Aakash Repeater Courses
Take Aakash iACST and get instant scholarship on coaching programs.
We have seen clothes hangers, sandwiches, traffic signboards, etc. Do you know which shape they represent? All these things are triangular. Triangles are fundamental geometric 2-D shapes characterised by three sides, three angles and a sum of 180 degrees for their interior angles. It is a polygon with the least number of sides. In this chapter, you will learn about different kind of triangles and their amazing properties. NCERT Solutions for Class 10 can help the students understand these concepts with clarity and make them efficient in solving problems involving triangles.
This Story also Contains
NCERT Solutions for Class 10 Maths provide clear and step-by-step solutions to the exercise problems in the NCERT textbook. Students who are in need of the quadratic equation solutions will find this article very useful. It covers questions from all the topics and will help you improve your speed and accuracy. These NCERT Solutions are trustworthy and reliable, as they are created by subject matter experts at Careers360, making them an essential resource for exam preparation. Get all solved exercises, full syllabus notes, and a free PDF from this NCERT article
Students who wish to access the NCERT solutions for class 10 chapter 6 can click on the link below to download the entire solution in PDF.
Below are the NCERT class 10 maths chapter 4 solutions for exercise questions
Triangles Class 10 NCERT Solutions Exercise: 6.1 Page number: 78 Total questions: 3 |
Take Aakash iACST and get instant scholarship on coaching programs.
Answer:
All circles are similar.
Since all the circles have a similar shape. They may have different radii, but the shape of all circles is the same.
Therefore, all circles are similar.
Answer:
All squares are similar.
Since all the squares have a similar shape. They may have a different side, but the shape of all squares is the same.
Therefore, all squares are similar.
Answer:
All equilateral triangles are similar.
Since all the equilateral triangles have a similar shape. They may have different sides, but the shape of all equilateral triangles is the same.
Therefore, all equilateral triangles are similar.
Answer:
Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.
Thus, (a) equal
(b) proportional
Q2 (i): Give two different examples of a pair of similar figures.
Answer:
The two different examples of a pair of similar figures are :
1. Two circles with different radii.
2. Two rectangles with different breadth and length.
Q2 (ii): Give two different examples of a pair of non-similar figures.
Answer:
The two different examples of a pair of non-similar figures are :
1. Rectangle and circle
2. A circle and a triangle.
Q3: State whether the following quadrilaterals are similar or not:
Answer:
Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. $1:2$, but their corresponding angles are not equal.
Triangles Class 10 NCERT Solutions Exercise: 6.2 Page number: 84-85 Total questions: 10 |
Q1: In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Answer:
(i)
Let EC be x
Given: DE || BC
By using the proportionality theorem, we get
$\frac{AD}{DB}=\frac{AE}{EC}$
$\Rightarrow \frac{1.5}{3}=\frac{1}{x}$
$\Rightarrow x=\frac{3}{1.5}=2\, cm$
$\therefore EC=2\, cm$
(ii)
Let AD be x
Given: DE || BC
By using the proportionality theorem, we get
$\frac{AD}{DB}=\frac{AE}{EC}$
$\Rightarrow \frac{x}{7.2}=\frac{1.8}{5.4}$
$\Rightarrow x=\frac{7.2}{3}=2.4\, cm$
$\therefore AD=2.4\, cm$
Answer:
Given :
PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
$\frac{PE}{EQ}=\frac{3.9}{3}=1.3\, cm$ and $\frac{PF}{FR}=\frac{3.6}{2.4}=1.5\, cm$
We have
$\frac{PE}{EQ} \neq \frac{PF}{FR}$
Hence, EF is not parallel to QR.
Answer:
Given :
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}\, cm$ and $\frac{PF}{FR}=\frac{8}{9}\, cm$
We have
$\frac{PE}{EQ} = \frac{PF}{FR}$
Hence, EF is parallel to QR.
Q2 (iii): E and F are points on the sides PQ and PR, respectively, of a triangle PQR. For each of the following cases, state whether EF || QR : PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Answer:
Given :
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
$\frac{PE}{PQ}=\frac{0.18}{1.28}=\frac{9}{64}\, cm$ and $\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}\, cm$
We have
$\frac{PE}{EQ} = \frac{PF}{FR}$
Hence, EF is parallel to QR.
Q3: In Fig. 6.18, if LM || CB and LN || CD, prove that $\frac{AM}{AB} = \frac{AN}{AD }$
Answer:
Given : LM || CB and LN || CD
To prove :
$\frac{AM}{AB} = \frac{AN}{AD }$
Since , LM || CB so we have
$\frac{AM}{AB}=\frac{AL}{AC}.............................................(1)$
Also, LN || CD
$\frac{AL}{AC}=\frac{AN}{AD}.............................................(2)$
From equations 1 and 2, we have
$\frac{AM}{AB} = \frac{AN}{AD }$
Hence proved.
Q4: In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC
Answer:
Given: DE || AC and DF || AE.
To prove :
$\frac{BF}{FE} = \frac{BE}{EC }$
Since , DE || AC so we have
$\frac{BD}{DA}=\frac{BE}{EC}.............................................1$
Also, DF || AE
$\frac{BD}{DA}=\frac{BF}{FE}.............................................2$
From equations 1 and 2, we have
$\frac{BF}{FE} = \frac{BE}{EC }$
Hence proved.
Q5: In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
Answer:
Given: DE || OQ and DF || OR.
To prove EF || QR.
Since DE || OQ so we have
$\frac{PE}{EQ}=\frac{PD}{DO}.............................................1$
Also, DF || OR
$\frac{PF}{FR}=\frac{PD}{DO}.............................................2$
From equations 1 and 2, we have
$\frac{PE}{EQ} = \frac{PF}{FR }$
Thus, EF || QR. (converse of basic proportionality theorem)
Hence proved.
Answer:
Given : AB || PQ and AC || PR
To prove: BC || QR
Since, AB || PQ so we have
$\frac{OA}{AP}=\frac{OB}{BQ}.............................................1$
Also, AC || PR
$\frac{OA}{AP}=\frac{OC}{CR}.............................................2$
From equations 1 and 2, we have
$\frac{OB}{BQ} = \frac{OC}{CR }$
Therefore, BC || QR. (converse basic proportionality theorem)
Hence proved.
Answer:
Let PQ be a line passing through the midpoint of line AB and parallel to line BC, intersecting line AC at point Q.
i.e. $PQ||BC$ and $AP=PB$ .
Using the basic proportionality theorem, we have
$\frac{AP}{PB}=\frac{AQ}{QC}..........................1$
Since $AP=PB$
$\frac{AQ}{QC}=\frac{1}{1}$
$\Rightarrow AQ=QC$
$\therefore$ Q is the midpoint of AC.
Answer:
Let P be the midpoint of line AB and Q be the midpoint of line AC.
PQ is the line joining midpoints P and Q of lines AB and AC, respectively.
i.e. $AQ=QC$ and $AP=PB$ .
we have,
$\frac{AP}{PB}=\frac{1}{1}..........................1$
$\frac{AQ}{QC}=\frac{1}{1}...................................2$
From equations 1 and 2, we get
$\frac{AQ}{QC}=\frac{AP}{PB}$
$\therefore$ By basic proportionality theorem, we have $PQ||BC$
Answer:
Draw a line EF passing through point O such that $EO||CD\, \, and\, \, FO||CD$
To prove :
$\frac{AO}{BO} = \frac{CO}{DO}$
In $\triangle ADC$ , we have $CD||EO$
So, by using the basic proportionality theorem,
$\frac{AE}{ED}=\frac{AO}{OC}........................................1$
In $\triangle ABD$ , we have $AB||EO$
So, by using the basic proportionality theorem,
$\frac{DE}{EA}=\frac{OD}{BO}........................................2$
Using equations 1 and 2, we get
$\frac{AO}{OC}=\frac{BO}{OD}$
$\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$
Hence proved.
Answer:
Draw a line EF passing through point O such that $EO||AB$
Given :
$\frac{AO}{BO} = \frac{CO}{DO}$
In $\triangle ABD$ , we have $AB||EO$
So, by using the basic proportionality theorem,
$\frac{AE}{ED}=\frac{BO}{DO}........................................1$
However, it is given that
$\frac{AO}{CO} = \frac{BO}{DO}..............................2$
Using equations 1 and 2, we get
$\frac{AE}{ED}=\frac{AO}{CO}$
$\Rightarrow EO||CD$ (By basic proportionality theorem)
$\Rightarrow AB||EO||CD$
$\Rightarrow AB||CD$
Therefore, ABCD is a trapezium.
Triangles Class 10 NCERT Solutions Exercise: 6.3 Page number: 94-97 Total questions: 16 |
Answer:
(i) $\angle A=\angle P=60 ^\circ$
$\angle B=\angle Q=80 ^\circ$
$\angle C=\angle R=40 ^\circ$
$\therefore \triangle ABC \sim \triangle PQR$ (By AAA)
So , $\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}$
(ii) As corresponding sides of both triangles are proportional.
$\therefore \triangle ABC \sim \triangle PQR$ (By SSS)
(iii) Given triangles are not similar because corresponding sides are not proportional.
(iv) $\triangle MNL \sim \triangle PQR$ by SAS similarity criteria.
(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides
(vi) In $\triangle DEF$ , we know that
$\angle D+\angle E+\angle F=180 ^\circ$
$\Rightarrow 70 ^\circ+80 ^\circ+\angle F=180 ^\circ$
$\Rightarrow 150 ^\circ+\angle F=180 ^\circ$
$\Rightarrow \angle F=180 ^\circ-150 ^\circ=30 ^\circ$
In $\triangle PQR$ , we know that
$\angle P+\angle Q+\angle R=180 ^\circ$
$\Rightarrow 30^{\circ}+80^{\circ}+\angle R=180^{\circ}$
$\Rightarrow 110 ^\circ+\angle R=180 ^\circ$
$\Rightarrow \angle R=180 ^\circ-110 ^\circ=70 ^\circ$
$\angle Q=\angle P=70 ^\circ$
$\angle E=\angle Q=80^{\circ}$
$\angle F=\angle R=30 ^\circ$
$\therefore \triangle DEF\sim \triangle PQR$ ( By AAA)
Q2: In Fig. 6.35, $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 ^\circ$ and $\angle CDO = 70 ^\circ$ . Find $\angle DOC , \angle DCO , \angle OAB$
Answer:
Given : $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 ^\circ$ and $\angle CDO = 70 ^\circ$
$\angle DOC+\angle BOC=180 ^{\circ} $ (DOB is a straight line)
$\Rightarrow \angle DOC+125 ^\circ=180 ^\circ$
$\Rightarrow \angle DOC=180 ^\circ-125 ^\circ$
$\Rightarrow \angle DOC=55 ^\circ$
In $\Delta ODC, $
$\angle DOC+\angle ODC+\angle DCO=180 ^\circ$
$\Rightarrow 55 ^\circ+ 70 ^\circ+\angle DCO=180 ^\circ$
$\Rightarrow \angle DCO+125 ^\circ=180 ^\circ$
$\Rightarrow \angle DCO=180 ^\circ-125 ^\circ$
$\Rightarrow \angle DCO=55 ^\circ$
Since, $\Delta ODC \sim \Delta OBA$, so
$\Rightarrow\angle OAB= \angle DCO=55 ^{\circ} $ (Corresponding angles are equal in similar triangles).
Answer:
In $\triangle DOC\, and\, \triangle BOA$ , we have
$\angle CDO=\angle ABO$ ( Alternate interior angles as $AB||CD$ )
$\angle DCO=\angle BAO$ ( Alternate interior angles as $AB||CD$ )
$\angle DOC=\angle BOA$ ( Vertically opposite angles are equal)
$\therefore \triangle DOC\, \sim \, \triangle BOA$ ( By AAA)
$\therefore \frac{DO}{BO}=\frac{OC}{OA}$ ( corresponding sides are equal)
$\Rightarrow \frac{OA }{OC} = \frac{OB }{OD }$
Hence proved.
Answer:
Given : $\frac{QR }{QS } = \frac{QT}{PR}$ and $\angle 1 = \angle 2$
To prove : $\Delta PQS \sim \Delta TQR$
In $\triangle PQR$ , $\angle PQR=\angle PRQ$
$\therefore PQ=PR$
$\frac{QR }{QS } = \frac{QT}{PR}$ (Given)
$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$
In $\Delta PQS\, and\, \Delta TQR$,
$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$
$\angle Q=\angle Q$ (Common)
$\Delta PQS \sim \Delta TQR$ (By SAS)
Answer:
Given : $\angle$ P = $\angle$ RTS
To prove RPQ ~ $\Delta$ RTS.
In $\Delta$ RPQ and $\Delta$ RTS,
$\angle$ P = $\angle$ RTS (Given)
$\angle$ R = $\angle$ R (common)
$\Delta$ RPQ ~ $\Delta$ RTS. (By AA)
Q6: In Fig. 6.37, if $\Delta$ ABE $\equiv$ $\Delta$ ACD, show that $\Delta$ ADE ~ $\Delta$ ABC.
Answer:
Given: $\triangle ABE \cong \triangle ACD$
To prove ADE ~ $\Delta$ ABC.
Since $\triangle ABE \cong \triangle ACD$
$AB=AC$ (By CPCT)
$AD=AE$ (By CPCT)
In $\Delta$ ADE and $\Delta$ ABC,
$\angle A=\angle A$ (Common)
and
$\frac{AD}{AB}=\frac{AE}{AC}$ ( $AB=AC$ and $AD=AE$ )
Therefore, $\Delta$ ADE ~ $\Delta$ ABC. ( By SAS criteria)
Q7 (i): In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta AEP \sim \Delta CDP$
Answer:
To prove : $\Delta AEP \sim \Delta CDP$
In $\Delta AEP \, \, and\, \, \Delta CDP$ ,
$\angle AEP=\angle CDP$ ( Both angles are right angle)
$\angle APE=\angle CPD$ (Vertically opposite angles )
$\Delta AEP \sim \Delta CDP$ ( By AA criterion)
Q7 (ii): In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta ABD \sim \Delta CBE$
Answer:
To prove : $\Delta ABD \sim \Delta CBE$
In $\Delta ABD \, \, and\, \, \Delta CBE$ ,
$\angle ADB=\angle CEB$ ( Both angles are right angle)
$\angle ABD=\angle CBE$ (Common )
$\Delta ABD \sim \Delta CBE$ ( By AA criterion)
Q7 (iii): In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta AEP \sim \Delta ADB$
Answer:
To prove : $\Delta AEP \sim \Delta ADB$
In $\Delta AEP \, \, \, and\, \, \Delta ADB$ ,
$\angle AEP=\angle ADB$ ( Both angles are right angle)
$\angle A=\angle A$ (Common )
$\Delta AEP \sim \Delta ADB$ ( By AA criterion)
Q7 (iv): In Fig. 6.38, altitudes AD and CE of $\Delta ABC$ intersect each other at the point P. Show that: $\Delta PDC \sim \Delta BEC$
Answer:
To prove : $\Delta PDC \sim \Delta BEC$
In $\Delta PDC \, \, and\, \, \, \Delta BEC$ ,
$\angle CDP=\angle CEB$ ( Both angles are right angle)
$\angle C=\angle C$ (Common )
$\Delta PDC \sim \Delta BEC$ ( By AA criterion)
Answer:
To prove : $\Delta ABE \sim \Delta CFB$
In $\Delta ABE \, \, \, and\, \, \Delta CFB$ ,
$\angle A=\angle C$ ( Opposite angles of a parallelogram are equal)
$\angle AEB=\angle CBF$ ( Alternate angles of AE||BC)
$\Delta ABE \sim \Delta CFB$ ( By AA criterion )
Q9 (i): In Fig. 6.39, ABC and AMP are two right triangles, right-angled at B and M, respectively. Prove that: $\Delta ABC \sim \Delta AMP$
Answer:
To prove : $\Delta ABC \sim \Delta AMP$
In $\Delta ABC \, \, and\, \, \Delta AMP$ ,
$\angle ABC=\angle AMP$ ( Each $90 ^\circ$ )
$\angle A=\angle A$ ( common)
$\Delta ABC \sim \Delta AMP$ ( By AA criterion )
Answer:
To prove :
$\frac{CA }{PA } = \frac{BC }{MP}$
In $\Delta ABC \, \, and\, \, \Delta AMP$ ,
$\angle ABC=\angle AMP$ ( Each $90 ^\circ$ )
$\angle A=\angle A$ ( common)
$\Delta ABC \sim \Delta AMP$ ( By AA criterion )
$\frac{CA }{PA } = \frac{BC }{MP}$ ( corresponding parts of similar triangles )
Hence proved.
Q10 (i): CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of $\Delta ABC\: \: and\: \: \Delta EGF$ respectively. If $\Delta ABC \sim \Delta EGF$ , show that: $\frac{CD}{GH} = \frac{AC}{FG}$
Answer:
To prove :
$\frac{CD}{GH} = \frac{AC}{FG}$
Given : $\Delta ABC \sim \Delta EGF$
$\angle A=\angle F,\angle B=\angle E\, \, and \, \, \angle ACB=\angle FGE,\angle ACB=\angle FGE$
$\therefore \angle ACD=\angle FGH$ ( CD and GH are bisectors of equal angles)
$\therefore \angle DCB=\angle HGE$ ( CD and GH are bisectors of equal angles)
In $\Delta ACD \, \, and\, \, \Delta FGH$
$\therefore \angle ACD=\angle FGH$ ( proved above)
$\angle A=\angle F$ ( proved above)
$\Delta ACD \sim \Delta FGH$ ( By AA criterion)
$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$
Hence proved.
Q10 (ii): CD and GH are respectively the bisectors of $\angle ABC \: \: and \: \: \angle EGF$ such that D and H lie on sides AB and FE of $\Delta ABC \: \: and \: \: \Delta EGF$ respectively. If $\Delta ABC \sim \Delta EGF$ , show that: $\Delta DCB \sim \Delta HGE$
Answer:
To prove : $\Delta DCB \sim \Delta HGE$
Given : $\Delta ABC \sim \Delta EGF$
In $\Delta DCB \,\, \, and\, \, \Delta HGE$ ,
$\therefore \angle DCB=\angle HGE$ ( CD and GH are bisectors of equal angles)
$\angle B=\angle E$ ( $\Delta ABC \sim \Delta EGF$ )
$\Delta DCB \sim \Delta HGE$ ( By AA criterion )
Q10 (iii): CD and GH are respectively the bisectors of $\angle ABC \: \: and \: \: \angle EGF$ such that D and H lie on sides AB and FE of $\Delta ABC \: \: and \: \: \Delta EGF$ respectively. If $\Delta ABC\sim \Delta EGF$ , show that: $\Delta DCA \sim \Delta HGF$
Answer:
To prove : $\Delta DCA \sim \Delta HGF$
Given : $\Delta ABC \sim \Delta EGF$
In $\Delta DCA \, \, \, and\, \, \Delta HGF$ ,
$\therefore \angle ACD=\angle FGH$ ( CD and GH are bisectors of equal angles)
$\angle A=\angle F$ ( $\Delta ABC \sim \Delta EGF$ )
$\Delta DCA \sim \Delta HGF$ ( By AA criterion )
Answer:
To prove : $\Delta ABD \sim \Delta ECF$
Given: ABC is an isosceles triangle.
$AB=AC \, \, and\, \, \angle B=\angle C$
In $\Delta ABD \, \, and\, \, \Delta ECF$ ,
$\angle ABD=\angle ECF$ ( $\angle ABD=\angle B=\angle C=\angle ECF$ )
$\angle ADB=\angle EFC$ ( Each $90 ^\circ$ )
$\Delta ABD \sim \Delta ECF$ ( By AA criterion)
Answer:
AD and PM are medians of triangles. So,
$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$
Given :
$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$
$\Rightarrow \frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}$
$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$
In $\triangle ABD\, and\, \triangle PQM,$
$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$
$\therefore \triangle ABD\sim \triangle PQM,$ (SSS similarity)
$\Rightarrow \angle ABD=\angle PQM$ ( Corresponding angles of similar triangles )
In $\triangle ABC\, and\, \triangle PQR,$
$\Rightarrow \angle ABD=\angle PQM$ (proved above)
$\frac{AB}{PQ}=\frac{BC}{QR}$
Therefore, $\Delta ABC \sim \Delta PQR$ . ( SAS similarity)
Answer:
In, $\triangle ADC \, \, and\, \, \triangle BAC,$
$\angle ADC = \angle BAC$ ( given )
$\angle ACD = \angle BCA$ (common )
$\triangle ADC \, \, \sim \, \, \triangle BAC,$ ( By AA rule)
$\frac{CA}{CB}=\frac{CD}{CA}$ ( corresponding sides of similar triangles )
$\Rightarrow CA^2=CB\times CD$
Answer:
$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$ (given)
Produce AD and PM to E and L such that AD=DE and PM=DE. Now,
join B to E, C to E, Q to L and R to L.
AD and PM are medians of a triangle; therefore
QM=MR and BD=DC
AD = DE (By construction)
PM=ML (By construction)
So, diagonals of ABEC bisecting each other at D, so ABEC is a parallelogram.
Similarly, PQLR is also a parallelogram.
Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR
$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$ (Given )
$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{2.AD}{2.PM}$
$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$
$\Delta ABE \sim \Delta PQL$ (SSS similarity)
$\angle BAE=\angle QPL$ ...................1 (Corresponding angles of similar triangles)
Similarity, $\triangle AEC=\triangle PLR$
$\angle CAE=\angle RPL$ ........................2
Adding equations 1 and 2,
$\angle BAE+\angle CAE=\angle QPL+\angle RPL$
$\angle CAB=\angle RPQ$ ............................3
In $\triangle ABC\, and\, \, \triangle PQR,$
$\frac{AB}{PQ}=\frac{AC}{PR}$ ( Given )
$\angle CAB=\angle RPQ$ ( From above equation 3)
$\triangle ABC\sim \triangle PQR$ ( SAS similarity)
Answer:
CD = pole
AB = tower
Shadow of pole = DF
Shadow of tower = BE
In $\triangle ABE\, \, and\, \triangle CDF,$
$\angle CDF=\angle ABE$ ( Each $90 ^\circ$ )
$\angle DCF=\angle BAE$ (Angle of sun at same place )
$\triangle ABE\, \, \sim \, \triangle CDF,$ (AA similarity)
$\frac{AB}{CD}=\frac{BE}{QL}$
$\Rightarrow \frac{AB}{6}=\frac{28}{4}$
$\Rightarrow AB=42$ cm
Hence, the height of the tower is 42 cm.
Answer:
$\Delta AB C \sim \Delta PQR$ ( Given )
$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$ ............... ....1( corresponding sides of similar triangles )
$\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R$ ....................................2
AD and PM are medians of a triangle. So,
$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$ ..........................................3
From equations 1 and 3, we have
$\frac{AB}{PQ}=\frac{BD}{QM}$ ...................................................................4
In $\triangle ABD\, and\, \triangle PQM,$
$\angle B=\angle Q$ (From equation 2)
$\frac{AB}{PQ}=\frac{BD}{QM}$ (From equation 4)
$\triangle ABD\, \sim \, \triangle PQM, $ (SAS similarity)
$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$
Here are the exercise-wise links for NCERT class 10, Chapter 6 Triangles:
Question:
In $\triangle XYZ$, points P, Q, and R are points on the sides XY, YZ, and XZ, respectively $\angle YXZ = 50^\circ$, XR = PR, ZR = QR, and $\angle RQZ = 80^\circ$. What is the value of $\angle PRQ$?
Answer:
In $\triangle XPR$
$\angle PXR = 50^\circ$
Since XR = PR, the triangle is isosceles.
$\angle$XPR = $\angle$PXR = 50$^\circ$
⇒ $\angle XRP = 180^\circ-50^\circ-50^\circ= 80^\circ$
Since ZR = QR, the triangle is isosceles.
Similarly,
$\angle RQZ = \angle RZQ = 80^\circ$
⇒ $\angle QRZ = 180^\circ- 80^\circ- 80^\circ = 20^\circ$
Now, $\angle PRQ = 180^\circ - \angle RQZ - \angle XPR$
$⇒\angle PRQ = 180^\circ-20^\circ- 80^\circ=80^\circ$
Hence, the correct answer is $80^\circ$.
Topics you will learn in NCERT Class 10 Maths Chapter 6 Triangles include:
Based on the sides, we have 3 triangles:
Based on the measurement of the angle, we have 3 triangles:
Two triangles are similar if they have the same ratio of corresponding sides and an equal pair of corresponding angles.
Criteria for Triangle Similarity:
Theorem 6.1: If a line is drawn parallel to one side of a triangle and it cuts the other two sides at different points, it will divide those two sides in the same ratio.
Theorem 6.2: If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle.
Theorem 6.3: If two triangles have their corresponding angles equal, then their corresponding sides are in the same ratio, which means the two triangles are similar.
Theorem 6.4: If the sides of one triangle are in the same ratio as the sides of another triangle, then their corresponding angles are also equal, which makes the two triangles similar.
Theorem 6.5: If one angle of a triangle is equal to one angle of another triangle and the sides around these angles are in the same ratio, then the two triangles are similar.
We at Careers360 compiled all the NCERT class 10 Maths solutions in one place for easy student reference. The following links will allow you to access them.
After completing the NCERT textbooks, students should practice exemplar exercises to gain a better understanding of the chapters and improve clarity. The following links will help students find exemplar exercises.
Here are some useful links for NCERT books and the NCERT syllabus for class 10:
Frequently Asked Questions (FAQs)
To prove the Pythagorean theorem using similar triangles, we can draw an altitude from the right angle to the hypotenuse, creating two smaller triangles that are similar to each other and the original triangle, then use the proportional sides of similar triangles to derive the theorem.
Congruent triangles are identical in both shape and size, while similar triangles have the same shape but can differ in size, with corresponding sides being proportional. These concepts are discussed in class 10 maths chapter 6 solutions, practice these to command the concepts.
Similar triangles have various real-life applications, including measuring inaccessible heights or distances, understanding how zoom works on cameras, and in fields like cartography, engineering, and construction.
Thales's Theorem or Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
NCERT Class 10 Maths Chapter 6, "Triangles," focuses on the similarity of triangles, including criteria for similarity (AA, SSS, SAS), and theorems like the Basic Proportionality Theorem and Pythagoras Theorem.
On Question asked by student community
Hello,
If you want to get your 10th marksheet online, you just need to visit an official website like https://www.cbse.gov.in/ or https://results.cbse.nic.in/ for the CBSE board, and for the state board, you can check their website and provide your roll number, security PIN provided by the school, and school code to download the result.
I hope it will clear your query!!
Hello
You asked about Class 10 sample paper board exam and most important questions. Practicing sample papers and previous year questions is one of the best ways to prepare for the board exam because it gives a clear idea of the exam pattern and types of questions asked. Schools and teachers usually recommend students to solve at least the last five years question papers along with model papers released by the board.
For Class 10 board exams, the most important areas are Mathematics, Science, Social Science, English, and Hindi or regional language. In Mathematics, questions from Algebra, Linear Equations, Geometry, Trigonometry, Statistics, and Probability are repeatedly seen. For Science, the key chapters are Chemical Reactions, Acids Bases and Salts, Metals and Non metals, Life Processes, Heredity, Light and Electricity. In Social Science, priority should be given to Nationalism, Resources and Development, Agriculture, Power Sharing, Democratic Politics, and Economics related topics. In English, focus on unseen passages, grammar exercises, and important writing tasks like letter writing and essays.
Follow these steps to access the SQPs and marking schemes:
Step 1: Visit https://cbseacademic.nic.in/
Step 2: Click on the link titled “CBSE Sample Papers 2026”
Step 3: A PDF will open with links to Class 10 and 12 sample papers
Step 4: Select your class (Class 10 or Class 12)
Step 5: Choose your subject
Step 6: Download both the sample paper and its marking scheme
If you are looking for Class 10 previous year question papers for 2026 preparation, you can easily access them through the links I’ll be attaching. These papers are very helpful because they give you a clear idea about the exam pattern, marking scheme, and the type of questions usually asked in board exams. Practicing these will not only improve your time management but also help you identify important chapters and commonly repeated questions.
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers-class-10
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers
Hello,
Yes, you can give the CBSE board exam in 2027.
If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.
Hope it helps !
Here are some strategies so u can do best in your board exams and get god score
1. Make a good and smart schedule
2. If u r from cbse board go through ncert books by heart
3. Solve pyqs of each subject
4. Do revision on daily basis
5. Practice on presentation and writing the answer .
6. Do your best and give exam with the best way possible all the best blud .
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE