NCERT Solutions For Class 10 Chapter – 6 Exercise – 6.4

NCERT Solution For Class 10 Math Chapter – 6 Exercise – 6.4

Q1. Let ∆ABC ∆DEF and their areas be 64 cm² and 121 cm² respectively If EF = 15.4 cm , find BC.

Solution:-

Area(ABC)/Area (DEF) = BC²/EF²

64/121 = BC²/EF²

BC/EF = 8/11

BC = 8/11 X EF

BC = 8/11 X 15.4 cm = 11.2 cm

Q2. Diagonals of 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:-

Ncert solutions class 10 chapter 6-29

In ∆AOB and ∆COD.

∟OAB = ∟OCD ( Alternate interior angles)

∟OBA = ∟ODC ( Alternate interior angles )

By AA similarity

∆AOB – ∆COD

So , Area ∆AOB/Area ∆ COD = (AB/CD)²

= (2/1)² { AB = 2CD}

= 4 :1

Q3. In figure , ABC and DBC are two triangles on the same base BC If AD intersects BC at O , Show that Area (ABC)/Area (DBC) = AO/DO

Solution:-

AL/DM = AO/DO (1) Ncert solutions class 10 chapter 6-31

Area (∆ABC)/Area (∆DBC) = 1/2 X (BC) X (AL)/1/2 X (BC) X (DM)

= AL/DM = AO/DO

Hence , Area (∆ABC)/Area(∆DBC) = AO/D0

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

Solution:-

Area of (∆ABC)/ Area of (∆PQR) = BC²/QR²

= BC²/QR² = 1

= BC²/QR²

BC = QR

AB = PQ and AC = PR

∆ABC ∆PQR [ SSS criterion of congruence]

Q5. 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:-

Ncert solutions class 10 chapter 6-34

Area ( ∆FBD) = Area (∆DEF) (I)

Area(∆AFE) = Area(∆DEF) (II)

And ,

Area(∆ABC) = Area (∆FBD) + Area (∆DEF) + Area ∆AFE) + Area (∆EDC) (IV)

From equation (I) and (ii) and (iii))

Hence , Area(∆DEF): Area(∆ABC) = 1:4

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

Solution:-

Area(∆ABC)/Area(∆DEF) = (AB²/DE²) …………………… (I)

AB/DE = BC/EF = CA/FD …………..(II)

AB/DE = 1/2BC /1/2EF = CD/FD

AB/DE = AM/DN ………….. (III)

∆ABC ∆DEF [ SAS similarity criterion]

AB/DE = AM/DN (III)

Hence proved.

Q7. Prove that the area of an equilateral triangle described on one side of square is equal to half the the area of the equilateral triangle described on one of its diagonals.

Solution:-

Ncert solutions class 10 chapter 6-37