NCERT Class 10 Maths Solutions Chapter - 6 Triangles, Ex - 6.4

Ex - 6.4

Question 1.  Let ABC ~ DEF their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Solution



Question 2.  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.

Solution


Since AB || CD
OAB = OCD            (Alternate interior angles)
OBA = ODC            (Alternate interior angles)
AOB = COD            (Vertically opposite angles)
Therefore AOB ~ COD        (By AAA rule)


Question 3.  In figure 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that 


Solution


Since ABC and DBC are one same base,
Therefore ratio between their areas will be as ratio of their heights.
Let us draw two perpendiculars AP and DM on line BC.

In APO and DMO,
APO = DMO = 90�
AOP = DOM          (vertically opposite angles)
OAP = ODM         (remaining angle) 
Therefore APO ~  DMO    (By AAA rule)


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

Solution

Let us assume two similar triangles as ABC ~ PQR


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

Solution


Since D and E are mid points of ABC


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


Let us assume two similar triangles as ABC ~ PQR. Let AD and PS be the   medians of these triangles.

A = P, B = Q, C = R
Since, AD and PS are medians


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 ABCD be a square of side a.
Therefore its diagonal      
Two desired equilateral triangles are formed as ABE and DBF
Side of an equilateral triangle ABE described on one of its side = a
Side of an equilateral triangle DBF described on one of its diagonal 
We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.


Question 8.  ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A)    2 : 1
(B)    1 : 2
(C)    4 : 1
(D)    1 : 4

Solution


We know that equilateral triangles are having all its angles as 60º and all its sides of same length. So, all equilateral triangles are similar to each other. So, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.
Let side of ABC = x
 Hence, (c)

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

If, two triangles are similar to each other, ratio between areas of these triangles will be equal to the square of the ratio between sides of these triangles.
Given that sides are in the ratio 4:9.


 
Hence, (d).

Ex-6.5