Similarity of Triangles — Exercise 16(B)

In the given figure, DE ∥ BC.

(i) Prove that ΔADE ∼ ΔABC.

(ii) Given that AD = $\dfrac{1}{2}$ DB, calculate DE, if BC = 4.5 cm.

(iii) Find $\dfrac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}}$.

(iv) Find $\dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}}$.

Answer

(i) Considering ΔADE and ΔABC,

∠BAC = ∠DAE [Common angles]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ ΔADE ∼ ΔABC (By A.A. axiom)

Hence, proved that ΔADE ∼ ΔABC.

(ii) Given,

$\Rightarrow AD = \dfrac{1}{2} BD \\[1em] \Rightarrow AD = \dfrac{1}{2}(AB - AD) \\[1em] \Rightarrow 2AD = AB - AD \\[1em] \Rightarrow 2AD + AD = AB \\[1em] \Rightarrow 3AD = AB \\[1em] \Rightarrow \dfrac{AD}{AB} = \dfrac{1}{3} \\[1em] \Rightarrow AD : AB = 1 : 3.$

Since triangles ADE and ABC are similar so, ratio of their corresponding sides will be equal.

$\therefore \dfrac{AD}{AB} = \dfrac{DE}{BC} \\[1em] \therefore \dfrac{1}{3} = \dfrac{DE}{4.5} \\[1em] \therefore DE = \dfrac{4.5}{3} \\[1em] \therefore DE = 1.5 \text{ cm}.$

Hence, the length of DE = 1.5 cm.

(iii) We know that,

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

$\Rightarrow \dfrac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}} = \dfrac{DE^2}{BC^2} \\[1em] = \Big(\dfrac{DE}{BC}\Big)^2 \\[1em] = \Big(\dfrac{1}{3}\Big)^2 \\[1em] = \dfrac{1}{9}.$

Hence, $\dfrac{\text{ar(ΔADE)}}{\text{ar(ΔABC)}} = \dfrac{1}{9}$.

(iv) From figure,

Area of trap.(BCED) = area(Δ ABC) - area(Δ ADE)

$\Rightarrow \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} = \dfrac{9}{1} \\[1em] \Rightarrow \text{ar(ΔABC)} = 9[\text{ar(ΔADE)}]\\[1em] \Rightarrow \text{ar(trap. BCED)} = 9[\text{ar(ΔADE)}] - \text{ar(ΔADE)} \\[1em] \Rightarrow \text{ar(trap. BCED)} = 8[\text{ar(ΔADE)}] \\[1em] \Rightarrow \dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \dfrac{1}{8}.$

Hence, $\dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \dfrac{1}{8}$.

Given that ΔABC ∼ ΔPQR.

(i) If ar(ΔABC) = 49 cm² and ar(ΔPQR) = 25 cm² and AB = 5.6 cm, find the length of PQ.

(ii) If ar(ΔABC) = 28 cm² and ar(ΔPQR) = 63 cm² and PR = 8.4 cm, find the length of AC.

(iii) If BC = 4 cm, QR = 5 cm and ar(ΔABC) = 32 cm² determine ar(ΔPQR).

Answer

(i) Given,

ΔABC ∼ ΔPQR

ar(ΔABC) = 49 cm²

ar(ΔPQR) = 25 cm²

AB = 5.6 cm

We know that,

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

$\therefore \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\dfrac{AB}{PQ}\Big)^2 \\[1em] \Rightarrow \dfrac{49}{25} = \Big(\dfrac{5.6}{PQ}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{49}{25}} = \dfrac{5.6}{PQ} \\[1em] \Rightarrow \dfrac{7}{5} = \dfrac{5.6}{PQ} \\[1em] \Rightarrow PQ = \dfrac{5 \times 5.6}{7}\\[1em] \Rightarrow PQ = \dfrac{28}{7}\\[1em] \Rightarrow PQ = 4 \text{ cm.}$

Hence, PQ = 4 cm.

(ii) Given,

ΔABC ∼ ΔPQR

ar(ΔABC) = 28 cm²

ar(ΔPQR) = 63 cm²

PR = 8.4 cm

We know that,

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

$\therefore \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\dfrac{AC}{PR}\Big)^2 \\[1em] \Rightarrow \dfrac{28}{63} = \Big(\dfrac{AC}{8.4}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{4}{9}} = \dfrac{AC}{8.4} \\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{AC}{8.4} \\[1em] \Rightarrow AC = \dfrac{2 \times 8.4}{3}\\[1em] \Rightarrow AC = \dfrac{16.8}{3}\\[1em] \Rightarrow AC = 5.6 \text{ cm.}$

Hence, AC = 5.6 cm.

(iii) Given,

ΔABC ∼ ΔPQR

BC = 4 cm

QR = 5 cm

ar(ΔABC) = 32 cm²

We know that,

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

$\therefore \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔPQR)}} = \Big(\dfrac{BC}{QR}\Big)^2 \\[1em] \Rightarrow \dfrac{32}{\text{ar(ΔPQR)}} = \Big(\dfrac{4}{5}\Big)^2 \\[1em] \Rightarrow \dfrac{32}{\text{ar(ΔPQR)}} = \dfrac{16}{25} \\[1em] \Rightarrow \text{ar(ΔPQR)} = \dfrac{32 \times 25}{16} \\[1em] \Rightarrow \text{ar(ΔPQR)} = 25 \times 2 \\[1em] \Rightarrow \text{ar(ΔPQR)} = 50 \text{ cm}^2.$

Hence, ar(ΔPQR) = 50 cm².

The areas of two similar triangles are 48 cm² and 75 cm² respectively. If the altitude of the first triangle is 3.6 cm, find the corresponding altitude of the other.

Answer

Let the length of altitude of other triangle be x cm

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding altitudes.

$\therefore \dfrac{\text{Area of first Δ}}{\text{Area of second Δ}} = \Big(\dfrac{\text{Altitude of first Δ}}{\text{Altitude of second Δ}}\Big)^2 \\[1em] \Rightarrow \dfrac{48}{75} = \Big(\dfrac{3.6}{x}\Big)^2 \\[1em] \Rightarrow \dfrac{16}{25} = \Big(\dfrac{3.6}{x}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{16}{25}} = \dfrac{3.6}{x} \\[1em] \Rightarrow \dfrac{4}{5} = \dfrac{3.6}{x} \\[1em] \Rightarrow x = \dfrac{18}{4} \\[1em] \Rightarrow x = 4.5 \text{ cm.}$

Hence, the length of altitude of other triangle = 4.5 cm.

In the given figure, AB ⟂ BC and DE ⟂ BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD.

Answer

Given,

AB ⟂ BC and DE ⟂ BC

Thus AB ∥ DE.

∠ABC = ∠DEC [Given]

∠ACB = ∠DCE [Common angle in both triangles]

∴ ΔABC ∼ ΔDEC (By A.A. axiom)

From figure,

DC = AC - AD = 24 - AD

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{DE}{AB} = \dfrac{DC}{AC} \\[1em] \Rightarrow \dfrac{3}{9} = \dfrac{24 - AD}{24} \\[1em] \Rightarrow 24 - AD = \dfrac{3 \times 24}{9} \\[1em] \Rightarrow 24 - AD = \dfrac{72}{9} \\[1em] \Rightarrow 24 - AD = 8 \\[1em] \Rightarrow AD = 24 - 8 \\[1em] \Rightarrow AD = 16 \text{ cm.}$

Hence, AD = 16 cm.

In the given figure, DE || BC. If DE = 4 cm, BC = 6 cm and ar(ΔADE) = 20 cm², find the area of ΔABC.

Answer

Considering ΔADE and ΔABC,

∠A = ∠A [Common angles]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ ΔADE ∼ ΔABC (By A.A. axiom)

We know that,

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

Let the area of ΔABC be x cm².

$\therefore \dfrac{\text{Area of ΔADE}}{\text{Area of ΔABC}} = \Big(\dfrac{{DE}}{{BC}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of ΔADE}}{\text{Area of ΔABC}} = \dfrac{{DE}^2}{{BC}^2} \\[1em] \Rightarrow \dfrac{20}{x} = \dfrac{4^2}{6^2} \\[1em] \Rightarrow \dfrac{20}{x} = \dfrac{16}{36} \\[1em] \Rightarrow x = \dfrac{20 \times 36}{16} \\[1em] \Rightarrow x = \dfrac{720}{16} \\[1em] \Rightarrow x = 45 \text{ cm}^2$

Hence, area of ΔABC = 45 cm².

In the given figure, LM ∥ BC. If AB = 6 cm, AL = 2 cm and AC = 9 cm, calculate :

(i) the length of CM,

(ii) Find the value of $\dfrac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}}$.

Answer

(i) Given,

AB = 6 cm

AL = 2 cm

LB = AB - AL = 6 - 2 = 4 cm

AC = AM + MC

AM = AC - MC

AM = 9 - MC

In ΔAML and ΔABC,

∠AML = ∠ACB [Corresponding angles are equal]

∠LAM = ∠BAC [Common angle]

ΔAML ∼ ΔABC [By AA similarity]

We know that,

Corresponding sides of similar triangles are proportional.

$\therefore \dfrac{AL}{LB} = \dfrac{AM}{MC} \\[1em] \Rightarrow \dfrac{2}{4} = \dfrac{9 - MC}{MC} \\[1em] \Rightarrow 2MC = 4(9 - MC) \\[1em] \Rightarrow 2MC = 36 - 4MC \\[1em] \Rightarrow 2MC + 4MC = 36 \\[1em] \Rightarrow 6MC = 36 \\[1em] \Rightarrow MC = \dfrac{36}{6} \\[1em] \Rightarrow MC = 6 \text{ cm.}$

Hence, CM = 6 cm.

(ii) Given

In ΔALM and ΔABC,

∠AML = ∠ACB [Corresponding angles are equal]

∠LAM = ∠BAC [Common angle]

ΔAML ∼ ΔABC [By AA similarity]

We know that,

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

$\therefore \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \Big(\dfrac{AL}{AB}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \Big(\dfrac{2}{6}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \dfrac{4}{36} \\[1em] \Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(ΔABC)}} = \dfrac{1}{9}.$

Let ar(ΔALM) = x, then ar(ΔABC) = 9x.

From figure,

ar(trap. LBCM) = ar(ΔABC) - ar(ΔALM)

= 9x - x

= 8x.

$\Rightarrow \dfrac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}} = \dfrac{x}{8x} = \dfrac{1}{8}$.

Hence, $\dfrac{\text{ar(ΔALM)}}{\text{ar(trap. LBCM)}} = \dfrac{1}{8}$.

In ΔABC, it is given that AB = 12 cm, ∠B = 90° and AC = 15 cm. If D and E are points on AB and AC respectively such that ∠AED = 90° and DE = 3 cm, prove that :

(i) ΔABC ∼ ΔAED.

(ii) ar(ΔAED) = 6 cm².

(iii) ar(quad BCED) : ar(ΔABC) = 8 : 9.

Answer

(i) Given,

∠ABC = ∠AED = 90° [Given]

∠BAC = ∠DAE [Common angle]

∴ ΔABC ∼ ΔAED (By A.A. axiom)

Hence, proved that ΔABC ∼ ΔAED.

(ii) ΔABC is right-angled triangle, applying pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ BC² = AC² - AB²

⇒ BC² = (15)² - (12)²

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = $\sqrt{81}$

⇒ BC = 9 cm

Area of ΔABC = $\dfrac{1}{2}$ × Base × height

= $\dfrac{1}{2}$ × 12 × 9

= 54 cm²

Since, ΔABC ∼ ΔAED,

We know that,

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

$\therefore \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{{ED}^2}{{BC}^2} \\[1em] \Rightarrow \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{3^2}{9^2} \\[1em] \Rightarrow \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{9}{81} \\[1em] \Rightarrow \dfrac{\text{ar(ΔAED)}}{\text{ar(ΔABC)}} = \dfrac{1}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = \text{ar(ΔABC)} \times \dfrac{1}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = \dfrac{54}{9} \\[1em] \Rightarrow \text{ar(ΔAED)} = 6 \text{ cm}^2$

Hence, proved that ar(ΔAED) = 6 cm².

(iii) From figure,

ar(quad. BCED) = ar(ΔABC) - ar(ΔAED)

ar(quad. BCED) = 54 - 6

ar(quad. BCED) = 48 cm².

$\Rightarrow \dfrac{\text{ar(quad. BCED)}}{\text{ar.(ΔABC)}} = \dfrac{48}{54} = \dfrac{8}{9}$.

Hence, proved that ar(quad BCED) : ar(ΔABC) = 8 : 9.

In the given figure, ∠PQR = ∠PST = 90°, PQ = 5 cm and PS = 2 cm.

(i) Prove that ΔPQR ∼ ΔPST.

(ii) Find area of ΔPQR : Area of quadrilateral SRQT.

Answer

(i) Considering ΔPQR and ΔPST.

∠P = ∠P [Common angles]

∠PQR = ∠PST [Both are equal to 90°]

∴ ΔPQR ∼ ΔPST (By A.A. axiom)

Hence, proved that ΔPQR ∼ ΔPST.

(ii) We know that,

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

$\Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPST}} = \dfrac{{PQ}^2}{{PS}^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPST}} = \dfrac{5^2}{2^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPQR} - \text{Area of SRQT}} = \dfrac{25}{4} \\[1em] \Rightarrow 4\text{ Area of ΔPQR} = 25 (\text{ Area of ΔPQR} - \text{Area of SRQT}) \\[1em] \Rightarrow 4\text{ Area of ΔPQR} = 25 \text{ Area of ΔPQR} - 25\text{Area of SRQT} \\[1em] \Rightarrow 25\text{Area of SRQT} = 25 \text{ Area of ΔPQR} - 4\text{ Area of ΔPQR}\\[1em] \Rightarrow 25\text{Area of SRQT} = 21 \text{ Area of ΔPQR} \\[1em] \Rightarrow \dfrac{\text{Area of ΔPQR}}{\text{Area of SRQT}} = \dfrac{25}{21}.$

Hence, area of ΔPQR : Area of quadrilateral SRQT = 25 : 21.

In a ΔPQR, L and M are two points on the base QR such that ∠LPQ = ∠QRP and ∠RPM = ∠QRP. Prove that

(i) ΔPQL ∼ ΔRPM.

(ii) QL × RM = PL × PM.

(iii) PQ² = QL × QR.

Answer

(i) In ΔPQL and ΔRPM

∠LPQ = ∠MRP [Given]

∠LQP = ∠RPM [Given]

∴ ΔPQL ∼ ΔRPM (y A.A. axiom)

Hence, proved that ΔPQL ∼ ΔRPM.

(ii) Since, ΔPQL ∼ ΔRPM and corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{QL}{PM} = \dfrac{PL}{RM} \\[1em] \Rightarrow QL \times RM = PL \times PM.$

Hence, proved that QL × RM = PL × PM.

(iii) In ΔPQL and ΔRQP

∠LPQ = ∠QRP [Given]

∠Q = ∠Q [Common]

∴ ΔPQL ∼ ΔRQP (By A.A. axiom)

Since, corresponding sides of similar triangle are proportional to each other.

$\therefore \dfrac{PQ}{RQ} = \dfrac{QL}{QP} \\[1em] \Rightarrow PQ^2 = QR \times QL.$

Hence, proved that PQ² = QR x QL.

In the adjoining figure, the medians BD and CE of a ΔABC meet at G. Prove that :

(i) ΔEGD ∼ ΔCGB.

(ii) BG = 2 × GD.

Answer

(i) Since, BD and CE are medians.

So, E is mid-point of AB and D is mid-point of AC.

By mid-point theorem,

ED ∥ BC and ED = $\dfrac{1}{2}$ BC [By mid-point theorem]

$\Rightarrow \dfrac{ED}{BC} = \dfrac{1}{2}$

$\Rightarrow \dfrac{BC}{ED} = 2$ .....(1)

In triangle EGD and BGC,

∠EGD = ∠BGC [Vertically opposite angles are equal]

∠DEG = ∠GCB [Alternate angles are equal]

∴ ΔEGD ∼ ΔCGB by (By A.A. axiom)

Hence, proved that ΔEGD ∼ ΔCGB.

(ii) Since, corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{BG}{GD} = \dfrac{BC}{ED} \\[1em] \Rightarrow \dfrac{BG}{GD} = 2 .....\text{(From 1)} \\[1em] \Rightarrow BG = 2GD.$

Hence, proved that BG = 2GD.

In the given diagram ∆ADB and ∆ACB are two right angled triangles with ∠ADB = ∠BCA = 90°. If AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm

(i) Prove that ∆APD ∼ ∆BPC.

(ii) Find the length of BD and PB

(iii) Hence, find the length of PA

(iv) Find area ∆APD : area ∆BPC

Answer

(i) In ∆APD and ∆BPC,

⇒ ∠APD = ∠BPC (Vertically opposite angles are equal)

⇒ ∠ADP = ∠BCP (Both equal to 90°)

Hence, proved that ∆APD ∼ ∆BPC.

(ii) In ∆ADB,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 10² = 6² + BD²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = $\sqrt{64}$ = 8 cm.

⇒ PB = BD - PD = 8 - 4.5 = 3.5 cm

Hence, BD = 8 cm and PB = 3.5 cm.

(iii) In ∆APD,

By pythagoras theorem,

⇒ AP² = AD² + DP²

⇒ AP² = 6² + (4.5)²

⇒ AP² = 36 + 20.25

⇒ AP² = 56.25

⇒ AP = $\sqrt{56.25}$ = 7.5 cm

Hence, length of AP = 7.5 cm.

(iv) We know that,

Ratio of area of similar triangles is equal to the square of the corresponding sides.

$\therefore \dfrac{\text{Area of △APD}}{\text{Area of △BPC}} = \dfrac{AD^2}{BC^2} \\[1em] = \dfrac{6^2}{(2.4)^2} \\[1em] = \dfrac{6 \times 6}{2.4 \times 2.4} \\[1em] = \dfrac{1 \times 1}{0.4 \times 0.4} \\[1em] = \dfrac{10 \times 10}{4 \times 4} \\[1em] = \dfrac{100}{16} \\[1em] = \dfrac{25}{4} \\[1em] = 25 : 4.$

Hence, area ∆APD : area ∆BPC = 25 : 4.

In the adjoining figure, PQRS is a parallelogram with PQ = 15 cm and RQ = 10 cm. If L is a point on RP such that RL : PL = 2 : 3 and QL produced meets RS at M and PS produced at N, find the lengths of PN and RM.

Answer

In ΔRLQ and ΔPLN,

⇒ ∠RLQ = ∠PLN [Vertically opposite angles are equal]

⇒ ∠LRQ = ∠LPN [Alternate angles are equal]

∴ ΔRLQ ∼ ΔPLN (By A.A. axiom)

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{RL}{LP} = \dfrac{RQ}{PN} \\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{10}{PN} \\[1em] \Rightarrow PN = \dfrac{30}{2} = 15 \text{ cm}.$

In ΔRLM and ΔPLQ,

⇒ ∠RLM = ∠PLQ [Vertically opposite angles are equal]

⇒ ∠LRM = ∠LPQ [Alternate angles are equal]

∴ ΔRLM ∼ ΔPLQ (By A.A. axiom)

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{RM}{PQ} = \dfrac{RL}{LP} \\[1em] \Rightarrow \dfrac{RM}{15} = \dfrac{2}{3} \\[1em] \Rightarrow RM = \dfrac{30}{3} \\[1em] \Rightarrow RM = 10 \text{ cm.}$

Hence, PN = 15 cm and RM = 10 cm.

In the given figure, ΔABC ∼ ΔPQR, AM and PN are altitudes, whereas AX and PY are medians. Prove that $\dfrac{AM}{PN} = \dfrac{AX}{PY}$.

Answer

Since ΔABC ∼ ΔPQR

So, their respective sides will be in proportion

$\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{BC}{QR}$

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

In ΔABM and ΔPQN,

∠ABM = ∠PQN [Since, ABC and PQR are similar]

∠AMB = ∠PNQ = 90° [Given ]

∴ ΔΑΒΜ ∼ ΔPQN by AA similarity

$\dfrac{AM}{PN} = \dfrac{AB}{PQ}$ .....(1)

Since, AX and PY are medians so they will divide their opposite sides.

BX = $\dfrac{BC}{2}$ and QY = $\dfrac{QR}{2}$

Therefore, we have:

$\dfrac{AB}{PQ} = \dfrac{BX}{QY}$

∠ABC = ∠PQR

So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal.

Hence, ∆ABX ∼ ∆PQY (by SAS similarity rule).So,

$\dfrac{AM}{PQ} = \dfrac{AX}{PY}$ .....(2)

From (1) and (2),

$\dfrac{AM}{PN} = \dfrac{AX}{PY}$

Hence, proved that $\dfrac{AM}{PN} = \dfrac{AX}{PY}$

In the given figure, BC ∥ DE, area (ΔABC) = 25 cm², area (trap. BCED) = 24 cm² and DE = 14 cm. Calculate the length of BC.

Answer

Area of ΔADE = Area of ΔABC + Area of trapezium BCED = 25 + 24 = 49 cm².

Given,

BC ∥ DE.

In ΔABC and ΔADE,

∠ABC = ∠ADE [Corresponding angles are equal]

∠ACB = ∠AED [Corresponding angles are equal]

∴ ΔABC ∼ ΔADE by (By A.A. axiom)

We know that,

The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$\therefore \dfrac{\text{Area of ΔABC}}{\text{Area of ΔADE}} = \dfrac{BC^2}{DE^2} \\[1em] \Rightarrow \dfrac{25}{49} = \dfrac{BC^2}{14^2} \\[1em] \Rightarrow BC^2 = \dfrac{25}{49} \times 196 \\[1em] \Rightarrow BC^2 = 100 \\[1em] \Rightarrow BC = 10 \text{ cm.}$

Hence, BC = 10 cm.

In the given figure, DE ∥ BC and DE : BC = 3 : 5. alculate ar(ΔADE) : ar(trap. BCED).

Answer

It is given that DE ∥ BC.

∠ADE = ∠ABC [Corresponding angles are equal]

∠AED = ∠ACB [Corresponding angles are equal]

∴ ΔADE ∼ ΔАВС (By A.A. axiom)

We know that,

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

$\dfrac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} = \dfrac{(BC)^2}{(DE)^2}$

Subtracting 1 from both sides, we get:

$\Rightarrow \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} - 1 = \dfrac{(5)^2}{(3)^2} - 1 \\[1em] \Rightarrow \dfrac{\text{ar(ΔABC)} - \text{ar(ΔADE)}}{\text{ar(ΔADE)}} = \dfrac{25}{9} - 1 \\[1em] \Rightarrow \dfrac{\text{ar(BCED)}}{\text{ar(ΔADE)}} = \dfrac{25 - 9}{9} \\[1em] \Rightarrow \dfrac{\text{ar(BCED)}}{\text{ar(ΔADE)}} = \dfrac{16}{9} \\[1em] \Rightarrow \dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \dfrac{9}{16}.$

Hence, ar(ΔADE) : ar(trap. BCED) = 9 : 16.

In ΔABC, D and E are mid-points of AB and AC respectively.
Find: ar(ΔADE) : ar(ΔABC).

Answer

We know that,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of the third side.

Given,

In Δ ABC,

D is mid-point of side AB and E is the mid-point of the side AC.

∴ DE ∥ BC and DE = $\dfrac{1}{2}$ BC

D is mid-point of side AB.

∴ AB = 2AD

Let us consider ΔADE and ΔABC

∠DAE = ∠BAC [Common angle]

∠ADE = ∠ABC [Corresponding angle are equal]

∴ ΔADE ∼ ΔABC by AA similarity.

$\Rightarrow \dfrac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \dfrac{AD^2}{AB^2} \\[1em] \Rightarrow \dfrac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \dfrac{AD^2}{(2AD)^2} \\[1em] \Rightarrow \dfrac{\text{area(ΔADE)}}{\text{area(ΔABC)}} = \dfrac{1}{4}.$

Hence, ar(ΔADE) : ar(ΔABC) = 1 : 4.

In ΔPQR, MN is parallel to QR and $\dfrac{PM}{QM} = \dfrac{2}{3}$.

(i) Find $\dfrac{MN}{QR}$.

(ii) Prove that ΔOMN and ΔORQ are similar.

(iii) Find: Area of ΔOMN : Area of ΔORQ.

Answer

(i) Considering ΔPMN and ΔPQR,

∠P = ∠P [Common angles]

∠PMN = ∠PQR [Corresponding angles are equal]

∴ ΔPMN ∼ ΔPQR by AA similarity.

Given,

$\dfrac{PM}{MQ} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{PM}{PQ - PM} = \dfrac{2}{3} \\[1em] \Rightarrow 3PM = 2(PQ - PM) \\[1em] \Rightarrow 3PM = 2PQ - 2PM \\[1em] \Rightarrow 3PM + 2PM = 2PQ \\[1em] \Rightarrow 5PM = 2PQ \\[1em] \Rightarrow \dfrac{PM}{PQ} = \dfrac{2}{5}.$

Since triangles are similar hence the ratio of the corresponding sides will be equal,

$\dfrac{MN}{QR} = \dfrac{PM}{PQ} = \dfrac{2}{5}$.

Hence, $\dfrac{MN}{QR} = \dfrac{2}{5}$

(ii) Considering ΔOMN and ΔORQ,

∠MON = ∠QOR (Vertically opposite angles are equal)

∠OMN = ∠ORQ (Alternate angles are equal)

Hence, by AA similarity ΔOMN ∼ ΔORQ.

(iii) We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \dfrac{MN^2}{QR^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \dfrac{2^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of ΔOMN}}{\text{Area of ΔORQ}} = \dfrac{4}{25} \\[1em]$

Hence, the ratio of the Area of ΔOMN : Area of ΔORQ = 4 : 25.

PQR is a triangle, S is a point on the side QR of ΔPQR such that ∠PSR = ∠QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.

(i) Prove ΔPQR ∼ ΔSPR.

(ii) Find the length of QR and PS.

(iii) Find $\dfrac{\text{area of ΔPQR}}{\text{area of ΔSPR}}$.

Answer

(i) In ΔPQR and ΔSPR,

⇒ ∠PSR = ∠QPR [Given ]

⇒ ∠PRQ = ∠PRS [Common angle]

∴ ΔPQR ∼ ΔSPR by AA similarity.

Hence, proved that ΔPQR ∼ ΔSPR.

(ii) Since, ΔPQR ∼ ΔSPR and corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{QR}{PR} = \dfrac{PR}{SR} \\[1em] \Rightarrow \dfrac{QR}{6} = \dfrac{6}{3} \\[1em] \Rightarrow \dfrac{QR}{6} = \dfrac{36}{3} = 12 \text{ cm.}$

Also,

$\Rightarrow \dfrac{PQ}{SP} = \dfrac{PR}{SR} \\[1em] \Rightarrow \dfrac{8}{SP} = \dfrac{6}{3} \\[1em] \Rightarrow SP = \dfrac{8 \times 6}{3} \\[1em] \Rightarrow SP = \dfrac{24}{3} = 4 \text{ cm.}$

Hence, QR = 12 cm and PS = 4 cm.

(iii) We know that,

Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides.

$\therefore \dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPQR}} = \Big(\dfrac{PQ}{SP}\Big)^2 \\[1em] = \Big(\dfrac{8}{4}\Big)^2 \\[1em] = (2)^2 \\[1em] = 4$

Hence, $\dfrac{\text{Area of ΔPQR}}{\text{Area of ΔPQR}} = \dfrac{4}{1}$

In ΔABC, ∠ABC = 90°, AB = 20 cm, AC = 25 cm, DE is perpendicular to AC such that ∠DEA = 90° and DE = 3 cm as shown in the given figure.

(a) Prove that ΔABC ~ ΔAED.

(b) Find the lengths of BC, AD and AE.

(c) If BCED represents a plot of land on a map whose actual area on ground is 576 m², then find the scale factor of the map.

Answer

(a) In ΔABC and ΔAED,

⇒ ∠ABC = ∠AED [Both = 90°]

⇒ ∠BAC = ∠DAE [Common angles]

∴ ΔABC ~ ΔAED (By AA similarity postulate)

Hence, proved that ΔABC ~ ΔAED.

(b) Given,

AB = 20 cm, AC = 25 cm, DE = 3 cm

In ΔABC,

By pythagoras theorem,

⇒ AB² + BC² = AC²

⇒ (20)² + BC² = (25)²

⇒ 400 + BC² = 625

⇒ BC² = 625 - 400

⇒ BC² = 225

⇒ BC = $\sqrt{225}$

⇒ BC = 15 cm.

We know that,

Since, corresponding sides of similar triangles are proportional we have :

$\dfrac{AB}{AE} = \dfrac{BC}{DE} = \dfrac{AC}{AD}$

Solving,

$\Rightarrow \dfrac{AB}{AE} = \dfrac{BC}{DE} \\[1em] \Rightarrow \dfrac{20}{AE} = \dfrac{15}{3} \\[1em] \Rightarrow \dfrac{20 \times 3}{15} = AE \\[1em] \Rightarrow AE = \dfrac{20}{5} \\[1em] \Rightarrow AE = 4\text{ cm}.$

Substituting values in $\dfrac{BC}{DE} = \dfrac{AC}{AD}$ we get :

$\Rightarrow \dfrac{BC}{DE} = \dfrac{AC}{AD} \\[1em] \Rightarrow \dfrac{15}{3} = \dfrac{25}{AD} \\[1em] \Rightarrow 5 = \dfrac{25}{AD} \\[1em] \Rightarrow AD = \dfrac{25}{5} \\[1em] \Rightarrow AD = 5\text{ cm}.$

Hence, BC = 15 cm, AE = 4 cm, AD = 5 cm.

(c) Given,

Area on ground = 576 m²

By formula,

Area of triangle = $\dfrac{1}{2}$ × base × height

Area of ΔABC = $\dfrac{1}{2}$ × AB × BC

= $\dfrac{1}{2} \times 20 \times 15$

= 150 cm².

Area of ΔAED = $\dfrac{1}{2}$ × AE × DE

$= \dfrac{1}{2} \times 4 \times 3$

$= \dfrac{1}{2} \times 12$

= 6 cm².

From figure,

⇒ Area of Quadrilateral (BCED) = Area of ΔABC - Area of ΔAED

= 150 - 6 = 144 cm².

⇒ Actual ground area = 576 m²

= 576 × 10000 cm² = 5760000 cm²

Let scale factor be k.

By formula,

k² = $\dfrac{\text{Area of BCED}}{\text{Actual area of BCED on ground}}$

Substituting values we get :

$\Rightarrow k^2 = \dfrac{144}{5760000} \\[1em] \Rightarrow k^2 = \dfrac{1}{40000} \\[1em] \Rightarrow k = \sqrt{\dfrac{1}{40000}}\\[1em] \Rightarrow k = \dfrac{1}{200}\\[1em]$

Hence, scale factor equals 1 : 200.