Jharkhand Board JAC Class 10 Maths Solutions Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.4

Question 1.

Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Solution :

ΔABC ~ ΔDEF

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 :

In trapezium ABCD, AB || CD and diagonals AC and BD intersect at O.

Then, in ΔAOB and ΔCOD.

∠OAB = ∠OCD (Alternate angles)

∠OBA = ∠ODC (Alternate angles)

∠AOB = ∠COD (Vertically opposite angles)

∴ By AAA criterion, ΔAOB ~ ΔCOD.

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

Solution :

Draw AM ⊥ BC and DN ⊥ BC.

Then, ar (ABC) = \(\frac{1}{2}\) × BC × AM and

ar (DBC) = \(\frac{1}{2}\) × BC × DN

In ΔAMO and ΔDNO

∠AMO = ∠DNO (Right angles)

∠AOM = ∠DON (Vertically opposite angles)

∴ By AA criterion, ΔAMO ~ ΔDNO.

∴ \(\frac{AM}{DN}=\frac{AO}{DO}\) ……………….(2)

From (1) and (2), ar (ABC) / ar (DBC) = \(\frac{AO}{DO}\)

Question 4.

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

Solution :

Given: ΔABC ~ ΔPQR and ar (ABC) = ar (PQR)

To prove : ΔABC ≅ ΔPQR

Proof: ΔABC ~ ΔPQR

∴ AB = PQ, BC = QR and CA = RP

∴ By SSS criterion for congruence of triangles, ΔABC = ΔPOR.

Question 5.

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC.

Solution :

In ΔABC, D, E and F are the mid-points of sides AB, BC and CA respectively.

Then, EF || AB and DE || AC

∴ EF || AD and DE || AF

∴ ADEF is a parallelogram.

∴ ∠A = ΔDEF

(Opposite angles of parallelogram)

Similarly, we can prove that ∠B = ∠EFD and ∠C = ∠EDF

Now, in ΔABC and ΔEFD,

∠A = ∠E, ∠B = ∠F and ∠C = ∠D

∴ By AAA criterion, ΔABC ~ ΔEFD.

∴ ar (DEF) / ar(ABC) = (\(\frac{EF}{AB}\))² ………………(1)

In ΔABC, E and F are the mid-points of BC and CA respectively.

∴ EF = \(\frac{1}{2}\)AB ………………(2)

(Alternately: ADEF is a parallelogram.

∴ EF = AD = \(\frac{1}{2}\)AB)

From (1) and (2),

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 :

Given: ΔABC ~ ΔPQR, AD and PM are medians of triangles ABC and PQR respectively.

Proof :

In ΔABC, AD is a median ∴ BD = \(\frac{1}{2}\)BC

In ΔPQR, PM is a median ∴ QM = \(\frac{1}{2}\)QR.

ΔABC ~ ΔPQR

∴ ∠B = ∠Q and \(\frac{AB}{PQ}=\frac{BC}{QR}\)

∴ ∠ABD = ∠PQM and \(\frac{AB}{PQ}=\frac{BD}{QM}\)

So, by SAS criterion ΔABD ~ ΔPQM.

∴ \(\frac{AB}{PQ}=\frac{AD}{PM}\) ……………(1)

Now, ΔABC ~ ΔPQR

∴ ar (ABC) / ar (PQR) = (\(\frac{AB}{PQ}\))²

∴ ar (ABC) / ar (PQR) = (\(\frac{AD}{PM}\))² [By (1)]

Question 7.

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

Solution :

ABCD is a square. PAB, is an equilateral triangle described on side AB and QAC is an equilateral triangle described on diagonal AC.

In ΔABC, ∠B = 90° and AB = BC (Properties of a square)

Then, AC² = AB² + BC² (Pythagoras theorem)

∴ AC² = AB² + AB²

∴ AC² = 2AB²

∴ \(\frac{\mathrm{AB}^2}{\mathrm{AC}^2}\) = \(\frac{1}{2}\)

∴ (\(\frac{18}{36}\))

ΔPAB is an equilateral triangle.

∴ ∠P = ∠A = ∠B = ∠60°

ΔQAC is an equilateral triangle.

∴ ∠Q = ∠A = ∠C = 60°

Thus, in ΔPAB and ΔQAC,

∠P = ∠Q and ∠A = ∠A and ∠B = ∠C

∴ By AAA criterion, ΔPAB ~ ΔQAC.

∴ ar (PAB) / ar(QAC) = (\(\frac{AB}{AC}\))²

∴ ar (PAB) / ar(QAC) = \(\frac{1}{2}\) [By (1)]

∴ ar (PAB) = \(\frac{1}{2}\)ar (QAC)

Thus, the area of an equilateral triangle described on a side of a square is half the area of an equilateral triangle described on one of its diagonals.

Tick the correct answer and justify:

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 :

The correct answer is (C) 4 : 1.

ΔABC and ΔBDE are equilateral triangles.

Hence, any of their correspondences is a similarity.

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 :

The correct answer is (D) 16 : 81.

By theorem 6.6.

Ratio of areas of two similar triangles

= (Ratio of their corresponding sides)²

= (4 : 9)²

= (\(\frac{4}{9}\))²

= 16 : 81