Triangles Test - 19

We know that, two triangles are congruent, if the sides and angles of one triangle are equal to the corresponding sides and angles of the other triangle.

Also, criterion for congruence of triangles are SAS (Side-Angle-Side), ASA (Angle-Side- Angle), SSS (Side-Side-Side) and RHS (right angle-hypotenuse-side).

So, SSA is not a criterion for congruence of triangles.

We know that, if ΔRST is congruent to ΔUVW i.e., ΔRST = ΔUVW, then sides of ΔRST fall on corresponding equal sides of ΔUVW and angles of ΔRST fall on corresponding equal angles of ΔUVW.

Here, given AB = QR, BC = PR and CA = PQ, which shows that AB covers QR, BC covers PR and CA covers PQ i.e., A correspond to Q, B correspond to R and C correspond to P.
or A↔Q,B↔R,C↔P

Under this correspondence,

ΔABC ≅ ΔQRP, so option (a) is incorrect,

or ΔBAC ≅ ΔRQP, so option (c) is incorrect,

or ΔCBA ≅ ΔPRQ, so option (b) is correct,

or ΔBCA ≅ ΔRPQ, so option (d) is incorrect.

Given that in triangle ABC, BC = AB which means ABC is an isosceles triangle.

We know the sum of angles in a triangle is 180°. Since it is an isosceles triangle the measure of two angles will be same. Let ∠A = x and ∠c = x

We know,

Sum of all angles = 180°

x + x + 80° = 180°

2x = 180°- 80°

2x =100°

x= \(\frac{100}{2}\) = 50°

Thus the measure of ∠A = 50°

Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively.

Let sides AB = 5 cm and CA = 1.5 cm

We know that, a closed figure formed by three intersecting lines (or sides) is called a triangle, if difference of two sides < third side and sum of two sides > third side

5 -1.5 < BC and 5 + 1.5 > BC

3.5 < BC and 6.5 > BC

Here, we see that options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

Given, In ΔABC and ΔDEF, AB = DF and ∠A = ∠D
We know that, two triangles will be congruent by SAS rule, if two sides and the included angle of one triangle are equal to the two sides and the included angle of other triangle.
AC = DE

In a given \(\triangle\)ABC we are given that the three angles are equal.

So, ∠A = ∠B = ∠C

According to the angle sum property of a triangle in  \(\triangle\)ABC

∠A + ∠B + ∠C = 180°

3∠A = 180°

∠A = \(\frac{180}{3}\)

∠A = 60°

∠A = ∠B = ∠C = 60°

Sum of the right angle in right angle triangle = 90°

Let the value of one acute angle = x

Sum of whole angle of triangle = 180°

x + x + 90° = 180°

2x + 90° = 180°

2x = 180° - 90°

2x = 90°

x = \(\frac{90^o}{2}\)

x = 45°

Value of two acute angles = 45°

Since we know that sum of 2 opposite interior angles is equal to exterior angle.

So, let one int. angle = x & other int. angle = x

So, x + x = 100°

2x = 100°

x = \(\frac{100^o}{2}\)

x = 50°

Therefore, each interior angle = 50°

Let, ABC be a triangle and AB, BC and AC produced to D, E and F respectively.

∠A + ∠B + ∠C = 180° (i)

∠CBD = ∠C + ∠A (Exterior angle theorem) (ii)

∠ACE = ∠A + ∠B (Exterior angle theorem) (iii)

∠BAF = ∠B + ∠C (Exterior angle theorem) (iv)

Adding (ii), (iii) and (iv) we get

∠CBD + ∠ACE + ∠BAF = 2∠A + 2∠B + 2∠C

∠CBD + ∠ACE + ∠BAF = 2 (∠A + ∠B + ∠C)

∠CBD + ∠ACE + ∠BAF = 2 x 180°

∠CBD + ∠ACE + ∠BAF = 360°

Thus, sum of all three exterior angles is 360°.

Given, AD perpendicular to BC

∠A = 100° 

In  ?ADB,

∠ADB + ∠B + ∠DAC = 180° 

90° + ∠B + \(\frac{1}{2}\)∠A = 180° 

∠B + \(\frac{1}{2}\) x 100° = 180° – 90° 

∠B + 50° = 90°

∠B = 90° - 50°

∠B = 40°

Let ABC is an acute angled triangle.

∠B = 90°

We know that, ∠A + ∠B + ∠C = 180° 

∠A + 90° + ∠C = 180° 

∠A + ∠C = 90° ..... (i)

In ΔAOC

∠AOC + ∠ACD + ∠CAD = 180°

∠AOC +  \(\frac{1}{2}\)∠C +  \(\frac{1}{2}\)∠A = 180°

∠AOC + \(\frac{1}{2}\)(∠A + ∠C) = 180° 

∠AOC + \(\frac{1}{2}\) x 90° = 180°   [From (i)]

∠AOC + 45° = 180°

∠AOC = 180° – 45° = 135° 

Thus, the angle at O between two bisectors is equal to 135°.

An exterior angle of a triangle is always greater than the opposite interior angles.

The sum of the acute angles of a right triangole has to be 90°.

x + 2x = 90°

3x = 90°

x = \(\frac{90^o}{3}\)

x = 30°