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)
The solutions of the fill ups are:
(i) Similar
(ii) Similar
(iii) Equilateral
(iv) (a) Equal
(b) Proportional
Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures
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
State whether the following quadrilaterals are similar or not:
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.
In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii)
(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
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
(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.
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
In Fig. 6.18, if LM || CB and LN || CD, prove that
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.
In Fig. 6.19, DE || AC and DF || AE. Prove that
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,
......(1)
In triangle BAE, DF is parallel to AE
Therefore,
By Basic proportionality theorem
...... (2)
From (1) and (2), we get
Hence, Proved.In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
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.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.
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.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)
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.
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)
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,
Hence, using basic proportionality theorem we get:
PQ parallel to BC
Hence, Proved.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that
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
The diagonals of a quadrilateral ABCD intersect each other at the point O such thatShow that ABCD is a trapezium
The quadrilateral ABCD is shown below, BD and AC are the diagonals.
Construction: Draw a line OE parallel to ABGiven: In ΔABD, OE is parallel to AB
To prove : ABCD is a trapeziumNow, 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
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
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:
(i) From the figure:
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Therefore, ΔABC ΔPQR [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(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)
In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB
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 ΔOBA
∠ OAB = ∠ OCD (Corresponding angles are equal in similar triangles)
Thus, ∠ OAB = 55°
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Show that
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.
In Fig. 6.36, and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.
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 theΔ PQS ∼ Δ TQR
Hence, Proved.
S and T are points on sides PR and QR of Δ PQR such that ∠P = ∠RTS. Show that Δ RPQ ~ Δ RTS
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 ΔRTS (By AAA similarity)
In Fig. 6.37, if Δ ABE ≅Δ ACD, show that
Δ ADE ~ Δ ABC.
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.
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
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity,
ΔAEP ΔCDP
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB (Each 90°)
∠ABD = ∠CBE (Common)
Hence, by using AA similarity,
ΔABD ΔCBE
(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (Each 90°)
∠PAE = ∠DAB (Common)
Hence, by using AA similarity,
ΔAEP ΔADB
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity,
ΔPDC ΔBEC
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
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 ΔCFB (By AA similarity)
Hence, Proved.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)
(i)
To Prove: Δ ABC ∼ Δ AMP
Given: In ΔABC and ΔAMP,
∠ABC = ∠AMP (Each 90°)
Proof:∠A = ∠A (Common)
∴ΔABC ΔAMP (By AA similarity)
Hence, Proved.
(ii) Δ 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
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
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.
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
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 ΔECF (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.
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.
To Prove: Δ ABC ∼ Δ PQR
Given:
Proof:
Median divides the opposite side
BD = and,
QM =
Now,
=
In Δ ABD and Δ PQM,
ΔABD ΔPQM (By SSS similarity)
∠ABD = ∠PQM (Corresponding angles of similar triangles)
In ΔABC and ΔPQR,
∠ABD = ∠PQM (Proved above)
ΔABC ΔPQR (By SAS similarity)
Hence, Proved.D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA2 = CB.CD
In ΔADC and ΔBAC,
∠ACD = ∠BCA (Common angle)
According to AA similarity, if two corresponding angles of two triangles are equal then the triangles are similar
ΔADC ΔBAC (By AA similarity)
We know that corresponding sides of similar triangles are in proportion
Hence in ΔADC and ΔBAC,
Hence Proved
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
To Prove: Δ ABC ∼ Δ PQR
Given:
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 ΔPQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∠BAE = ∠QPL ..... (i)
Similarly, it can be proved that
ΔAEC ΔPLR 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 ΔPQR (By SAS similarity criterion)
Hence, Proved.
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
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
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 ΔCDF (AAA similarity)
AB = 42 m
Height of the Tower = 42 m
If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that
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 ΔPQM (Since two sides are proportional and one angle is equal then by SAS similarity)
Hence, Proved
Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC
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
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
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
In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
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 ΔDMO (By AA similarity)
Hence, proved.
If the areas of two similar triangles are equal, prove that they are congruent
Let us consider two similar triangles as ΔABC ΔPQR (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 ΔPQR (By SSS rule)
Hence, Proved.
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
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.
=
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians
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,
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
Let ABCD be a square of side a
To Prove = Area of ΔABE = 1/2 Area of ΔADBTwo 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.
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
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
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
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)
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
(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
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM×MR
⇒ 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
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
(i) In ΔADB and ΔCAB, we have
∠DAB = ∠ACB (Each of 90o)
∠ABD = ∠CBA (Common angle)
Therefore,
ΔADB ΔCAB (AA similarity)
AB2 = CB x BD
(ii) Let ∠CAB = x
In ΔCBA,
∠CBA + ∠CAB + ∠ACB = 180o∠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 ΔCAD (By AAA similarity)
AC2 = DC * BC
(iii) In ΔDCA and ΔDAB, we have
∠DCA = ∠DAB (Each 90o)
∠CDA = ∠ADB (Common angle)
Therefore,
ΔDCA ΔDAB (By AA similarity)
AD2 = BD x CD
ABC is an isosceles triangle right angled at C. Prove that
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.ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
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 sidesTherefore, the given triangle is a right-angled triangle
Hence, Proved.
ABC is an equilateral triangle of side 2a. Find each of its altitudes
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
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
In Rhombus ABCD,
AB, BC, CD and AD are the sides of the rhombus.
BD and AC are the diagonals.
To prove: AB2 + BC2 + CD2 +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,=(AC)2 + (BD)2
Hence Sum of squares of sides of a rhombus equals to sum of squares of diagonals of rhombus.
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)
(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 + AF2 ..................eq(i)
Similarly, in ΔBOD,
OB2 = OD2 + BD2 ..................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)AF2 + BD2 + EC2 = AE2 + CD2 + BF2
Hence, Proved
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
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
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?
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
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?
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.
Applying Pythagoras theorem,
Distances between planes is 300√61 km
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
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
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
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
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 + BD2 .................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.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
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 = and DB =
Putting these values in (iii), we get
AC2 – = AB2 –
16AC2 – BC2 = 16AB2 – 9BC2
16AB2 – 16AC2 = 8BC2
2AB2 = 2AC2 + BC2
In an equilateral triangle ABC, D is a point on side BC such that Prove that 9 AD2 = 7 AB2
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.
AM2 = AB2 - BM2
or AM2 = a2 - 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)2 = (altitude)2 + (base)2
Applying Pythagoras theorem in ΔADM, we obtain
AD2 = AM2 + DM2
Now, a = AB or a2 = AB2
36 AD2 = 28 AB2
9 AD2 = 7 AB2
Hence, Proved
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
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)
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°
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
ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B
∠B = 90°
Hence, the correct answer is (C)
In Fig. 6.56, PS is the bisector of ∠QPR of Δ PQR. Prove that
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
In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that:
(i) (ii)
(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
In Fig. 6.58, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that
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)]
In Fig. 6.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that
To Prove: AC2 = 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 = AB2 – 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 = AC2 [ DC = BC - BD]
AC2 = AB2– BD2 + BC2 + BD2 -2BC x BD
AC2 = AB2 + BC2 - 2BC x BD
Hence, Proved.
In Fig. 6.60, AD is a median of a triangle ABC and AM⊥ BC. Prove that:
(i)
(ii)
(iii)
(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
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides
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 ABBD2 = 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.
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
(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
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
(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
In Fig. 6.63, D is a point on side BC of Δ ABC such thatProve that AD is the bisector of ∠ BAC
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?
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
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 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