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Triangles Class 10th Mathematics CBSE Solution

Class 10th Mathematics CBSE Solution
Exercise 6.1
  1. Fill in the blanks using the correct word given in brackets: (i) All circles are…
  2. Give two different examples of pair of similar figures and non similar figures…
  3. State whether the following quadrilaterals are similar or not:
Exercise 6.2
  1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)…
  2. E and F are points on the sides PQ and PR respectively of a PQR. For each of the…
  3. In Fig. 6.18, if LM || CB and LN || CD, prove that
  4. In Fig. 6.19, DE || AC and DF || AE. Prove that bf/fe = be/ec
  5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
  6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB…
  7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of…
  8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides…
  9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at…
  10. The diagonals of a quadrilateral ABCD intersect each other at the point O such…
Exercise 6.3
  1. State which pairs of triangles in Fig. are similar.Write the similarity…
  2. In Fig. 6.35, ODC ~ OBA, BOC = 125 and CDO = 70. Find DOC, DCO and OAB…
  3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at…
  4. In Fig. 6.36, qr/qs = qt/pr and 1 = 2. Show that PQS ~ TQR.
  5. S and T are points on sides PR and QR of PQR such that P = RTS. Show that RPQ ~…
  6. In Fig. 6.37, if ABE ACD, show that ADE ~ ABC.
  7. In Fig. 6.38, altitudes AD and CE of ABCintersect each other at the point P.…
  8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects…
  9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M…
  10. CD and GH are respectively the bisectors of ACB and EGF in such a way that D…
  11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC…
  12. Sides AB and BC and median AD of a triangle ABC are respectively proportional…
  13. D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that…
  14. Sides AB and AC and median AD of a triangle ABC are respectively proportional…
  15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the…
  16. If AD and PM are medians of triangles ABC and PQR, respectively where ABC ~…
Exercise 6.4
  1. Let ABC ~ DEF and their areas be, respectively, 64 cm^2 and 121 cm^2 . If EF =…
  2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.…
  3. In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD…
  4. If the areas of two similar triangles are equal, prove that they are congruent…
  5. D, E and F are respectively the mid-points of sides AB, BC and CA of ABC. Find…
  6. Prove that the ratio of the areas of two similar triangles is equal to the…
  7. Prove that the area of an equilateral triangle described on one side of a square…
  8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC.…
  9. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles…
Exercise 6.5
  1. Sides of triangles are given below. Determine which of them are right triangles.…
  2. PQR is a triangle right angled at P and M is a point on QR such that PM QR. Show…
  3. In Fig. 6.53, ABD is a triangle right angled at Aand AC BD. Show that: (i) AB^2…
  4. ABC is an isosceles triangle right angled at C. Prove that ab^2 = 2ac^2…
  5. ABC is an isosceles triangle with AC = BC. If ab^2 = 2ac^2 prove that ABC is a…
  6. ABC is an equilateral triangle of side 2a. Find each of its altitudes…
  7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum…
  8. In Fig. 6.54, O is a point in the interior of a triangle ABC, OD BC, OE AC and…
  9. A ladder 10 m long reaches a window 8 m above theground. Find the distance of…
  10. A wire attached to a vertical pole of height 18 m is 24 m long and has a stack…
  11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per…
  12. Two poles of heights 6 m and 11 m stand on aplane ground. If the distance…
  13. D and E are points on the sides CA and CB respectively of a triangle ABC right…
  14. The perpendicular from A on side BC of aABC intersects BC at D such that DB = 3…
  15. In an equilateral triangle ABC, D is a point on side BC such that bd = 1/3 bc…
  16. In an equilateral triangle, prove that three times the square of one side is…
  17. Tick the correct answer and justify: InABC, AB = 63 cm, AC = 12 cm and BC = 6…
Exercise 6.6
  1. In Fig. 6.56, PS is the bisector of QPR of PQR. Prove that qs/sr = pq/pr…
  2. In Fig. 6.57, D is a point on hypotenuse AC of ABC, DM BC and DN AB. Prove that:…
  3. In Fig. 6.58, ABC is a triangle in which ABC 90 and AD CB produced. Prove that…
  4. In Fig. 6.59, ABC is a triangle in which ABC 90 and AD BC. Prove that ac^2 =…
  5. In Fig. 6.60, AD is a median of a triangle ABC and AM BC. Prove that: (i) ac^2 =…
  6. Prove that the sum of the squares of the diagonals of parallelogram is equal to…
  7. In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove…
  8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point…
  9. In Fig. 6.63, D is a point on side BC of ABC such that bd/cd = ab/ac Prove that…
  10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above…

Exercise 6.1
Question 1.

Fill in the blanks using the correct word given in brackets:

(i) All circles are __________. (congruent, similar)

(ii) All squares are ___________. (similar, congruent)

(iii) All ____________ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are _________ and (b) their corresponding sides are ___________. (equal, proportional)


Answer:

The solutions of the fill ups are:

(i) Similar


(ii) Similar


(iii) Equilateral


(iv) (a) Equal


(b) Proportional



Question 2.

Give two different examples of pair of

(i) Similar figures

(ii) Non-similar figures


Answer:

Two examples of similar figures are:

(i) Two equilateral triangles with sides 1 cm and 2 cm



(ii) Two squares with sides 1 cm and 2 cm



Now two examples of non-similar figures are:


(i) Trapezium and square



(ii) Triangle and parallelogram




Question 3.

State whether the following quadrilaterals are similar or not:



Answer:

The given quadrilateral PQRS and ABCD are not similar because though their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.




Exercise 6.2
Question 1.

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



Answer:

(i) Let us take EC = x cm

Given: DE || BC

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Now, using basic proportionality theorem, we get:


 = 



x = 


x = 2 cm


Hence, EC = 2 cm


(ii) Let us take AD = x cm


Given: DE || BC

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Now, using basic proportionality theorem, we get




x = 


Hence, AD = 2.4 cm


Question 2.

E and F are points on the sides PQ and PR respectively of a Δ PQR. 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 = 0.36 cm


Answer:

(i)

Given :
PE = 3.9 cm,

EQ = 3 cm,

PF = 3.6 cm,

FR = 2.4 cm


Now we know,

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR






Therefore, EF is not parallel to QR


(ii)

We know that,
Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR



Hence,


Therefore, EF is parallel to QR


(iii)

In this we know that,

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

So, if the lines EF and QR are to be parallel, then ratio PE:EQ should be proportional to PF:PR




Hence,



EF is parallel to QR


Question 3.

In Fig. 6.18, if LM || CB and LN || CD, prove that




Answer:



Given: LM ll CB and LN ll CD
From the given figure:
In ΔALM and ΔABC

LM || CB
Proportionality theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion

Now, using basic proportionality theorem that the corresponding sides will have same proportional lengths, we get:


 (i)



And in ΔALN and ΔACD corresponding sides will be proportional


Therefore,


 (ii)


From (i) and (ii), we obtain



Hence, Proved.


Question 4.

In Fig. 6.19, DE || AC and DF || AE. Prove that





Answer:





Given: 

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides 
into two distinct points, then the line divides those sides in proportion.

In triangle ABC, DE is parallel to AC

Therefore,

By Basic proportionality theorem

 ......(1)

In triangle BAE, DF is parallel to AE


Therefore,

By Basic proportionality theorem

 ...... (2)

From (1) and (2), we get

Hence, Proved.
 
Question 5.

In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.


Answer:

To Prove: EF ll QR

Given: In triangle POQ, DE parallel to OQ

Proof:

In triangle POQ, DE parallel to OQ

Hence,

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

 (Basic proportionality theorem) (i)

Now,



Hence,


 (Basic proportionality theorem) (ii)


From (i) and (ii), we get



Therefore,


EF is parallel to QR (Converse of basic proportionality theorem)

Hence, Proved.
 
Question 6.

In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.Show that BC || QR.


Answer:

To Prove: BC ll QR

Given that in triangle POQ, AB parallel to PQ

Hence,

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

 (Basic proportionality theorem)


Now, 


Therefore,

Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Using Basic proportionality theorem, we get:



From above equations, we get



BC is parallel to QR (By the converse of Basic proportionality theorem)

Hence, Proved.
 
Question 7.

Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX)


Answer:

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ is parallel to BC.


To Prove: PQ bisects AC
Given: PQ ll BC and PQ bisects AB
Proof: 
According to Theorem 6.1: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.
Now, using basic proportionality theorem, we get




[As AP = PB coz P is the mid-point of AB]

AQ = QC


Or, Q is the mid-point of AC
Hence proved.


Question 8.

Using Theorem 6.2, 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:


To Prove: PQ ll BC
Given: P and Q are midpoints of AB and AC
Proof: 
Let us take the given figure in which PQ is a line segment which joins the mid-points P and Q of line AB and AC respectively

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


We observe that,



And,



Therefore,



Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Hence, using basic proportionality theorem we get:


PQ parallel to BC
Hence, Proved.


Question 9.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that 


Answer:

The figure is given below:


Given: ABCD is a trapezium
AB ll CD
Diagonals intersect at O
To Prove = 

Construction: Construct a line EF through point O, such that EF is parallel to CD.


Proof:

In ΔADC, EO is parallel to CD
According to basic propotionality theorem, if a side is drawn parallel to any side of the triangle then the corresponding sides formed are propotional

Now, using basic proportionality theorem in ΔABD and ΔADC, we obtain

In ΔABD, OE is parallel to AB

So, using basic proportionality theorem in ΔEOD and ΔABD, we get


From (i) and (ii), we get

Therefore by cross multiplying we get,


Hence, Proved


Question 10.

The diagonals of a quadrilateral ABCD intersect each other at the point O such thatShow that ABCD is a trapezium


Answer:

The quadrilateral ABCD is shown below, BD and AC are the diagonals.

Construction: Draw a line OE parallel to AB

Given: In ΔABD, OE is parallel to AB

To prove : ABCD is a trapezium
According to basic proportionality theorem, if in a triangle another line is drawn parallel to any side of triangle, 
then the sides so obtain are proportional to each other.

Now, using basic proportionality theorem in ΔDOE and ΔABD, we obtain


 ...(i)


It is given that,


 ...(ii)


From (i) and (ii), we get


 ...(iii)
Now for ABCD to be a trapezium AB has to be parallel of CD

Now From the figure we can see that If eq(iii) exists then,

EO || DC (By the converse of basic proportionality theorem)

Now if,

⇒ AB || OE || DC

Then it is clear that

⇒ AB || CD

Thus the opposite sides are parallel and therefore it is a trapezium.

Hence,

ABCD is a trapezium



Exercise 6.3
Question 1.

State which pairs of triangles in Fig. 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:













Answer:

(i) From the figure:

∠A = ∠P = 60°


∠B = ∠Q = 80°


∠C = ∠R = 40°


Therefore, ΔABC Î”PQR [By AAA similarity]

Now corresponding sides of triangles will be propotional,



(ii)
From the triangle,



Hence the corresponding sides are propotional

Thus the corresponding angles will be equal

The triangles ABC and QRP are similar to each other by SSS similarity


(iii) The given triangles are not similar because the corresponding sides are not proportional


(iv) In triangle MNL and QPR, we have

∠M = ∠Q = 70o

but 

Therefore, MNL and QPR are not similar.

(v) In triangle ABC and DEF, we have


AB = 2.5, BC = 3


∠A = 80o


EF = 6


DF = 5


∠F = 80o



And, 


∠B ≠ ∠F


Hence, triangle ABC and DEF are not similar


(vi) In triangle DEF, we have


∠D + ∠E + ∠F = 180o (Sum of angles of triangle)


70o + 80o + ∠F = 180o


∠F = 30o


In PQR, we have


∠P + ∠Q + ∠R = 180o


∠P + 80o + 30o = 180o


∠P = 70o


In triangle DEF and PQR, we have


∠D = ∠P = 70o


∠F = ∠Q = 80o


∠F = ∠R = 30o


Hence, ΔDEF ~ ΔPQR (AAA similarity)


Question 2.

In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB


Answer:

From the figure,

We see, DOB is a straight line

∠ DOC + ∠ COB = 180° (angles on a straight line form a supplementary pair)

∠ DOC = 180° - 125°

∠ DOC = 55°

Now, In ΔDOC,

∠ DCO + ∠ CDO + ∠ DOC = 180°

(Sum of the measures of the angles of a triangle is 180°)

∠ DCO + 70° + 55° = 180°

∠ DCO = 55°

It is given that ΔODC Î”OBA

∠ OAB = ∠ OCD (Corresponding angles are equal in similar triangles)


Thus,  OAB = 55°


Question 3.

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Show that 


Answer:


In ΔDOC and ΔBOA,

∠CDO = ∠ABO (Alternate interior angles as AB || CD)


∠DCO = ∠BAO (Alternate interior angles as AB || CD)


∠DOC = ∠BOA (Vertically opposite angles)


Therefore,

ΔDOC ∼ ΔBOA [ BY AAA similarity]

Now in similar triangles, the ratio of corresponding sides are proportional to each other. Therefore,

 ......... (Corresponding sides are proportional)

or,


Hence, Proved.


Question 4.

In Fig. 6.36,  and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.



Answer:



To Prove: Δ PQS ∼ Δ TQR

Given: In ΔPQR,

∠PQR = ∠PRQ

Proof: 

As∠PQR = ∠PRQ

PQ = PR [sides opposite to equal angles are equal] (i)

Given,

 by (1)


In Δ PQS and Δ TQR, we get

∠Q = ∠Q


Therefore,

By SAS similarity Rule which states that Triangles are similar if two sides in one triangle are in the 
same proportion to the corresponding sides in the other, and the included angle are equal. 

Δ PQS ∼ Δ TQR 

Hence, Proved.


Question 5.

S and T are points on sides PR and QR of Δ PQR such that ∠P = ∠RTS. Show that Δ RPQ ~ Δ RTS


Answer:


In ΔRPQ and ΔRST,

∠ RTS = ∠ QPS (Given)

∠ R = ∠ R (Common to both the triangles)
If two angles of two triangles are equal, third angle will also be equal. As the sum of interior angles of triangle is constant and is 180°

∴ ΔRPQ Î”RTS (By AAA similarity)


Question 6.

In Fig. 6.37, if Δ ABE ≅Δ ACD, show that
Δ ADE ~ Δ ABC.



Answer:

To Prove: Δ ADE ∼ Δ ABC

Given: ΔABE ≅ ΔACD

Proof: 

ΔABE ≅ ΔACD

∴ AB = AC (By CPCT) (i)

And,

AD = AE (By CPCT) (ii)



In ΔADE and ΔABC,


Dividing equation (ii) by (i)


∠A = ∠A (Common)

SAS Similarity: Triangles are similar if two sides in one triangle are in the same proportion to the corresponding sides in the other, and the included angle are equal. 

Therefore,

ΔADEΔABC (By SAS similarity)

Hence, Proved.


Question 7.

In Fig. 6.38, altitudes AD and CE of Δ ABCintersect each other at the point P. Showthat:

(i) Δ AEP ~ Δ CDP

(ii) Δ ABD ~ Δ CBE

(iii) Δ AEP ~ Δ ADB

(iv) Δ PDC ~ Δ BEC



Answer:

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)


∠APE = ∠CPD (Vertically opposite angles)


Hence, by using AA similarity,


ΔAEP  Î”CDP


(ii) In ΔABD and ΔCBE,


∠ADB = ∠CEB (Each 90°)


∠ABD = ∠CBE (Common)


Hence, by using AA similarity,


ΔABD  Î”CBE


(iii) In ΔAEP and ΔADB,


∠AEP = ∠ADB (Each 90°)


∠PAE = ∠DAB (Common)


Hence, by using AA similarity,


ΔAEP  Î”ADB


(iv) In ΔPDC and ΔBEC,


∠PDC = ∠BEC (Each 90°)


∠PCD = ∠BCE (Common angle)


Hence, by using AA similarity,


ΔPDC  Î”BEC



Question 8.

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB


Answer:



To Prove: Δ ABE ∼ Δ CFB
Given: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. As shown in the figure.
Proof:
In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram are equal)

∠AEB = ∠CBF (Alternate interior angles are equal because AE || BC)

Therefore,

ΔABE  Î”CFB (By AA similarity)

Hence, Proved.
 
Question 9.

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) Δ ABC ~ Δ AMP

(ii) 



Answer:

(i) 

To Prove: Δ ABC ∼ Δ AMP

Given: In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

Proof:

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ΔABC  Î”AMP (By AA similarity)

Hence, Proved.


(ii) Î” ABC ∼ Δ AMP

Now we get that,

Similarity Theorem - If the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar. And the converse is also true, so we have


Hence, Proved.
 
Question 10.

CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF in such a way that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that:
(i) 

(ii) Δ DCB ~ Δ HGE

(iii) Δ DCA ~ Δ HGF


Answer:


Given, Δ ABC ∼ Δ FEG …..eq(1)


⇒ corresponding angles of similar triangles


⇒ ∠ BAC = ∠ EFG ….eq(2)


And ∠ ABC = ∠ FEG …….eq(3)


⇒ ∠ ACB = ∠ FGE


⇒ 


⇒ ∠ ACD = ∠ FGH and ∠ BCD = ∠ EGH ……eq(4)


Consider Δ ACD and Δ FGH


⇒ From eq(2) we have


⇒ ∠ DAC = ∠ HFG


⇒ From eq(4) we have


⇒ ∠ ACD = ∠ EGH


Also, ∠ ADC = ∠ FGH


⇒ If the 2 angle of triangle are equal to the 2 angle of another triangle, then by angle sum property of triangle 3rd angle will also be equal.


⇒ by AAA similarity we have in two triangles if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles are similar.


∴ Δ ADC ∼ Δ FHG


⇒ By Converse proportionality theorem


⇒ 


Consider Δ DCB and Δ HGE


From eq(3) we have


⇒ ∠ DBC = ∠ HEG


⇒ From eq(4) we have


⇒ ∠ BCD = ∠ FGH


Also, ∠ BDC = ∠ EHG


∴ Δ DCB ∼ ΔHGE


Hence proved.


Question 11.

In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC,prove that Δ ABD ~ Δ ECF


Answer:



To Prove: Δ ABD ∼ Δ ECF


Given: ABC is an isosceles triangle, AD is perpendicular to BC

BC is produced to E and EF is perpendicular to AC

Proof: 


Given that ABC is an isosceles triangle

AB = AC

⇒ ∠ABD = ∠ECF


In ΔABD and ΔECF,


∠ADB = ∠EFC (Each 90°)


∠ABD = ∠ECF (Proved above)

Therefore,

ΔABD  Î”ECF (By using AA similarity criterion)

AA Criterion: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

Hence, Proved.


Question 12.

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of
Δ PQR (see Fig. 6.41). Show that
Δ ABC ~ Δ PQR.


Answer:


To Prove: Δ ABC ∼ Δ PQR

Given:


Proof:
Median divides the opposite side

BD =  and,


QM = 


Now, 

Multiplying and dividing by 2, we get,

 = 



In Δ ABD and Δ PQM,


Side-Side-Side (SSS) Similarity Theorem - If the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar.

ΔABD  Î”PQM (By SSS similarity)


∠ABD = ∠PQM (Corresponding angles of similar triangles)


In ΔABC and ΔPQR,


∠ABD = ∠PQM (Proved above)


The SAS Similarity Theorem states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar.

ΔABC  Î”PQR (By SAS similarity)

Hence, Proved.
 
Question 13.

D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA2 = CB.CD 


Answer:

In ΔADC and ΔBAC,



To Prove : CA2 = CB . CD

Given: ∠ADC = ∠BAC 

Proof: Now In ΔADC and ΔBAC,

∠ADC = ∠BAC 

∠ACD = ∠BCA (Common angle)

According to AA similarity, if two corresponding angles of two triangles are equal then the triangles are similar

ΔADC  Î”BAC (By AA similarity)


We know that corresponding sides of similar triangles are in proportion


Hence in ΔADC and ΔBAC,





Hence Proved


Question 14.

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 Δ ABC ~ Δ PQR


Answer:

To Prove: Δ ABC ∼ Δ PQR
Given:

Proof: 


Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. 
Then, join B to E, C to

E, Q to L, and R to L

We know that medians divide opposite sides.

Hence, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC,

Diagonals AE and BC bisect each other at point D.

Therefore,

Quadrilateral ABEC is a parallelogram.

AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given in the question that,



ΔABE  Î”PQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∠BAE = ∠QPL ..... (i)

Similarly, it can be proved that

ΔAEC  Î”PLR and

∠CAE = ∠RPL ..... (ii)

Adding equation (i) and (ii), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒∠CAB = ∠RPQ .... (iii)

In ΔABC and ΔPQR,

 (Given)

∠CAB = ∠RPQ [Using equation (iii)]

ΔABC  Î”PQR (By SAS similarity criterion)
Hence, Proved.


Question 15.

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


Answer:

Let AB and CD be a tower and a pole respectively

And, the shadow of BE and DF be the shadow of AB and CD respectively


To find: AB

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle


Therefore,


∠DCF = ∠BAE


And,


∠DFC = ∠BEA


∠CDF = ∠ABE (Tower and pole are vertical to the ground)


ΔABE Î”CDF (AAA similarity)


Hence, By the properties of similar triangles that if two triangles are similar, their corresponding sides will be proportional.


AB = 42 m
Height of the Tower = 42 m


Question 16.

If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that 


Answer:

It is given that ΔABC is similar to ΔPQR




Given: Δ ABC ~ Δ PQR

AD and PM are medians 
We know that the corresponding sides of similar triangles are in proportion

 ..........eq(i)

And also the corresponding angles are equal

∠A = ∠P

∠B = ∠Q

∠C = ∠R ..........eq(ii)


Since AD and PM are medians, they divide their opposite sides in two equal parts


BD =  and,


QM =  .......eq(iii)


From (i) and (iii), we get


(iv)


In ΔABD and ΔPQM,


∠B = ∠Q [Using (ii)]


 [Using (iv)]


ΔABD  Î”PQM (Since two sides are proportional and one angle is equal then by SAS similarity)



Hence, Proved



Exercise 6.4
Question 1.

Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC


Answer:

It is given that,

Δ ABC ~ Δ DEF


Therefore,


Given:


EF = 15.4 cm


ar (ΔABC) = 64cm2


ar (ΔDEF) = 121 cm2


Hence,



Taking square root on both of the sides


BC = (8 × 15.4) / 11


BC = 8 × 1.4 = 11.2 cm


Question 2.

Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD


Answer:

Since AB || CD,

∴∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

ΔAOB ~ΔCOD (By AAA similarity)
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles. If two similar triangles have a scale factor of a : b, then the ratio of their areas is a2 : b2.

Since, AB = 2 CD (Given)

Therefore,

Therefore, the ratio of the areas of triangles AOB and COD is 4:1


Question 3.

In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that


Answer:


Construction: Draw two perpendiculars AP and DM on line BC and AB

Proof:

Area of a triangle = 1/2 x Base x Height

Therefore,



In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)


∠AOP = ∠DOM (Vertically opposite angles)


∴ ΔAPO  Î”DMO (By AA similarity)


As we know in similar triangles the sides are proportional to each other.



Hence, proved.


Question 4.

If the areas of two similar triangles are equal, prove that they are congruent


Answer:


Let us consider two similar triangles as ΔABC  Î”PQR (Given)
When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangles. 
If two similar triangles have a scale factor of a : b, then the ratio of their areas is a2 : b2.


Given that,
ar(Δ ABC) = ar(Δ PQR)

Therefore putting in equation (i) we get,

AB = PQ


BC = QR


And,


AC = PR


Therefore,


ΔABC Î”PQR (By SSS rule)
Hence, Proved.


Question 5.

D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas of ΔDEF and Δ ABC


Answer:


Given: D, E and F are the mid points of sides AB, BC and CA respectively.

Because D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC,

Midpoint Theorem: The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side.
Therefore, From mid-point theorem,

DE || AC and DE =  AC [1]

DF || BC and DF =  BC [2]

EF || AB and EF =  AB [3]


From [1], [2] and [3]


By SSS Similarity Criterion, we can see that

ΔDEF ∼ ΔABC

Theorem: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

 = 


Question 6.

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians


Answer:


Let us assume two similar triangles as ΔABC ~ ΔPQR.

Let AD and PS be the medians of these triangles

Then, because ΔABC ~ΔPQR

 .....(i)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R .......(ii)

Since AD and PS are medians,

BD = DC =BC/2

And, QS = SR =QR/2

Equation (i) becomes,

 .......(iii)

In ΔABD and ΔPQS,

∠B = ∠Q [From (ii)]

And,

 [From (iii)]

ΔABD ~ ΔPQS (SAS similarity)

Therefore, it can be said that

.......(iv)

From (i) and (iv), we get

And hence,


Question 7.

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals


Answer:

Let ABCD be a square of side a

To Prove = Area of ΔABE = 1/2 Area of ΔADB
Proof: 
Let the side of square = a
To find the length of the diagonal of the square,
Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
Applying pythagoras theorem in ΔADB
AD2 + AB2 = BD2

a2 + a2 = BD2
BD = √2 a 

Two desired equilateral triangles are formed as ΔABE and ΔDBF

Side of an equilateral triangle, ΔABE, described on one of its sides = a

Side of an equilateral triangle, ΔDBF, described on one of its diagonals = 



Therefore, Area of equilateral triangle on one of the side of the square is half of the area of equilateral triangle on diagonal.


Question 8.

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
A. 2 : 1 B. 1 : 2

C. 4 : 1 D. 1 : 4


Answer:



Given: D is mid point of BC

We know:

Equilateral triangles have all its angles as 60° and all its sides are of the same length. Therefore, all equilateral triangles are similar to each other.


Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.


Let side of ΔABC = x

Therefore,


Side of ΔBDE = 


Therefore,


Hence, the correct answer is C


Question 9.

Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

A. 2 : 3 B. 4 : 9

C. 81 : 16 D. 16 : 81


Answer:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of theratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9


Therefore,


Ratio between areas of these triangles =2



Hence, the correct answer is (D)




Exercise 6.5
Question 1.

Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm


Answer:

(i) Given: sides of the triangle are 7 cm, 24 cm, and 25 cm

Squaring the lengths of these sides, we get: 49, 576, and 625.


49 + 576 = 625


Or, 72 + 242 = 252


The sides of the given triangle satisfy Pythagoras theorem


Hence, it is a right triangle


We know that the longest side of a right triangle is the hypotenuse


Therefore, the length of the hypotenuse of this triangle is 25 cm


(ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm


Squaring the lengths of these sides, we will obtain 9, 64, and 36


However, 9 + 36 ≠ 64


Or, 32 + 62 ≠ 82


Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side


Therefore, the given triangle is not satisfying Pythagoras theorem


Hence, it is not a right triangle


(iii)Given that sides are 50 cm, 80 cm, and 100 cm.


Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.


And, 2500 + 6400 ≠ 10000


Or, 502 + 802 ≠ 1002


Now, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side


Therefore, the given triangle is not satisfying Pythagoras theorem


Hence, it is not a right triangle


(iv)Given: Sides are 13 cm, 12 cm, and 5 cm


Squaring the lengths of these sides, we get 169, 144, and 25.


Clearly, 144 +25 = 169


Or, 122 + 52 = 132


The sides of the given triangle are satisfying Pythagoras theorem


Therefore, it is a right triangle


We know that the longest side of a right triangle is the hypotenuse


Therefore, the length of the hypotenuse of this triangle is 13 cm


Question 2.

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM×MR


Answer: 


⇒ Let ∠MPR = x


⇒ In Δ MPR, ∠MRP = 180-90-x


⇒ ∠MRP = 90-x


Similarly in Δ MPQ,


∠MPQ = 90-∠MPR = 90-x


⇒ ∠MQP = 180-90-(90-x)


⇒ ∠MQP = x


In Δ QMP and Δ PMR


⇒ ∠MPQ = ∠MRP


⇒ ∠PMQ = ∠RMP


⇒ ∠MQP = ∠MPR


⇒ Δ QMP ∼ Δ PMR


⇒  = 


⇒ PM2 = MR × QM


Hence proved


Question 3.

In Fig. 6.53, ABD is a triangle right angled at Aand AC ⊥ BD. Show that:

(i) AB2 = BC . BD

(ii) AC2 = BC . DC

(iii) AD2 = BD . CD



Answer:

(i) In ΔADB and ΔCAB, we have

∠DAB = ∠ACB (Each of 90o)


∠ABD = ∠CBA (Common angle)


Therefore,


ΔADB Î”CAB (AA similarity)



AB2 = CB x BD


(ii) Let ∠CAB = x


In ΔCBA,

∠CBA + ∠CAB + ∠ACB = 180o
As, ∠ACB = 90o, we have

∠CBA = 180o – 90o – x


∠CBA = 90o – x


Similarly, in ΔCAD


∠CAD = 90o - ∠CBA


= 90o – x


∠CDA = 180o – 90o – (90o – x)


∠CDA = x


In triangle CBA and CAD, we have


∠CBA = ∠CAD


∠CAB = ∠CDA


∠ACB = ∠DCA (Each 90o)


Therefore,


ΔCBA Î”CAD (By AAA similarity)



AC2 = DC * BC


(iii) In ΔDCA and ΔDAB, we have


∠DCA = ∠DAB (Each 90o)


∠CDA = ∠ADB (Common angle)


Therefore,


ΔDCA Î”DAB (By AA similarity)


AD2 = BD x CD


Question 4.

ABC is an isosceles triangle right angled at C. Prove that 


Answer:


To Prove: 2 AC2 = AB2

Given: ΔABC is an isosceles triangle

Proof:

AC = CB ( Two sides of an isosceles triangle are equal, as the side opposite to right angle is largest, rest of the two sides are equal)

Pythagoras Theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Using Pythagoras theorem in ΔABC (i.e., right-angled at point C), we get

AC2 + CB2 = AB2


AC2 + AC2 = AB2 (AC = CB)

So,

2AC2 = AB2

Hence, Proved.
 
Question 5.

ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.


Answer:

To Prove: ABC is a right angled triangle
Given: AB2 = 2AC2

Now 2 AC2 can be split into two parts

AB2 = AC2 + AC2

in an isosceles triangle ABC two sides are equal, and it is given that AC = BC. So,

AB2 = AC2 + BC2 (As, AC = BC)

Now According to pythagoras theorem, in a right angled triangle, square of one side equals to the sum of squares of other two sides
And clearly above equation satisfies it. Thus, the equation satisfies pythagoras theorem and the triangle should be right angled for that.

Therefore, the given triangle is a right-angled triangle
Hence, Proved.


Question 6.

ABC is an equilateral triangle of side 2a. Find each of its altitudes


Answer:

Let AD be the altitude of the given equilateral triangle, ΔABC

We know that altitude bisects the opposite side


BD = DC = a



In triangle ADB,


∠ADB = 90O


Using Pythagoras theorem, we get


AD2 + DB2 = AB2


AD2 + a2 = (2a)2


AD2 + a2 = 4a2


AD2 = 3a2


AD = a


In an equilateral triangle, all the altitudes are equal in length.


Hence, the length of each altitude will be 



Question 7.

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


Answer:

In Rhombus ABCD, 
AB, BC, CD and AD are the sides of the rhombus.
BD and AC are the diagonals.

To prove: AB2 + BC2 + CD+AD2 = AC2 + BD2
Proof:
The figure is shownbelow:

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain


AB2 = AO2 + OB2 .....................eq(i)


BC2 = BO2 + OC2 .......................eq (ii)


CD2 = CO2 + OD2 ....................eq(iii)


AD2 = AO2 + OD2.....................eq (iv)


Now after adding all equations, we get,


AB2 + BC2 + CD2 + AD2 = 2 (AO2 + OB2 + OC2 + OD2)

Diagonals of a rhombus bisect each other, 
Thus AO = AC/2, OB = BD/2, OC=AC/2, and OD= BD/2


=(AC)2 + (BD)2

Hence Sum of squares of sides of a rhombus equals to sum of squares of diagonals of rhombus.


Question 8.

In Fig. 6.54, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that


(i) 

(ii) 


Answer:

(i)

To Prove : OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + EC2
Given: OD, OE and OF are perpendiculars on sides BC, AC and AB respectively
Construction : Join OA, OB and OC
Now according to pythagoras theorem, In a right angled triangle,
(hypotenuse)2 = (altitude)2 + (base)2

Applying Pythagoras theorem in ΔAOF, we obtain

OA2 = OF2 + AF..................eq(i)

Similarly, in ΔBOD,

OB2 = OD2 + BD..................eq(ii)

Similarly, in ΔCOE,

OC2 = OE2 + EC2 .................eq(iii)


Adding these equations, we get


OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
Rearranging the equations we get,

OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + EC2

Hence, Proved


(ii)

To Prove: AF2 + BD2 + EC2 = AE2 + CD2 + BF2
From the above given result from (i),

AF2 + BD2 + EC2 = (AO2 – OE2) + (OC2 – OD2) + (OB2 – OF2)

and from eq(i), (ii) and (iii)
AO2 - OE2 = AE2 , OC2 - OD2 = CD2 , OB2 - OF2 - BF2
Putting these values in above equation we get,

AF2 + BD2 + EC2 = AE2 + CD2 + BF2
Hence, Proved


Question 9.

A ladder 10 m long reaches a window 8 m above theground. Find the distance of the foot of the ladder from base of the wall


Answer:

Let OA be the wall and AB be the ladder


By Pythagoras theorem,


AB2 = OA2 + BO2


(10)2 = (8)2 + OB2


100 = 64 + OB2


OB2 = 36


OB = 6 cm


Therefore, the distance of the foot of the ladder from the base of the wall is 6 m



Question 10.

A wire attached to a vertical pole of height 18 m is 24 m long and has a stack attached to the other end. How far from the base of the pole should the stack be driven so that the wire will be taut?


Answer:

To find: OA

Let OB be the pole and AB be the wire

By Pythagoras theorem,

Pythagoras Theorem : the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.


AB2 = OB2 + OA2


(24)2 = (18)2 + OA2


OA2 = (576 – 324)


OA2 = 252


OA = 6√7 m


Question 11.

An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes afterhours?


Answer:



we know, 

Distance = speed × time

Distance traveled by the plane flying towards north in = 1,000 x

= 1,500 km


Similarly, distance traveled by the plane flying towards west in  = 1,200 x


= 1,800 km


Let these distances be represented by OA and OB respectively.


Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Applying Pythagoras theorem,




Distances between planes is 300√61 km


Question 12.

Two poles of heights 6 m and 11 m stand on aplane ground. If the distance between the feetof the poles is 12 m, find the distance between their tops


Answer:

Let CD and AB be the poles of height 11 m and 6 m

Therefore, CP = 11 - 6 = 5 m


From the figure, it can be observed that AP = 12m


Applying Pythagoras theorem for ΔAPC, we obtain



AP2 + PC2 = AC2


(12)2 + (5)2 = AC2


AC2 = (144 + 25)


AC2 = 169


AC = 13 m


Therefore, the distance between their tops is 13 m



Question 13.

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2


Answer:

To Prove: AE2 + BD2 = AB2 + DE2

Given: D and E are midpoints of AD and CB and ABC is right angled at C
Applying Pythagoras theorem in ΔACE, we obtain

Pythagoras theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

AC2 + CE2 = AE2 ..........eqn(i)

Applying Pythagoras theorem in triangle BCD, we get

BC2 + CD2 = BD2 .........eqn(ii)


Adding equations (i) and (ii), we get


AC2 + CE2 + BC2 + CD2 = AE2 + BD.................eqn (iii)


Applying Pythagoras theorem in triangle CDE, we get


DE2 = CD2 + CE2


Applying Pythagoras in triangle ABC, we get


AB2 = AC2 + CB2


Putting these values in eqn(iii), we get


DE2 + AB2 = AE2 + BD2

Hence, Proved.
 
Question 14.

The perpendicular from A on side BC of aΔABC intersects BC at D such that DB = 3 CD(see Fig. 6.55). Prove that 2AB2 = 2AC2 + BC2


Answer:

We have two right angled triangles now ΔACD and ΔABD

Applying Pythagoras theorem for ΔACD, we obtain

AC2 = AD2 + DC2

AD2 = AC2 – DC2 ......................eq(i)

Applying Pythagoras theorem in ΔABD, we obtain

AB2 = AD2 + DB2

AD2 = AB2 – DB2 ........................eq(ii)

Now we can see from equation i and equation ii that LHS is same. Thus,

From (i) and (ii), we get

AC2 – DC2 = AB2 – DB2 (iii)

It is given that 3DC = DB

Therefore,
DC + DB = BC

DC + 3DC = BC
4 DC = BC ........................eq(iv)

and also,
DC = DB/3
putting this in eq (iii)
DB = 
So,

DC =  and DB = 

Putting these values in (iii), we get


AC2 – = AB2 – 


16AC2 – BC2 = 16AB2 – 9BC2


16AB2 – 16AC2 = 8BC2


2AB2 = 2AC2 + BC2


Question 15.

In an equilateral triangle ABC, D is a point on side BC such that Prove that 9 AD= 7 AB2


Answer:

The figure is given below:

Given: BD = BC/3

To Prove: 9 AD2 = 7 AB2

Proof:
Let the side of the equilateral triangle be a, and AM be the altitude of ΔABC

BM = MC = BC/2 = a/2 [ Altitude of an equilateral triangle bisect the side]

And, then, in ΔABM, by pythagoras theorem we write,
Pythagoras Theorem : Square of the Hypotenuse equals to the sum of the squares of other two sides.

AM= AB- BM2

or AM= a- a2/4



BD = a/3 [ BC = a]

DM = BM – BD

= a/2 – a/3

= a/6


According to pythagoras theorem in a right angled triangle,
(hypotenuse)= (altitude)2 + (base)2

Applying Pythagoras theorem in ΔADM, we obtain

AD2 = AM2 + DM2

Now, a = AB or a= AB2

36 AD= 28 AB2

9 AD2 = 7 AB
Hence, Proved


Question 16.

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes


Answer:

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC


To Prove : 4 × (Square of altitude) = 3 × (Square of one side)
Proof:
Altitude of equilateral triangle divides the side in two equal parts. Therefore,


Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Applying Pythagoras theorem in ΔABE, we obtain

AB2 = AE2 + BE2

4 × (Square of altitude) = 3 × (Square of one side)


Question 17.

Tick the correct answer and justify:
In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm, the angle B is:

A. 120° B. 60°

C. 90° D. 45°


Answer:


Given that, AB = 6 cm,

AC = 12 cm,

And BC = 6 cm


It can be observed that


AB2 = 108

AC2 = 144

And, BC2 = 36


AB2 +BC2 = AC2

Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B

∠B = 90°


Hence, the correct answer is (C)



Exercise 6.6
Question 1.

In Fig. 6.56, PS is the bisector of ∠QPR of Δ PQR. Prove that


Answer:

Construct a line segment RT parallel to SP which intersects the extended line segment QP at point T

Given: PS is the angle bisector of ∠QPR.

Proof:

∠QPS = ∠SPR (i)

By construction,

∠SPR = ∠PRT (As PS || TR, By interior alternate angles) (ii)

∠QPS = ∠QTR (As PS || TR, By interior alternate angles) (iii)

Using these equations, we get:

∠PRT = ∠QTR

PT = PR

By construction,

PS || TR



Basic Proportionality Theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.



By using basic proportionality theorem for ΔQTR,



Hence, Proved


Question 2.

In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that:
(i)  (ii) 


Answer:

(i) To Prove: DM2 = DN. MC

Construction:join DB

We have, DN || CB,

DM || AB,

And ∠B = 90° (Given)

As opposite sides are parallel and equal and also each angle is 90°, DMBN is a rectangle

DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC

∠CDB = 90°

Now from the figure we can say that

∠2 + ∠3 = 90° ........eq(i)

In ΔCDM,

∠1 + ∠2 + ∠DMC = 180° [ Sum of angles of a triangle = 180°]

∠1 + ∠2 = 90° ...........eq(ii)


In Δ DMB,

∠3 + ∠DMB + ∠4 = 180° [ Sum of angles of a triangle = 180°]

⇒∠3 + ∠4 = 90° ........eq(iii)


From (i) and (ii), we get

∠1 = ∠3

From (i) and (iii), we get

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

ΔDCM similar to ΔBDM (AA similarity)

(AA Similarity : When you have two triangles where one is a smaller version of the other, you are looking at two similar triangles.)

BM/DM = DM/MC

Cross multiplying we get,

DN/DM = DM/MC (BM = DN)

DM2 = DN × MC


Hence, Proved.


(ii) 
To Prove: DN2= AN x DM

In right triangle DBN,

∠5 + ∠7 = 90° (iv)

In right triangle DAN,

∠6 + ∠8 = 90° (v)

D is the foot of the perpendicular drawn from B to AC

∠ADB = 90°

∠5 + ∠6 = 90° (vi)

From equation (iv) and (vi), we obtain

∠6 = ∠7

From equation (v) and (vi), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

Hence,

ΔDNA similar to ΔBND (AA similarity criterion)

( AA similarity Criterion: When you have two triangles where one is a smaller version of the other, you are looking at two similar triangles.)

AN/DN = DN/NB

DN2 = AN x NB

DN2= AN x DM (As NB = DM)

Hence, Proved


Question 3.

In Fig. 6.58, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that


Answer:

Using Pythagoras theorem in ΔADB, we get:

AB2 = AD2 + DB2 (i)

Applying Pythagoras theorem in ΔACD, we obtain

AC2 = AD2 + DC2

AC2 = AD2 + (DB + BC)2

AC2 = AD2 + DB2 + BC2 + 2DB x BC

AC2 = AB2 + BC2+ 2DB x BC [Using equation (i)]


Question 4.

In Fig. 6.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that


Answer:



To ProveAC2 = AB2 + BC2 - 2BC x BD

Given: AD is Perpendicular on BC and angle ABC < 90°

Proof: 

Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Applying Pythagoras theorem in Δ ADB, we obtain

AD2 + DB2 = AB2

AD2 = AB– DB2 .....eq(i)

Applying Pythagoras theorem in Δ ADC, we obtain

AD2 + DC2 = AC2

AB2– BD2 + DC2 = AC2 [Using equation (i)]

AB2– BD2 + (BC - BD)2 = AC[ DC = BC - BD]

AC2 = AB2– BD2 + BC2 + BD2 -2BC x BD

AC2 = AB2 + BC2 - 2BC x BD

Hence, Proved.


Question 5.

In Fig. 6.60, AD is a median of a triangle ABC and AM⊥ BC. Prove that:



(i) 

(ii) 

(iii) 


Answer:

(i) Using, Pythagoras theorem in ΔAMD, we get

AM2 + MD2 = AD2 (i)


Applying Pythagoras theorem in ΔAMC, we obtain


AM2 + MC2 = AC2


AM2 + (MD + DC)2 = AC2


(AM2 + MD2) + DC2 + 2MD.DC = AC2


AD2 + DC2 + 2MD.DC = AC2[Using equation (i)]


Using, DC = BC/2 we get


AD2 + (BC/2)2 + 2MD * (BC/2) = AC2


AD2 + (BC/2)2 + MD * BC = AC2


(ii) Using Pythagoras theorem in ΔABM, we obtain


AB2 = AM2 + MB2


= (AD2– DM2) + MB2


= (AD2– DM2) + (BD - MD)2


= AD2– DM2 + BD2 + MD2 - 2BD × MD


= AD2 + BD2 - 2BD × MD


= AD2 + (BC/2)2 – 2 (BC/2) * MD


= AD2 + (BC/2)2 – BC * MD


(iii)Using Pythagoras theorem in ΔABM, we obtain


AM2 + MB2 = AB2 (1)


Applying Pythagoras theorem in ΔAMC, we obtain


AM2 + MC2 = AC2(2)


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


2AM2 + MB2 + MC2 = AB2 + AC2


2AM2 + (BD - DM)2 + (MD + DC)2 = AB2 + AC2


2AM2+BD2 + DM2 - 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2


2AM2 + 2MD2 + BD2 + DC2 + 2MD ( - BD + DC) = AB2 + AC2


2 (AM2 + MD2) + (BC/2)2 + (BC/2)2 + 2MD (-BC/2 + BC/2) = AB2 + AC2


2AD2 + BC2/2 = AB2 + AC2



Question 6.

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


Answer:

ABCD is a parallelogram in which AB = CD and AD = BC

Perpendicular AN is drawn on DC and perpendicular DM is drawn on AB extend up to M


In Δ AMD,


AD2 = DM2 + AM2 ...............eq(i)


In Δ BMD,


BD2 = DM2 + (AM + AB)2


Or,

(AM + AB)2 = AM2 + AB2 + 2 AM x AB

BD2 = DM2 + AM2 + AB2 + 2AM x AB ..................eq(ii)


Substituting the value of AM2 from (i) in (ii), we get


BD2 = AD2 + AB2 + 2 x AM x AB .............eq(iii)


In Δ AND,


AD2 = AN2 + DN2 ......................eq(iv)


In Δ ANC,


AC2 = AN2 + (DC – DN)2


Or,


AC2 = AN2 + DN2 + DC2 – 2 x DC x DN ............eq(v)


Substituting the value of AD2 from (iv) in (v), we get


AC2 = AD2 + DC2 – 2 x DC x DN ...............eq(vi)


We also have,


AM = DN and AB = CD


Substituting these values in (vi), we get


AC2 = AD2 + DC2 – 2 x AM x AB .................eq(vii)


Adding (iii) and (vii), we get


AC2 + BD2 = AD2 + AB2 + 2 x AM x AB + AD2 + DC2 – 2 x AM x AB


Or,


AC2 + BD2 = AB2 + BC2 + DC2 + AD2


Hence, proved.


Question 7.

In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that:

(i) Δ APC ~ Δ DPB

(ii) AP . PB = CP . DP



Answer:

(i) In triangle APC and DPB,

∠CAP = ∠BDP (Angles on the same side of a chord are equal)


∠APC = ∠DPB (Opposite angles)


Hence,


Δ APC ~ Δ DPB (By AAA similarity)


(ii) Since, the two triangles are similar


Hence,



Or,


AP * PB = CP * DP


Hence, proved



Question 8.

In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P(when produced) outside the circle. Prove that

(i) Δ PAC ~ Δ PDB

(ii) PA . PB = PC . PD



Answer:

(i) In triangle PAC and PDB

∠PAC + ∠CAB = 180o (Linear pair)


∠CAB + ∠BDC = 180O (Opposite angles of a cyclic quadrilateral are supplementary)


Hence,


∠PAC = ∠PDB


Similarly,


∠PCA = ∠PBD


Hence,


Δ PAC ~ Δ PDB


(ii) Since the two triangles are similar, so



Or,


PA * PB = PC * PD


Hence, proved



Question 9.

In Fig. 6.63, D is a point on side BC of Δ ABC such thatProve that AD is the bisector of ∠ BAC


Answer: 
To Prove : AD bisects ∠BAC

Now from ΔABD and ΔADC,
As it is given that

And, AD is common to both triangles,
Therefore,
ΔABD ∼ ΔADC (BY SSS theorem)
Now by similarity ∠BAD = ∠DAC
Hence, AD must be the bisector of the angle BAC

Question 10.

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip ofthe rod. Assuming that her string(from the tip of her rod to the fly) is taut,how much string does she have out(see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?


Answer:



As per the question:

AD = 1.8 m
BD = 2.4 m
CD = 1.2 m
speed of string when she pulls in = 5 cm per second


To find: Length of string, AB and 


In triangle ABD, length of string i.e. AB can be calculated as follows [By Pythagoras theorem]

AB2 = AD2 + BD2

= (1.8)2 + (2.4)2
= 3.24 + 5.76
= 9


Or, AB = 3 m



Let us assume that the string reaches at point M after 12 seconds

Now, To find: Distance of fly, from the girl.


Length of string pulled in, after 1 second = 5 cm

Length of string pulled in, after 12 seconds = 5 * 12 = 60 cm

= 0.6 m [As, 1 m = 100 cm]

Remaining length, AM = 3 – 0.6 = 2.4 m

In triangle AMD, we can find MD by using Pythagoras theorem,

MD2 = AM2 – AD2
= 2.42 – 1.82
= 5.76 – 3.24
= 2.52 m


Or, MD = 1.58 m


Also,

Horizontal distance between the girl and the fly = CD + MD
= 1.2 + 1.58 = 2.78 m


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