NCERT Solutions for Class 10 Maths Chapter 6: Triangles - Exercise 6.4

EXERCISE NO: 6.4

1. Let \[\Delta \mathbf{ABC\sim}\Delta \mathbf{DEF}\] and their areas be, respectively, $\mathbf{64}\ \mathbf{c}{{\mathbf{m}}^{\mathbf{2}}}$ and$\mathbf{121}\,\,\mathbf{c}{{\mathbf{m}}^{\mathbf{2}}}$. If $\mathbf{EF}=\mathbf{15}.\mathbf{4}$ cm, find $\mathbf{BC}$.

Ans: We are provided that, \[\Delta ABC\text{ }\sim \Delta DEF\].

Therefore, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DEF \right)}={{\left( \frac{AB}{DE} \right)}^{2}}={{\left( \frac{BC}{EF} \right)}^{2}}=\left( \frac{AC}{DF} \right)$

Also, it is provided that,

$EF=15.4$ cm,

$area\left( \Delta ABC \right)=64\ c{{m}^{2}}$,

$area\left( \Delta DEF \right)=121\,c{{m}^{2}}$.

Thus, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DEF \right)}={{\left( \frac{BC}{EF} \right)}^{2}}$

Therefore, $\left( \frac{64}{121} \right)=\frac{B{{C}^{2}}}{15.4}$

$\Rightarrow \frac{BC}{15.4}=\frac{8}{11}$ cm

$\Rightarrow BC=\frac{8\times 15.4}{11}=11.2$ cm

2. Diagonals of a trapezium $\mathbf{ABCD}$ with $\mathbf{AB}||\mathbf{DC}$ intersect each other at the point $\mathbf{O}$. If \[\mathbf{AB}=\mathbf{2CD}\], find the ratio of the areas of triangles $\mathbf{AOB}$ and $\mathbf{COD}$.

Ans: 

It is provided that, \[AB||CD\].

Therefore, \[\angle OAB=\angle OCD\] and \[\angle OBA=\angle ODC\] (since, these are alternate interior angles)

Now, for the triangles \[\Delta AOB\] and \[\Delta COD\], we have

\[\angle AOB=\angle COD\] (Since, these are vertically opposite angles)

Similarly,

\[\angle OAB=\angle OCD\] 

\[\angle OBA=\angle ODC\]

Hence, \[\Delta AOB\sim \Delta COD\] (According to the AAA similarity property)

3. In the following figure, $\mathbf{ABC}$ and $\mathbf{DBC}$ are two triangles on the same base $\mathbf{BC}$. If $\mathbf{AD}$ intersects $\mathbf{BC}$ at $\mathbf{O}$, show that $\frac{\mathbf{area}\left( \mathbf{\Delta ABC} \right)}{\mathbf{area}\left( \mathbf{\Delta DBC} \right)}=\frac{\mathbf{AO}}{\mathbf{DO}}$. 

Ans: On the line$BC$, sketch two perpendiculars, \[AP\] and \[DM\].

Recall that, area of a triangle $=\frac{1}{2}\times Base\times height$

Therefore, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DBC \right)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times DM}=\frac{AP}{DM}$

For the triangle, \[\Delta APO\] and \[\Delta DMO\], we have

\[\angle APO=\angle DMO\] (both the angles are of ${{90}^{\circ }}$)

\[\angle AOP=\angle DOM\] (These angles are vertically opposite)

Thus, \[\Delta APO\sim \Delta DMO\] (According to the $AA$ similarity property)

Therefore, $\frac{AP}{DM}=\frac{AO}{DO}$

Hence, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DBC \right)}=\frac{AO}{DO}$

4. If the areas of two similar triangles are equal, prove that they are congruent.

Ans:

Suppose that, \[\Delta ABC\sim \Delta PQR\] are two similar triangles.

Then, we have

$\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}={{\left( \frac{AB}{PQ} \right)}^{2}}={{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}$                        …… (i)

Also, it is provided that, $area\left( \Delta ABC \right)=area\left( \Delta PQR \right)$.

This implies, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}=1$                                           …… (ii)

Equating the equations (i) and (ii), it yields

$1={{\left( \frac{AB}{PQ} \right)}^{2}}={{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}$

Thus, \[AB=PQ\] , \[BC=QR\], and \[AC=PR\].

Hence,\[\Delta ABC\cong \Delta PQR\] (According to the $SSS$ congruence property)

5. $\mathbf{D},\,\,\mathbf{E}$ and $\mathbf{F}$ are respectively the mid-points of sides $\mathbf{AB},\,\,\mathbf{BC}$ and $\mathbf{CA}$ of $\mathbf{\Delta }\,\mathbf{ABC}$. Find the ratio of the area of $\mathbf{\Delta DEF}$ and $\mathbf{\Delta ABC}$.

Ans:

It is given that, $D$ is the mid-point of $AB$ and $E$ is the mid-point of $BC$ in the triangle $\Delta ABC$.

Therefore, $DE||AC$ and $DE=\frac{1}{2}AC$.

Again, for the triangles $\Delta BED$ and $\Delta BCA$, we have

$\angle BED=\angle BCA$ (corresponding angles)

Similarly, 

$\angle BDE=\angle BAC$ and 

$\angle EBA=\angle CBA$.

Therefore, $\Delta BED\sim \Delta BCA$ (According to the $AAA$ similarity property)

Thus, we have

$\frac{area\left( \Delta BED \right)}{area\left( \Delta BCA \right)}={{\left( \frac{DE}{AC} \right)}^{2}}$

$\Rightarrow \frac{area\left( \Delta BED \right)}{area\left( \Delta BCA \right)}=\frac{1}{4}$

$\Rightarrow area\left( \Delta BED \right)=\frac{1}{4}area\left( \Delta BCA \right)$

In a similar way, 

$area\left( \Delta CFE \right)=\frac{1}{4}area\left( \Delta CBA \right)$ and

$area\left( \Delta ADF \right)=\frac{1}{4}area\left( \Delta ABA \right)$.

Again, 

$area\left( \Delta DEF \right)=area\left( \Delta ABC \right)-\left[ area\left( \Delta BED \right)+area\left( \Delta CFE \right)+area\left( \Delta ADF \right) \right]$

$\Rightarrow area\left( \Delta DEF \right)=area\left( \Delta ABC \right)-\frac{3}{4}area\left( \Delta ABC \right)$

$\Rightarrow \frac{area\left( \Delta DEF \right)}{area\left( \Delta ABC \right)}=\frac{1}{4}$

6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding mediAns:

Ans:

Suppose that, $\Delta ABC\sim \Delta PQR$ are two similar triangles. Also suppose that, $AD$, $PS$ are the medians of the triangle $\Delta ABC$ and $\Delta PQR$ respectively.

Therefore, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$                        …… (i)

$\angle A=\angle P$, $\angle B=\angle Q$, and $\angle C=\angle R$         …… (ii)

Also, we have

$BD=DC=\frac{BC}{2}$,                                           …… (iii)

$QS=SR=\frac{QR}{2}$                                             ……. (iv)

From the equations (i), (ii), and (iii), we obtain

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}$                                          …… (v)

For the triangles $\Delta ABD$ and $\Delta PQS$, we have

$\angle B=\angle Q$ [by equation (ii)]

Also, $\frac{AB}{PQ}=\frac{BD}{QS}$ [by equation (v)]

Therefore, $\Delta ABD\sim \Delta PQS$ (According to the $SAS$ similarity property)

Thus, we can write that,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}$                                           …… (vi)

So, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}={{\left( \frac{AB}{PQ} \right)}^{2}}{{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}$.

Then, the equation (i) and (vi) together implies that,

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}$

Thus, we obtain

$\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}={{\left( \frac{AD}{PS} \right)}^{2}}$. 

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Ans: 

In the above figure, \[ABCD\] is a square of side $a$ units.

Then, its diagonal $=\sqrt{2a}$ units.

Suppose that, \[\Delta ABE\] and \[\Delta DBF\] are the two equilateral triangles.

Also, side of the equilateral triangle, \[\Delta ABE\] described on one of its sides $=\sqrt{2a}$ units; and side of the equilateral triangle, \[\Delta DBF\] described on one of its diagonals.

Now, recall that, each of angles of an equilateral triangle is ${{90}^{\circ }}$ and length of all of its sides are the same. So, all equilateral triangles are always similar to each other. Thus, the square of the ratio between the sides of these triangles will be equal to the ratio between their areas.

Hence, $\frac{area\text{ of }\Delta \text{ABE}}{area\text{ of }\Delta \text{DBF}}={{\left( \frac{a}{\sqrt{2}a} \right)}^{2}}=\frac{1}{2}$.

That is, area of $\Delta ABE=\frac{1}{2}\times $ area of $\Delta DBF$.

So, the required result is proved.

8. $\mathbf{ABC}$ and $\mathbf{BDE}$ are two equilateral triangles such that $\mathbf{D}$ is the mid-point of $\mathbf{BC}$. Ratio of the area of triangles $\mathbf{ABC}$ and $\mathbf{BDE}$ is

(a) $\mathbf{2}:\mathbf{1}$

(b) $\mathbf{1}:\mathbf{2}$

(c) $\mathbf{4}:\mathbf{1}$

(d) $\mathbf{1}:\mathbf{4}$

Ans:

Recall that, each of the angles of an equilateral is of ${{90}^{\circ }}$ and length all the sides of it are of equal. So, all equilateral triangles are similar to one another.

Thus, the square of the ratio between the sides of these triangles will be equal to the ratio between their areas.

Now, suppose that, sides of $\Delta ABC$ is of $x$ units.

Then, sides of $\Delta BDE$  is of $\frac{x}{2}$ units.

So, $\frac{area\text{ of  }\Delta \text{ABC}}{area\text{ of }\Delta \text{BDE}}={{\left( \frac{x}{\frac{x}{2}} \right)}^{2}}=\frac{4}{1}$

Hence, the ratio between areas of triangles $ABC$ and $BDE$ is $4:1$.

That is, the correct answer is the option (c).

9. Sides of two similar triangles are in the ratio $\mathbf{4}:\mathbf{9}$. Areas of these triangles are in the ratio 

(a) $\mathbf{2}:\mathbf{3}$

(b) $\mathbf{4}:\mathbf{9}$

(c) $\mathbf{81}:\mathbf{16}$

(d) $\mathbf{16}:\mathbf{81}$

Ans: Recall that, for similar triangles, the square of the ratio between the sides of these triangles will be equal to the ratio between their areas.

We are provided that the ratio of the sides of two similar triangles is $4:9$.

So, areas of these triangles will be in the following ratio 

$={{\left( \frac{4}{9} \right)}^{2}}=\frac{16}{81}$.

That is, $16:81$.

Thus, option (d) is the correct answer.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

Opting for the NCERT solutions for Ex 6.4 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.4 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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