# RD Sharma Solutions Class 10 Triangles 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.

Solution:

(i)

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we get,

$\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{BC}{EF})^{2}$ $\frac{16}{25} = (\frac{2.3}{EF})^{2}$ $\frac{4}{5} = \frac{2.3}{EF}$

Therefore, EF = 2.875 cm

(ii) $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AB}{DE})^{2}$ $\frac{9}{64} = (\frac{AB}{DE})^{2}$ $\frac{3}{8} = \frac{AB}{5.1}$

Therefore, AB = 1.9125 cm

(iii) $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AC}{DF})^{2}$ $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{19}{8})^{2}$ $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{361}{64})$

Therefore, the ratio of the areas of the two triangles are 361:64

(iv) $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AB}{DE})^{2}$ $\frac{36}{64} = (\frac{AB}{DE})^{2}$ $\frac{6}{8} = \frac{AB}{6.2}$

Therefore, AB = 4.65 cm

(v) $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{AB}{DE})^{2}$ $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{1.2}{1.4})^{2}$ $\frac{ar \Delta ABC}{ar \Delta DEF} = (\frac{36}{49})$

Therefore, the ratio of the areas of the two triangles are 36: 49

1. 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 $\Delta APQ$.

Solution:

Given:

$\Delta ACB$ is similar to $\Delta APQ$

BC = 10 cm

PQ = 5 cm

BA = 6.5 cm

AP = 2.8 cm

To Find: CA, AQ and that the area of $\Delta ACB$: Area of $\Delta APQ$

(i) CA and AQ

Given: $\Delta ACB$ – $\Delta APQ$

Since, for any two similar triangles the sides are proportional.

We know that:

$\frac{AB}{AQ} = \frac{BC}{PQ} = \frac{AC}{AP}$ $\frac{AB}{AQ} = \frac{BC}{PQ}$ $\frac{6.5}{AQ} = \frac{10}{5}$

i.e., AQ = 3.25 cm

Similarly,

$\frac{BC}{PQ} = \frac{CA}{AP}$ $\frac{CA}{2.8} = \frac{10}{5}$

i.e., CA = 5.6 cm

(ii) Area of $\Delta ACB$: Area of $\Delta APQ$

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we get,

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

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

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

= $\frac{4}{1}$

Therefore, the ratio is 4:1

1. 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?

Solution:

Given: The areas of two similar triangles are 81cm2 and 49cm2.

To find: The ratio of their corresponding heights and the ratio of their corresponding medians

(i) The ratio of their corresponding heights.

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes, we get,

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

Taking square root on both sides, we get

$\frac{9}{7} = \frac{altitude \; 1}{altitude \; 2}$

Therefore, the ratio= Altitude 1: altitude 2 = 9: 7

(ii) The ratio of their corresponding medians.

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their medians, we get,

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

Taking square root on both sides, we get

$\frac{9}{7} = \frac{median \; 1}{median \; 2}$

Therefore, the ratio = Median 1: median 2 = 9: 7

1. 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.

Solution:

Given:

The area of two similar triangles is 169cm2 and 121cm2.

The longest side of the larger triangle is 26cm.

To find: the longest side of the smaller triangle

Since, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we get,

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

Taking square roots of LHS and RHS, we get,

$\frac{13}{11} = \frac{side \; of \; the \; larger \; triangle\;}{side \; of \; the \; smaller \; triangle\;}$ $\frac{13}{11} = \frac{ 26 }{side \; of \; the \; smaller \; triangle\;}$

So, we get the side of the smaller triangle = $\frac{11 \; \times \;26}{13}$ = 22 cm

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

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

Solution:

Given:

The areas of two similar triangles are 25cm2 and 36cm2.

Altitude of first triangle = 2.4cm.

To find= the altitude of the other triangle

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes, we have,

$\frac{ar\;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{altitude \; 1}{altitude \; 2})^{2}$ $\frac{ 25 }{ 36 } = (\frac{ 2.4 }{altitude \; 2})^{2}$

Taking square root of both LHS and RHS, we get

$\frac{ 5 }{ 6 } = \frac{ 2.4 }{altitude \; 2}$

Altitude 2 = 2.88 cm

Therefore, the altitude of the other is 2.88 cm.

1. 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 $\Delta ABC$.

Solution:

Given:

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

To find= Ratio of areas of triangle

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes, we get,

$\frac{ar\;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{altitude \; 1}{altitude \; 2})^{2}$ $\frac{ar \;(triangle \; 1)}{ar \;(triangle \; 2)} = (\frac{ 6 }{ 9 })^{2}$ $\frac{ar \;(triangle \; 1)}{ar \;(triangle \; 2)} = \frac{ 36 }{ 81}$ $\frac{ar \;(triangle \; 1)}{ar \;(triangle \; 2)} = \frac{ 4 }{ 9 }$

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

Therefore, the ratio of the areas of two triangles = 4: 9.

1. 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 $\Delta ANC \, and \, \Delta ABC$.

Solution:

Given:

In $\Delta ABC$,

$\angle A = 90^{\circ}$

AN $\perp$ BC

BC= 12 cm

AC = 5 cm.

To find: Ratio of the $\Delta ANC \; to \;\Delta ABC$.

In $\Delta ANC \; and \;\Delta ABC$,

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

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

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

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we get,

$\frac{Ar ( \Delta ANC )}{Ar( \Delta ABC )} = (\frac{ AC }{ BC })^{2}$ $\frac{Ar ( \Delta ANC )}{Ar( \Delta ABC )} = (\frac{ 5 cm }{ 12 cm })^{2}$ $\frac{Ar ( \Delta ANC )}{Ar( \Delta ABC )} = \frac{ 25 }{ 144 }$

Therefore, the ratio= 25:144

1. 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.

Solution:

From figure,

DE $\parallel$ BC.

In $\Delta ADE \; and \;\Delta ABC$

We know that,

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

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

Hence, $\Delta ADE \; and \;\Delta ABC$ (AA Similarity)

(i) Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have,

$\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{DE^{ 2 }}{BC^{ 2 }}$ $\frac{16}{Ar (\Delta ABC)} = \frac{4^{ 2 }}{6^{ 2 }}$ $Ar (\Delta ABC) = \frac{6^{ 2 }\; \times \;16}{4^{ 2 }}$ $Ar ( \Delta ABC)$ = 36 cm2

(ii) Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides, we have,

$\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{DE^{ 2 }}{BC^{ 2 }}$ $\frac{25}{Ar(\Delta ABC)} = \frac{4^{ 2 }}{8^{ 2 }}$ $Ar (\Delta ABC) = \frac{8^{ 2 }\; \times \; 25}{4^{ 2 }}$ $Ar ( \Delta ABC)$ = 100 cm2

(iii) According to the question,

$\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{DE^{ 2 }}{BC^{ 2 }}$ $\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{3^{ 2 }}{5^{ 2 }}$ $\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{ 9 }{ 25 }$

Let the area of $\Delta ADE$ = 9x sq units

area of $\Delta ABC$ = 25x sq units

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

1. In $\Delta ABC$ , D and E are the mid- points of AB and AC respectively. Find the ratio of the areas $\Delta ADE$ and $\Delta ABC$ .

Solution:

Given:

In $\Delta ABC$, D and E are the midpoints of AB and AC respectively.

To find= Ratio of the areas of $\Delta ADE \; : \;\Delta ABC$

Given: D and E are the midpoints of AB and AC respectively.

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

Also, DE = $\frac{1}{2}$BC

From $\Delta ADE \; and \;\Delta ABC$ $\angle ADE = \angle B$(Corresponding angles)

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

Thus, $\Delta ADE \; – \;\Delta ABC$ (AA Similarity)

Since the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides, we get,

$\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{AD^{ 2 }}{AB^{ 2 }}$ $\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{1^{ 2 }}{2^{ 2 }}$ $\frac{Ar (\Delta ADE)}{Ar (\Delta ABC)} = \frac{ 1 }{ 4 }$

Therefore, the ratio of the areas $\Delta ADE$:$\Delta ABC[/latex=1:4 1. 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. Solution: Given: The area of the two similar triangles is \(\ 100cm^{2}$ and $\ 49cm^{2}$. And the altitude of the bigger triangle is 5cm

To find: The corresponding altitude of the other triangle

Since the ratio of the areas of the two similar triangles is equal to the ratio of squares of their corresponding altitudes, we get,

$\frac{ar(bigger \, triangle 1)}{Ar(triangle2)}$=$(\frac{altitude \, of \, the \, bigger \, triangle1}{altitude 2})^{2}$ $(\frac{100}{49})$=$(\frac{5}{altitude 2})^{2}$

Squaring LHS and RHS,

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

Therefore, Altitude 2=3.5cm

1. 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.

Solution:

Given : the area of the two triangles is $\ 121cm^{2}$ and $\ 64cm^{2}$respectively and the median of the first triangle is 12.1cm

To find= the corresponding medians of the other triangle

Since the ratio of the areas of the two similar triangles are equal to the ratio of the squares of their medians, we have,

$(\frac{ar(triangle1)}{ar(triangle2)}$=$(\frac{median1}{median2})^{2}$ $(\frac{121}{64}$=$(\frac{12.1}{median2})^{2}$

Taking the square roots on both LHS and RHS.

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

Therefore, Median 2=8.8cm

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

Solution:

Given : the area of the two similar $\Delta ABC$=$20cm^{2}$ and $\Delta DEF$ =$45cm^{2}$ and AB=5cm

To Find= DE

Since the ratio of the areas of the two similar triangles is equal to the ratio of the squares of their corresponding sides, we have,

$\frac{Ar\Delta ABC}{Ar\Delta DEF}=(\frac{AB}{DE})^{2}$ $\frac{20}{45}$=$\frac{5}{DE}^{2}$ $\frac{20}{45}$=$\frac{5}{DE}$ ${DE}^{2}$=$\frac{25\times45}{20}$ ${DE}^{2}$=$\frac{225}{4}$

Therefore, DE=7.5cm

1. 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 $\frac{ BP }{ AB }$.

Solution:

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

To find : $\frac{BP}{AB}$

According to the question,

$PQ\parallel BC$ and

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

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

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

We know that,

$PQ\parallel BC$ 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)

Since 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, we have,

$\frac{Ar\Delta APQ}{Ar\Delta ABC}=(\frac{AP}{AB})^{2}$ $\frac{Ar\Delta APQ}{2Ar\Delta ABC}=(\frac{AP}{AB})^{2}$ $\frac{1}{2}=\frac({AP}{AB})^{2}$ $\sqrt{\frac{1}{2}}=\frac({AP}{AB})$

AB=$\sqrt{2AP}$

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

1. 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.

Solution:

Given: ratio of the areas of triangles ABC and PQR = 9:16. Also, BC=4.5cm

To find: QR

Since the ratio of the areas of the two similar triangles is equal to the ratio of the squares of their corresponding sides, we have,

$\frac{Ar\Delta ABC}{Ar\Delta PQR}=(\frac{BC}{QR})^{2}$ $\frac{9}{16}$ =$(\frac{4.5}{QR})^{2}$ $\frac{3}{4}$=$\frac{4.5}{QR}$

QR=$\frac{18}{3}$ = 6cm

Therefore, QR = 6cm

1. 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 $\Delta ABC$.

Solution:

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 Prove= 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, we know that, if a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side.

$PQ\parallel BC$

Then,

$\Delta ABC$ 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)

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

Hence Proved.

1.  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 $\Delta BDE$.

Solution:

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

To find = $\frac{\Delta ABC}{\Delta BDE}$

According to the question,

In $\Delta ABC$ and $\Delta BDE$,

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

$\angle DBE$ =$\angle ABC$ $\Delta ABC \sim \Delta BDE$

Since the ratio of the two similar triangles is equal to the ratio of the squares of their corresponding sides, we get

Then,

$\frac{Ar\Delta ABC}{Ar\Delta BDE}=(\frac{AB}{BD})^{2}$ $(\frac{AB+DA}{BD})^{2}$ $(\frac{3x+2x}{2x})^{2}$ $\frac{Ar\Delta ABC}{Ar\Delta BDE}=(\frac{25}{4})$

Therefore, areas of $\Delta ABC$ : $\Delta BDE$= 25:4

1. 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 $\Delta BDE$.

Solution:

Given: D is the midpoint of the side BC in $\Delta ABC$, where  $\Delta ABC$ and $\Delta BDE$ are equilateral triangles.

To find =$\frac{Ar\Delta ABC}{Ar\Delta BDE}$

According to the question,

In $\Delta ABC$, $\Delta BDE$ $\Delta ABC \sim \Delta BDE$ (AAA similarity)

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

Since the ratio of the areas of the two similar triangles is equal to the ratio of squares of their corresponding sides, we have,

DC=x, and BD= x

Then,

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

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

=$(\frac{1x+1x}{1x})^{2}$ $\frac{Ar\Delta ABC}{Ar\Delta BDE}= 4:1$

Therefore, areas of $\Delta ABC$ : $\Delta BDE$=4:1

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

Solution:

Given: Areas of two isosceles triangles having equal vertical angles are in the ratio 36: 25.

To find: Ratio of the corresponding heights of the triangles given

Assume that $\Delta ABC$ and $\Delta PQR$ are two isosceles triangles with $\angle A = \angle P$.

Then,

$\frac{AB}{AC} = \frac{PQ}{PR}$

In $\Delta ABC$ and $\Delta PQR$,

$\angle A = \angle P$ $\frac{AB}{AC} = \frac{PQ}{PR}$

∴ $\Delta ABC$ – $\Delta PQR$ (SAS similarity)

Let AD be the altitude of $\Delta ABC$ and PS be the altitudes of $\Delta PQR$.

Since the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes, we get,

$\frac{ar \; \Delta ABC}{ar \; \Delta PQR} = (\frac{AD}{PS})^{2}$ $\frac{ 36 }{ 25 } = (\frac{AD}{PS})^{2}$ $\frac{ AD }{ PS } = \frac{ 6 }{ 5 }$

Therefore, the ratio of the corresponding heights of the triangles given in the question is 6: 5.

1. In the given figure. $\Delta ABC$ and $\Delta DBC$ are on the same base BC. If AD and BC intersect at O, Prove that

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

Solution:

Given: AD and BC intersect at O and $\Delta ABC$ and $\Delta DBC$ lie on the same BC.

To Prove: $\frac{Ar\Delta ABC}{Ar\Delta DBC}=\frac{AO}{DO}$ $AL \perp BC and DM \perp BC$

According to the question,

From $\Delta ALO$ and $\Delta ALO$

We get,

$\angle ALO$ = $\angle DMO$ =$90^{\circ}$ $\angle AOL$ =$\angle DOM$ (vertically opposite angles)

Hence,

$\Delta ALO \sim\Delta DMO$ $\frac{AL}{DM} = \frac{AO}{DO}$ $\frac{Ar\Delta ABC}{Ar\Delta BCD} = \frac{\frac{1}{2}\times BC\times AL}{\frac{1}{2}\times BC\times DM}$

=$\frac{AL}{DM}$

=$\frac{AO}{DO}$

Hence Proved.

1. 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) $\frac{Area\: of\; \left (\Delta AOD \right ) }{Area\: of\; \left (\Delta COD \right )}$

Solution:

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

From $\Delta AOB and \Delta COD$

(i) According to the question,

We get,

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

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

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

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

1. a) Since the ratio of areas of two triangles is equal to the ratio of squares of their corresponding sides, we get,
$\frac{Ar\Delta AOB}{Ar\Delta COD}=(\frac{AO}{CO})^{2}$

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

1. b) Since the ratio of two similar triangles is equal to the ratio of their corresponding

sides, we get,

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

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

1. 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.

Solution:

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.

According to the question,

From $\Delta APQ$ and $\Delta ABC$ $\angle APQ =\angle B$ (corresponding angles)

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

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

Since the ratio of areas of the two similar triangles is equal to the ratio of the squares of their corresponding sides, we get,

$\frac{Ar\Delta APQ}{Ar\Delta ABC}=(\frac{AP}{AB})^{2}$ $\frac{Ar\Delta APQ}{Ar\Delta ABC}= \frac{1x^{2}}{(1x+2x)^{2}}$ $\frac{Ar\Delta APQ}{Ar\Delta ABC}= \frac{1}{9}$

Let the Area of $\Delta APQ$ = 1x sq. units

Let the Area of $\Delta ABC$ = 9x sq. units

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

Ar[trapBCED]=9x-1x

Ar[trapBCED]=8x sq units

Therefore,

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