1. Draw a rectangle of area 12 sq. CmAnswer:

Length x breadth = 12 sq.cmSo sides may be = 1 x 12

= 2 x 6= 3 x 4

= x 24

So we can draw 4 rectangles with area 12 sq.cm1)

2)

3)

4)

in the fourth figure breadth is cm. x 24 = 12 sq cm

2) Draw a rectangle with area 12 sq.cm and one side 8 cm

Answer :

Here only one rectangle is possible.

3) Draw a triangle of area 12 sq.cm

Answer :

As before we can draw many triangles with area 12 sq.cm

4) Draw triangle with area 12 sq cm and one side 8cm

Answer: in the case of rectangle we draw only one rectangle. But here we can draw many triangles.

In the case of rectangle with one side 8cm and area 12 sqcm, only one rectangle appears. But in the case of triangle with one side 8cm and area 12sq cm, we can draw many of them. The reason is the following law.

A) Draw a triangle of sides 4 centimetres, 5 centimetres and 6 centimetres. Draw an isosceles triangle with the same base and area.

Here red triangle is an isosceles triangle and have the same area that of the blue triangle, because both have same base and altitude.

B ) Can you draw another triangle of the same base ( 6cm ) and area (12 sq.cm) with the left side 7centimetres?Answer : yes

In this figure red triangle have sides 6cm , 4cm, 5 cm and one side of the blue triangle is 7 cm.Both triangles have same base and altitude. Therefore area is 12 sq cm

Problems in page number 7 :-

Answer:

in this figure Red triangle have side 6 cm, 3 cm, 4 cm. Blue, Green, Yellow triangles are all right triangles and have same base and height of the triangle ABC and therefore have same area.

Answer :

AR =7 CMGreen, Red, Blue and yellow triangles all have same base and altitude. < BAQ = 90, < BCQ = 60, < ACR = 30 degrees.

Answer :

Red and green triangles have samearea and vertices of green triangleare on the circle

Answer :

As shown above we can draw 4 triangles with area 12 sq cm but only one with area 24 sq cm

Answer:

In the figure BPC, BQC, ACR, ACP, ABQ, ABR, ABP ( ACBP is a parallelogram ) ACQ, BCR total 9 triangles with the area of blue triange.

Triangles within the the parallelogram are congruent to blue triangles. There are 3 such triangles.

Answer :

< ADC + < BDC = 180 degree

Therefore AB is a strait line.

A parallel line is drawn through CNow Blue and Green triangles have same base and height and therefore their areas are equal.

Answer :

if x and y or x and z are fixed we can draw two triangles with same area. If z and y are kept fixed then only one triangle can draw.

Quadrilateral and triangle

A ) Explain how to find the area of a quadrilateral given below by transforming it in to a triangle.

Answer:

Step 1: Name the quadrilateral as ABCD and draw a diagonal BDStep 2: Extend the side AB Step 3: Draw a parallel line through C to BD which meet at E

Step 4: Now join ED

Then triangle AED (shaded triangle ) has the same area that of the quadrilateral ABCD

( In page no 15 Problems 1 , 2, 3 are for drawing. Draw the pictures in your note book and show me )

Answer:

To find Area of yellow region in the figure

Yellow part is divided into twotrapeziums. Trapezium EFGD andTrapezium AHFE

Area of trapezium =

Area of Trapezium EFGD =

= 2sq cm

Area of Trapezium AHFE = = 5 sq cm

Area of yellow part = 5 + 2 = 7 sq cmtotal area of rectangle ABCD = length X breadth = 5 X 3 = 15 sq cm

Area of Green Part = 15 – 7 = 8 sq cm

Triangle division

Answer :

Areas of blue , yellow, green triangles

Area of Blue triangle =

Area of Yellow triangle =

Area of Green triangle =

Ratio of areas of Blue, Yellow, Green triangles = : :

and h are same in the ratiio

Therefore ratio of areas = a : b : c ,

Answer:

CP divides AB equally and p is the mid point. Therefore Area of APC = Area of triangle BPC

Similarly AO and BO divides APC and BPC into equal parts respectively.

Therefore all four triangles in the picture have same area.

Ie Area of all small triangles are area of triangle ABC .

Answer:

Given that P is the mid point of AB. That is area of APC = area of BPCand given that CO : OP = 2: 1 This means ratio of areas of AOC and AOP = 2 : 1and ratio of areas of BOC to area of BOP is also = 2:1If area of APO = x, then area of AOC = 2xGiven P is mid point ofAB, So area of APO = Area of BPOie We can take area of BPO =x If area of BPO = x then area of BOC = 2xThat is area of green triangle = 2x All triangles have same area.

One triangle has area that of ABC

Answer:Let ACB is the angle and CO is the bisector. OP and OQ are are perpendiculars to the arms of the angle from a point on the bisector.

Now construct another pictures such a way that AC = BC

ie ABC becomes an isosceles triangle and D isthe midpoint of AB

This means AD : BD = 1 : 1

Area of ADC : Area of BDC = 1: 1

Area of ADC = Area od BDC

ie Area and bases of these triangles are equal.

That means altitudes PD = QDie Every perpendicular from the bisector hassame length.

Answer:

(i) Let < ACE = x0.But < ACE and < BDC are corresponding angles and therefore are equal, ie < BDC = x0.

BCD is an isosceles triangle and base angles are equal.

Therefore < BDC = < CBD =x0.

< BCE and < CBD are alternate interior angles and therefore equal.Ie < BCE = X0.We have proved < ACE = x0.= < BCE That means CE is the bisector of < ACB

(ii) Firstly draw line AB = 8cm . Complete the triangle, ABC with sides 4 cm and 5cm.

Extend AC to D such that CD =5 cm. Join BD. Draw CE parallel to BD, ie AE : EB =4:5

(iii) In this case we cannot construct a triangle with sides 8 cm, 3cm, 4cm.

Keeping the ratio 3:4, we take the measures of other sides as 6cm and 8 cm, 6:8 = 3:4

R divides PQ = 8cm in the ratio 3:4

Answer:

In the figure ratioof Area of CEDto Area of CEB = 30 : 40 = 3 : 4

ie ED : BE = 3 : 4

Areas of triangles, AED and AEB are also in the same ratio 3 : 4

Let total area of AED and AEB = x sq cm

ie

= 140 sq cm

Area of AEB = = 80 sq cm

Area of quadrilateral = 60 +30+40+ 140 = 270 sq cm

Answer :

Let Area of green triangle = x sq cmArea of yellow triangle =y sq cmArea of red triangle = z sq cm

Given AB II CD ACD and BCD have same base and lies between same parallel lines.

Therefore their areas are equal.

Ie x + y = x = z

ie y = z

ie Area of yellow triangle = Area of red triangle

Answer : In the above problem we proved yellow and red triangles have same areaTherefore Area of red triangle = 10 sq cmRatio of areas of yellow triangle and red triangle = 1 : 2Same ratio for blue and red trianglesie Area of blue triangle =5 sq cm

Total area of the Trapezium = 10 + 20+ 10 +5 = 45 sq cm

Answer:As in the above cases area of yellow and red triangles are equal and let it be x sq cm.

Ratio of areas 4: x = x : 9

product of extremes = product of middles

ie

ie x2 = 36

ie x = 6

Area of trapezium = 9 +4+6+6 = 25 sq cm

End