**Exercise 4.6**

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**1. Triangles ABC and DEF are similar.**

**(i) If area of (\(\Delta ABC\)) = 16 cm ^{2 }, area (\(\Delta DEF\)) = 25 cm^{2 }and BC = 2.3 cm, find EF.**

**(ii) If area (\(\Delta ABC\)) = 9 cm ^{2 }, area (\(\Delta DEF\)) = 64 cm^{2 }and DE = 5.1 cm, find AB.**

**(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles.**

**(iv) If area of (\(\Delta ABC\)) = 36 cm ^{2 }, area (\(\Delta DEF\)) = 64 cm^{2 }and DE = 6.2 cm, find AB.**

**(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the area of two triangles.**

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**Answer:**

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

\(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{BC}{EF})^{2}\)

\(\frac{16}{25} = (\frac{2.3}{EF})^{2}\)

\(\frac{4}{5} = \frac{2.3}{EF}\)

EF = 2.875 cm

**(ii)** \(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AB}{DE})^{2}\)

\(\frac{9}{64} = (\frac{AB}{DE})^{2}\)

\(\frac{3}{8} = \frac{AB}{5.1}\)

AB = 1.9125 cm

**(iii)** \(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AC}{DF})^{2}\)

\(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{19}{8})^{2}\)

\(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{361}{64})\)

**(iv)** \(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AB}{DE})^{2}\)

\(\frac{36}{64} = (\frac{AB}{DE})^{2}\)

\(\frac{6}{8} = \frac{AB}{6.2}\)

AB = 4.65 cm

**(v)** \(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AB}{DE})^{2}\)

\(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{1.2}{1.4})^{2}\)

\(\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{36}{49})\)

**2. In the fig 4.178, \(\Delta ACB\) is similar to \(\Delta APQ\). If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm, AP = 2.8 cm, find CA and AQ. Also, find the Area of \(\Delta ACB\): Area of \(\Delta APQ\).**

**Answer:**

Given: \(\Delta ACB\)

BC = 10 cm

PQ = 5 cm

BA = 6.5 cm

AP = 2.8 cm

**Find:**

**(1) CA and AQ**

**(2) Area of \(\Delta ACB\): Area of \(\Delta APQ\)**

**(1)** It is given that \(\Delta ACB\)

We know that for any two similar triangles the sides are proportional. Hence

\(\frac{AB}{AQ} = \frac{BC}{PQ} = \frac{AC}{AP}\)

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

\(\frac{6.5}{AQ} = \frac{10}{5}\)

AQ = 3.25 cm

Similarly,

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

\(\frac{CA}{2.8} = \frac{10}{5}\)

CA = 5.6 cm

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

\(\frac{ar \Delta ACQ}{ar \Delta APQ} = (\frac{BC}{PQ})^{2}\)

= \( (\frac{10}{5})^{2}\)

= \( (\frac{2}{1})^{2}\)

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

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**3. The areas of two similar triangles are 81 cmÂ² and 49 cmÂ² respectively. Find the ration of their corresponding heights. What is the ratio of their corresponding medians?**

**Answer:**

Given: The area of two similar triangles is 81cm^{2} and 49cm^{2} respectively.

To find:

**(1)** The ratio of their corresponding heights.

**(2)** The ratio of their corresponding medians.

**(1)** We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

\(\frac{ar\;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{altitude \; 1}{altitude \; 2})^{2}\)

\(\frac{81}{49} = (\frac{altitude \; 1}{altitude \; 2})^{2}\)

Taking square root on both sides, we get

\(\frac{9}{7} = \frac{altitude \; 1}{altitude \; 2}\)

Altitude 1: altitude 2 = 9: 7

**(2)** We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their medians.

\(\frac{ar\;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{median \; 1}{median \; 2})^{2}\)

\(\frac{81}{49} = (\frac{median \; 1}{median \; 2})^{2}\)

Taking square root on both sides, we get

\(\frac{9}{7} = \frac{median \; 1}{median \; 2}\)

Median 1: median 2 = 9: 7

**4. The areas of two similar triangles are 169 ****cm ^{2}and 121 cm^{2}respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.**

**Answer:**

Given:

The area of two similar triangles is 169cm2 and 121cm2 respectively. The longest side of the larger triangle is 26cm.

To find:

Longest side of the smaller triangle

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

\(\frac{ar \;(larger \; triangle)}{ar \;(smaller \;triangle)} = (\frac{side \; of \; the \; larger \; triangle\;}{side \; of \; the \; smaller \; triangle\;})^{2}\)

\(\frac{169}{121} = (\frac{side \; of \; the \; larger \; triangle\;}{side \; of \; the \; smaller \; triangle\;})^{2}\)

Taking square root on both sides, we get

\(\frac{13}{11} = \frac{side \; of \; the \; larger \; triangle\;}{side \; of \; the \; smaller \; triangle\;}\)

\(\frac{13}{11} = \frac{ 26 }{side \; of \; the \; smaller \; triangle\;}\)

Side of the smaller triangle = \(\frac{11 \; \times \;26}{13}\)

Hence, the longest side of the smaller triangle is 22 cm.

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**5. The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.**

**Answer:**

Given:

The area of two similar triangles is 25cm^{2} and 36cm^{2} respectively. If the altitude of first triangle 2.4cm.

To find:

The altitude of the other triangle

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

\(\frac{ar\;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{altitude \; 1}{altitude \; 2})^{2}\)

\(\frac{ 25 }{ 36 } = (\frac{ 2.4 }{altitude \; 2})^{2}\)

Taking square root on both sides, we get

\(\frac{ 5 }{ 6 } = \frac{ 2.4 }{altitude \; 2}\)

Altitude 2 = 2.88 cm

Hence, the corresponding altitude of the other is 2.88 cm.

**6. ABC is a triangle in which ****âˆ **** A = 90Â°, AN âŠ¥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the areas of \(\Delta ANC\) and \(\Delta ABC\).**

**Answer:**

Given:

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively.

To find:

Ratio of areas of triangle

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

\(\frac{ar\;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{altitude \; 1}{altitude \; 2})^{2}\)

\(\frac{ar \;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{ 6 }{ 9 })^{2}\)

\(\frac{ar \;(triangle \; 1)}{ar \;(triangle \; 2)} = \frac{ 36 }{ 81}\)

\(\frac{ar \;(triangle \; 1)}{ar \;(triangle \; 2)} = \frac{ 4 }{ 9 }\)

ar (triangle 1): ar (triangle 2) = 4: 9

Hence, the ratio of the areas of two triangles is 4: 9.

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**7. ABC is a triangle in which \(\angle A ^{\circ}, \, AN \perp BC\), BC = 12 cm and AC = 5 cm. Find the ratio of the areas of \(\Delta ANC \, and \, \Delta ABC\).**

**Answer:**

Given:

In \(\Delta ABC\)

To find:

Ratio of the triangles \(\Delta ANC \; and \;\Delta ABC\)

In \(\Delta ANC \; and \;\Delta ABC\)

\(\angle ACN = \angle ACB\)

\(\angle A = \angle ANC\)

Therefore, \(\Delta ANC \; – \;\Delta ABC\)

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

Therefore,

\(\frac{Ar ( \Delta ANC )}{Ar( \Delta ABC )} = (\frac{ AC }{ BC })^{2}\)

\(\frac{Ar ( \Delta ANC )}{Ar( \Delta ABC )} = (\frac{ 5 cm }{ 12 cm })^{2}\)

\(\frac{Ar ( \Delta ANC )}{Ar( \Delta ABC )} = \frac{ 25 }{ 144 }\)

**8. In Fig, DE || BC**

**(i) If DE = 4m, BC = 6 cm and Area \(\left ( \Delta ADE \right ) = 16 \, cm^{2}\), find the area of \(\Delta ABC\).**

**(ii) If DE = 4cm, BC = 8 cm and Area \(\left ( \Delta ADE \right ) = 25 \, cm^{2}\), find the area of \(\Delta ABC\).**

**(iii) If DE : BC = 3 : 5. Calculate the ratio of the areas of \(\Delta ADE\) and the trapezium BCED.**

**Answer:**

In the given figure, we have DE \(\parallel\)

In \(\Delta ADE \; and \;\Delta ABC\)

\(\angle ADE = \angle B\)

\(\angle DAE = \angle BAC\)

So, \(\Delta ADE \; and \;\Delta ABC\)

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

Hence,

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{DE^{ 2 }}{BC^{ 2 }}\)

\(\frac{16}{Ar (\Delta ABC)} = \frac{4^{ 2 }}{6^{ 2 }}\)

\(Ar (\Delta ABC) = \frac{6^{ 2 }\; \times \;16}{4^{ 2 }}\)

\(Ar ( \Delta ABC)\)^{2}

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

Hence,

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{DE^{ 2 }}{BC^{ 2 }}\)

\(\frac{25}{Ar(\Delta ABC)} = \frac{4^{ 2 }}{8^{ 2 }}\)

\(Ar (\Delta ABC) = \frac{8^{ 2 }\; \times \; 25}{4^{ 2 }}\)

\(Ar ( \Delta ABC)\)^{2}

(iii) We know that

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{DE^{ 2 }}{BC^{ 2 }}\)

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{3^{ 2 }}{5^{ 2 }}\)

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{ 9 }{ 25 }\)

Let the area of \(\Delta ADE\)

area of \(\Delta ABC\)

Now, \(\frac{Ar \left ( \Delta ADE \right )}{ Ar \left ( trap BCED \right )} = \frac{9x}{16x}\)

\(\frac{Ar \left ( \Delta ADE \right )}{ Ar \left ( trap BCED \right )} = \frac{9}{16}\)

**9. In \(\Delta ABC\) , D and E are the mid- points of AB and AC respectively. Find the ratio of the areas \(\Delta ADE\) and \(\Delta ABC\) . **

**Answer:**

Given:

In \(\Delta ABC\)

To find:

Ratio of the areas of \(\Delta ADE \; and \;\Delta ABC\)

It is given that D and E are the midpoints of AB and AC respectively.

Therefore, DE II BC (Converse of mid-point theorem)

Also, DE = \(\frac{1}{2}\)

In \(\Delta ADE \; and \;\Delta ABC\)

\(\angle ADE = \angle B\)

\(\angle DAE = \angle BAC\)

So, \(\Delta ADE \; – \;\Delta ABC\)

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

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{AD^{ 2 }}{AB^{ 2 }}\)

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{1^{ 2 }}{2^{ 2 }}\)

\(\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{ 1 }{ 4 }\)

**10. The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude of the bigger triangles is 5 cm, find the corresponding altitude of the other.**

**Answer: **

Given: the area of the two similar triangles is \(\ 100cm^{2}\)

To find: their corresponding altitude of the other triangle

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

\(\frac{ar(bigger \, triangle 1)}{Ar(triangle2)}\)

\((\frac{100}{49})\)

Taking squares on both the sides

\((\frac{10}{7}\)

Altitude 2=3.5cm

**11. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.**

**Answer:**

Given : the area of the two triangles is \(\ 121cm^{2}\)

To find the corresponding medians of the other triangle

We know that ratio of the areas of the two similar triangles are equal to the ratio of the squares of their merdians

\((\frac{ar(triangle1)}{ar(triangle2)}\)

\((\frac{121}{64}\)

Taking the squareroot on the both sides

\((\frac{11}{8}\)

Median2=8.8cm

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**12. If \(\Delta ABC \sim \Delta DEF\) such that AB = 5cm, area (\(\Delta ABC \)) = 20 cm2 and area (\(\Delta DEF \)) = 45 cm2, determine DE.**

**Answer:**

Given : the area of the two similar \(\Delta ABC\)

To measure of DE

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

\(\frac{Ar\Delta ABC}{Ar\Delta DEF}=(\frac{AB}{DE})^{2}\)

\(\frac{20}{45}\)

\(\frac{20}{45}\)

\({DE}^{2}\)

\({DE}^{2}\)

DE=7.5cm

**13. In \(\Delta ABC \), PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides \(\Delta ABC \) into twoÂ parts equal in area. Find \(\frac{ BP }{ AB }\).**

**Answer:**

Given: in \(\Delta ABC\)

To find : \(\frac{BP}{AB}\)

We have \(PQ\parallel BC\)

Ar(\(\Delta APQ\)

Ar(\(\Delta APQ\)

2(Ar(\(\Delta APQ\)

Now \(PQ\parallel BC\)

In \(\Delta ABC\)

\(\angle APQ =\angle B\)

\(\angle PAQ =\angle BAC\)

In \(\Delta ABC \sim\Delta APQ\)

We know that the ratio of the areas of the two similar triangles is used and is equal to the ratio of their squares of the corresponding sides.

Hence

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}=(\frac{AP}{AB})^{2}\)

\(\frac{Ar\Delta APQ}{2Ar\Delta ABC}=(\frac{AP}{AB})^{2}\)

\(\frac{1}{2}=\frac({AP}{AB})^{2}\)

\(\sqrt{\frac{1}{2}}=\frac({AP}{AB})\)

AB=\(\sqrt{2AP}\)

AB=\(\sqrt{2(AB-BP}\)

\(\sqrt{2BP}\)

\(\frac{BP}{AB}\)

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**14. The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.**

**Answer: **

Given: the areas of the two similar triangles ABC and PQR are in the ratio 9:16. BC=4.5cm

To find: Length of QR

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

\(\frac{Ar\Delta ABC}{Ar\Delta PQR}=(\frac{BC}{QR})^{2}\)

\(\frac{9}{16}\)

\(\frac{3}{4}\)

QR=\(\frac{18}{3}\)

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**15. ABC is a triangle and PQ is a straight line meeting AB and P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 m, prove that area of \(\Delta APQ \) is one – sixteenth of the area of \(\Delta ABC \).**

**Answer: **

Given : in \(\Delta ABC\)

To find Ar(\(\Delta APQ\)

In \(\Delta ABC\)

\(\frac{AP}{PB}\)

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

According to converse of basic proportional theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence,

\(PQ\parallel BC\)

Hence in \(\Delta ABC\)

\(\angle APQ\)

\(\angle PAQ\)

\(\Delta ABC \sim \Delta APQ\)

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}=\frac({AP}{AB})^{2}\)

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}=(\frac{AP}{(AB+BP)})^{2}\)

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}=(\frac{1}{4})^{2}\)

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}=(\frac{1}{16})\)

**Â **

**16. Â If D is a point on the side AB of \(\Delta ABC \)Â such that AD : DB = 3 : 2 and E is a point on BC such that DE || AC. Find the ratio of areas of \(\Delta ABCÂ \) and \(\Delta BDE \).**

**Answer: **

Given In \(\Delta ABC\)

To find

\(\frac{\Delta ABC}{\Delta BDE}\)

In \(\Delta ABC\)

\(\angle BDE\)

\(\angle DBE\)

\(\Delta ABC \sim \Delta BDE\)

We know that the ratio of the two similar triangles is equal to the ratio of the squares of their corresponding sides

Let AD=2x and BD =3x

Hence

\(\frac{Ar\Delta ABC}{Ar\Delta BDE}=(\frac{AB}{BD})^{2}\)

\((\frac{AB+DA}{BD})^{2}\)

\((\frac{3x+2x}{2x})^{2}\)

\(\frac{Ar\Delta ABC}{Ar\Delta BDE}=(\frac{25}{4})\)

**17. If \(\Delta ABC \) and \(\Delta BDE \) are equilateral triangles, where D is the midpoint of BC, find the ratio of areas of \(\Delta ABC \)Â and \(\Delta BDE \)Â .**

**Answer: **

Given InÂ \(\Delta ABC\)

To find \(\frac{Ar\Delta ABC}{Ar\Delta BDE} \)

In \(\Delta ABC\)

\(\Delta ABC Â \sim \Delta BDE\)

Since D is the mid point of BC, BD : DC=1

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

Let DC=x, and BD= x

Hence

\(\frac{Ar\Delta ABC}{Ar\Delta BDE}=(\frac{BC}{BD})^{2}\)

=\((\frac{BD+DA}{DC})^{2}\)

=\((\frac{1x+1x}{1x})^{2}\)

\(\frac{Ar\Delta ABC}{Ar\Delta BDE}= 4:1 \)

**18.Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights. **

**Answer:**

Given:

Two isosceles triangles have equal vertical angles and their areas are in the ratio of 36: 25.

To find:

Ratio of their corresponding heights

Suppose \(\Delta ABC \)

Therefore,

\(\frac{AB}{AC} = \frac{PQ}{PR}\)

In \(\Delta ABC\)

\(\angle A = \angle P\)

\(\frac{AB}{AC} = \frac{PQ}{PR}\)

âˆ´ \(\Delta ABC\)

Let AD and PS be the altitudes of \(\Delta ABC\)

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

\(\frac{ar \; \Delta ABC}{ar \; \Delta PQR} = (\frac{AD}{PS})^{2}\)

\(\frac{ 36 }{ 25 } = (\frac{AD}{PS})^{2}\)

\(\frac{ AD }{ PS } = \frac{ 6 }{ 5 }\)

Hence, the ratio of their corresponding heights is 6: 5.

**19. In the given figure. \(\Delta ABC\) and \(\Delta DBC\) are on the same base BC. If AD and BC intersect at O, Prove that**

**Â \(\frac{Area\: of\; \left (\Delta ABC \right ) }{Area\: of\; \left (\Delta DBC \right )} = \frac{AO}{DO}\)**

**Â **

**Answer:**

Given \(\Delta ABC\)

Prove that :\(\frac{Ar\Delta ABC}{Ar\Delta DBC}=\frac{AO}{DO}\)

\(AL \perp BC and DM \perp BC\)

Now, in \(\Delta ALO\)

\(\angle ALO\)

\(\angle AOL\)

ThereforeÂ \(\Delta ALO \sim\Delta DMO\)

âˆ´ \(\frac{AL}{DM} = \frac{AO}{DO}\)

\(\frac{Ar\Delta ABC}{Ar\Delta BCD} = \frac{\frac{1}{2}\times BC\times AL}{\frac{1}{2}\times BC\times DM}\)

=\(\frac{AL}{DM}\)

=\(\frac{AO}{DO}\)

**20. ABCD is a trapezium in which AB || CD. The DiagonalÂ AC and BC intersect at O. Prove that :**

**(i) Â \(\Delta AOB ~ \Delta COD \)**

**(ii) If OA = 6 cm, OC = 8 cm,**

**Find:**

**(a) \(\frac{Area\: of\; \left (\Delta AOB \right ) }{Area\: of\; \left (\Delta COD \right )}\)**

**(b) \(\frac{Area\: of\; \left (\Delta AOD \right ) }{Area\: of\; \left (\Delta COD \right )}\)**

**Â **

**Answer: **Given ABCD is the trapezium which \(AB\parallel CD\)

The diagonals AC and BD intersect at o.

To prove:

(i) \(\Delta AOB \sim\Delta COD\)

(ii) If OA = 6 cm, OC = 8 cm

To find :

(a)\(\frac{Ar\Delta AOB}{Ar\Delta COD}\)

(b)\(\frac{Ar\Delta AOD}{Ar\Delta COD}\)

Construction : Draw a line MN passing through O and parallel to AB and CD

Now in \(\Delta AOB and \Delta COD \)

(i) Now in \(\angle OAB\)

(ii) \(\angle OBA\)

\(\angle AOB\)

\(\Delta AOB \sim\Delta COD\)

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

\(\frac{Ar\Delta AOB}{Ar\Delta COD}=(\frac{AO}{CO})^{2}\)

=\((\frac{6}{8})^{2}\)

\(\frac{Ar\Delta AOB}{Ar\Delta COD}\)

b) We know that the ratio of two similar triangles is equal to the artio of their corresponding

sides.

\(\frac{Ar\Delta AOB}{Ar\Delta COD}=(\frac{AO}{CO})^{2}\)

=\((\frac{6 cm}{8 cm})^{2}\)

**21. In \(\Delta ABC\), P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of \(\Delta APQ \) and trapezium BPQC.**

**Answer: **Â Given : In \(\Delta ABC\)

To find : The ratio of the areas of \(\Delta APQ\)

In \(\Delta APQ\)

\(\angle APQ =\angle B\)

\(\angle PAQ =\angle BAC\)

So, \(\Delta APQ \sim\Delta ABC\)

We know that the ratio of areas of the twosimilar triangles is equal to the ratio of the squares of their corresponding sides.

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}=(\frac{AP}{AB})^{2}\)

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}= \frac{1x^{2}}{(1x+2x)^{2}}\)

\(\frac{Ar\Delta APQ}{Ar\Delta ABC}= \frac{1}{9}\)

Let Area of \(\Delta APQ\)

Ar[trapBCED]=Ar(\(\Delta ABC\)

=9x-1x

=8x sq units

Now,

\(\frac{Ar\Delta APQ}{Ar( trapBCED)} \)