Q.1Â Â Â State which pairs of triangle in figure, 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 : Sol.
(i) In s ABC and PQR, we observe that
, and
Therefore by AAA criterion of similarity,
(ii) In s ABC and PQR, we observe that
Therefore by SSS criterion of similarity,
(iii) In s LMP and DEF, we observe that the ratio of the sides of these triangles are not equal.
So, the two triangles are not similar.
(iv) In s MNL and QPR, we observe that
But,
Therefore These two triangles are not similar as they do not satisfy SAS criterion of similarity.
(v) In s ABC and FDE, we have
But, [Because AC is not given]
So, by SAS criterion of similarity, these two triangless are not similar.
(vi) In s DEF and PQR
[Because ]
So, by AAA criterion of similarity,
Q.2Â Â Â Â Â In figures, , and . Find , and .
Sol.
Since BD is a line and OC is a ray on it,
Therefore
In , we have
It is given that
,
and
Hence, ,
and
Q.3Â Â Â Â Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that .
Sol.
Given : ABCD is a trapezium in which AB || DC.
To prove :
Proof : In OAB and OCD, we have
[Vert. opp. ]
[Alternate ]
and [Alternate ]
Therefore by AAA criterion of similarity,
Hence,
[Because in case of two similar triangles, the ratio of their corresponding sides are equal]
Q.4Â Â Â Â Â In figure, and . Show that
Sol.
We have,
.............. (1)
Also, [Given]
PR = PQ .............. (2) [Because sides opp. to equal are equal]
From (1) and (2), we get
........... (3)
Therefore in PQS and TQR, we have
and
Therefore by SAS criterion of similarity,
.
Q.5Â Â Â Â Â S and T are points on sides PR and QR of such that . Show that .
Sol.
In RPQ and RTS, we have
[Given]
[Common]
Therefore by AA criterion of similarity,
Q.6Â Â Â Â Â In figure, if , show that .
Sol.
It is given that
Therefore AB = AC [Because corresponding parts of congruent triangles are equal]
and, AE = AD
.......... (1)
Therefore In ADE and ABC, we have
[Because of (1)]
and, [Common]
Thus, by SAS criterion of similarity,
Q.7Â Â Â Â Â Â In figure, altitudes AD and CE of intersect each other at the point P, Show that :
Â Â Â Â Â Â Â Â Â Â Â Â (i)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (iii)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (iv)
Sol.
(i)Â Â In AEP and CDP, we have
Â Â Â Â Â Â [Because and ]
Â Â Â Â Â Â [Vert. opp. ]
Â Â Â Â Â Â Therefore by AA - criterion of similarity, we have
Â Â Â Â Â Â
(ii)Â In ABD and CBE, we have
Â Â Â Â Â Â [Common angle]
Â Â Â Â Â Â
Â Â Â Â Â Â Therefore by AA-criterion of similarity, we have
Â Â Â Â Â Â
(iii)Â In AEP and ADB, we have
Â Â Â Â Â Â Â Â [Becauese and ]
Â Â Â Â Â Â Â Â [Common angle]
Â Â Â Â Â Â Â Â Therefore by AA-criterion of similarity, we have
Â Â Â Â Â Â Â Â
(iv)Â Â In PDC and BEC, we have
Â Â Â Â Â Â Â Â [Because and ]
Â Â Â Â Â Â Â Â [Common angle]
Â Â Â Â Â Â Â Â Therefore AA-criterion of similarity, we have
Â Â Â Â Â Â Â Â
Q.8Â Â Â Â Â E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that .
Sol.
In ABE and CFB, we have
[Alt. ]
[Opp. of a ||gm]
Therefore by AA-criterion of similarity, we have
Q.9Â Â Â Â Â In figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that :
Â Â Â Â Â Â Â Â Â Â Â Â Â Â (i)Â
Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii)
Sol.
(i)Â Â In ABC and AMP, we have
Â Â Â Â Â Â [Given]
Â Â Â Â Â Â and, [Common ]
Â Â Â Â Â Â Therefore by AA, criterion of similarity, we have
Â Â Â Â Â Â
(ii)Â We have, [As proved above]
Â Â Â Â Â Â Â
Q.10Â Â Â Â CD and GH are respectively the bisectors of and such that D and H lie on sides AB and FE of and respectively. If , show that :
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (i)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (ii)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (iii)
Sol.
We have,
.............. (1)
and,
and .............. (2)
[Because CD and GH are bisectors of and respectively.
Therefore In DCA and HGF, we have
[From (1)]
[From (2)]
Therefore by AA-criterion of similarity, we have
which proves the (iii) part.
We have, [As proved above]d
which proves the (i) part.
In DCB and HGE, we have
[From (2)]
[Because ]
Therefore by AA-criterion of similarity, we have
which proves, the (ii) part.
Q.11Â Â Â In figure, E is a point on side CB produced of an isosceles triangle ABC withÂ AB = AC. If and , prove that .
Sol.
Here, is isosceles with AB = AC
In ABD and ECF, we have
[Because ]
[Because and ]
Therefore by AA-criterion of similarity,
.
Q.12Â Â Â Â Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of (see figure). Show that .
Sol.
Given : AD is the median of ABC and PM is the median of PQR such that
To prove :
Proof : [Given]
and, [Given]
Also, [Given]
By SSS-criterion of similarity, we have
[Similar have corresponding ]
and, [Given]
Therefore by SAS-criterion of similarity, we have
Q.13Â Â Â D is a point on the side BC of a triangle ABC such that . Show that .
Sol.
In ABC and DAC, we have
and
Therefore by AA-criterion of similarity, we have
Q.14Â Â Â 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 .
Sol.Â Â Â Â Â Â Â Same as question 12.
Q.15Â Â Â A 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.
Sol.
Let AB be the vertical pole and AC be the shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF. Let DE = x metres.
We have, AB = 6 m, AC = 4 cm and DF = 28 m
In ABC and DEF, we have
,
and
[Because each is the angular elevation of the sun]
Therefore by AA-criterion of similarity,
Hence, the height of the tower is 42 metres.
Q.16Â Â Â If AD and PM are medians of triangles ABC and PQR respectively, where , prove that
Sol.
Given : AD and PM are the medians of ABC and PQR respectively, where
To prove :
Proof : In ABD and PQM, we have
[Because ]
Â [Since AD and PM are the median of BC and QR respectively]
Therefore by SAS - criterion of similarity
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