RD Sharma Solutions Class 10 Triangles Exercise 4.6

RD Sharma Solutions Class 10 Chapter 4 Exercise 4.6

RD Sharma Class 10 Solutions Chapter 4 Ex 4.6 PDF Free Download

Exercise 4.6

 

1. Triangles ABC and DEF are similar.

(i) If area of (\(\Delta ABC\)) = 16 cm2 , area (\(\Delta DEF\)) = 25 cm2 and BC = 2.3 cm, find EF.

(ii) If area (\(\Delta ABC\)) = 9 cm2 , area (\(\Delta DEF\)) = 64 cm2 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 cm2 , area (\(\Delta DEF\)) = 64 cm2 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.

 

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: \(\Delta ACB\) is similar to \(\Delta APQ\)

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

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.

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

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

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

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

 

 

3. The areas of two similar triangles are 81 cm² and 49 cm² 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 81cm2 and 49cm2 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 cm2and 121 cm2respectively. 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 = \(\frac{11 \; \times \;26}{13}\) = 22 cm

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

 

 

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 25cm2 and 36cm2 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 A = 90°, AN ⊥ 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.

 

 

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 \(\Delta ABC\), \(\angle A = 90^{\circ}\), AN \(\perp\) BC, BC= 12 cm and AC = 5 cm.

To find:
Ratio of the triangles \(\Delta ANC \; and \;\Delta ABC\).

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

\(\angle ACN = \angle ACB\) (Common)

\(\angle A = \angle ANC\) (\(90^{\circ}\))

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

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 \(\parallel\) BC.

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

\(\angle ADE = \angle B\) (Corresponding angles)

\(\angle DAE = \angle BAC\) (Common)

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

 

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

Hence,

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

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

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

\(Ar ( \Delta ABC)\) = 36 cm2

 

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

Hence,

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

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

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

\(Ar ( \Delta ABC)\) = 100 cm2

 

(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 \(\Delta ADE\) = 9x sq units

area of \(\Delta ABC\) = 25x sq units

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 \(\Delta ABC\), D and E are the midpoints of AB and AC respectively.

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

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

\(\angle ADE = \angle B\)(Corresponding angles)

\(\angle DAE = \angle BAC\) (common)

So, \(\Delta ADE \; – \;\Delta ABC\) (AA Similarity)

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 \(\ 100cm^{2}\) and \(\ 49cm^{2}\) respectively. If the altitude of the bigger triangle is 5cm

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.

\(\frac{ar(bigger \, triangle 1)}{Ar(triangle2)}\)=\((\frac{altitude \, of \, the \, bigger \, triangle1}{altitude 2})^{2}\)

\((\frac{100}{49})\)=\((\frac{5}{altitude 2})^{2}\)

Taking squares on both the sides

\((\frac{10}{7}\)=\(( \frac{5}{altitude 2})\)

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 \(\ 121cm^{2}\) and \(\ 64cm^{2}\)respectively. If the merdian of the first triangle is 12.1cm

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

\((\frac{ar(triangle1)}{ar(triangle2)}\)=\((\frac{median1}{median2})^{2}\)

\((\frac{121}{64}\)=\((\frac{12.1}{median2})^{2}\)

Taking the squareroot on the both sides

\((\frac{11}{8}\)=\((\frac{12.1}{median2})\)

Median2=8.8cm

 

 

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 \(\Delta ABC\)=\(20cm^{2}\) and \(\Delta DEF\) =\(45cm^{2}\) and AB=5cm

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.

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

 

\(\frac{20}{45}\)=\(\frac{5}{DE}^{2}\)

\(\frac{20}{45}\)=\(\frac{5}{DE}\)

\({DE}^{2}\)=\(\frac{25\times45}{20}\)

\({DE}^{2}\)=\(\frac{225}{4}\)

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 two  parts equal in area. Find \(\frac{ BP }{ AB }\).

 

Answer:

Given: in \(\Delta ABC\), PQ is a line segment intersecting AB at P and AC at  such that \(PQ\parallel BC\) and PQ divides \(\Delta ABC\) in two parts equal in area

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

We have \(PQ\parallel BC\) and

Ar(\(\Delta APQ\)) = Ar(quad BPQC)

Ar(\(\Delta APQ\)) +Ar(\(\Delta APQ\)) =Ar(quad BPQC) +Ar(\(\Delta APQ\))

 

2(Ar(\(\Delta APQ\)) =Ar(\(\Delta ABC\))

Now \(PQ\parallel BC\) and BA is a transversal

In \(\Delta ABC\) and \(\Delta APQ\))

\(\angle APQ =\angle B\)       (corresponding angles)

\(\angle PAQ =\angle BAC\)    (common)

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

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

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

\(\frac{BP}{AB}\)=\(\frac{\sqrt{2}-1}{\sqrt{2}}\)

 

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.

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

\(\frac{9}{16}\) =\((\frac{4.5}{QR})^{2}\)

\(\frac{3}{4}\)=\(\frac{4.5}{QR}\)

 

QR=\(\frac{18}{3}\) = 6cm

 

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 \(\Delta ABC\), PQ is a line segment intersecting AB at P and AC at Q. AP = 1cm , PB = 3cm, AQ= 1.5 cm and QC= 4.5cm

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

In \(\Delta ABC\)

\(\frac{AP}{PB}\)=\(\frac{AQ}{QC}\)

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

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 \(\Delta ABC\) and \(\Delta APQ\)

\(\angle APQ\) =\(\angle B\) (corresponding angles)

\(\angle PAQ\)=\(\angle BAC\) (common)

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

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

 

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

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

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

 

 

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

 

Answer:

Given In \(\Delta ABC\), D is appoint on the side AB such that AD:DB=3:2. E is a point on side BC such that \(DE\parallel AC\)

To find

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

In \(\Delta ABC\),\(\Delta BDE\),

\(\angle BDE\) =\(\angle A\) (corresponding angles)

\(\angle DBE\) =\(\angle ABC\)

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

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 \(\Delta ABC \)  and \(\Delta BDE \)  .

 

Answer:

Given In  \(\Delta ABC\), \(\Delta BDE\) are equilateral triangles. D is the point of BC.

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

In \(\Delta ABC\), \(\Delta BDE\)

\(\Delta ABC  \sim \Delta BDE\) (AAA criteria of similarity all angles of the equilateral triangles are equal)

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 \(\Delta ABC \) and \( \Delta PQR \) are two isosceles triangles with \(\angle A = \angle P\).

Therefore,

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

In \(\Delta ABC\) and \(\Delta PQR\),

\(\angle A = \angle P\)

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

\(\Delta ABC\)\(\Delta PQR\) (SAS similarity)

 

Let AD and PS be the altitudes of \(\Delta ABC\) and \(\Delta PQR\), respectively.

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

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

 

Answer:

Given \(\Delta ABC\) and \(\Delta DBC\) are on the same BC. AD and BC intersect at O.

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

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

Now, in \(\Delta ALO\) and \(\Delta ALO\)we have

\(\angle ALO\) = \(\angle DMO\) =\(90^{\circ}\)

\(\angle AOL\) =\(\angle DOM\) (vertically opposite angles)

Therefore  \(\Delta ALO \sim\Delta DMO\)

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

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

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

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

 

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

 

(i)  \(\Delta AOB ~ \Delta COD \)

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

Find:

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

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

 

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 \(\angle OAB\) =\(\angle OCD\)  (Alternate angles)

(ii) \(\angle OBA\) =\(\angle ODC\)                (Alternate angles)

\(\angle AOB\) =\(\angle COD\) (vertically opposite angle)

\(\Delta AOB \sim\Delta COD\) (A.A.Acrieteria)

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

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

 

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

\(\frac{Ar\Delta AOB}{Ar\Delta COD}\)=\((\frac{6}{8})^{2}\)

 

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

sides.

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

 

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

 

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:  Given : In \(\Delta ABC\) , P divides the side AB such that AP: PB =1:2, Q is a point on AC on such that PQ \(\parallel\) BC

To find : The ratio of the areas of \(\Delta APQ\) and the trapezium BPQC.

In \(\Delta APQ\)  and \(\Delta ABC\)

\(\angle APQ =\angle B\) (corresponding angles)

\(\angle PAQ =\angle BAC\) (common)

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

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 \(\Delta APQ\)  =1 sq. units and Area of \(\Delta ABC\)=9x sq.units

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

=9x-1x

=8x sq units

Now,

\(\frac{Ar\Delta APQ}{Ar( trapBCED)} \) =\(\frac{x sq units}{8x sq units}\) =\(\frac{1}{8}\)

 

 


Practise This Question

Find the zeroes of the following quadratic polynomial:
6x2+x5