Triangles — Class 10 Mathematics
Mathematics · Chapter 6
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Chapter summary
Key ideas & terms
- Similar triangles
- Triangles with equal corresponding angles and proportional sides.
- Basic Proportionality Theorem
- A line parallel to one side divides the other two sides proportionally.
- AA criterion
- Two triangles are similar if two pairs of angles are equal.
- Corresponding sides
- Sides opposite to equal angles in similar triangles.
Extra questions & answers
State the Basic Proportionality Theorem (Thales’ theorem).
If a line is drawn parallel to one side of a triangle to intersect the other two sides, it divides those sides in the same ratio.
State the AA similarity criterion.
If two angles of one triangle are equal to two angles of another, the triangles are similar.
When are two triangles said to be similar?
When their corresponding angles are equal and their corresponding sides are in the same ratio (proportion).
State the Basic Proportionality Theorem (Thales’ theorem).
If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, it divides those two sides in the same ratio (refer to the figure in your NCERT textbook).
Name the criteria used to prove two triangles similar.
The AA (or AAA), SSS and SAS similarity criteria.
State the AA similarity criterion.
If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
How are the areas of two similar triangles related to their sides?
The ratio of the areas of two similar triangles equals the square of the ratio of any pair of corresponding sides.
In a triangle, a line parallel to BC meets AB at D and AC at E. If
By the Basic Proportionality Theorem,
If two similar triangles have corresponding sides in the ratio
The ratio of areas is the square of the ratio of sides, i.e.
Give one real-life use of similar triangles.
Finding inaccessible heights and distances, such as the height of a tower or tree using the length of its shadow and a known object’s shadow.
EXERCISE 6.1
1. Fill in the blanks using the correct word given in brackets :
(i) All circles are ______. (congruent, similar)
(ii) All squares are ______. (similar, congruent)
(iii) All ______ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ______ and (b) their corresponding sides are ______. (equal, proportional)
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.
2. Give two different examples of pair of
(i) similar figures. (ii) non-similar figures.
(i) Two pairs of similar figures are two circles of different radii and two squares of different side lengths.
(ii) Two pairs of non-similar figures are a square and a rectangle whose adjacent sides are unequal, and an equilateral triangle and an isosceles triangle that is not equilateral.
3. State whether the following quadrilaterals are similar or not:
The quadrilaterals are not similar. Although the corresponding sides are proportional, the first quadrilateral is not a square because its angles are not all right angles, whereas every angle of the second quadrilateral is 90
∘
. Thus, their corresponding angles are not equal.
EXERCISE 6.2
1. In Fig. 6.17, (i) and (ii), DE
parallelBC. Find EC in (i) and AD in (ii).
(i) By the Basic Proportionality Theorem,
Therefore,
so EC=2
textcm.
(ii) Again,
Thus,
which gives AD=2.4
textcm.
2. E and F are points on the sides PQ and PR respectively of a
trianglePQR. For each of the following cases, state whether EF
parallelQR :
(i) PE=3.9
textcm,EQ=3
textcm,PF=3.6
textcm and FR=2.4
textcm
(ii) PE=4
textcm,QE=4.5
textcm,PF=8
textcm and RF=9
textcm
(iii) PQ=1.28
textcm,PR=2.56
textcm,PE=0.18
textcm and PF=0.36
textcm
(i)
The ratios are unequal, so EF
not
parallelQR.
(ii)
The ratios are equal, so EF
parallelQR.
(iii) EQ=1.28−0.18=1.10
textcm and FR=2.56−0.36=2.20
textcm. Hence,
and
Therefore, EF
parallelQR.
3. In Fig. 6.18, if LM
parallelCB and LN
parallelCD, prove that
In
triangleABC, LM
parallelCB. Therefore, by the Basic Proportionality Theorem,
In
triangleADC, LN
parallelCD. Hence,
From (1) and (2),
4. In Fig. 6.19, DE
parallelAC and DF
parallelAE. Prove that
In
triangleBAE, DF
parallelAE. Therefore,
In
triangleBAC, DE
parallelAC. Therefore,
Combining (1) and (2),
5. In Fig. 6.20, DE
parallelOQ and DF
parallelOR. Show that EF
parallelQR.
In
trianglePOQ, DE
parallelOQ, so
In
trianglePOR, DF
parallelOR, so
Thus,
By the converse of the Basic Proportionality Theorem in
trianglePQR, EF
parallelQR.
6. In Fig. 6.21, A,B and C are points on OP,OQ and OR respectively such that AB
parallelPQ and AC
parallelPR. Show that BC
parallelQR.
Since AB
parallelPQ, in
triangleOPQ,
Since AC
parallelPR, in
triangleOPR,
Therefore,
By the converse of the Basic Proportionality Theorem in
triangleOQR, BC
parallelQR.
7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Let D be the midpoint of AB in
triangleABC, and let a line through D, parallel to BC, meet AC at E. Since DE
parallelBC, Theorem 6.1 gives
As D is the midpoint of AB, AD=DB, so
Hence,
which means AE=EC. Therefore, E is the midpoint of AC, so the line bisects the third side.
8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Let D and E be the midpoints of AB and AC, respectively, in
triangleABC. Then
Therefore,
Thus, D and E divide AB and AC in the same ratio. By Theorem 6.2, DE
parallelBC.
9. ABCD is a trapezium in which AB
parallelDC and its diagonals intersect each other at the point O. Show that
Consider
triangleAOB and
triangleCOD. Since AB
parallelDC,
and
by alternate interior angles. Also,
as they are vertically opposite angles. Hence,
Therefore, corresponding sides are proportional:
Cross-multiplying gives AO
cdotDO=BO
cdotCO, and therefore
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that
Show that ABCD is a trapezium.
The given relation can be written as
so
In
triangleAOB and
triangleCOD,
because they are vertically opposite angles. The two sides including these angles are proportional, so
by the SAS similarity criterion. Hence,
These are alternate interior angles, and therefore AB
parallelDC. Thus, ABCD is a trapezium.
EXERCISE 6.3
1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
(i) The corresponding angles are equal, so by the AAA criterion,
(ii) The corresponding sides satisfy
Therefore, by the SSS criterion,
(iii) The side lengths are not proportional because
are not equal. Hence, the triangles are not similar.
(iv)
and the included angles are both 70
∘
. Therefore, by the SAS criterion,
(v) The available proportional sides do not include the equal 80
∘
angles, so no similarity criterion is satisfied. The triangles are not similar.
(vi) The angles of the first triangle are 70
∘
,80
∘
,30
∘
, corresponding to the angles of the second triangle. Therefore, by the AAA criterion,
2. In Fig. 6.35,
triangleODC
sim
triangleOBA,
angleBOC=125
∘
and
angleCDO=70
∘
. Find
angleDOC,
angleDCO and
angleOAB.
Since D,O,B are collinear,
angleDOC and
angleBOC form a linear pair. Thus,
In
triangleODC,
Since
triangleODC
sim
triangleOBA, the angle at C corresponds to the angle at A. Hence,
3. Diagonals AC and BD of a trapezium ABCD with AB
parallelDC intersect each other at the point O. Using a similarity criterion for two triangles, show that
Consider
triangleAOB and
triangleCOD. Since AB
parallelDC,
and
by alternate interior angles. Also,
as vertically opposite angles. Therefore,
by the AA criterion. Hence, corresponding sides are proportional:
4. In Fig. 6.36,
and
angle1=
angle2. Show that
trianglePQS
sim
triangleTQR.
Since P lies on QT and S lies on QR,
The given equality of
angle1 and
angle2 gives
Using the given proportional relation and the configuration in the figure, the sides including these equal angles are proportional. Therefore, by the SAS similarity criterion,
5. S and T are points on sides PR and QR of
trianglePQR such that
angleP=
angleRTS. Show that
triangleRPQ
sim
triangleRTS.
Since S lies on PR and T lies on QR, the angle at R is common in direction:
It is also given that
Thus, two corresponding angles are equal. Therefore, by the AA similarity criterion,
6. In Fig. 6.37, if
triangleABE
cong
triangleACD, show that
triangleADE
sim
triangleABC.
From
triangleABE
cong
triangleACD, corresponding sides are equal:
Therefore,
Also,
because D lies on AB and E lies on AC. Thus, the sides including the equal angle are proportional. By the SAS criterion,
7. In Fig. 6.38, altitudes AD and CE of
triangleABC intersect each other at the point P. Show that:
(i)
triangleAEP
sim
triangleCDP
(ii)
triangleABD
sim
triangleCBE
(iii)
triangleAEP
sim
triangleADB
(iv)
trianglePDC
sim
triangleBEC
(i) Since CE
perpAB and AD
perpBC,
Also,
angleAPE=
angleCPD as vertically opposite angles. Hence,
by AA.
(ii)
Also, since D lies on BC and E lies on AB,
Therefore,
by AA.
(iii)
Moreover, AP lies on AD and AE lies on AB, so
Hence,
by AA.
(iv)
Also, PC lies on CE and CD lies on CB, so
Thus,
by AA.
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
triangleABE
sim
triangleCFB.
In parallelogram ABCD,
Since F lies on CD, CF
parallelAB. Since E lies on AD produced, AE
parallelBC. Therefore,
and
Thus, by the AA similarity criterion,
9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i)
triangleABC
sim
triangleAMP
(ii)
(i) Since the triangles are right angled,
Also, the angle at A is common because AM lies along AC and AP lies along AB:
Therefore, by AA,
(ii) Corresponding sides of similar triangles are proportional. The hypotenuse CA corresponds to PA, and BC corresponds to MP. Hence,
10. CD and GH are respectively the bisectors of
angleACB and
angleEGF such that D and H lie on sides AB and FE of
triangleABC and
triangleEFG respectively. If
triangleABC
sim
triangleFEG, show that:
(i)
(ii)
triangleDCB
sim
triangleHGE
(iii)
triangleDCA
sim
triangleHGF
Since
triangleABC
sim
triangleFEG,
Because CD and GH bisect the corresponding angles,
and
(ii) Also, as D lies on AB and H lies on FE,
Therefore, by AA,
(iii) Similarly,
Together with
angleDCA=
angleHGF, this gives
by AA.
(i) From
triangleDCA
sim
triangleHGF, corresponding sides are proportional, so
11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB=AC. If AD
perpBC and EF
perpAC, prove that
triangleABD
sim
triangleECF.
Since AD
perpBC,
Since EF
perpAC,
As AB=AC, the base angles of the isosceles triangle are equal:
Because D lies on BC, E,C,B are collinear, and F lies on AC,
Thus, two corresponding angles are equal. By AA,
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of
trianglePQR (see Fig. 6.41). Show that
triangleABC
sim
trianglePQR.
Let
Since AD and PM are medians,
so BD=kQM.
By Apollonius' theorem in
triangleABC,
Substituting AD=kPM, BD=kQM, and AB=kPQ,
By Apollonius' theorem in
trianglePQR,
Therefore,
so AC=kPR. Hence,
Thus, by the SSS similarity criterion,
13. D is a point on the side BC of a triangle ABC such that
angleADC=
angleBAC. Show that CA
2
=CB
cdotCD.
Since D lies on BC,
It is given that
Therefore, by AA,
Thus, corresponding sides are proportional:
Cross-multiplying,
Hence,
14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that
triangleABC
sim
trianglePQR.
Let
Since AD and PM are medians, BD=DC and QM=MR.
Using Apollonius' theorem in
triangleABC,
Substituting AB=kPQ, AC=kPR, and AD=kPM,
By Apollonius' theorem in
trianglePQR,
Therefore,
so BC=kQR. Hence,
By the SSS similarity criterion,
15. A vertical pole of length 6
textm casts a shadow 4
textm long on the ground and at the same time a tower casts a shadow 28
textm long. Find the height of the tower.
At the same time, the sun's rays make the same angle with the ground, so the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow. Let the tower's height be h
textm. Then
Therefore,
The height of the tower is 42
textm.
16. If AD and PM are medians of triangles ABC and PQR, respectively where
triangleABC
sim
trianglePQR, prove that
Since
triangleABC
sim
trianglePQR,
As AD and PM are medians,
Hence, from (1),
Also,
because D lies on BC, M lies on QR, and corresponding angles of the original similar triangles are equal. From (1) and (2),
Therefore, by SAS similarity,
Corresponding sides are proportional, so
Thus,