{"id":624130,"date":"2023-09-01T06:56:36","date_gmt":"2023-09-01T06:56:36","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=624130"},"modified":"2023-09-01T06:57:15","modified_gmt":"2023-09-01T06:57:15","slug":"kc-sinha-exercise-5-3-mathematics-solution-class-10-chapter-5-triangles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-5-3-mathematics-solution-class-10-chapter-5-triangles\/","title":{"rendered":"KC Sinha: Exercise 5.3 – Mathematics Solution Class 10 Chapter 5 Triangles"},"content":{"rendered":"\n\n\n\n\n

Question 1 <\/h4>\n\n\n\n

State which of the following pairs of triangles are similar. Write the similarity criterion used and write the pairs of similar triangles in symbolic form (all lengths of sides are in cm).<\/strong><\/p>\n\n\n\n

(i) <\/strong><\/a><\/p>\n\n\n\n

(ii) <\/strong><\/a>
(iii) <\/strong><\/a><\/p>\n\n\n\n

(iv) <\/strong><\/a><\/p>\n\n\n\n

(v) <\/strong><\/a><\/p>\n\n\n\n

(vi) <\/strong><\/a>

<\/strong>
(vii) <\/strong><\/a>
<\/strong><\/p>\n\n\n\n

Sol :
(i) In \u0394ABC,
\u2220A = 70\u00b0 and \u2220B = 50\u00b0
And we know that, sum of the angles = 180\u00b0
\u21d2\u2220A + \u2220B +\u2220C = 180\u00b0
\u21d270\u00b0 + 50\u00b0 +\u2220C = 180\u00b0
\u21d2 120\u00b0 +\u2220C = 180\u00b0
\u21d2 \u2220C = 60\u00b0
<\/p>\n\n\n\n

And In \u0394DEF
\u2220F = 70\u00b0 and \u2220E = 50\u00b0
And we know that, sum of the angles = 180\u00b0
\u21d2\u2220D + \u2220E +\u2220F = 180\u00b0
\u21d2\u2220D + 50\u00b0 + 70\u00b0 = 180\u00b0
\u21d2 \u2220D = 60\u00b0
Yes,\u0394ABC ~ \u0394DEF [by AAA similarity criterion]
<\/p>\n\n\n\n

(ii) In \u0394ABC and \u0394PQR
Here,[Math Processing Error], [Math Processing Error], [Math Processing Error]
<\/p>\n\n\n\n

As,[Math Processing Error]
So, \u0394ABC ~ \u0394PQR [by SSS similarity criterion]
<\/p>\n\n\n\n

(iii) In\u0394MNL and \u0394PQR
\u2220NML = \u2220PQR = 70\u00b0<\/p>\n\n\n\n

[Math Processing Error]
[Math Processing Error]
and [Math Processing Error]
[Math Processing Error]
<\/p>\n\n\n\n

No, the two triangles are not similar.
<\/p>\n\n\n\n

(iv) In \u0394PQR and \u0394LMN
\u2220PQR = \u2220LNM = 50\u00b0
[Math Processing Error]
and [Math Processing Error]
[Math Processing Error]
\u2234\u0394PQR ~ \u0394LMN [by SAS similarity criterion]
<\/p>\n\n\n\n

(v) In \u0394LMP and \u0394DEF
Here,[Math Processing Error] <\/p>\n\n\n\n

[Math Processing Error], [Math Processing Error], [Math Processing Error]<\/p>\n\n\n\n

As[Math Processing Error]
So, no two triangles are not similar
<\/p>\n\n\n\n

(vi) In \u0394ABC and \u0394PQR
\u2220A = \u2220Q = 85\u00b0
\u2220B = \u2220P = 60\u00b0
and \u2220C =\u2220R = 35\u00b0
So, \u0394PQR ~ \u0394LMN [by AAA similarity]
<\/p>\n\n\n\n

(vii) In \u0394ABC and \u0394PQR
Here,[Math Processing Error],[Math Processing Error], [Math Processing Error]
As,[Math Processing Error]
So, \u0394ABC ~ \u0394PQR [by SSS similarity criterion]<\/p>\n\n\n\n

\n\n\n\n

Question 2 <\/h4>\n\n\n\n

If diagonals AC and BD of trapezium ABCD with AB || CD intersect each other at 0 and AB= 18 cm, DC = 30 cm, OB =y cm, OD= 10 cm, find y.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: ABCD is a trapezium with AB || CD
and diagonals AB and CD intersecting at O
To find: y
Firstly, we prove that \u0394OAB ~ \u0394ODC
Let \u0394OAB and \u0394ODC
\u2220AOB = \u2220COD [vertically opposite angles]
\u2220OBA = \u2220ODC [\u2235AB || CD with BD as transversal.
alternate angles are equal]
\u2220OAB = \u2220OCD [\u2235AB || CD with BD as transversal.
alternate angles are equal]
<\/p>\n\n\n\n

\u2234 \u0394OAB ~ \u0394ODC [by AAA similarity]
Since triangles are similar. Hence corresponding sides are proportional.
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2 y = 6cm<\/p>\n\n\n\n

\n\n\n\n

Question 3 <\/h4>\n\n\n\n

In the given figure BC = 5 cm, AC = 5.5 cm and AB= 4.6 cm. P and Q are points on AB and AC respectively such that PQ || BC. If PQ = 2.5 cm, find other sides of \u0394APQ.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: PQ || BC
To find: AP and AQ
Since, PQ || BC, AB is transversal, then,
\u0394APQ = \u0394ABC [by corresponding angles]
Since, PQ || BC, AC is transversal, then,
\u0394APQ = \u0394ABC [by corresponding angles]
In \u0394APQ and\u0394ABC
\u2220APQ = \u2220ABC
\u2220AQP = \u2220ACB
<\/p>\n\n\n\n

\u2234 \u0394APQ \u2245 \u0394ABC [by AAA similarity]
Since, the corresponding sides of similar triangles are proportional
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2AP = 2.3
<\/p>\n\n\n\n

Now, taking [Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2AQ = 2.75
Therefore, AP = 2.3cm and AQ = 2.75cm<\/p>\n\n\n\n

\n\n\n\n

Question 4 <\/h4>\n\n\n\n

In the given figure \u0394ABR ~ <\/strong>\u0394PQR, if PQ = 30 cm, AR = 45 cm, AP = 72 cm and QR = 42 cm, find PR and BR.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: \u0394ABR ~ \u0394PQR
As, \u0394ABR and \u0394PQR are similar
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2BR = 70cm
and PR = AP \u2013 AR = 72 \u2013 45 = 27cm<\/p>\n\n\n\n

\n\n\n\n

Question 5<\/h4>\n\n\n\n

In the given figure, QA and PB are perpendiculars to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm, find AQ.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Let us take \u0394OAQ and \u0394OBP
\u2220AOQ = \u2220BOP (vertically opposite angles)
\u2220OAQ = \u2220OBP (each 90\u00b0)
\u2234 \u0394OAQ ~\u0394OBP (by AA similarity criterion)
Given: AO = 10 cm, BO = 6 cm and PB = 9 cm
As, \u0394OAQ ~\u0394OBP
[Math Processing Error]
[Math Processing Error]
\u21d2AQ = 15cm<\/p>\n\n\n\n

\n\n\n\n

Question 6 <\/h4>\n\n\n\n

In the given figure \u0394ACB ~ \u0394APQ, if BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP= 2.8 cm, Find CA and AQ.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: \u0394ACB ~ \u0394APQ
As, \u0394ACB and \u0394APQ are similar
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
Taking [Math Processing Error]
<\/p>\n\n\n\n

\u21d2CA = 5.6cm
Now, taking [Math Processing Error]
\u21d2AQ =3.25cm<\/p>\n\n\n\n

\n\n\n\n

Question 7 <\/h4>\n\n\n\n

In the given figure, XY || BC. Find the length of XY, given BC = 6 cm.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: XY || BC
To find: XY
Since, XY || BC, AB is transversal, then,
\u0394AXY = \u0394ABC [by corresponding angles]
Since, XY || BC, AC is transversal, then,
\u0394AYX = \u0394ABC [by corresponding angles]
In \u0394AXY and\u0394ABC
\u2220AXY = \u2220ABC
\u2220AYX = \u2220ACB
\u2234 \u0394AXY \u2245 \u0394ABC [by AA similarity]
Since, triangles are similar, hence corresponding sides will be proportional
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2XY = 1.5
Therefore, XY= 1.5cm<\/p>\n\n\n\n

\n\n\n\n

Question 8 <\/h4>\n\n\n\n

The perimeters of two similar triangles, ABC and PQR (\u0394ABC~ \u0394PQR) are respectively 72 cm and 48 cm. If PQ = 20 cm, find AB.<\/strong><\/p>\n\n\n\n

Sol :
Given: \u0394ABC~\u0394PQR, PQ =20cm
And perimeter of \u0394ABC and\u0394PQR are 72cm and 48cm respectively.
As, \u0394ABC ~\u0394PQR
[Math Processing Error] (corresponding sides are proportional)
[Math Processing Error][Math Processing Error]
[Math Processing Error]<\/p>\n\n\n\n

[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2AB = 30cm<\/p>\n\n\n\n

\n\n\n\n

Question 9 <\/h4>\n\n\n\n

In the given figure, if PQ || RS, prove that \u0394POQ ~ \u0394 SOR.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: PQ || RS
To Prove: \u0394POQ ~\u0394SOR
Let us take\u0394POQ and\u0394SOR
\u2220OPQ = \u2220OSR (as PQ || RS, Alternate angles)
\u2220POQ = \u2220ROS (vertically opposite angles)
\u2220OQP = \u2220ORS (as PQ || RS, Alternate angles)
\u2234 \u0394POQ ~\u0394SOR (by AAA similarity criterion)
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 10 <\/h4>\n\n\n\n

In the given figure, if \u2220A = \u2220C, then prove that \u0394 AOB ~ \u0394 COD<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: \u2220A = \u2220C
To Prove: \u0394AOB ~\u0394COD
Let us take \u0394AOB and\u0394COD
\u2220A = \u2220C (given)
\u2220AOB = \u2220COD (vertically opposite angles)
\u2234 \u0394AOB ~\u0394COD (by AA similarity criterion)
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 11 <\/h4>\n\n\n\n

In the given figure DB \u22a5 BC, DE \u22a5 AB and AC \u22a5 BC, prove that \u0394BDE ~\u0394ABC.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
We have, DB \u4e04 BC and AC \u4e04 BC
\u2220B + \u2220C = 90\u00b0 + 90\u00b0
\u21d2 \u2220B + \u2220C = 180\u00b0
\u2234 BD || AC
\u21d2\u2220EBD = \u2220CAB (alternate angles)
Let us take \u0394BDE and\u0394ABC
\u2220BED = \u2220ACB (each 90\u00b0)
\u2220EBD = \u2220CAB (alternate angles)
\u2234 \u0394BDE ~\u0394ABC (by AA similarity criterion)
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 12 <\/h4>\n\n\n\n

In the given figure, <\/strong>\u22201 =\u22202 and [Math Processing Error], prove that \u0394ACB ~ \u0394DCE<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :<\/p>\n\n\n\n

We have, [Math Processing Error]
[Math Processing Error]
[Math Processing Error] (\u2235, BD = DC as \u22201 = \u22202) \u2026(i)
Also, \u22201 = \u22202
i.e. \u2220DBC = \u2220ACB
\u2234 \u0394ACB ~ \u0394DCE (by SAS similarity criterion)
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 13 <\/h4>\n\n\n\n

In an isosceles \u0394ABC with AC = BC, the base AB is produced both ways to P and Q such that AP x BQ = AC2<\/sup>. Prove that : \u0394ACP ~ \u0394BQC<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given ABC is an isosceles triangle and AC = BC
\u2235 AC = BC
\u21d2\u2220CAB = \u2220CBA
\u21d2180\u00b0 \u2013 \u2220CAB = 180\u00b0 \u2013 \u2220CBA
\u21d2 \u2220CAP = \u2220CBQ
Also, AP x BQ = AC2<\/sup>
[Math Processing Error]
[Math Processing Error] (\u2235 AC =BC)<\/p>\n\n\n\n

Thus, by SAS similarity, we get
\u0394ACP ~ \u0394BQC
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 14 <\/h4>\n\n\n\n

In the given figure, find \u2220P.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
From the figure,
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
<\/p>\n\n\n\n

Hence, [Math Processing Error]
<\/p>\n\n\n\n

Now it can be seen that both the triangles are similar as the corresponding sides are propotional.
From the figure we can see that,
\u2220 P = \u2220 C
From \u0394ABC,
\u2220 A + \u2220 B + \u2220 C = 180\u00b0
60\u00b0 + 80\u00b0 + \u2220 C = 180\u00b0
\u2220 C = 180\u00b0 \u2013 140\u00b0
\u2220 C = 40\u00b0
Hence, \u2220 P = 40\u00b0<\/p>\n\n\n\n

\n\n\n\n

Question 15 <\/h4>\n\n\n\n

P and Q are points on the sides AB and AC respectively of a \u0394ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, show that BC = 3 PQ.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Here, [Math Processing Error]
and [Math Processing Error]
[Math Processing Error]<\/p>\n\n\n\n

\u2234 PQ || BC [by converse of basic proportionality theorem]
Now, take \u0394APQ and \u0394ABC
\u2220APQ = \u2220ABC (corresponding angles)
\u2220AQP = \u2220ACB (corresponding angles)
<\/p>\n\n\n\n

\u2234 \u0394APQ ~ \u0394ABC (by AA similarity criterion)
<\/p>\n\n\n\n

Since, triangles are similar, hence corresponding sides will be proportional
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2BC = 3PQ
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 16 <\/h4>\n\n\n\n

P and Q are respectively the points on the sides AB and AC of a \u0394ABC. If AP = 2 cm, PB = 6 cm, AQ = 3 cm and QC = 9, Prove that BC = 4PQ.<\/strong><\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Here, [Math Processing Error]
and [Math Processing Error]
[Math Processing Error]<\/p>\n\n\n\n

\u2234 PQ || BC [by the converse of basic proportionality theorem]
Now, take \u0394APQ and \u0394ABC
\u2220APQ = \u2220ABC (corresponding angles)
\u2220AQP = \u2220ACB (corresponding angles)
\u2234 \u0394APQ ~ \u0394ABC (by AA similarity criterion)
<\/p>\n\n\n\n

Since, triangles are similar, hence corresponding sides will be proportional
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2BC = 4PQ
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 17 <\/h4>\n\n\n\n

In the given figure, [Math Processing Error] and AB = 5 cm. Find the value of DC.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
In \u0394AOB and \u0394COD,
\u2220 AOB = \u2220COD (Vertically opposite angles)
 [Math Processing Error](given)
Therefore according to SAS similarity criterion,
\u2234 \u0394AOB ~ \u0394COD
<\/p>\n\n\n\n

Since, triangles are similar, hence corresponding sides will be proportional
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2 DC = 10cm<\/p>\n\n\n\n

\n\n\n\n

Question 18 <\/h4>\n\n\n\n

In the given figure, OA .OB = OC.OD, show that: \u2220A = \u2220C and \u2220B = \u2220D.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: OA \u00d7 OB = OC \u00d7 OD
To Prove: \u2220A = \u2220C and \u2220B = \u2220D
Now, OA .OB = OC.OD
 [Math Processing Error]\u2026(i)
<\/p>\n\n\n\n

In \u25b3AOD and \u25b3COB
[Math Processing Error] (from (i))
\u2220AOD = \u2220COB (vertically opposite angles)
\u2234 \u25b3AOD ~ \u25b3COB (by SAS similarity criterion)
We know that if two triangles are similar then their corresponding angles are equal.
\u21d2 \u2220A = \u2220C and \u2220B = \u2220D
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 19 <\/h4>\n\n\n\n

In the given figure, CM and RN are respectively the medians of \u0394ABC and \u0394PQR. If \u0394ABC ~ \u0394PQR, prove that:<\/strong><\/p>\n\n\n\n

(i) \u0394AMC ~ \u0394PNR<\/strong>
(ii) <\/strong>[Math Processing Error]<\/strong>
(iii) \u0394CMB ~ \u0394RNQ<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: CM is the median of \u25b3ABC and RN is the median of \u25b3PQR
Also, \u25b3ABC ~ \u25b3PQR
To Prove: (i) \u25b3AMC ~\u25b3PNR
CM is median of \u25b3ABC
<\/p>\n\n\n\n

So, [Math Processing Error] \u2026(1)
<\/p>\n\n\n\n

Similarly, RN is the median of \u25b3PQR
So, [Math Processing Error] \u2026(2)
<\/p>\n\n\n\n

Given \u25b3ABC ~\u25b3PQR
[Math Processing Error]
(corresponding sides of similar triangle are proportional)
[Math Processing Error]
[Math Processing Error] (from (1) and (2))<\/p>\n\n\n\n

[Math Processing Error] \u2026(3)
<\/p>\n\n\n\n

Also, since \u25b3ABC ~\u25b3PQR
\u2220A = \u2220P \u2026(4)
(corresponding angles of similar triangles are equal)
<\/p>\n\n\n\n

In \u25b3AMC and \u25b3PNR<\/p>\n\n\n\n

\u2220A = \u2220P (from (4))
[Math Processing Error] (from (3))
\u2234 \u25b3AMC ~\u25b3PNR (by SAS similarity)
Hence Proved
<\/p>\n\n\n\n

(ii)To Prove:[Math Processing Error]
In part (i), we proved that \u25b3AMC ~\u25b3PNR
So, [Math Processing Error]
(corresponding sides of a similar triangle are proportional)
Therefore, [Math Processing Error]
[Math Processing Error]
[Math Processing Error]
Hence Proved
<\/p>\n\n\n\n

(iii) \u25b3CMB ~\u25b3RNQ
Given \u25b3ABC ~\u25b3PQR
[Math Processing Error]
(corresponding sides of similar triangle are proportional)
[Math Processing Error]
[Math Processing Error] (from (1) and (2))
[Math Processing Error] \u2026(5)
Also, since \u25b3ABC ~\u25b3PQR
\u2220B = \u2220Q \u2026(6)
(corresponding angles of similar triangles are equal)
In \u25b3CMB and \u25b3RNQ
\u2220B = \u2220Q (from (6))
[Math Processing Error] (from (5))
\u2234 \u25b3CMB ~\u25b3RNQ (by SAS similarity)
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 20 <\/h4>\n\n\n\n

In the adjoining figure, the diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Show that DF x FE = BF x FA.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: ABCD is a parallelogram
To Prove: DF x FE = BF x FA
In \u25b3AFD and \u25b3BFE
\u22201 = \u22202 (alternate angles)
\u22203 = \u22204 (vertically opposite angles)
\u2234 \u25b3AFD ~\u25b3BFE (by AA similarity criterion)
So, [Math Processing Error]
(corresponding sides of similar triangle are proportional)
[Math Processing Error]
\u21d2 DF x FE = BF x FA
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 21 <\/h4>\n\n\n\n

In the given figure, DEFG is a square and \u2220BAC is a right angle. Show that DE2<\/sup>= BD x EC.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: DEFG is a square and \u2220BAC = 90\u00b0
To Prove: DE2<\/sup> = BD \u00d7 EC.
In \u25b3AGF and \u25b3DBG
\u2220GAF = \u2220BDG [each 90\u00b0]
\u2220AGF = \u2220DBG
[corresponding angles because GF|| BC and AB is the transversal]
\u2234\u25b3AFG ~ \u25b3DBG [by AA Similarity Criterion] \u2026(1)
In \u25b3AGF and \u25b3EFC
\u2220GAF = \u2220CEF [each 90\u00b0]
\u2220AFG = \u2220ECF
[corresponding angles because GF|| BC and AC is the transversal]
\u2234\u25b3AGF ~\u25b3EFC [by AA Similarity Criterion] \u2026(2)
From equation (1) and (2), we have
\u25b3DBG ~ \u25b3EFC
Since, the triangle is similar. Hence corresponding sides are proportional
[Math Processing Error]
[Math Processing Error] [\u2235DEFG is a square]
\u21d2DE2<\/sup> = BD \u00d7 EC
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 22 <\/h4>\n\n\n\n

In the given figure, ABD is a right angled triangle being right angled at A and AD \u22a5 BC. Show that:<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

(i) AB2<\/sup>= BC.BD<\/strong>
(ii) AC2<\/sup> = BC. DC<\/strong>
(iii) AB. AC. = BC. AD<\/strong>
Sol :
(i) In \u0394DAB and \u0394ACB
\u2220ADB = \u2220CAB [each 90\u00b0]
\u2220DAB = \u2220CAB [common angle]
\u2234 \u25b3DAB ~\u25b3ACB [by AA similarity]
<\/p>\n\n\n\n

Since the triangles are similar, hence corresponding sides are in proportional.
[Math Processing Error]
\u21d2 AB2<\/sup>= BC\u00d7BD
<\/p>\n\n\n\n

(ii) In \u25b3ACB and \u25b3DAC<\/p>\n\n\n\n

\u2220CAB = \u2220ADC [each 90\u00b0]
\u2220CAB = \u2220CAD [common angle]
\u2234 \u25b3ACB ~\u25b3DAC [by AA similarity]
Since the triangles are similar, hence corresponding sides are in proportional.
[Math Processing Error]
\u21d2 AC2<\/sup> = BC. DC
<\/p>\n\n\n\n

(iii) In part (i) we proved that \u25b3DAB ~\u25b3ACB
[Math Processing Error]
\u21d2 AB \u00d7 AC = BC \u00d7 AD
Hence Proved<\/p>\n\n\n\n

\n\n\n\n

Question 23 <\/h4>\n\n\n\n

In the given figure, \u2220ABC = 90\u00b0 and BD \u22a5 AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: \u2220ABC = 90\u00b0 and BD\u4e04 AC
and AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm
To find: BC
Firstly, we have to show that \u25b3ABC ~\u25b3BDC
Let \u25b3ABC and \u25b3BDC
\u2220ABC = \u2220BDC [each 90\u00b0]
\u2220ACB = \u2220BCD [common angle]
\u2234 \u25b3ABC ~\u25b3BDC [by AA similarity criterion]
Since, triangles are similar, hence corresponding sides are proportional.
[Math Processing Error]
[Math Processing Error]
[Math Processing Error]
\u21d2 BC = 8.1cm<\/p>\n\n\n\n

\n\n\n\n

Question 24 <\/h4>\n\n\n\n

In the given figure, \u2220CAB =90\u00b0 and AD \u22a5 BC. Show that \u0394BDA ~ \u0394BAC. If AC = 75 cm, AB = 1 cm and BC = 1.25 cm, find AD.<\/strong><\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Sol :
Given: \u2220CAB =90\u00b0 and AD\u4e04 BC
and AC = 75 cm, AB = 1 cm and BC = 1.25 cm
Now, In \u25b3ADB and \u25b3CAB<\/p>\n\n\n\n

\u2220ADB = \u2220CAB [each 90\u00b0]
\u2220ABD = \u2220CBA [common angle]
\u2234 \u25b3ADB ~\u25b3CAB [by AA similarity]
Since the triangles are similar, hence corresponding sides are in proportional.
[Math Processing Error]
[Math Processing Error]
\u21d2 AD = 60cm<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 10 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1  State which of the following pairs of triangles are similar. Write the similarity criterion used and write the pairs of similar triangles in symbolic form (all lengths of sides are in cm). (i)  (ii) (iii)  (iv)  (v)  (vi)  (vii)  Sol :(i) In \u0394ABC,\u2220A = 70\u00b0 and \u2220B = 50\u00b0And we know that, sum of the […]<\/p>\n","protected":false},"author":302,"featured_media":624114,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[24],"tags":[],"boards":[],"class_list":["post-624130","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-10","entry"],"yoast_head":"\nKC Sinha: Exercise 5.3 - Mathematics Solution Class 10 Chapter 5 Triangles - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1  State which of the following pairs of triangles are similar. 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