Free The Triangle and Its Properties 01 Practice Test - 7th grade 

The sum of all the angles of a triangle is ${180}^{\circ }$​.

Therefore,
$\text{Third angle}={180}^{\circ }-({50}^{\circ }+{70}^{\circ })={180}^{\circ }-\left({120}^{\circ }\right)={60}^{\circ }$​

In a right angled triangle, square of the hypotenuse (the side opposite to the right angle) is equal to the sum of the squares of the other two sides. This is Pythagoras' theorem.​

In an isosceles triangle, both median and altitude are same.

The sum of all the three angles of a triangle is ${180}^{\circ }$.

We know that obtuse angle is an angle which is greater than ${90}^{\circ }$ and less than ${180}^{\circ }$
Hence a triangle cannot have two obtuse angles.




 

The sum of 2 obtuse angles will be greater than (90 + 90), i.e. greater than ${180}^{\circ }$.

Since the sum of 2 angles of the triangle is more than ${180}^{\circ }$, the sum of three angles will be more than ${180}^{\circ }$ for sure.

This is not possible as the sum of 3 angles of a triangle is fixed i.e. ${180}^{\circ }$ and  cannot exceed this limit.

Thus, a triangle cannot have 2 obtuse angles.

Sum of lengths of any two sides of a triangle is always greater than the third side. Hence, the given statement is false.

The six elements of a triangle are 3 sides and 3 angles. Hence, $3\times 3=9$.

Given $\angle$C = 60^o and BC = AC,
Therefore $\angle A=\angle B$  (base angles are equal in an isosceles triangle)
$\therefore$  by angle sum property of triangle, $\angle A=\angle B={60}^o$

We know that,  $\angle$x +$\angle$y = $\angle$A +$\angle$B  (Exterior angle property)

$\therefore \angle A+\angle B+\angle x+\angle y={60}^{\circ }+{60}^{\circ }+{120}^{\circ }={240}^{\circ }$

$\angle$C = ${60}^{\circ }$​ and BC = AC,
By angle sum property of triangle, $\angle$A = $\angle$B = ${60}^{\circ }$​

Also,
$\angle ACB+\angle x+\angle y={180}^0$  [ BCD is straight line]
$⟹{60}^{\circ }+2y+y={180}^{\circ }$​ 
$⟹{60}^{\circ }+3y={180}^{\circ }$
$⟹3y={180}^{\circ }-{60}^{\circ }$​
$⟹y={40}^{\circ }$

Hence, $\angle B+y={60}^{\circ }+{40}^{\circ }={100}^{\circ }$​

      $\angle ACB+\angle ABC=\angle DAC$
                  [Exterior angle property]
     $\angle ACB+{60}^{\circ }={150}^{\circ }\angle ACB={90}^{\circ }\therefore \text{Reflex}\angle ACB={360}^{\circ }-{90}^{\circ }={270}^{\circ }=3\times {90}^{\circ }=3\text{right angles}$