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

Q2 Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Answer:

Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.

AB = 2 CD ( Given )

In $\triangle AOB\, and\, \triangle COD,$

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

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

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

$\therefore \triangle AOB\, \sim \, \triangle COD$ (AAA similarity)

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{AB^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{(2CD)^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4.CD^2}{CD^2}$

$\Rightarrow \frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4}{1}$

$\Rightarrow ar(\triangle AOB)=ar(\triangle COD)=4:1$

Q3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that $\frac{ar (ABC)}{ar ( DBC )} = \frac{AO}{DO}$

Answer:

Let DM and AP be perpendicular on BC.

$area\,\,of\,\,triangle=\frac{1}{2}\times base\times perpendicular$

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

In $\triangle APO\, and\, \triangle DMO,$

$\angle APO=\angle DMO$ (Each $90 ^\circ$ )

$\angle AOP=\angle MOD$ (Vertically opposite angles)

$\triangle APO\, \sim \, \triangle DMO,$ (AA similarity)

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

Since

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

$\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{AP}{ MD}=\frac{AO}{DO}$

Q4 If the areas of two similar triangles are equal, prove that they are congruent.

Answer:

Let $\triangle ABC\, \sim \, \triangle DEF,$ , therefore,

$ar(\triangle ABC\,) = \,ar( \triangle DEF)$ (Given )

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}................................1$

$\therefore \frac{ar(\triangle ABC)}{ar(\triangle DEF)}=1$

$\Rightarrow \frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}=1$

$AB=DE$

$BC=EF$

$AC=DF$

$\triangle ABC\, \cong \, \triangle DEF$ (SSS )

Q5 D, E, and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$ . Find the ratio of the areas of $\Delta DEF \: \:and \: \: \Delta ABC$

Answer:

D, E, and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$ . ( Given )

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

In $\Delta BED \: \:and \: \: \Delta ABC$ ,

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

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

$\Delta BED \: \:\sim \: \: \Delta ABC$ (By AA)

$\frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{DE^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{(\frac{1}{2}AC)^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{1}{4}$

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times ar(\triangle ABC)$

Let ${ar(\triangle ABC)$ be x.

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times x$

Similarly,

$\Rightarrow ar(\triangle CEF)=\frac{1}{4}\times x$ and $\Rightarrow ar(\triangle ADF)=\frac{1}{4}\times x$

$ar(\triangle ABC)=ar(\triangle ADF)+ar(\triangle BED)+ar(\triangle CEF)+ar(\triangle DEF)$

$\Rightarrow x=\frac{x}{4}+\frac{x}{4}+\frac{x}{4}+ar(\triangle DEF)$

$\Rightarrow x=\frac{3x}{4}+ar(\triangle DEF)$

$\Rightarrow x-\frac{3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{4x-3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{x}{4}=ar(\triangle DEF)$

$\frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{\frac{x}{4}}{x}$

$\Rightarrow \frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{1}{4}$

Q6 Proves that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer:

Let AD and PS be medians of both similar triangles.

$\triangle ABC\sim \triangle PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}............................1$

$\angle A=\angle P,\angle B=\angle Q,\angle \angle C=\angle R..................2$

$BD=CD=\frac{1}{2}BC\, \, and\, QS=SR=\frac{1}{2}QR$

Purring these value in 1,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}..........................3$

In $\triangle ABD\, and\, \triangle PQS,$

$\angle B=\angle Q$ (proved above)

$\frac{AB}{PQ}=\frac{BD}{QS}$ (proved above)

$\triangle ABD\, \sim \triangle PQS$ (SAS )

Therefore,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}................4$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{AC^2}{PR^2}$

From 1 and 4, we get

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

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AD^2}{PS^2}$

Q7 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.

Answer:

Let ABCD be a square of side units.

Therefore, diagonal = $\sqrt{2}a$

Triangles form on the side and diagonal are $\triangle$ ABE and $\triangle$ DEF, respectively.

Length of each side of triangle ABE = a units

Length of each side of triangle DEF = $\sqrt{2}a$ units

Both the triangles are equilateral triangles with each angle of $60 ^\circ$ .

$\triangle ABE\sim \triangle DBF$ ( By AAA)

Using area theorem,

$\frac{ar(\triangle ABC)}{ar(\triangle DBF)}=(\frac{a}{\sqrt{2}a})^2=\frac{1}{2}$

Q8 Tick the correct answer and justify : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is

(A) 2: 1 (B) 1: 2 (C) 4 : 1 (D) 1: 4

Answer:

Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

All angles of the triangle are $60 ^\circ$ .

$\triangle$ ABC $\sim \triangle$ BDE (By AAA)

Let AB=BC=CA = x

then EB=BD=ED= $\frac{x}{2}$

$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=(\frac{x}{\frac{x}{2}})^2=\frac{4}{1}$

Option C is correct.

Q9 Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

(A) 2 : 3 (B) 4: 9 (C) 81: 16 (D) 16: 81

Answer:

Sides of two similar triangles are in the ratio 4: 9.

Let triangles be ABC and DEF.

We know that

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{4^2}{9^2}=\frac{16}{81}$

Option D is correct.