Chapter 6: Triangles

False

According to the question,

Let us assume that,

A = 25 cm

B = 5 cm

C = 24 cm

Now, Using Pythagoras Theorem,

We have,

A² = B² + C²

B² + C² = (5)² + (24)²

B² + C² = 25 + 576

B² + C² = 601

A² = 600

600 ≠ 601

A² ≠ B² + C²

Since the sides does not satisfy the property of Pythagoras theorem, triangle with sides 25cm, 5cm and 24cm is not a right triangle.

False

We know that,

Corresponding angles are equal in similar triangles.

So, we get,

∠D = ∠R

∠E = ∠P

∠F = ∠Q

True

According to the question,

PQ = 12.5 cm

PA = 5 cm

BR = 6 cm

PB = 4 cm

Then,

QA = QP – PA = 12.5 – 5 = 7.5 cm

So,

PA/AQ = 5/7.5 = 50/75 = 2/3 … (i)

PB/BR = 4/6 = 2/3 … (ii)

Form Equations (i) and (ii).

PA/AQ = PB/BR

We know that, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Therefore,

AB || QR.

True

In ∆PBC and ∆PDE,

∠BPC = ∠EPD [vertically opposite angles]

PB/PD = 5/10 = ½ … (i)

PC/PE = 6/12 = ½ … (ii)

From equation (i) and (ii),

We get,

PB/PD = PC/PE

Since, ∠BPC of ∆PBC = ∠EPD of ∆PDE and the sides including these.

Then, by SAS similarity criteria

∆PBC ∼ ∆PDE

We know that,

Sum of the three angles of a triangle = 180°.

Then, from ∆PQR,

∠P + ∠Q + ∠R = 180°

55° + 25° + ∠R = 180°

So, we get,

∠R = 180° – (55° + 25°) = 180°- 80° = 100°

Similarly, from ∆TSM,

∠T + ∠S + ∠M = 180°

∠T + ∠25° + 100° = 180°

So, we get,

∠T = 180°- (∠25° + 100°)

∠T = 180° – 125° = 55°

In ∆PQR and ∆TSM,

We have,

∠P = ∠T,

∠Q = ∠S

∠R = ∠M

Hence, ∆PQR ∼ ∆TSM

Since, all corresponding angles are equal,

∆QPR is similar to ∆TSM,

False

Two quadrilaterals cannot be similar, if only their corresponding angles are equal