆ†ABC, D and E are points on the sides AB and AC respectively.

DE || BC

AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm

Consider the âˆ†ABC,

Given, DE || BC

So, AD/AB = AE/AC = DE/BC

2.4/AB = 2/5

AB = (2.4 x 5)/2

AB = 12/2

AB = 6 cm

Then, consider AE/AC = DE/BC

3.2/AC = 2/5

AC = (3.2 x 5)/2

AC = 16/2

AC = 8 cm

Hence, BD = AB â€“ AD

= 6 â€“ 2.4

= 3.6 cm

CE = AC â€“ AE

= 8 â€“ 3.2

= 4.8 cm

6. In a âˆ†ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.

Solution:-

From the question it is given that,

In a âˆ†ABC, D and E are points on the sides AB and AC respectively.

AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm

Consider the âˆ†ABC,

EC = AC â€“ AE

= 8.8 â€“ 3.3

= 5.5 cm

= 57/95

By dividing both numerator and denominator by 19 we get,

= 3/5

AE/EC = 3.3/5.5

= 33/55

By dividing both numerator and denominator by 11 we get,

= 3/5

Therefore, DE || BC

7. If the areas of two similar triangles are 360 cmÂ² and 250 cmÂ² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.

Solution:-

From the question it is given that, the areas of two similar triangles are 360 cmÂ² and 250 cmÂ².

one side of the first triangle is 8 cm

So, PQR and XYZ are two similar triangles,

So, let us assume area of âˆ†PQR = 360 cm2, QR = 8 cm

And area of âˆ†XYZ = 250 cm2

Assume YZ = a

We know that, area of âˆ†PQR/area of âˆ†XYZ = QR2/yz2

360/250 = (8)2/a2

360/250 = 64/a2

By cross multiplication we get,

a2 = (250 x 64)/360

a2 = 400/9

a = âˆš(400/9)

a = 20/3

a =

Therefore, the length of the corresponding side of the second triangle YZ =

8. In the adjoining figure, D is a point on BC such that âˆ ABD = âˆ CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, findÂ

(i) BCÂ

(ii) DCÂ

(iii) area of âˆ†ACD : area of âˆ†BCA.

Solution:-

From the question it is given that,

AB = 5 cm, AC = 3 cm and AD = 4 cm

Now, consider the âˆ†ABC and âˆ†ACD

âˆ C = âˆ C â€¦ [common angle for both triangles]

âˆ ABC = âˆ CAD â€¦ [from the question]

So, âˆ†ABC ~ âˆ†ACD

Then, AB/AD = BC/AC = AC/DC

5/4 = BC/3

BC = (5 x 3)/4

BC = 15/4

BC = 3.75 cm

5/4 = 3/DC

DC = (3 x 4)/5

DC = 12/5

Dc = 2.4 cm

(iii) Consider the âˆ†ABC and âˆ†ACD

âˆ CAD = âˆ ABC â€¦ [from the question]

âˆ ACD = âˆ ACB â€¦ [common angle for both triangle]

Therefore, âˆ†ACD ~ âˆ†ABC

Then, area of âˆ†ACD/area of âˆ†ABC = AD2/AB2

= 42/52

= 16/25

Therefore, area of âˆ†ACD : area of âˆ†BCA is 16: 25.

9. In the adjoining figure, the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of âˆ†AOE : area of parallelogram ABCD.Â

Solution:-

From the given figure,

The diagonals of a parallelogram intersect at O.

OE is drawn parallel to CB to meet AB at E.

In the figure four triangles have equal area.

So, area of âˆ†OAB = Â¼ area of parallelogram ABCD

Then, O is midpoint of AC of âˆ†ABC and DE || CB

E is also midpoint of AB

Therefore, OE is the median of âˆ†AOB

Area of âˆ†AOE = Â½ area of âˆ†AOB

= Â½ Ã— Â¼ area of parallelogram ABCD

= 1/8 area of parallelogram ABCD

So, area of âˆ†AOE/area of parallelogram ABCD = 1/8

Therefore, area of âˆ†AOE: area of parallelogram ABCD is 1: 8.

10. In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of âˆ†AOB and âˆ†COD.

Solution:-

From the question it is given that, ABCD is a trapezium in which AB || DC. If 2AB = 3DC.

So, 2AB = 3DC

AB/DC = 3/2

Now, consider âˆ†AOB and âˆ†COD

âˆ AOB = âˆ COD â€¦ [because vertically opposite angles are equal]

âˆ OAB = âˆ OCD â€¦ [because alternate angles are equal]

Therefore, âˆ†AOB ~ âˆ†COD â€¦ [from AA axiom]

Then, area of âˆ†AOB/area of âˆ†COD = AB2/DC2

= 32/22

= 9/4

Therefore, the ratio of the areas of âˆ†AOB and âˆ†COD is 9: 4

11. In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that

Solution:-

From the question it is given that,

ABCD is a parallelogram. E is mid-point of BC.

DE meets the diagonal AC at O.

(i) Now consider the âˆ†AOD and âˆ†EDC,

âˆ AOD = âˆ EOC â€¦ [because Vertically opposite angles are equal]

âˆ OAD = âˆ OCB â€¦ [because alternate angles are equal]

Therefore, âˆ†AOD ~ âˆ†EOC

Then, OA/OC = DO/OE = AD/EC = 2EC/EC

OA/OC = DO/OE = 2/1

Therefore, OA: OC = 2: 1

(ii) From (i) we proved that, âˆ†AOD ~ âˆ†EOC

So, area of âˆ†OEC/area of âˆ†AOD = OE2/DO2

area of âˆ†OEC/area of âˆ†AOD = 12/22

area of âˆ†OEC/area of âˆ†AOD = Â¼

Therefore, area of âˆ†OEC: area of âˆ†AOD is 1: 4.

13. A model of a ship is made to a scale of 1: 250 calculate:

(i) The length of the ship, if the length of model is 1.6 m.

(ii) The area of the deck of the ship, if the area of the deck of model is 2.4 m2.

(iii) The volume of the model, if the volume of the ship is 1 km3.

Solution:-

From the question it is given that, a model of a ship is made to a scale of 1 : 250

(i) Given, the length of the model is 1.6 m

Then, length of the ship = (1.6 Ã— 250)/1

= 400 m

(ii) Given, the area of the deck of the ship is 2.4 mÂ²

Then, area of deck of the model = 2.4 Ã— (1/250)2

= 1,50,000 m2 = 4 m2

(iii) Given, the volume of the model is 1 km3

Then, Volume of ship = (1/2503) Ã— 1 km3

= 1/(250)3 Ã— 10003

= 43

= 64 m3

Therefore, volume of ship is 64 m3.