MATHEMATICS

STANDARD IX

Part – 2

2013-2014

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WELCOME

SIMILAR TRIANGLES

Congruency of two triangles

If three sides of a triangle are equal to the three sides of another triangle, then these triangles are are congruent.

If two sides of a triangle and their included angle are equal to two sides of another triangle and their included angle, then these triangles are congruent.

Just because two sides and some angles of a triangle are equal to two sides and some angle of another triangle, the two triangles need not be congruent.

If one side and the two angles on it of a triangle are equal to one side and the two angles on it of another triangle then these triangles are congruent.

If the hypotenuse and one another side of a right angled triangle are equal to the hypotenuse and one other Side of another right angled triangle, then these two triangles are congruent.

If all the angles of a triangle are equal to the angles of another triangle, then all the pairs of sides opposite equal angles have the same ratio.

c b z y a x

x/a = y/b = z/c

This can be shortened a bit more :

If all the angles of a triangle are equal to the angles of another triangle, then the sides opposite equal angles are proportional.

Can you find out the proportional sides in the following figures?

A Pa) 500 700 500 700

B C Q

Ans : A = Q ; C = R ; A = P The pairs of proportional sides AB,PQ ; AC,PR ; BC,QR

700

800 800 300

E F Q R

Ans: E = Q F = R D = P

The pairs of proportional sides:

EF, QR ; DF, PR ; DE, PQ

D P

In the figures below, we have ABC and some other triangles with the same angle. Write against each, the names of the equal angles and the lengths of the equal sides.

A K

B C (i) L M

K = A L = B M = ………….. KL = ………………. LM = 12 cm MK = ………………………………….

Ans : K = A , L= B , M= C

BC/LM = AC/MK = AB/KL ( If all the angles of a triangle are equal to the angles of

another triangle, then the sides opposite equal angles are proportional)

BC = 6 cm, AB = 4 cm, AC =5 cm, LM = 12 cm

Therefore BC/LM = ½

That is, BC/LM = AC/MK = AB/KL = ½

AC/MK = ½

5/MK = ½

MK = 10 cm

AB/KL = ½

4/KL = ½

KL = 8 CM

SAME ANGLES

Give a triangle, there are several ways to draw another one with the same angles, but of different size. Look at this triangle: A

B CSuppose we want to enlarge it without altering the

angles. We can extend the left and right sides as much as we want ant this won’t change the top angle. A B C D E

We get a triangle, however we close the extended sides: A

If the two bottom angles are also to be equal, how should we draw the bottom line? The bottom line should be parallel to the line just above. A

B C

Can you prove that if the lines at the bottom are parallel, then the triangles would have equal angles. Ans : Angle B and angle D are corresponding angles formed, when

B C

AD cuts the parallel lines BC and DE. B = D (When a pair of parallel lines is cut by a third lines, each pair of corresponding angles are equal)

C and E are corresponding angles formed, when AE cuts the parallel lines BC and DE. C = E (When a pair of parallel lines is cut by a third line, each pair of corresponding angles are equal) A is a common angle to both ABC and ADE have equal angles.

Irrational problem

If in two triangles withn the same three angles, the ratio of one pair of sides opposite equal angles can be expressed in terms of natural numbers, then we can show that the other pairs of sides also have the same ratio, by dividing the triangles into smaller ones, as we have shown.

But there are instances where the ratio of sides cannot be expressed in terms of rational numbers . For example, draw an isosceless right angled triangle with the lenghts of the perpendicular sides 1 and another isosceless right angled

triangle with the lengths of the perpendicular sides 2 .

2 2

1 2 1 2

The angles of both triangles are 450 , 450 ,900. But the

ratio of sides opposite equal angles is 1: 2 .However small we divide one of the perpendicular

sides of one triangle, we cannot completely fill with it, perpendicular side of the other triangle. The same is true for the hypotenuse also.

So, the method of dissection of the triangles to prove equality of ratios will not work in cases such as this.

From this we have understood that the instances where the ratio of the sides opposite angles cannot be expressed in natural numbers, then we can’t prove that the other sides also have the same ratio dividing the triangle.

REVERSE QUESTION

If the sides of two triangles can be so paired that the ratio of the lengths of BC sides paired are all the same, then the pair of angles opposite each of these pairs of sides are equal .

We can also state this in a more concise form:

If the sides of a triangle are proportional to the sides of another triangle, then the angles opposite such sides are equal.

c b z y z y

Thus the three sides of PQR are equal to the three sides of XYZ and so these triangles are congruent. So, the angles opposite their pairs of equal sides are also equal: X = P , Y= Q , Z = R P = A, Q = B , R = C X = A , Y = B, Z = C

a x x

A P X

B C Q R Y Z

Triangle speciality

If the angles of a triangle are equal to angles of another triangle, then their sides are

proportional; on the other hand, if the sides of two triangles are proportional, then they have the

same angles. Among polygons, only triangles have this pecularity.

For example, all angles of a square and a rectangle which is not a square, are right

angles; but the sides are not proportional.

On the other hand, a square and a rhombus which is not a square have proportional

sides; but the angles are not equal.

similarity

We saw that if the angles of a triangle are all equal to the angles of

another triangle, then the sides of the two triangles are proportional; and on the

other hand, if the lenghths of the sides of a triangle are proportional to the

lengths of the sides of another triangle, then the angles of one triangle are equal

to the angles of the other.

In ABC shown below, A is a right angle

Draw the perpendicular from A to BC . Now we get two small right angled triangles also.

B C PWhat can we say about the angles of these?Let’s write B = X0 for convenience. A

B C PThen we get C = (90-x) 0 from the right angled triangle ABC.

?

B C

A

A

(One angle is right angle, Third angle = 180-(90+x)= 90-x)

Similarly, using the right angled triangles ABP and ACP, we can write other angles in terms of x.

Thus the angles of the triangles ABP and ACP are 900 , X0 and (90-x)0. So , they are similar.The angles of our original triangle ABC are also these. So, this triangle is also similar to ABP and ACP.Thus in a right angled triangle, the perpendicular to the hypotenuse from the opposite vertex divides, it into two right angled triangles, which are similar to each other; they are also similarly to the original triangle.

X0 (90-x)0

(90-x)0

B P C

A

If two sides of a triangle are proportional to two sides of another triangle and if their included angles are equal, then the triangles are similar.

c b y z y

X = P , Y = Q , Z = R P = A , Q = B , R = C X = A , Y = B , Z = C

B a C Q x R Y x Z

P X

A

Similarity and Congruence

If two triangles are congruent, then they are also similar. (Congruent triangles

have the same angles; also the ratio of each pair of sides opposite equal angles

is 1:1)

But two similar triangles may not be congruent.

Look at a comparison of these two concepts :

If two triangles have their sides proportional, then they are similar; they are

congruent only if sides are equal.

If two triangles have two pairs of angles equal, then they are similar; they are

congruent only if two pairs of angles and their common side are equal.

If two triangles have two pairs of sides proportional and their included angles

equal, then they are smilar; they are congruent only if two pairs of sides are

equal and their included angles are equal.

? The circles shown below have the same centre O.

Prove that OAB and OPQ are similar.

P

Q

A

O

B

Ans: Consider OAB and OPQ. O is common to both the triangles.

Therefore AOB = POB ……………….(1)In OAB, OA= OB(Radii of the circle)That is, OA/OB = 1 …………………..(2)In OPQ, OP=OQ (Radii of the circle)That is, OP/OQ = 1 ……………………(3)

From (2) and (3), OA/OB= OP/OQ ……………..(4)

If a pair of sides are proportional and their included angles are equal, then also the triangles are similar.

From (1) and (4), OAB and OPQ are similar triangles.

Prove that in a triangle, a line dividing two sides proportionally is parallel to the third side.

Ans : In the figure, line PQ cuts the sides AB and AC of ABC proportionally.

A P Q B C i.e., AP/PB = AQ/QC ……………(1)

We want to prove that the line PQ is parallel to BCFrom the data given,We can say PB/AP = QC/AQAdding 1 to both sides 1 + PB/AP = 1+ QC/AQNow 1+ PB/AP = (AP + PB)/AP = AB/AP1 + QC/AQ = (AQ+QC)/AQ = AC/AQTherefore AB/AP = AC/AQNoq the sides AB and AC of ABC and AP and AQ of APQ are proportional and the included angles of both the triangle are A itself.Therefore the triangles are similar.

Since the angles opposite to the proportional sides of the simi9lar triangles are equal APQ = ABC

Since they are corresponding angles, PQ || BC

END