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Exercise 6.2 - Chapter 6 Triangles class 10 ncert solutions Maths - SaraNextGen [2024]


Question 1:

In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2645/Chapter%206_html_m4a074616.jpg

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2645/Chapter%206_html_m1605a652.jpg

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2645/Chapter%206_html_2d18db46.jpg

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2645/Chapter%206_html_2ce4782b.gif

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2645/Chapter%206_html_4f044872.jpg

Let AD = cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2645/Chapter%206_html_21dedbbf.gif

 

Question 2:

E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Answer:

(i)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2648/Chapter%206_html_5985f6b6.jpg

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2648/Chapter%206_html_mbc1563a.gif

(ii)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2648/Chapter%206_html_7cd946c3.jpg

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2648/Chapter%206_html_e81cbb4.gif

(iii)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2648/Chapter%206_html_6168fbb3.jpg

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2648/Chapter%206_html_m1053f8c9.gif

 

Question 3:

In the following figure, if LM || CB and LN || CD, prove that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2652/Chapter%206_html_m6c7892ec.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2652/Chapter%206_html_m210d666d.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2652/Chapter%206_html_m210d666d.jpg

In the given figure, LM || CB

By using basic proportionality theorem, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2652/Chapter%206_html_6d420a0c.gif

 

Question 4:

In the following figure, DE || AC and DF || AE. Prove that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2657/Chapter%206_html_5e9b45b3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2657/Chapter%206_html_m507899cc.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2657/Chapter%206_html_m5d61c865.jpg

In ΔABC, DE || AC

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2657/Chapter%206_html_m692270fb.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2657/Chapter%206_html_c51663f.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2657/Chapter%206_html_m238b1cd7.gif

 

Question 5:

In the following figure, DE || OQ and DF || OR, show that EF || QR.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2661/Chapter%206_html_1b4d2ac5.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2661/Chapter%206_html_m13b99bda.jpg

In Δ POQ, DE || OQ

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2661/Chapter%206_html_2ccff3f5.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2661/Chapter%206_html_m22e15ec8.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2661/Chapter%206_html_5d125a72.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2661/Chapter%206_html_m31b5d561.jpg

 

Question 6:

In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2666/Chapter%206_html_m2e4875cb.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2666/Chapter%206_html_m5f6dd6b8.jpg

In Δ POQ, AB || PQ

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2666/Chapter%206_html_593f6bf4.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2666/Chapter%206_html_491b8dcc.jpg

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2666/Chapter%206_html_m689851b5.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2666/Chapter%206_html_m3165d71a.jpg

 

Question 7:

Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer:

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/55555555555%2013_00_19(4).png

Consider the given figure in which l is a line  drawn through the mid-point P of line segment AB meeting AC at Q, such that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2671/Chapter%206_html_53bbca23.gif.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2671/Chapter%206_html_707fad93.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2671/Chapter%206_html_m422f3a06.gif

Or, Q is the mid-point of AC.

 

Question 8:

Using Converse of basic proportionality theorem, 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).

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2675/Chapter%206_html_5b91dee3.jpg

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

i.e., AP = PB and AQ = QC

It can be observed that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2675/Chapter%206_html_28e25231.gif

Hence, by using basic proportionality theorem, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2675/Chapter%206_html_6e6f9c5.gif

 

Question 9:

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_161fd90b.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_720dfdf5.jpg

Draw a line EF through point O, such that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_716b4dc3.gif

In ΔADC,https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_6769a1ad.gif

By using basic proportionality theorem, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_2403acb1.gif

In ΔABD, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_m9aa01c8.gif

So, by using basic proportionality theorem, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_m562c7b7e.gif

From equations (1) and (2), we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2678/Chapter%206_html_5ae9d7ca.gif

 

Question 10:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2680/Chapter%206_html_161fd90b.gifShow that ABCD is a trapezium.

Answer:

Let us consider the following figure for the given Question.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2680/Chapter%206_html_mb6c545b.jpg

Draw a line OE || AB

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2680/Chapter%206_html_m3e30a2a2.jpg

In ΔABD, OE || AB

By using basic proportionality theorem, we obtain

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2680/Chapter%206_html_m6a38d519.gif

However, it is given that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2680/Chapter%206_html_1681a126.gif

⇒ EO || DC [By the converse of basic proportionality theorem]

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.

Also Read : Exercise-6.3-Chapter-6-Triangles-class-10-ncert-solutions-Maths

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