Regular engagement with S Chand Class 10 Maths Solutions ICSE Chapter 12 Similar Triangles Ex 12(d) can boost students confidence in the subject.
S Chand Class 10 ICSE Maths Solutions Chapter 12 Similar Triangles Ex 12(d)
Question 1.
In the figure, in ∆PQR if XY || QR, PX = 1 cm, XQ = 3 cm, YR = 4.5 cm and QR = 9 cm, find (i) PY and (ii) XY. Further, if the area of the ∆PXY is A cm², find in terms of A, (iii) the area of the triangle PQR (iv) the area of the figure XYRQ.
Solution:
In ∆PQR,
XY || QR
PX = 1 cm, XQ = 3 cm, YR = 4.5 cm and QR = 9 cm
Let PY = x cm and XY = y cm
∵ XY || QR
∴ ∆PXY ~ ∆PQR
∴ \(\frac{P X}{P Q}=\frac{P Y}{P R}=\frac{X Y}{Q R}\)
⇒ \(\frac{1}{1+3}=\frac{x}{x+4.5}=\frac{y}{9}\)
⇒ \(\frac{1}{4}=\frac{x}{x+4.5}=\frac{y}{9}\)
\(\frac{x}{x+4.5}=\frac{1}{4}\) ⇒ 4x = x + 4.5
⇒ 4x – x = 4.5 ⇒ 3x = 4.5
(i) ⇒ x = \(\frac { 4.5 }{ 3 }\) = 1.5 cm
and \(\frac { y }{ 9 }\) = \(\frac { 1 }{ 4 }\) = 4y = 9 ⇒ \(\frac { 9 }{ 4 }\) = 2.25 cm
(ii) ∴ PY = 1.5 cm and XY = 2.25 cm
(iii) ∵ ∆PXY ~ ∆PQR (proved)
∴ \(\frac { 1 }{ 2 }\)
(Areas of the similar triangles are proportional to the square of their corresponding sides)
⇒ \(\frac { 1 }{ 2 }\)
(∵ Area of APXY = A cm²)
∴ Area APQR = 16A cm²
(iv) Now area of figure XYRQ = area of ∆PQR – area of ∆PXY
= 16A – A = 15A cm²
Question 2.
The figure shows two isosceles similar triangles. If PQ and BC are not parallel and PC = 4cm, AQ = 3cm, QB = 12cm and BC = 15cm, calculate :
(i) the length of AP.
(ii) the ratio of the areas of triangle APQ
Solution:
In the figure two triangles are similar
i.e. ∆ABC ~ ∆APQ, PQ is not parallel
To BC, PC = 4cm, AQ = 3cm, QB = 12cm, BC = 15cm
Let AP = x, and PQ = y
Then \(\frac{\mathrm{AQ}}{\mathrm{AC}}=\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}\)
⇒ \(\frac{3}{x+4}=\frac{x}{3+12}=\frac{y}{15}\)
⇒ \(\frac{3}{x+4}=\frac{x}{15}\) ⇒ x² + 4x = 45 (By cross multiplication)
⇒ x² + 4x – 45 = 0
⇒ x² + 9x – 5x – 45 = 0
⇒ x (x + 9) – 5 (x + 9) = 0
⇒ (x + 9) (x – 5) = 0
Either x + 9 = 0, then x = – 9 which is not possible being negative
or x – 5 = 0, then x = 5
∴ AP = 5
and \(\frac{3}{x+4}=\frac{y}{15} \Rightarrow \frac{3}{5+4}=\frac{y}{15}\)
⇒ y = \(\frac { 3×15 }{ 2 }\) = 5
∵ ∆APQ ~ ∆ABC
∴ \(\frac{\text { area } \triangle \mathrm{APQ}}{\text { area } \triangle \mathrm{ABC}}=\frac{\mathrm{PQ}^2}{\mathrm{BC}^2}\)
{Areas of two similar triangles are proportional to the square of their corresponding sides}
\(\frac{(5)^2}{(15)^2}=\frac{25}{225}=\frac{1}{9}\)
Ratio between the areas of ∆APQ and ∆ABC = 1 : 9
Question 3.
Given that ∆s ABC, DEF are similar. Find
(a) the ratio of the area of ∆ABC to the area of ∆DEF if AB = 2 and DE = 4.
(b) \(\frac { AB }{ DE }\), if ∆ABC : ADEF = 16 : 25.
Solution:
∆ABC ~ ADEF
AB = 2, DE = 4
Question 4.
In a ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point on AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.
Solution:
In ∆ABC,
P is a point on AB such that
AP : PB = 1 : 2
and Q is a point on AC such that PQ || BC
∵ PQ || BC
∴ ∆APQ ~ ∆ABC
∵ \(\frac{\text { area of } \triangle \mathrm{APQ}}{\text { area of } \triangle \mathrm{ABC}}=\frac{\mathrm{AP}^2}{\mathrm{AB}^2}\)
(Areas of similar triangles are proportional to the squares of their corresponding sides)
= \(\frac{(1)^2}{(1+2)^2}=\frac{(1)^2}{(3)^2}=\frac{1}{9}\)
⇒ 9 areas of ∆APQ = area of ∆ABC
Subtracting area of ∆APQ from both sides area of ∆ABC – area of ∆APQ = 9 area of ∆APQ – area of ∆APQ
⇒ area of trap. BPQC = 8 area of ∆APQ
⇒ \(\frac{\text { area of } \triangle \mathrm{APQ}}{\text { area of trap. } \mathrm{BPQC}}=\frac{1}{8}\)
∴ area of ∆APQ : area of trap. BPQC = 1 : 8
Question 5.
The areas of two similar triangles ABC and DEF are 36 cm² and 81 cm² respectively. If EF = 6.9 cm, determine BC.
Solution:
Area of ∆ABC = 36 cm²
and area of ADEF = 81 cm²
EF = 6.9 cm
∵ ∆ABC ~ ADEF
∴ \(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEF}}=\frac{\mathrm{BC}^2}{E F^2}\)
(Areas of two similar triangles are in the ratio of the squares on their corresponding sides)
⇒ \(\frac{36}{81}=\frac{(\mathrm{BC})^2}{(6.9)^2} \Rightarrow \frac{(6)^2}{(9)^2}=\frac{(\mathrm{BC})^2}{(6.9)^2}\)
⇒ \(\frac{\mathrm{BC}}{6.9}=\frac{6}{9} \Rightarrow \mathrm{BC}=\frac{6.9 \times 6}{9}\)
⇒ BC = 4.6 cm
Question 6.
In the figure, DE || BC. AD = 3 cm, DB = 2 cm, area of ∆ABC = 10 cm². Find the area of ∆ADE.
Solution:
In ∆ADE, BC || DE
AD = 3 cm, DB = 2 cm
∴ AB = AD – DB = 3 – 2 = 1 cm
Area of ∆ABC = 10 cm
∵ In ∆ADE; BC || DE
∴ ∆ABC ~ ∆ADE
∴ \(\frac{\text { area } \triangle \mathrm{ABC}}{\text { area } \triangle \mathrm{ADE}}=\frac{\mathrm{AB}^2}{\mathrm{AD}^2}\)
{Areas of two similar triangles are proportional to the square of their corresponding sides}
⇒ \(\frac{10}{\text { area } \triangle \mathrm{ADE}}=\frac{(1)^2}{(3)^2}=\frac{1}{9}\)
⇒ area ∆ADE =10 x 9 = 90 cm²
Question 7.
In the figure, the angles PRS and PQR are similar PS = 2 cm and PR = 3 cm. If the area of the triangle PRS is 2 cm², calculate the area of ∆PQR.
Solution:
In ∆PRS and ∆PQR,
∠PRS – ∠PQR (given)
∠RPS – ∠RPQ (common)
∆PRS – ∆PQR (AA axiom)
∴ \(\frac{\text { area of } \triangle P R S}{\text { area of } \triangle P Q R}=\frac{P^2}{P^2}\)
{Areas of two similar triangles are proportional to the squares of their corresponding sides}
⇒ \(\frac{2}{\text { area of } \triangle \mathrm{PQR}}=\frac{(2)^2}{(3)^2}=\frac{4}{9}\)
⇒ area of ∆PQR = \(\frac{2 \times 9}{4}=\frac{9}{2}\) = 4.5 cm²
Question 8.
In the figure, DE || BC and AD : DB = 5 : 4.
Find \(\frac{\text { Area }(\triangle \mathrm{ADE})}{\text { Area }(\triangle \mathrm{ABC})}\)
Solution:
Question 9.
In the figure, in ∆PQR, PQ = 8 cm, PR = 10 cm and ∠Q = 90°. A and B are points on sides PQ and PR respectively such that AB = 2 cm and ∠ABP = 90°. Find :
(i) the area of APAB
(ii) the area of quad. AQRB : area of ∆PQR.
Solution:
In ∆PAB and ∆PQR,
∠ABP = ∠Q (each 90°)
∠P = ∠P (common)
∴ ∆PAB ~ ∆PQR (AA axiom)
In right ∆PQR, PQ = 8 cm and PR = 10 cm But PR² = PQ² + QR² (Pythagoras theorem)
⇒ (10)² = (8)² + (QR)²
⇒ 100 = 64 + (QR)²
⇒ QR² = 100 – 64 = 36 = (6)²
∴ QR = 6 cm
∵ ∆PAB ~ ∆PQR
∴ \(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{PB}}{\mathrm{PQ}} \Rightarrow \frac{2}{6}=\frac{\mathrm{PB}}{8}\)
⇒ PB = \(\frac { 2 }{ 6 }\) x 8 = \(\frac { 8 }{ 3 }\) cm
∴ Area of right ∆PAB = \(\frac { 1 }{ 2 }\) AB x PB
= \(\frac { 1 }{ 2 }\) x 2 x \(\frac { 8 }{ 3 }\) = \(\frac { 8 }{ 3 }\) cm²
Again ∆PAB ~ ∆PQR
∴ \(\frac { area (∆PAB) }{ area(∆PQR) }\) = \(\frac{\mathrm{AB}^2}{\mathrm{QR}^2}\)
{Areas of similar triangles are proportional to the squares of their corresponding sides}
= \(\frac{(2)^2}{(6)^2}=\frac{4}{36}=\frac{1}{9}\)
∴ area (∆PQR) = 9 area (∆PAB) … (i)
Subtracting area (∆PAB) from both sides, area (∆PQR) – are (∆PAB) = 9 area (∆PAB) – area (∆PAB)
⇒ area quad. AQRB = 8 area (∆PAB)
⇒ area (quad. AQRB) 8 area (∆PAB)
\(\frac{\text { area }(\text { quad. } A Q R B)}{\text { area }(\triangle \mathrm{PQR})}=\frac{8 \text { area }(\triangle \mathrm{PAB})}{9 \text { area }(\triangle \mathrm{PAB})}=\frac{8}{9}\) (from (i))
∴ Area quad. AQRB : area (∆PQR) = 8 : 9
Question 10.
The areas of two similar triangles ABC and PQR are 64 sq. cm and 121 sq. cm respectively. If QR = 15.4 cm, find BC.
Solution:
∆ABC ~ ∆PQR
Area (∆ABC) = 64 cm
and area (APQR) = 121 cm
QR = 15.4 cm
∵ Triangles are similar
∴ \(\frac { area (∆ABC) }{ area(∆PQR) }\) = \(\frac{\mathrm{BC}^2}{\mathrm{QR}^2}\)
{Areas of similar triangles are similar to the squares of their corresponding sides}
⇒ \(\frac{64}{121}=\frac{(\mathrm{BC})^2}{(15.4)^2} \Rightarrow \frac{(8)^2}{(11)^2}=\frac{(\mathrm{BC})^2}{(15.4)^2}\)
⇒ \(\frac{\mathrm{BC}}{15.4}=\frac{8}{11}\)
⇒ BC = \(\frac { 8 }{ 11 }\) x 15.4
⇒ BC = 11.2 cm
Question 11.
D, E, F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC. Determine the ratio of the areas of the triangles DEF and ABC.
Solution:
D, E and F are the mid-points of the sides
BC, CA and AB of ∆ABC
DE, EF and FD are joined
∵ E and F are mid-points of AC and AB
∴ EF || BC and = \(\frac { 1 }{ 2 }\)BC
⇒ \(\frac { EF }{ BC }\) = \(\frac { 1 }{ 2 }\) … (i)
Similarly D and E are the mid-point of BC and AC
∵ DE || AB and = \(\frac { 1 }{ 2 }\)AB
⇒ \(\frac { DE }{ AB }\) = \(\frac { 1 }{ 2 }\) … (ii)
and D and F are mid-points of BC and AB
∴ DF || AC and = \(\frac { 1 }{ 2 }\) AC
⇒ \(\frac { DF }{ AC }\) = \(\frac { 1 }{ 2 }\) … (iii)
From (i), (ii) and (iii)
\(\frac{\mathrm{EF}}{\mathrm{BC}}=\frac{\mathrm{DE}}{\mathrm{AB}}=\frac{\mathrm{DF}}{\mathrm{AC}}\left(\text { each }=\frac{1}{2}\right)\)
∆DEF ~ ∆ABC
∴ \(\frac{\text { area } \triangle \mathrm{DEF}}{\text { area } \triangle \mathrm{ABC}}=\frac{\mathrm{EF}^2}{\mathrm{BC}^2}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
∴ area ADEF : area ∆ABC = 1 : 4
Question 12.
ABCD is a trapezium in which AB = 2 DC and AB || DC. If AC and BD intersect at O, prove that area (∆AQB) = 4 area (ACOD).
Solution:
ABCD is a trapezium in which AB || DC and
AB = 2DC
AC and BD intersect each other at O
In ∆AOB and ∆COD,
∠AOB = ∠COD
(vertically opposite anlges)
∠ABO = ∠ODC (alternate angles)
∴ ∆AOB ~ ∆COD (AAA axiom)
∵ ∵ \(\frac { area (∆AOB) }{ area(∆COD) }\) = \(\frac{\mathrm{AB}^2}{\mathrm{DC}^2}=\frac{(2 \mathrm{DC})^2}{\mathrm{DC}^2}\)
\(\frac{4 \mathrm{DC}^2}{\mathrm{DC}^2}=\frac{4}{1}\)
∴ area (∆AOQ) = 4 area (ACOD)
Hence proved.
Question 13.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Let ∆ABC and ADEF are similar
and area (∆ABC) = area (ADEF)
Similarly AB = DE and AC = DF
Hence ∆ABC ≅ ∆DEF
(S.S.S. axiom of congruency)
Hence proved.
Question 14.
Prove that the ratio of corresponding altitudes of two similar triangles is equal to the ratio of their corresponding sides.
Solution:
In ∆ABC and ∆DEF,
AP ⊥ BC and DQ ⊥ EF
∵ ∆ABC ~ ADEF
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}\) and ∠B = ∠E
Now in ∆ABP and ∆DEQ,
∠B = ∠E (proved)
∠P = ∠Q (each 90°)
∴ ∆ABP ~ ∆DEQ (AA axiom)
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AP}}{\mathrm{DQ}}\)
But \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}\) (proved)
∴ \(\frac{\mathrm{AP}}{\mathrm{DQ}}=\frac{\mathrm{BC}}{\mathrm{EF}}\)
Hence proved.
Question 15.
ABC is a triangle, PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides ∆ABC into two of equal parts equal area. Find \(\frac { BP }{ AB }\).
Solution:
In ∆ABC, P and Q are the points on AB and AC such that
and area (∆APQ) = area (quad, PBCQ)
⇒ area (∆APQ) + area (∆APQ) = area (quad. PBCQ) + area (∆APQ)
⇒ 2 area (∆PQ) = \(\frac { 1 }{ 2 }\) area (∆ABC)
⇒ \(\frac { area (∆APQ) }{ area(∆ABC) }\) = \(\frac { 1 }{ 2 }\)
∵ PQ || BC
∴ ∆APQ ~ ∆ABC
∴ \(\frac { area (∆APQ) }{ area(∆ABC) }\) = \(\frac{A P^2}{A B^2}=\frac{1}{2}\)
∴ \(\frac{A P^2}{A B^2}=\frac{1}{2} \Rightarrow \frac{A P}{A B}=\frac{1}{\sqrt{2}}\)
\(\frac{A B-B P}{A B}=\frac{1}{\sqrt{2}}\)
⇒ \(\frac{A B}{A B}-\frac{B P}{A B}=\frac{1}{\sqrt{2}}\)
1 – \(\frac{B P}{A B}=\frac{1}{\sqrt{2}}\)
⇒ \(\frac{\mathrm{BP}}{\mathrm{AB}}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}\)
Hence \(\frac{\mathrm{BP}}{\mathrm{AB}}=\frac{\sqrt{2}-1}{\sqrt{2}}\)