Areas of Similar Triangles NCERT Exercise 6.4 Class Ten Mathematics

Triangle

Exercise 6.4

Question 1: Let Δ ABC ∼ Δ DEF and their areas be, respectively, 64 sq cm and 121 sq m. If EF = 15.4 cm, find BC.

Solution: In Δ ABC ∼ Δ DEF;

`text(ar ABC)/text(ar DEF)=(BC^2)/(EF^2)`

Or, `(BC^2)/(15.4^2)=(64)/(121)`

Or, `(BC)/(15.4)=(8)/(11)`

Or, `BC=11.2 cm`


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

similar triangles exercise solution

Solution: In triangles AOB and COD;

∠ AOB = ∠ COD (Opposite angles)

∠ OAC = ∠ OCD (Alternate angles)

Hence; Δ AOB ∼ Δ COD

Hence;

`text(ar AOB)/text(ar COD)=(AB^2)/(CD^2)`

`=((2CD)^2)/(CD^2)=(4CD^2)/(CD)^2=4:1`


Question 3: In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that `text(ar ABC)/text(ar DBC)=(AO)/(DO)`

similar triangles exercise solution

Solution: Let us draw altitudes AM and DN on BC; respectively from A and D

`text(ar ABC)/text(ar DBC)=(1/2xxBCxxAM)/(1/2xxBCxxDN)`

`=(AM)/(DN)`

In ΔAMO and ΔDNO;

∠ AMO = ∠ DNO (Right angle)

∠ AOM = ∠ DON (Opposite angles)

Hence; ΔAMO ∼ ΔDNO

Hence;

`(AM)/(DN)=(AO)/(DO)`

Or, `text(ar ABC)/text(ar DBC)=(AO)/(DO)`

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

Solution: Let us take two triangles ABC and PQR with equal areas.

Then, we have;

`text(ar ABC)/text(ar PQR)=1/1`

In this case;

`(AB^2)/(PQ^2)=(AC^2)/(PR^2)=1/1`

Or, `(AB)/(PQ)=(AC)/(PQ)=1`

Hence; the triangles are congruent.


Question 5: D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas of Δ DEF and Δ ABC.

similar triangles exercise solution

Solution: Since D, E and F are mid points of AB, BC and AC

Hence; ΔBAC ∼ΔDFE

So,

`(DF)/(BC)=(EF)/(AB)=(DE)/(AC)=1/2`

So,

`text(ar DEF)/text(ar ABC)=1^2/2^2=1/4`

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: In case of two similar triangles ABC and PQR;

`text(ar ABC)/text(ar PQR)=(AB^2)/(PQ^2)=(AC^2)/(PR^2)`

Let us assume AD and PM are the medians of these two triangles.

Then;

`(AB^2)/(PQ^2)=(AC^2)/(PR^2)=(AD^2)/(PM^2)`

Or, `text(ar ABC)/text(ar PQR)=(AD^2)/(PM^2)`

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: Let us take a square with side ‘a’

Then the diagonal of square will be `a\sqrt2`

Area of equilateral triangle with side ‘a’

`=sqrt3/4a^2`

Area of equilateral triangle with side `a\sqrt2`

`=sqrt3/4(a\sqrt2)^2`

Ratio of two areas can be given as follows:

`(sqrt3/4xx\a^2)/(sqrt3/4xx2a^2)=1/2`

Proved

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

  1. 2 : 1
  2. 1 : 2
  3. 4 : 1
  4. 1: 4

Solution: (c) 4 : 1

For explanation; refer to question 5

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

  1. 2 : 3
  2. 4 : 9
  3. 81 : 16
  4. 16: 81

Solution: (d) 16 : 91

Note: Areas are in duplicate ratio of corresponding sides in case of similar figures.



Copyright © excellup 2014