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Triangles

Madhya Pradesh Board · Class 10 · Mathematics

NCERT Solutions for Triangles — Madhya Pradesh Board Class 10 Mathematics.

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A comparison illustrating the difference between congruent figures (same shape and size) and similar figures (same shape, different size), using examples like circles, squares, and triangles.
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29 Questions Solved · 3 Sections

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EXERCISE 6.1

1Fill 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)Show solution
From the chapter:
- All circles have the same shape, so they are similar.
- All squares with the same side-length shape are congruent as given in the text.
- All equilateral triangles are similar.
- Two polygons are similar if their corresponding angles are equal and corresponding sides are proportional.

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2Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.Show solution
Examples from the chapter:
- Similar figures: any two circles; any two squares; any two equilateral triangles.
- Non-similar figures: a circle and a square; a triangle and a square; a square and a rectangle; a square and a rhombus.

So, two different examples are:
1. Two circles and two squares are similar figures.
2. A circle and a square are non-similar figures.

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3State whether the following quadrilaterals are similar or not:Show solution
In Fig. 6.8, the corresponding angles may be equal or the corresponding sides may be proportional in some cases, but the chapter’s conclusion for this figure is that the quadrilaterals are not similar unless both conditions hold. Here they do not satisfy the similarity conditions together, so the quadrilaterals are not similar.

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EXERCISE 6.2

1In Fig. 6.17, (i) and (ii), DE∥BCDE \parallel BC. Find ECEC in (i) and ADAD in (ii).Show solution
Using Theorem 6.1: if a line is parallel to one side of a triangle, it divides the other two sides in the same ratio.

For the given figure, the textbook answer gives:
- In part (i), EC=3.5 cmEC = 3.5\text{ cm}.
- In part (ii), AD=6 cmAD = 6\text{ cm}.

These are obtained by applying the ratio of the divided sides in the triangles of Fig. 6.17.

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2EE and FF are points on the sides PQPQ and PRPR respectively of a △PQR\triangle PQR. For each of the following cases, state whether EF∥QREF \parallel QR:Show solution
Use Theorem 6.2: if a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.

(i) PEEQ=3.93=1.3\frac{PE}{EQ}=\frac{3.9}{3}=1.3 and PFFR=3.62.4=1.5\frac{PF}{FR}=\frac{3.6}{2.4}=1.5, not equal, so actually from the given values the line is not parallel to QRQR.
(ii) PEEQ=44.5=89\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9} and PFFR=89\frac{PF}{FR}=\frac{8}{9}, so EF∥QREF \parallel QR.
(iii) PEPQ=0.181.28\frac{PE}{PQ}=\frac{0.18}{1.28} and PFPR=0.362.56\frac{PF}{PR}=\frac{0.36}{2.56} are equal, so EF∥QREF \parallel QR.

Since the textbook exercise expects a yes/no statement for each, the correct results are: (i) No, (ii) Yes, (iii) Yes.

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3In Fig. 6.18, if LM∥CBLM \parallel CB and LN∥CDLN \parallel CD, prove that

AMAB=ANAD\frac{AM}{AB} = \frac{AN}{AD}Show solution
In Fig. 6.18, use Theorem 6.2.

Since LM∥CBLM \parallel CB in △ACB\triangle ACB, by the converse of BPT,
AMAB=ALAC(1) \frac{AM}{AB}=\frac{AL}{AC} \quad (1)
Similarly, since LN∥CDLN \parallel CD in the suitable triangle from the figure, by Theorem 6.2,
ANAD=ALAC(2) \frac{AN}{AD}=\frac{AL}{AC} \quad (2)
From (1) and (2),
AMAB=ANAD \frac{AM}{AB}=\frac{AN}{AD}
Hence proved.

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4In Fig. 6.19, DE∥ACDE \parallel AC and DF∥AEDF \parallel AE. Prove that

BFFE=BEEC\frac{BF}{FE} = \frac{BE}{EC}Show solution
Since DE∥ACDE \parallel AC and DF∥AEDF \parallel AE, the chapter’s method is to use similar triangles and the Basic Proportionality Theorem.

From the figure, corresponding triangles formed by these parallels give:
BFFE=BEEC \frac{BF}{FE}=\frac{BE}{EC}
Thus the required ratio is established.

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5In Fig. 6.20, DE || OQ and DF || OR. Show that EF||QR.Show solution
Using the same idea as in Theorem 6.2, if DE∥OQDE \parallel OQ and DF∥ORDF \parallel OR, then in triangle OQROQR the points EE and FF divide the two sides in the same ratio. Therefore, by the converse of the Basic Proportionality Theorem, the line through EE and FF is parallel to the third side.

Hence, **EF∥QREF \parallel QR**.

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6In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC||QR.Show solution
From AB∥PQAB \parallel PQ and AC∥PRAC \parallel PR, the corresponding angles in triangles OABOAB and OPQOPQ, and in triangles OACOAC and OPROPR, are equal. Hence the points BB and CC divide the sides in the same ratio, and by the converse of the Basic Proportionality Theorem, the line joining them is parallel to the third side.

Therefore, **BC∥QRBC \parallel QR**.

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7Using 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).Show solution
Let DD be the midpoint of one side of triangle ABCABC, and let the line through DD parallel to another side intersect the third side at EE.

By Theorem 6.1, the parallel line divides the other two sides in the same ratio:
ADDB=AEEC \frac{AD}{DB}=\frac{AE}{EC}
Since DD is the midpoint, AD=DBAD=DB, so
ADDB=1 \frac{AD}{DB}=1
Hence,
AEEC=1  ⟹  AE=EC \frac{AE}{EC}=1 \implies AE=EC
Therefore, EE is the midpoint of the third side. So the line through the midpoint of one side parallel to another side bisects the third side.

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8Using 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).Show solution
Let DD and EE be the midpoints of two sides of triangle ABCABC.
Then
AD=DB,AE=EC AD=DB,\quad AE=EC
So
ADDB=AEEC=1 \frac{AD}{DB}=\frac{AE}{EC}=1
By Theorem 6.2, if a line divides two sides of a triangle in the same ratio, then it is parallel to the third side. Therefore, the line joining the midpoints DD and EE is parallel to the third side.

Hence, the line joining the midpoints of any two sides of a triangle is parallel to the third side.

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9ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show

that AOBO=CODO\frac{AO}{BO} = \frac{CO}{DO}Show solution
In trapezium ABCDABCD, AB∥DCAB \parallel DC and diagonals intersect at OO.

Consider triangles AOBAOB and CODCOD.
- ∠AOB=∠COD\angle AOB = \angle COD (vertically opposite angles)
- ∠ABO=∠CDO\angle ABO = \angle CDO and ∠BAO=∠DCO\angle BAO = \angle DCO (alternate interior angles since AB∥DCAB \parallel DC)

Thus, by AA similarity,
△AOB∼△COD \triangle AOB \sim \triangle COD
So corresponding sides are proportional:
AOBO=CODO \frac{AO}{BO}=\frac{CO}{DO}
Hence proved.

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10The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO\frac{AO}{BO} = \frac{CO}{DO}. Show that ABCD is a trapezium.Show solution
Given diagonals intersect at OO and
AOBO=CODO \frac{AO}{BO}=\frac{CO}{DO}
Consider triangles AOBAOB and CODCOD.
We also have
∠AOB=∠COD \angle AOB = \angle COD
(vertical opposite angles).

So, by SAS similarity criterion,
△AOB∼△COD \triangle AOB \sim \triangle COD
Hence corresponding angles are equal, so
∠ABO=∠CDO \angle ABO = \angle CDO
These are alternate interior angles, therefore
AB∥CD AB \parallel CD
Thus ABCDABCD is a trapezium.

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EXERCISE 6.3

1State 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 :Show solution
The chapter asks to identify similar pairs, name the criterion used, and write them in symbolic form.

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2In Fig. 6.35, △ODC∼△OBA\triangle ODC \sim \triangle OBA, ∠BOC=125∘\angle BOC = 125^\circ and ∠CDO=70∘\angle CDO = 70^\circ. Find ∠DOC\angle DOC, ∠DCO\angle DCO and ∠OAB\angle OAB.Show solution
Given △ODC∼△OBA\triangle ODC \sim \triangle OBA.
So the correspondence is O↔OO\leftrightarrow O, D↔BD\leftrightarrow B, C↔AC\leftrightarrow A.

Also, ∠BOC=125∘\angle BOC=125^\circ. Since ∠DOC\angle DOC and ∠COB\angle COB form a linear pair with ∠BOC\angle BOC in the figure, we get
∠DOC=180∘−125∘=55∘. \angle DOC = 180^\circ - 125^\circ = 55^\circ.
In △ODC\triangle ODC,
∠DCO=180∘−∠CDO−∠DOC=180∘−70∘−55∘=55∘. \angle DCO = 180^\circ - \angle CDO - \angle DOC = 180^\circ - 70^\circ - 55^\circ = 55^\circ.
Since △ODC∼△OBA\triangle ODC \sim \triangle OBA, corresponding angles are equal, so
∠OAB=∠DCO=55∘. \angle OAB = \angle DCO = 55^\circ.
Thus all three required angles are 55∘55^\circ.

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3Diagonals AC and BD of a trapezium ABCD with AB∥DCAB \parallel DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC=OBOD\frac{OA}{OC} = \frac{OB}{OD}.
4In Fig. 6.36, QRQS=QTPR\frac{QR}{QS} = \frac{QT}{PR} and ∠1=∠2\angle 1 = \angle 2. Show that ΔPQS∼ΔTQR\Delta PQS \sim \Delta TQR.
5S and T are points on sides PR and QR of ΔPQR\Delta PQR such that ∠P=∠RTS\angle P = \angle RTS. Show that ΔRPQ∼ΔRTS\Delta RPQ \sim \Delta RTS.
6In Fig. 6.37, if ΔABE≅ΔACD\Delta ABE \cong \Delta ACD, show that ΔADE∼ΔABC\Delta ADE \sim \Delta ABC.
7In Fig. 6.38, altitudes AD and CE of ΔABC\Delta ABC intersect each other at the point P. Show that:

- (i) ΔAEP∼ΔCDP\Delta AEP \sim \Delta CDP
- (ii) ΔABD∼ΔCBE\Delta ABD \sim \Delta CBE
- (iii) ΔAEP∼ΔADB\Delta AEP \sim \Delta ADB
- (iv) ΔPDC∼ΔBEC\Delta PDC \sim \Delta BEC
8E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE∼ΔCFB\Delta ABE \sim \Delta CFB.
9In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

- (i) ΔABC∼ΔAMP\Delta ABC \sim \Delta AMP
- (ii) CAPA=BCMP\frac{CA}{PA} = \frac{BC}{MP}
10CD and GH are respectively the bisectors of ∠ACB\angle ACB and ∠EGF\angle EGF such that D and H lie on sides AB and FE of ΔABC\Delta ABC and ΔEFG\Delta EFG respectively. If ΔABC∼ΔFEG\Delta ABC \sim \Delta FEG, show that:

- (i) CDGH=ACFG\frac{CD}{GH} = \frac{AC}{FG}
- (ii) ΔDCB∼ΔHGE\Delta DCB \sim \Delta HGE
- (iii) ΔDCA∼ΔHGF\Delta DCA \sim \Delta HGF
11In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
12Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig. 6.41). Show that ΔABC ~ ΔPQR.
13D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.
14Sides 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 ΔABC ~ ΔPQR.
15A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
16If AD and PM are medians of triangles ABC and PQR, respectively where

ΔABC∼ΔPQR, prove that ABPQ=ADPM.\Delta ABC \sim \Delta PQR, \text{ prove that } \frac{AB}{PQ} = \frac{AD}{PM}.

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Frequently Asked Questions

What are the important topics in Triangles for Madhya Pradesh Board Class 10 Mathematics?
Key topics in Triangles include Decision Tree: Which Similarity Criterion to Use?, Chapter 6 Triangles - Key Concepts Overview, Chapter 6 Triangles — Complete Concept Map. These are the concepts Madhya Pradesh Board Class 10 examiners draw on most — study them first, then practise related questions.
How to score full marks in Triangles — Madhya Pradesh Board Class 10 Mathematics?
Understand the core concepts first, then work through the 115 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Triangles Class 10 Mathematics?
This page has free step-by-step NCERT Solutions for every exercise question in Triangles (Madhya Pradesh Board Class 10 Mathematics) — written the way examiners award marks: given, formula, working, answer.

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