Chapter 6: Triangles

According to the question,

In ∆PQR,

PR² = QR² and QM⊥PR

Using Pythagoras theorem, we have,

PR² = PQ² + QR²

∆PQR is right angled triangle at Q.

From ∆QMR and ∆PMQ, we have,

∠M = ∠M

∠MQR = ∠QPM [= 90°-∠R]

So, using the AAA similarity criteria,

We have,

∆QMR ∼ ∆PMQ

Also, we know that,

Area of triangles = ½ × base × height

So, by property of area of similar triangles,

QM² = PM × RM

Hence proved.

According to the question,

DE || AB

Using basic proportionality theorem,

CD/AD = CE/BE

∴ If a line is drawn parallel to one side of a triangle such that it intersects the other sides at distinct points, then, the other two sides are divided in the same ratio.

Hence, we can conclude that, the line drawn is equal to the third side of the triangle.

(x + 3) (3x + 4) = x (3x + 19)

3x² + 4x + 9x + 12 = 3x² + 19x

19x – 13x = 12

6x = 12

∴ x = 12/6 = 2

According to the question,

∆ NSQ ≅ ∆MTR

∠1 = ∠2

Since,

∆NSQ = ∆MTR

So,

SQ = TR ….(i)

Also,

∠1 = ∠2 ⇒ PT = PS….(ii)

From Equation (i) and (ii).

PS/SQ = PT/TR

⇒ ST || QR

By converse of basic proportionality theorem, If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.

∴ ∠1 = PQR

And

∠2 = ∠PRQ

In ∆PTS and ∆PRQ.

∠P = ∠P [Common angles]

∠1 = ∠PQR (proved)

∠2 = ∠PRQ (proved)

∴ ∆PTS – ∆PRQ

Hence proved.

According to the question,

PQRS is a trapezium in which PQ‖RS and PQ = 3RS

PQ/RS = 3/1 = 3 …(i)

In ∆POQ and ∆ROS,

∠SOR = ∠QOP [vertically opposite angles]

∠SRP = ∠RPQ [alternate angles]

∴ ∆POQ ∼ ∆ROS [by AAA similarity criterion]

By property of area of similar triangle,

Therefore, the required ratio = 9:1.

According to the question,

AC and PQ intersect each other at the point O and AB||DC.

From ∆AOP and ∆COQ,

∠AOP = ∠COQ [Since they are vertically opposite angles]

∠APO = ∠CQO [since, AB||DC and PQ is transversal, the angles are alternate angles]

∴ ∆AOP ∼ ∆COQ [using AAA similarity criterion]

Then, since, corresponding sides are proportional

We have,

OA/OC = AP/CQ

OA × CQ = OC × AP

Hence Proved.

Let ABC be an equilateral triangle of side 8 cm

AB = BC = CA = 8 cm. (all sides of an equilateral triangle is equal)

Draw altitude AD which is perpendicular to BC.

Then, D is the mid-point of BC.

∴ BD = CD = ½

BC = 8/2 = 4 cm

Now, by Pythagoras theorem

AB² = AD² + BD²

⇒ (8)² = AD² + (4)²

⇒ 64 = AD² + 16

⇒ AD = √48 = 4√3 cm.

Hence, altitude of an equilateral triangle is 4√3 cm.

According to the question,

AB = 4 cm,

DE = 6 cm

EF = 9 cm

FD = 12 cm

Also,

∆ABC ∼ ∆DEF

We have,

By taking first two terms, we have

And by taking last two terms, we have,

Now,

Perimeter of ∆ABC = AB + BC + AC

= 4 + 6 + 8 = 18 cm

Thus, the perimeter of the triangle is 18 cm.