Isosceles Right Triangle: Formulas, Pythagoras Theorem and Area

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Isosceles right triangle is a two dimensional three sided figure in which one angle measures 90°, and the other two angles measure 45° each. In an isosceles right triangle, Hypotenuse is given by formula H=B\(\sqrt{2}\), the area is given by B2/2, and perimeter is given by 2B+H. [H-hypotenuse, B-any of the other two sides].

Key Takeaways: Triangle, Isosceles triangle, Isosceles right triangle, Hypotenuse, Area, Perimeter.


Triangle

In geometry, a triangle is a two-dimensional figure with three sides and three angles.

Triangle

The summation of the three angles of any triangle is always 180°.

The sum of the two sides of any triangle (in isosceles right triangle too) is always greater than the third side.


Types of Triangles

According to the measure of angles in a triangle, the triangles can be: 

  • Acute 
  • Obtuse 
  • Right

Types of Triangles

According to the length of sides in a triangle, triangles can be: 

According to the length of sides in a triangle

According to both angles and sides in a triangle, the triangle can be:

  • Right equilateral 
  • Right isosceles 
  • Obtuse Scalene

According to both angles and sides in a triangle

Also Read:


Isosceles triangle

A triangle in which the length of two of its sides is equal is called an isosceles triangle. Since the length of two sides of the isosceles triangle is equal, the corresponding angles are equal to each other too. 

Isosceles triangle

Here, side AC=BC; and angle ∠CAB= ∠CBA

The unequal side (AB) of an isosceles triangle is usually referred to as the 'base' of the triangle.


Right triangle

Right angle triangle is a triangle in which one angle is 90°. The longest side of the right triangle is the hypotenuse and the other two sides are height and the base. The opposite side (BC) of the right angle, whose ends are touched by arms of the angle (AB & AC), is the hypotenuse. 

Right triangle

AC ⊥ AB

∠A=90º

Also Read:


Isosceles right triangle

A triangle in which one angle measures 90°, and the other two angles measure 45° each is an isosceles right triangle. In an isosceles right triangle (figure below), ∠A and ∠C measure 45° each, and ∠B measures 90°. The sides AB and AC are equal. This type of triangle is also known as a 45-90-45 triangle

AC, the side opposite of ∠B, is the hypotenuse. And AB or AC can be taken as height or base

Isosceles right triangle

Also Read: Quadrilateral

Formulas

Given below are the formulas to construct a triangle which includes:

  1. Pythagoras theorem
  2. Area
  3. Perimeter

Pythagoras Theorem

Pythagoras theorem, which applies to any right-angle triangle, also applies to isosceles right triangles. 

In an isosceles right triangle, the length of base (BC) is equal to length of height (AB). If base (BC) is taken as ‘B’, then AB=BC=’B’. Let the hypotenuse be taken as ‘H’

Pythagoras theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the square of the other two sides.

(Hypotenuse)2 = (Side)2 + (Side)2

H2=B2+B2

H2=2B2
H=B\(\sqrt{2}\)

Area

The area of a triangle is half of the base times height.

Area = ½ × b × h              

Here, b-base; h-height

This applies to right isosceles triangles also. 

As stated above, in an isosceles right-triangle the length of base (BC) is equal to length of height (AB). If, base (BC) is taken as ‘B’, then AB=BC=’B’

Area of right isosceles triangle= ½ ×B×B

= B2/2

Perimeter

The Sum of all sides of a triangle is the perimeter of that triangle. 

In the right isosceles triangle, since two sides (Base BC and Height AB) are same and taken as ‘B’ each. And the hypotenuse is taken as ‘H’. 

The perimeter= AB+BC+AC

= B+B+H

=2B+H

Also Read: 


Things to Remember

  • Triangles are polygons with the least number of sides; since any figure with two sides only forms an angle.
  • In the right triangle two sides are perpendicular to each other; in the right isosceles triangle these two sides are of equal length.
  • The difference between any two sides is smaller than the 
  • third side-AB-BC<AC
  • The concept of right isosceles triangles is used by architects. Even Egyptians used these to design pyramids.

Sample Question

Ques: Find the length of the other two sides of the isosceles right triangle given below: (2 marks)

Find the length of the other two sides of the isosceles right triangle given below:

Ans: We know the length of the hypotenuse is \(\sqrt{2}\), we need to determine the lengths of the other two sides. 

Because the length of the hypotenuse is \(\sqrt{2}\) times the length of the other sides, we can construct by using Pythagoras theorem for isosceles right triangle- H2=2B2

H2=2B2

H=B2

B=H\(\sqrt{2}\)

x=\(\frac{\sqrt{8}}{\sqrt{2}}\)

x=\(\sqrt{4}\)=2

So, the length of the other two sides is 2 units each. 

Ques: The lengths of sides of an isosceles right triangle measure x=14cm. Find its perimeter. (2 marks)

The lengths of sides of an isosceles right triangle measure x=14cm. Find its perimeter

Ans: For an isosceles right triangle,
H=B\(\sqrt{2}\)

H= x\(\sqrt{2}\)

H=14\(\sqrt{2}\) cm

Now the perimeter= 2B+H

P= (142) +19.80

P=47.8cm

So, the perimeter of the given triangle is 47.8cm.

Ques: A triangle has two angles are 450 and two sides equal to 8. What is the perimeter of this triangle? (2 marks)

Ans: When a triangle has two angles equal to 450, it must be a 450:450:900 isosceles right triangle.

Since two sides are 8. 

The hypotenuse= 8\(\sqrt{2}\) [using H=B\(\sqrt{2}\), obtained from Pythagoras theorem]

Perimeter= 2B+H= 8+8+8\(\sqrt{2}\) =16+ 8\(\sqrt{2}\)

So, the perimeter is 16+ 8\(\sqrt{2}\) units.

Ques: An isosceles right triangle has hypotenuse of h=18\(\sqrt{2}\) inches.  Find its area. (3 marks)

An isosceles right triangle has hypotenuse of h=18?2 inches.  Find its area.

Ans: To calculate the triangle's area, we need to find the lengths of its sides. Hypotenuse is given

H=B\(\sqrt{2}\)

B=\(\frac{H}{\sqrt{2}}\)

B=\(\frac{18 \sqrt{2}}{\sqrt{2}}\)

B=18 inches

Two sides of the right isosceles triangles are 18 inches each.

Now, area of right isosceles triangles:

Area= B2/2

=182/2

Area =162 inch2

Ques: What are the different kinds of triangle? (3 marks)

Ans: Triangles can be classified according to the measure of angles, of sides, and of both sides, angles taken together-

According to measure of angles in triangle, the triangles can be- 1) Acute 2) Obtuse 3) Right

According to length of sides in triangle, triangles can be- 1) Equilateral 2) Isosceles 3) Scalene

According to both angles and sides in triangle, triangle can be- 1) Right equilateral 2) Right isosceles 3) Obtuse Scalene

Ques: The equal sides of an isosceles right triangle are 8 units each. Find the area. (2 marks)

Ans: For an isosceles right triangle, 

the area formula = B2/2 

Here, B = 8 units

Thus, Area = 82/2 = 32 square units

Thus, the area of the isosceles right triangle is 32 sq units.

Ques: The perimeter of an isosceles right triangle is 10 + 5\(\sqrt{2}\) units. If the hypotenuse measures 5\(\sqrt{2}\) units then, find the measure of the other sides. (2 marks)

Ans: For an isosceles right triangle, the perimeter formula is given by 2B+H 

Here, H = 5\(\sqrt{2}\)units, Perimeter (P)= 10 + 5\(\sqrt{2}\) units

By using the formula,

P= 2B+ H 

10 + 5\(\sqrt{2}\)=2B+ 5\(\sqrt{2}\)

B=5 

Hence, the length of the other side is 5 units each.

CBSE X Related Questions

  • 1.
    Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]


      • 2.

        In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15^\circ$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$.


          • 3.

            From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
            Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$


              • 4.
                Find the zeroes of the polynomial: \[ q(x) = 8x^2 - 2x - 3 \] Hence, find a polynomial whose zeroes are 2 less than the zeroes of \(q(x)\)


                  • 5.
                    In a right triangle ABC, right-angled at A, if $\sin B = \dfrac{1}{4}$, then the value of $\sec B$ is

                      • 4
                      • $\dfrac{\sqrt{15}}{4}$
                      • $\sqrt{15}$
                      • $\dfrac{4}{\sqrt{15}}$

                    • 6.

                      In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.

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