Area of Triangle


 
 
Concept Explanation
 

Area of Triangle

Area of Triangle: Area of a triangle is equal to half of the product of base and height

A right triangle is the one in which the measure of any one of the interior angles is 90 degrees. It is to be noted here that since the sum of interior angles in a triangle is 180 degrees, only 1 of the 3 angles can be a right angle. If the other two angles are equal, that is 45 degrees each, the triangle is called an isosceles right angled triangle. However, if the other two angles are unequal, it is a scalene right angled triangle.

Area of Triangle = frac{1}{2}times basetimes height

Perimeter of a right triangle = a + b + c = Sum of three sides

Where a, b and c are the measure of its three sides.

Pythagoras Theorem defines the relationship between the three sides of a right angled triangle. Thus, if the measure of two of the three sides of a right triangle is given, we can use the Pythagoras Theorem to find out the third side. Hypotenuse^2=Perpendicular^2+Base^2

Example: Find The Area Of A Right Triangle as given: AB = 4cm and AC = 3cm

Solution:  using Pythagoras theorem,

       large (AB)^{2}=(AB)^{2}+(BC)^{2}

      large (BC)^{2}=(AC)^{2}-(AB)^{2}

     large (BC)^{2}=25-16

      large (BC)^{2}=9

     large BC=sqrt{9}

      BC = 3 cm

Area of triangle=  large =frac{1}{2}times AB times BC

 dpi{100} large =frac{1}{2}times 4 times 3=6cm^2

Theorem 1     The area of a triangle is half the product of any its sides and the corresponding altitude.

Given        Delta ABC   in which AL is the altitude to the side BC.

To prove    ar(Delta ABC)=frac{1}{2}(BCtimes AL)

Construction  Through C and A draw  CD || BA and  AD || BC  respectively, intersecting each other at D.

Proof     By construction we have, BA || CD and AD || BC    RightarrowABCD is a parallelogram as both opposite pair of sides are parallel to each other.

Since AC  is a diagonal of parallelogram ABCD

therefore;;;;ar(Delta ABC)=frac{1}{2}ar(parallel ^{gm} ABCD)

Rightarrow ;;;ar (Delta ABC)=frac{1}{2}(BCtimes AL)      

Beacuse   is the base and  AL is the corresponding   altitude of  parallelogram ABCD.

Hence Proved.

Given        Delta ABC   in which AL is the altitude to the side BC.

To prove    ar(Delta ABC)=frac{1}{2}(BCtimes AL)

Construction  Through C and A draw  CDparallel BA   and   ADparallel BC  respectively, intersecting each other at D.

Proof     We have,

                BAparallel CD                                  [ By construction]

and,        ADparallel BC                                 [ By construction]

therefore;;;; BCDA is a parallelogram.

Since AC  is a diagonal of  parallel ^{gm} BCDA.

therefore;;;;ar(Delta ABC)=frac{1}{2}ar(parallel ^{gm} BCDA)

Rightarrow ;;;ar (Delta ABC)=frac{1}{2}(BCtimes AL)                        [ because BC  is the base and  AL is the corresponding

                                                                                                        altitude of  parallel ^{gm}BCDA ]

[ Fig. required  (pg. no. 15.7)  xxxxx]

Sample Questions
(More Questions for each concept available in Login)
Question : 1

In the given right triangle right angled at A, AD = 6 cm , DB=10 cm. the find out the Area of triangle DAB.

Right Option : A
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Question : 2

A ____________ of a triangle divides it into two triangle of equal area .

Right Option : B
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Question : 3

In In ; Delta ABC ;and; Delta ;PQR,;AB=PQ,;BC=QR; and ;AC;=;PR. ;If large ar(Delta ABC);=;20 ;sq.; units,then ;ar(Delta PQR) is; equal;to

 

Right Option : B
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