Parallel Lines

Let the ends of our 8cm line be A and B.

Method-1:

Let us divide the given line into five equal parts as below.

Here if we consider AD, DB then we can notice that AD = 2parts and DB = 3 parts

Hence

Method-2: (by using the ratios)

Then the point dividing our line from end A will be

And it will be= 8-3.2 = 4.8cm from end B.

Method -1:

Given ratio of sides is 3:4.

Let one side be 3 , then other side would be 4

Therefore, total perimeter = 3 + 4 + 3 + 4 = 14

Here our perimeter is less than the required perimeter by 15-14 = 1

Step-1: using circle(centre B, radius BC) extend the base with side length (BC) as shown below

Step-2: take any slant line below base which will have a more length of 0.5cm than the base length(AE). ( here the extra length is half the difference between required perimeter and present perimeter)

Step-3: now join EG and draw a parallel line to EG through B. this meets AG at I.

Step-4: now take AI and IG as lengths of sides of required rectangle and form a rectangle as shown below:

Hence the bottom rectangle KJIA, is our required rectangle with perimeter 15.

Method-2:

Let ‘a’ and ‘b’ be the length and width of rectangle.

Then given

Then we know that perimeter will be 2(a + b)

Therefore, b = 4.286cm &

For equilateral triangle, all the sides are equal. Let ‘x’ be the length of each side.

Given perimeter = 10cm

x + x + x = 10cm

3x = 10cm

Let each part be ‘x’.

Then sides will be 3x, 4x, 5x respectively.

Given perimeter = 10cm

3x + 4x + 5x = 10cm

12x = 10cm

Then sides are

Let each part be ‘x’.

Then sides will be 2x, 3x, 4x respectively.

Given perimeter = 10cm

2x + 3x + 4x = 10cm

9x = 10cm

Then the sides are

Here AB, CD are parallel, then draw a line parallel to AB or CD through P.

Then,

We know that ,

Any three parallel lines will cut any two lines into pieces whose lengths are in the same ratio.

Therefore,

PA×PD = PB×PC

For verification, we can take a small trapezium and prove it as follows:

Here 3.8×2.12 = 7.98 = 1.9×4.25 approximately

Let a right angled triangle DEF,

H and G are mid points of DF, DE respectively

Here by pythagorous,

Since H and G are mid-points of DF, DE respectively. We can have

By pythagorous theorem,

Hence length of third side of larger triangle is 6cm

And 3cm, 4cm, 5cm are lengths of sides of smaller triangle.

i) We know that,

In any triangle, line drawn parallel to one side, passing through mid-point of another side will also meet the third side at its mid-point.

Therefore,

Let AD be ‘x’, then DC = x. similarly AE = y then EB = y

By pythagorus theorem for bigger triangle ABC,

----1

Similarly for smaller triangle AED,

-----2

From 1 & 2, we have

Hence proved.

ii)

We have the above following data.

And from pythagorous theorem,

We have

Hence the distance between mid-point of hypotenuse to vertex B (or) A (or) C is x

Hence proved.

iii) We know that

circumcentre is a point from which all the vertices are at equal distances.

In the above case, we have proved the mid-point of hypotenuse is equidistant from all the three vertices. Therefore mid-point of hypotenuse is the circumcentre of the given triangle.

Hence proved.

i) Here we have AD, PQ, BC as three parallel lines.

We know that

Any three parallel lines will cut any two lines into pieces whose lengths are in the same ratio.

For triangle ABC,

Therefore, --------1

Similarly, for triangle ACD,

--------2

From 1 & 2,

Hence proved

ii) From previous part we have,

On inverting the numerators and denominators,

On adding one on both sides of equation,

From figure, on simplifying,

On inverting the numerators and denominators,

Hence proved.

Given ,

Here AC and BD are diagonals and they bisect each other as we know.

Therefore we can say that AC is the median and passes through mid-poin of DB.

We know that

In any triangle, all the medians intersect at a single point and that point divides each median in the ratio 2:1 measured from vertex.

Therefore from triangle ABD,

We have

-------1

Similarly, from triangle BCD,

We have

; ----2

From 1 & 2

AG:GI:IC = AG:(GH + HI):IC

AG:GI:IC = 2:2:2 = 1:1:1

Hence the lines from vertices to mid-points of opposite side will divide the diagonal into three equal parts as shown above.

Hence proved.

Given ABCD is a quadrilateral. And EFGH be the mid-points of the above quadrilateral.

Then,

We know that,

If the mid-points of adjacent sides of a triangle are joined, then the line joining those mid-points will be parallel to the third side and half of its length.

Therefore, from triangle’s ABD and BCD,

EF//DB and HG//DB

⇒ EF//HG

Similarly, from triangle ACD and triangle ABC,

FG//AC and EH//AC

⇒ FG//EH

Since EF//HG ,

And FG//EH ,

Therefore EFGH forms a parallelogram for any type of quadrilateral taken.

Hence proved.

If original quadrilateral was a rectangle then,

Then the new quadrilateral formed would be a rhombus as shown in figure, due to

Equal sides are generated because of equal opposite lengths of original quadrilateral.

Therefore,

And diagonals are perpendicular to each other. Hence EFGH forms a rhombus.

If original quadrilateral was a square then,

Then the new quadrilateral formed would be a square as shown in figure, due to

Equal and perpendicular adjacent sides are generated because of equal lengths of original quadrilateral. Even new diagonals are equal and perpendicular.

Therefore,

And diagonals are equal and perpendicular to each other. Hence EFGH forms a square.

Here AB, CD are parallel, then draw a line parallel to AB or CD through P.

Then,

We know that ,

Any three parallel lines will cut any two lines into pieces whose lengths are in the same ratio.

Therefore, if we consider QR, AP, line through B parallel to QR, then it divides the BA and BP in the same ratio.

----1

Similarly, if we consider QR,AP, line through B parallel to QR, then it divides the BA and BP in the same ratio.

--------2

From 1 & 2, we have,

Here AB, CD are parallel, then draw a line parallel to AB or CD through P.

Then,

We know that ,

Any three parallel lines will cut any two lines into pieces whose lengths are in the same ratio.

Therefore,