X similar triangle solves extra score guide questions

CBSE TEST PAPER MATHEMATICS (Class-10) SIMILAR TRIANGLE

1. In an equilateral Δ ABC, the side BC is trisected at D. Prove that 9AD2 = 7AB2

2.  P and Q are points on sides AB and AC respectively, of ΔABC. If AP = 3 cm,PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.

3. The image of a tree on the film of a camera is of length 35 mm, the distance from the lens to the film is 42 mm and the distance from the lens to the tree is 6 m. How tall is the portion of the tree being photographed?

4. . Prove that in any triangle the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median, which bisects the third side.

5.  If a straight line is drawn parallel to one side of a triangle intersecting the othertwo sides, then it divides the two sides in the same ratio.

6. If ABC is an obtuse angled triangle, obtuse angled at B and if AD ^ CB Prove that  AC²=AB²+ BC²+2 BC x BD

7. If a straight line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

8. ABCD is a quadrilateral with AB =AD. If AE and AF are internal bisectors of ΔABC, D and E are points on AB and AC respectively such that AD/ DB = AEC/EC and ΔABC is isosceles.

9. In a ΔABC, points D, E and F are taken on the sides AB, BC and CA respectively such that DE IIAC and FE II AB.

10. . Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.

11. If a perpendicular is drawn from the vertex of a right angled triangle to its hypotenuse, then the triangles on each side of the perpendicular are similar to the whole triangle.

12. A man sees the top of a tower in a mirror which is at a distance of 87.6 m from the tower. The mirror is on the ground, facing upward. The man is 0.4 m away from the mirror, and the distance of his eye level from the ground is 1.5 m. How tall is the tower? (The foot of man, the mirror and the foot of the tower lie along a straight line).

13. In a right Δ ABC, right angled at C, P and Q are points of the sides CA and CB respectively, which divide these sides in the ratio 2: 1. Prove that

(I) 9AQ²= 9AC²+4BC² (II) 9 BP²= 9 BC²+ 4AC² (III) 9 (AQ²+BP²) = 13AB²

14. ABC is a triangle. PQ is the line segment intersecting AB in P and AC in Q such that PQ parallel to BC and divides Δ ABC into two parts equal in area. Find BP: AB.

15. P and Q are the mid points on the sides CA and CB respectively of triangle ABC right angled at C. Prove that4(AQ²+BP²) = 5 AB²

CBSE NCERT MATH 10th Chapter 6 Triangles Test Paper

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cbse-math-triangles NCERT optional question solved

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Given that ABCD be a parallelogram.

Construction : We have to Draw AF and DE perpendicular on side DC and on extended side AB to use Pythagoras theorem

In ΔDEA, DE2 + EA2 = DA2 … (i)

Similarly, In ΔDEB,

DE2 + EB2 = DB2

DE2 + (EA + AB)2 = DB2 ( EB =EA + AB)

DE2 + EA2 + AB2 + 2EA × AB == DB2

(DE2 + EA2) + AB2 + 2EA × AB = DB2

DA2 + AB2 + 2EA × AB = DB2 … (ii)( Using -----1)

In ΔAFC, AC2 = AF2 + FC2

= AF2 + (DC − FD)2

= AF2 + DC2 + FD2 − 2DC × FD

= (AF2 + FD2) + DC2 − 2DC × FD

=AC2 = AD2 + DC2 − 2DC × FD ( using AD2 = AF2 + FD2)… (iii)

Now,given that  ABCD is a parallelogram, AB = CD … (iv) And, BC = AD … (v)

Look In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF -Alternate angles )

AD = AD (Common)

∴ ΔEAD cong. ΔFDA (AAS congruence criterion)

⇒ EA = DF … (vi)
Adding equations (i) and (iii), we obtain

DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2

DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2

BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2

[Using equations (iv) and (vi)]

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

Q. In Fig. is the perpendicular bisector of the line segment DE, FA perpendiculat OB and F E intersects OB at the point C. Prove that 1/ OA + 1/OB = 1/OC

CBSE Maths X Triangles – Criteria for Similarity of Triangles

Math Adda


10th math Converse of Thales theorem


10th math Triangles Thales theorem and angle bisector theorem


10th math SAS similarity of triangle


10th SSS similarity for two triangles are similar

10th AA- similarity for two triangles are similar

10th math SAS similarity of triangle

Math Adda

Statement: Two triangles are similar to each other when one angle of a triangle equal to an angle of other triangle and sides making these angles are proportional.

Proof: We are given DABC and DPQR such that  < A = <P

And AB / PQ = AC / PR


We have to prove. DABC ~ DPQR

To prove this: On the sides PQ and QR of DPQR, take points M and N such that 

AB = PM and AC = PN Join MN

Now 

AB / PQ = AC / PR
PM / PQ = PN / PR [Because AB = PM and AC = PN] 

Thus, MR || QR [Converse of Thales theorem]

So, <1 =<Q And <2 = <R [corresponding angles]

Therefore, DPMN ~ DPQR [A. A Similarity]

So, PM / PQ = MN / QR = PN / PR --------------- (1) [Sides of similar triangles are proportional]

Now, in DABC and DPMN

AB = PM [we have constructed]

<A = <P [Given]

AC = PN [we have constructed]

Thus, DABC@ DPMN

Thus, <A = <P, <B = <M and <C = <N

So DABC ~ DPQR [Because DABC @ DPMN and DPMN ~ DPQR]

This condition of similarity is known as SAS similarity.

10th SSS similarity for two triangles are similar

Two triangles are similar when their corresponding sides are proportional.

Let us prove this condition as a theorem, with the help of Thales Theorem.

Statement: Two triangles are similar to each other when their corresponding sides are proportional.

Proof: We are given DABC and DPQR such that

AB / PQ = BC / QR = AC / PR

We have to prove D ABC ~ DPQR

To prove this: On the sides PQ and QR of DPQR take points M and N such that

AB = PM and AC = PN. Join MN
Now,
 AB / PQ = AC / PR [Given]
i.e. PM / PQ = PN / PR [Because AB = PM and AC = PN]

Thus, MN || QR [Converse of Thales theorem]

So, <1 = <Q and <2 = <R [corresponding angles]

Therefore, DPMN ~ DPQR [AA similarity]

So, PM / PQ = MN / QR = PN / PR ----------------- (1) [Sides of similar triangles are proportional]

Since AB / PQ = PM / PQ [Because AB = PM]

And AB / PQ = BC / QR [Given]

Therefore,

PM / PQ = BC / QR ----------------- (2)

From (1) and (2)

MN / QR = BC / QR

MN =BC

Now, in D ABC and DPMN

AB =PM [We have constructed]

BC = MN [Proved above]

AC = PN [We have constructed]

Thus, DABC@D PMN [SSS congruency]

Thus, < A = <P, <B = <M and <C = vN

So DABC ~ DPQR [Because DABC @ D PMN and D PMN ~ D PQR]

We represent this condition of similarity as side - side similarity or SSS similarity

Two triangles are similar when one angle of a triangle is equal to an angle of other triangle and the sides making these angles are proportional.

Let us prove this as a theorem with the help of Thales theorem.

AA similarity Conditions under which two Triangles are Similar

Math Adda    Conditions under which two Triangles are Similar



Two triangles are similar when their corresponding angles are equal.

Let us prove these conditions as a theorem with the help of Thales Theorem.

Statement: Two triangles are similar if their corresponding angles are equal.

Proof: we are given D ABC and D PQR such that

< A = <P 
<B = <Q
<C = <R

We have to prove: DABC ~ DPQR

To prove this: On the sides PQ and PR of DPQR take points M and N such that 

AB = PM and AC = PN. Join MN.

Now, in DABC and DPMN

AB = PM [We have constructed] 
<A = <P [Given] 
AC = PN [We have constructed] 

Therefore DABC ~ DPMN [SAS congruency] 

This means
<PMN = <B [c. p. c. t]
But <B = <Q [Given] 

Therefore, 
<PMN = <Q
These are corresponding angles.

Thus, MN || QR

Now, 

PM / MQ = NR / PN {Thales theorom}

i.e. MQ / PM = NR / PN

i.e. MQ / PM + 1 = NR / PN +1

i.e. MQ + PM / PM = NR + PN/ PN

i.e. PQ / PM = PN / PR

i.e. PM / PQ = PR / PN

Since PM = AB and PN = AC

Therefore

AB / PQ = AC / PR

Similarly we can prove

AB / PQ = BC / QR

Hence AB / PQ = BC / QR = AC / PR, we already have been given that <A = <P, <B = <Q and <C = <F.

This way both conditions of similarity are fulfilled.

Thus, DABC ~ DPQR when their corresponding angles are equal.

We denote this condition as 

Angle - Angle - Angle similarity or AAA similarity.

Note: When two angles of a triangle are equal to two corresponding angles of another triangle then their remaining angles are also equal to each other. This AAA similarity is also written as AA similarity

10th math Converse of Thales theorem

Math Adda

If a line divides any two sides of a triangles proportionally then the line is parallel to the third side

Statement: If a line divides any two sides of a triangles proportionally (in same ratio), then the line is parallel to the third side.

Proof: We are given ABC 

AD / BD = AE / CE

We have to prove: DE is parallel to BC

Let DE's not parallel to BC then an another line DE' is parallel to BC.

Now

AD / BD = AE / CE [Given]

And AD / BD = AE' / CE' [Thales Theorem]

Therefore AE / CE = AE' / CE'

i.e. AE / CE + 1 = AE' / CE' + 1

i.e. AE + CE / CE = AE' + CE' / CE' 

i.e. CE = CE'

But this is not possible until E and E' is coincident.

Thus, our assumption is wrong and DE is parallel to BC.

10th math Triangles Thales theorem and angle bisector theorem

CBSE MATH Similarity of Triangles FOR X

MATH ADDA "10TH MATHS BY JSUNIL": Chapter 12: Trigonometry -Trigonometry Identities

MATH ADDA "10TH MATHS BY JSUNIL": Chapter 12: Trigonometry -Trigonometry Identities: "CBSE Maths eBooks Let us take any acute angle taking point P on OB we draw perpendicular PQ on OAB. Thus we have a right angled triangle O..."