Area of Right triangle

                      

  Area of Right triangle:-

Today in this article I want to discuss with you the topic area of right triangle,to start my discussion let us know,

 

**What is a right triangle?

Answer:- A triangle whose any one angle measures 90° is called right triangle.

In the picture,
    ABC is a right triangle,whose <B=90°

Now we will learn 

***what is the area of right triangle?

Answer:- The area of right triangle

can be defined the total inside space covered by the three sides of the triangle.
Again,
we can explain
 The area of right triangle = ½ × base×perpendicular/height

Problems  related on area of right triangle:-

Question(1):- The base of a right triangle is 4cm and its height is 6cm, find its area.

Answer:-

        

∵ Base of right triangle(b)= 4cm 
and height/perpendicular of right triangle(p) = 6cm
∴ area of right triangle = ½ ×b ×p
                                     = ½ ×4×6
                                     = (2×6)m²
                                     = 12m²

**Question(2):- The area of a right triangle is 15 cm² and its perpendicular is 3 cm, find its base.

Answer:- 

1st method,

∵The area of right triangle(A)=15cm²
               perpendicular(p) = 3 cm
   ∴ base of the triangle(b)= 2×A/p
                                         = (2×15)/3
                                         = 30/3
                                         = 10cm

2nd method,

∵The area of right triangle(A)=15cm²

               perpendicular(p) = 3 cm
Let the base of the right triangle =b cm
 Now, According to question,
 ½ × b × p = 15
  =>½ × b × 3 =15
  => b × 3 = 15 × 2
  => b × 3 = 30
   => b = 30/3
   => b =10
∴ Base of the right triangle(b) = 10 cm

  

Question(3):- The area of a right triangle is
20 cm² and its base 5 cm find its height/perpendicular.

Answer:- 

1st method:-

       ∵ Area of right triangle(A) = 20cm²
         Base of right triangle(b) = 5 cm
 ∴ height/perpendicular(h/p)= 2×A/b
                                       = 2× 20/5
                                      = 40/5
                                      = 8 cm

 2nd method:-

        ∵ Area of right triangle(A) = 20cm²

         Base of right triangle(b) = 5 cm
                  Let, perpendicular = p cm
         Now,according to question,
          ½ × b × p = 20
        =>½ ×5×p = 20
        =>5×p = 2×20
        =>5×p = 40
        => p = 40/5
        =>p  = 8
   ∴ perpendicular = 8 cm
When we say right triangle there must be mention the famous law of Pythagorus.

So, I am explaining 'Pythagorus law':-

"Pythagorus law states that sum of the squares of base and perpendicular of a right triangle is equal to the square of its hypotenuse"

Application of Pythagorus law:-

Question(1):-Perpendicular drawn from the vertex A of Δ ABC intersect BC at D, such that DB=3CD, Prove that 2AB²= 2AC² +BC²

Answer:- 

                       

In the picture, AD⊥BC in Δ ABC is drawn,which intersects BC at point D,such that DB=3CD ∴ BC=DB+CD 
                      =>BC =3CD+CD
                      =>BC = 4CD

 To prove:- 2AB²=2AC²+BC²

    

Proof:- In right triangle ADB,

Applying Pythagorus law, we find
∵ AB²= AD²+ BD² [∵<ADB = 90°]
=>2AB²=2AD²+2BD²—---------(1)[∵multiplying both sides by 2]
Again,in right triangle ADC,
Applying Pythagorus law, we find
∵ AC² = AD² + CD²  [∵ <ADC =90°]
=>2AC²=2AD²+2CD²—----(2)[∵multiplying both sides by 2]

Now equation(1) -equation(2) we get,
2AB²-2AC² = 2BD²- 2CD²
=>2AB² - 2AC² = 2(3CD)² - 2CD² [∵ putting DB =3CD]
 =>2AB² - 2AC² = 18CD² - 2CD²
 =>2AB² - 2AC² = 16CD²
 =>2AB² - 2AC² = (4CD)²
=>2AB² - 2AC² = (BC)² [∵putting 4CD= BC]
=>2AB² - 2AC² = BC²    

=>2AB² = 2AC² +BC²

                                    Proved

More Question:-

Question(2):- In Δ ABC, <C=90°, D and E are two points on the sides CA and CB,Prove that, AE² + BD² = AB² + DE²

Answer:-   

In the picture, ABC is a Δ, whose <C =90° Dand E are two points on the sides CA and CB,

To prove:-AE² + BD² = AB² + DE²

Proof:- ∵ In right triangle ABC,

         we get,
       ∵ AB² = AC² + BC²—-----(i)
       => AE² = AC² + CE²—---(ii)
        =>BD² = BC² +CD²—----(iii)
        =>DE² = CE² + CD²—---(iv)

Click Now

[Note:- Students are advised 1st realise the above process how to solve area problems of right triangle then try similar type of different problems]

FREQUENTLY ASKED QUESTIONS:-

What is the four walls formula of a cube?

Answer: The four walls area formula of a cube = 4l² where, l = length of each face of the cube.

what is the total surface area formula of a cube ?

Answer:- The total surface area formula of a cube = 6ll² where, l= length of each face of the cube.

what is the ceiling area formula of a house?

Answer :- The ceiling area formula of a house = l × b where, l = length of the house and b = breadth of the house.

What is four walls area formula of a cuboid ?

Answer :- The four walls formula of a cuboid = 2(l+b)h where, l = length of the cuboid, b = breadth of the cuboid and h = height of the cuboid.

What is curved surface area formula of a cylinder?

Answer:-The curved surface area formula of a cylinder = 2πrh where, r = radius of the cylinder and h = height of the cylinder.

What is curved surface area formula of a cone?

Answer :- Curved surface area formula of a cone = πrl where, r = radius of the cube, and l = slent height of the cone.

Share this post:-

Please give your valueable feedback:--

Previous

Next