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Learn Exercise 6.4 with Free Lessons & Tips

Question 1:

Let $\triangle ABC\sim \triangle DEF$ and their areas be, respectively, $64 cm^{2}$ and $121cm^{2}$ . If EF = 15.4 cm, find BC.

Solution 1:

Given $\Delta ABC\sim \Delta DEF$

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

According to the Question,

EF=15.4 cm

$ar\left ( \Delta ABC \right )=64cm^{2}$

$ar(\Delta DEF)=121cm^{2}$

$\therefore \frac{ar\Delta ABC}{ar\Delta DEF}=\left ( \frac{BC}{EF} \right )^{2}$

$\left ( \frac{64 cm^{2}}{121cm^{2}} \right )=\frac{BC^{2}}{(15.4cm^{2})}$

$\frac{BC}{15.4}=\left ( \frac{8}{11} \right ) cm$

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

Comments

Tasneem

Question 2:

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

Solution 2:

Since AB || CD,

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ ΔAOB ∼ ΔCOD (By AAA similarity criterion)

$\therefore \frac{ar\Delta AOB}{ar\Delta COD}=\left ( \frac{AB}{CD} \right )^{2}$

Since AB= 2CD

$\therefore \frac{ar\Delta AOB}{ar\Delta COD}=\left ( \frac{2CD}{CD} \right )^{2}=\frac{4}{1}=4:1$

Comments

Tasneem

Question 3:

In the figure, ABC and DBC are two triangles on the same base BC.

If AD intersects BC at O, show that:

$\frac{ar(ABC)}{ar(DBC)}=\frac{AO}{DO}$

Solution 3:

Let us draw two perpendiculars AP and DM on line BC.\

We know that area of a triangle =

$\therefore \frac{ar\Delta ABC}{ar\Delta DBC}=\left ( \frac{\frac{1}{2}BC\times AP}{\frac{1}{2}BC\times DM} \right )=\frac{AP}{DM}$

n ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ∼ ΔDMO (By AA similarity criterion)

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

$\therefore \frac{ar\Delta ABC}{ar\Delta DBC}=\frac{AO}{DO}$

Comments

Tasneem

Question 4:

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

Solution 4:

Let us assume two similar triangles as ΔABC ∼ ΔPQR.

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

Given that $ar \Delta ABC\sim ar\Delta PQR$

$\therefore \frac{ar\Delta ABC}{ar\Delta PQR}=1$

Substituting in equation (i)

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

$\Rightarrow AB=PQ, BC=QR, AC=PR$

$\therefore\Delta ABC \cong \Delta PQR$ (by SSS congruence criterion)

Comments

Tasneem

Question 5:

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

Solution 5:

D and E are the mid-points of ΔABC.

$\therefore DE\parallel AC$ and $DE=\frac{1}{2}AC$

In $\Delta BED$ and $\Delta BCA,$

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

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

$\angle EBD=\angle CBA$ (Common angles)

$\therefore \Delta BED\sim \Delta BCA$

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

$\therefore \frac{ar\Delta BED}{ar\Delta BCA}=\left ( \frac{1}{4} \right )$

$ar\Delta BED=\frac{1}{4}ar(\Delta BCA)$

Similarly, $ar\Delta CFE=\frac{1}{4}ar(\Delta CBA) and ar\Delta ADF=\frac{1}{4}ar(\Delta ABC)$

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

On substitution, we get

$ar(\Delta DEF)=ar(\Delta ABC)-\frac{3}{4}ar \Delta ABC= \frac{1}{4}ar\Delta ABC$

$\therefore \frac{ar\Delta DEF}{ar\Delta ABC}=\left ( \frac{1}{4} \right )$

Comments

Tasneem

Question 6:

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

Solution 6:

Let us assume two similar triangles as ΔABC ∼ ΔPQR. Let AD and PS be the medians of these triangles.

ΔABC ∼ ΔPQR

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}..............(i)$

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PS are medians,

∴ BD = DC =

And, QS = SR =

Equation (1) becomes

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}..............(iii)$

In ΔABD and ΔPQS,

∠B = ∠Q [Using equation (2)]

And, $\frac{AB}{PQ}=\frac{BD}{QS}$ [Using equation (3)]

∴ ΔABD ∼ ΔPQS (SAS similarity criterion)

Therefore, it can be said that

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}..............(iv)$

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

From equations (1) and (4), we may find that

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

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

Comments

Tasneem

Question 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

Solution 7:

Let ABCD be a square of side a.

Therefore, its diagonal

Two desired equilateral triangles are formed as ΔABE and ΔDBF.

Side of an equilateral triangle, ΔABE, described on one of its sides = a

Side of an equilateral triangle, ΔDBF, described on one of its diagonals

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

$\frac{Area of \Delta ABC}{Area of \Delta DBF}=\left ( \frac{a}{\sqrt{2}a} \right )^{2}=\frac{1}{2}$

Comments

Tasneem

Question 8:

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

Solution 8:

Let side of ΔABC = x

Therefore, side of

$\frac{Area of \Delta ABC}{Area of \Delta BDE}=\left ( \frac{x}{\frac{x}{2}} \right )^{2}=\frac{4}{1}$

Hence, the correct answer is (C).

Comments

Tasneem

Question 9:

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

Solution 9:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangles = $\left ( \frac{4}{9}^{2} \right )=\frac{16}{81}$

Hence, the correct answer is (D).

Comments

Tasneem

All Exercises - Chapter 6 - Triangles

Exercise 6.1

Exercise 6.2

Exercise 6.3

Exercise 6.4

Exercise 6.5

Exercise 6.6

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Learn Exercise 6.4 with Free Lessons & Tips

All Exercises - Chapter 6 - Triangles

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