Chapter 6: Triangles

According to the question,

∠A = ∠C,

AB = 6 cm, BP = 15 cm,

AP = 12 cm

CP = 4 cm

From ∆APB and ∆CPD,

∠A = ∠C

∠APB = ∠CPD [vertically opposite angles]

∴ By AAA similarity criteria,

∆APD ∼ ∆CPD

Using basic proportionality theorem,

Considering AP/CP = PB/PD, we get,

Considering, AP/CP = AB/CD

Therefore,

Length of PD = 5 cm

Length of CD = 2 cm

According to the question,

∆ABC ∼ ∆EDF

From property of similar triangle,

We know that, corresponding sides of ∆ABC and ∆EDF are in the same ratio.

AB/ED = AC/EF = BC/DF ….(i)

According to the question,

AB = 5cm, AC = 7cm

DF = 15cm and DE = 12cm

Substituting these values in Equation (i), we get,

Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm

Let a ∆ABC in which a line DE parallel to BC intersects AB at D and AC at E.

To prove DE divides the two sides in the same ratio.

AD/DB = AE/EC

Construction:

Join BE, CD

Draw EF ⊥ AB and DG ⊥ AC.

We know that,

Area of triangle = ½ × base × height

Then,

Since,

∆BDE and ∆DEC lie between the same parallel DE and BC and are on the same base DE.

We have,

area (∆BDE) = area(∆DEC) …..(iii)

From Equation (i), (ii) and (iii),

We get,

AD/DB = AE/EC

Hence proved.

According to the question,

PQRS is a parallelogram,

Therefore, PQ || SR and PS || QR.

Also given, AB || PS.

To prove:

OC || SR

From ∆OPS and OAB,

PS||AB

∠POS = ∠AOB [common angle]

∠OSP = ∠OBA [corresponding angles]

∆OPS ∼ ∆OAB [by AAA similarity criteria]

Then,

Using basic proportionality theorem,

We get,

PS/AB = OS/OB …(i)

From ∆CQR and ∆CAB,

QR || PS || AB

∠QCR = ∠ACB [common angle]

∠CRQ = ∠CBA [corresponding angles]

∆CQR ∼ ∆CAB

Then, by basic proportionality theorem

From Equation (i) and (ii),

Subtracting 1 from L.H.S and R.H.S, we get,

SR || OC [By converse of basic proportionality theorem]

Hence proved.

Let the length of the ladder = AC = 5 m

Let the height of the wall on which ladder is placed = BC = 4m.

From right angled ∆EBD,

Using the Pythagoras Theorem,

ED² = EB² + BD²

(5)² = (EB)² + (14)² [ BD = 1.4]

25 = (EB)² + 1.96

(EB)² = 25 –1.96 = 23.04

EB = √23.04 = 4.8

Now, we have,

EC = EB – BC = 4.8 – 4 = 0.8

Hence, the top of the ladder would slide upwards on the wall by a distance of 0.8 m.

According to the question,

AC⊥CB,

AC = 2x km,

CB = 2 (x + 7) km and AB = 26 km

Thus, we get ∆ ACB right angled at C.

Now, from ∆ACB,

Using Pythagoras Theorem,

AB² = AC² + BC²

⇒ (26)² = (2x)² + {2(x + 7)}²

⇒ 676 = 4x² + 4(x² + 196 + 14x)

⇒ 676 = 4x² + 4x² + 196 + 56x

⇒ 676 = 8² + 56x + 196

⇒ 8x² + 56x – 480 = 0

Dividing the equation by 8, we get,

x² + 7x – 60 = 0

x² + 12x – 5x – 60 = 0

x(x + 12)-5(x + 12) = 0

(x + 12)(x – 5) = 0

∴ x = -12 or x = 5

Since the distance can’t be negative, we neglect x = -12

∴ x = 5

Now,

AC = 2x = 10km

BC = 2(x + 7) = 2(5 + 7) = 24 km

Thus, the distance covered to city B from city A via city C = AC + BC

AC + BC = 10 + 24

= 34 km

Distance covered to city B from city A after the highway was constructed = BA = 26 km

Therefore, the distance saved = 34 – 26 = 8 km.

Let MN = 18 m be the flag pole and its shadow be LM = 9.6 m.

The distance of the top of the pole, N from the far end, L of the shadow is LN.

In right angled ∆LMN,

LN² = LM² + MN² [by Pythagoras theorem]

⇒ LN² = (9.6)² + (18)²

⇒ LN² = 9.216 + 324

⇒ LN² = 416.16

∴ LN = √416.16 = 20.4 m

Hence, the required distance is 20.4 m

To help students master the concepts in the chapter easily, we have offered the free NCERT Exemplar for Class 10 Maths Chapter 6 – Triangles PDF here, which can be downloaded easily. Also, they can make use of exemplar books, notes, question papers and 10th Class Maths NCERT Solutions to get a good grasp of the concepts of each chapter.

Students are advised to check sample papers and previous years’ questions and solve them to get an idea of questions asked from the chapter Triangles in the Class 10 board exam. Also, practise 10th standard Exemplar Solutions for Maths for all chapters. Download BYJU’S – The Learning App and get personalised video lessons to learn about maths concepts, such as triangles.