Class 10 Mathematics Note

Area of Triangle and Quadrilateral

Perimeter and Area of Geometrical figures

Rectangle:


 The perimeter of rectangle = 2(l + b). 
Area of rectangle = l × b; (l and b are the length and breadth of rectangle)
 Diagonal of rectangle = $\sqrt {{{\rm{l}}^2}{\rm{\: }} + {\rm{\: }}{{\rm{b}}^2}} $


Square:


Perimeter of square = 4 × S. 
 Area of square = S × S. 
 Diagonal of square = S√2; (S is the side of square) 


Triangle: 

 Perimeter of triangle = (a + b + c); (a, b, c are 3 sides of a triangle) 
 Area of triangle = $\sqrt {{\rm{s}}\left( {{\rm{s\: }} - {\rm{\: a}}} \right)\left( {{\rm{s\: }} - {\rm{\: b}}} \right)\left( {{\rm{s\: }} - {\rm{\: c}}} \right)} ;$ (s is the semi-perimeter of triangle) 
 S = $\frac{1}{2}$ (a + b + c) 
 Area of triangle = $frac{1}{2}$× b × h; (b base , h height) 
 Area of an equilateral triangle = $\frac{{{{\rm{a}}^2}\sqrt 3 }}{4};$ (a is the side of triangle) 


Parallelogram: 

 Perimeter of parallelogram = 2 (sum of adjacent sides) 
 Area of parallelogram = base × height 


Rhombus: 


 Area of rhombus = base × height 
 Area of rhombus = $frac{1}{2}$× length of one diagonal × length of other diagonal
 Perimeter of rhombus = 4 × side


Trapezium: 


 Area of trapezium =$frac{1}{2}$(sum of parallel sides) × (perpendicular distance between them) 
 = $frac{1}{2}$☐(p1 + p2) × h (p1, pare two parallel sides) 


Circle: 


 Circumference of circle = 2πr
= πd
Where, π = 3.14 or π = 22/7
r is the radius of circle 
 d is the diameter of circle
 Area of circle = πr2
Area of ring = Area of outer circle - Area of inner circle.

 

Theorems

Parallelograms standing on the same base and between the same parallel lines are equal in area.

 

 

 Given: ☐PQRS and ☐AQRB are standing on the same base and between the same paralleled lines

 TO prove: ☐PQRS = ☐AQRB

 

S. NO

 Statement

Reason

1.

2.

3.

4.

5.

6.

In Δ PQAand ΔSRB

∠ PAQ=∠ AQR =∠ SRB(A)

∠ APQ=∠ SRQ =∠ RSB(A)

AQ= BR (S)

Δ PQA≃ΔSRB

Δ PQA+ AQRS=AQRS +ΔSRB

☐PQRS = ☐AQRB

 

 

Being an alternate angleand opposite angle of parallelogram

Being an alternate angleand opposite angle of parallelogram

Opposite side of the parallelogram

 BYA.A. Saxiom

 Adding common part

 From 5

 

 

 

 

 

The area of the triangle is equal to the area of the parallelograms, both standing on the same base and between the sane parallel lines

 

 

Given:Δ TQR and☐PQRS are standing on the same base and between the same parallel linesPT and RQ

TO prove: Δ TQR == $frac{1}{2}$☐ PQRS

Construction:join QU

S.No

Statement

Reason

1.

 

2.

 

3.

 

4.

 

RQUTis a parallelogram

 

Δ TQR =$frac{1}{2}$☐TRQU

 

☐TRQU = ☐QRSP

 

Δ TQR = $frac{1}{2}$☐TRQU

By constructionQU || TR

 

The diagonal bisects the parallelogram.

 

Parallelogram standing on the same base and between the same parallel lines are equal in area

 

 From 2 and 3

 

 

Examples 1

 ABCD and PQRD are the two parallelograms

Prove that:

☐ABCD =☐PQRD

Given: ABCD and PQRD are parallelogram

To prove:☐ABCD = ☐PQRD

S.NO

Statement

Reason

1.

2.

3.

 

 

 

Δ DPC=$frac{1}{2}$☐ ABCD

Δ DPC=$frac{1}{2}$☐PQRD

☐ ABCD = ☐PQRD

 

 

1. Triangle and the parallelogram standing on the same base and between the same parallel lines.

2. Triangle and the parallelogram standing on the same base and between the same parallel lines.

From2 and 3

 

 

 

 

 

Examples 2

In the given figure, ABCD is the parallelogram X is the any point within it .Prove that he sum of the area of XCD and XCB is equal to half of the area of ABCD .

Given: ABCD is a parallelogram X is at any point in centre.

To prove:Δ XCD + Δ XAB =$\frac{1}{2}{\rm{\: }}$parallelogram ABCD

Construction: Draw MN || AB

S.NO

Statement

Reason

1.

2.

3.

4.

MNCD is a parallelogram

ABNM is a parallelogram

Δ XCD =$frac{1}{2}$MNCD

Δ XAB= $frac{1}{2}$ABNM

 

AB ||MNand AB ||CD

AB ||MNand AB ||CD

Triangle and the parallelogram standing on the same base and between same parallel lines

Triangle and the parallelogram standing on the same base and between same parallel lines

 


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