Similarity (With Applications to Maps and Models) Archives

The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their: areas.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

The ratio between the altitudes of two similar triangles is same as the ratio between their sides. So, The ratio between the areas of two similar triangles is same as the square of the ratio between...

The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their: (i) medians. (ii) perimeters.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

The ratio between the altitudes of two similar triangles is same as the ratio between their sides. So, (i) The ratio between the medians of two similar triangles is same as the ratio between their...

Two similar triangles are equal in area. Prove that the triangles are congruent.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Let’s consider two similar triangles as \[\blacktriangle ABC\text{ }\sim\text{ }\blacktriangle PQR\] So, \[Ar\left( \blacktriangle ABC \right)/\text{ }Ar\left( \blacktriangle PQR \right)\] \[=\text{...

In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that: (i) EF = FB, (ii) AG: GD = 2: 1

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) In \[\vartriangle BFD\text{ }and\text{ }\vartriangle BEC,\] \[\angle BFD\text{ }=\angle BEC\] [Corresponding angles] \[\angle FBD\text{ }=\angle EBC\] [Common] Hence, \[\vartriangle...

In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle QRP is similar to triangle ABC.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: In \[\vartriangle ABC,\text{ }as\text{ }PR\text{ }||\text{ }BC\text{ }by\text{ }BPT\]we have \[AP/PB\text{ }=\text{ }AR/RC\] And, in \[\vartriangle PAR\text{ }and\text{ }\vartriangle...

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm. Find the area of ΔACD: area of ΔABC

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: As, \[\vartriangle ACD\text{ }\sim\text{ }\vartriangle BCA\] We have, \[Ar\left( \vartriangle ACD \right)/\text{ }Ar\left( \vartriangle BCA \right)\] \[=\text{ }A{{D}^{2}}/\text{...

In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm. (i) Prove that ΔACD is similar to ΔBCA. (ii) Find BC and CD

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) In \[\vartriangle ACD\text{ }and\text{ }\vartriangle BCA\] \[\angle DAC\text{ }=\angle ABC\] [Given] \[\angle ACD\text{ }=\angle BCA\][Common angles] Hence, \[\vartriangle ACD\text{...

In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate: (i) EF (ii) AC

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) In \[\vartriangle PCD\text{ }and\text{ }\vartriangle PEF\] \[\angle CPD\text{ }=\angle EPF\][Vertically opposite angles] \[\angle DCE\text{ }=\angle FEP\][As DC || EF, alternate...

In the following figure, DE || AC and DC || AP. Prove that: BE/EC = BC/CP.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: Given, \[DE\text{ }||\text{ }AC\] So, \[BE/EC\text{ }=\text{ }BD/DA\text{ }\left[ By\text{ }BPT \right]\] And, \[DC\text{ }||\text{ }AP\] So, \[BC/CP\text{ }=\text{ }BD/DA\text{ }\left[...

In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown. Prove that: AB/BC = PQ/QR

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: Let join \[AR\]such that it intersects \[BQ\]at point \[X.\] In \[\vartriangle ACR,\text{ }BX\text{ }||\text{ }CR\] By\[BPT\], we have \[AB/BC\text{ }=\text{ }AX/XR\text{ }\ldots \text{...

In the following figure, ∠AXY = ∠AYX. If BX/AX = CY/AY, show that triangle ABC is isosceles.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[\angle AXY\text{ }=~\angle AYX\] So, \[AX\text{ }=\text{ }AY\][Sides opposite to equal angles are equal.] Also, from BPT we have \[BX/AX\text{ }=\text{...

Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: AB/PQ = AD/PM

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle PQR\] And, \[AD\text{ }and\text{ }PM\]are the angle bisectors. So, \[\angle BAD\text{ }=\angle QPM\]...

Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that: AB/PQ = AD/PM.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle PQR\] So, \[\angle ABC\text{ }=\angle PQR\] i.e. \[\angle ABD\text{ }=\angle PQM\] Also, \[\angle...

Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: AB/PQ = AD/PM.

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle PQR\] \[AD\text{ }and\text{ }PM\]are the medians, so \[BD\text{ }=\text{ }DC\text{ }and\text{ }QM\text{...

In the following figure, AB, CD and EF are perpendicular to the straight line BDF. If AB = x and; CD = z unit and EF = y unit, prove that: 1/x + 1/y = 1/z

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: In \[\Delta \text{ }FDC\text{ }and\text{ }\Delta \text{ }FBA,\] \[\angle FDC\text{ }=\angle FBA\text{ }\left[ As\text{ }DC\text{ }||\text{ }AB \right]\] \[\angle DFC\text{ }=\angle BFA\]...

In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm, Calculate: PE

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: We already have, \[\vartriangle AEB\text{ }\sim\text{ }\vartriangle FEC\] So, \[AE/FE\text{ }=\text{ }BE/CE\] \[=\text{ }AB/FC\] \[AE/FE\text{ }=\text{ }9/13.5\] Or, \[\left( AF\text{...

In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm, Calculate: (i) EC (ii) AF

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) In \[\Delta \text{ }AEB\text{ }and\text{ }\Delta \text{ }FEC,\] \[\angle AEB\text{ }=\angle FEC\] [Vertically opposite angles] \[\angle BAE\text{ }=\angle CFE\] [Since, AB||DC] Hence,...

In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm. Find: XY

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[XY\text{ }||\text{ }BC\] So, In \[\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC\] \[\angle AXY\text{ }=\angle ABC\] [Corresponding angles]...

In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm. Find: (i) AY/YC (ii) YC/AC

Class 10, Exercise 15E, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[XY\text{ }||\text{ }BC\] So, In \[\Delta \text{ }AXY\text{ }and\text{ }\Delta \text{ }ABC\] \[\angle AXY\text{ }=\angle ABC\][Corresponding angles]...

A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find: (i) OA, if OA’ = 6 cm (ii) OC’, if OC = 21 cm Also, state the value of: (a) OB’/OB (b) C’A’/CA

Class 10, Exercise 15D, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(i)\[OA\text{ }=\text{ }6\text{ }cm\] So, \[OA\text{ }\left( 3 \right)\text{ }=\text{ }OA\] \[OA\text{ }\left( 3 \right)\text{ }=\text{ }6\] Or, \[OA\text{ }=\text{ }2\text{ }cm\] (ii) \[OC\text{...

A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find: (i) A’B’, if AB = 4 cm. (ii) BC, if B’C’ = 15 cm.

Class 10, Exercise 15D, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

According to the given question, \[\Delta \text{ }ABC\]is enlarged and the scale factor \[m\text{ }=\text{ }3\]to the \[\Delta \text{ }ABC\] (i) \[AB\text{ }=\text{ }4\text{ }cm\] So, \[AB\left( 3...

A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate: (i) the length of M’ N’, if MN = 8 cm. (ii) the length of LM, if L’ M’ = 5.4 cm.

Class 10, Exercise 15D, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

According to the given question, \[\Delta \text{ }LMN\] has been reduced by a scale factor \[m\text{ }=\text{ }0.8\text{ }to\text{ }\Delta \text{ }LMN\] (i) \[MN\text{ }=\text{ }8\text{ }cm\] So,...

A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’ Calculate: (i) the length of AB, if A’ B’ = 6 cm. (ii) the length of C’ A’ if CA = 4 cm.

Class 10, Exercise 15D, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Given that, \[\Delta \text{ }ABC\]has been enlarged by scale factor \[m\text{ }of\text{ }2.5\text{ }to\text{ }\Delta \text{ }ABC\] (i) \[AB\text{ }=\text{ }6\text{ }cm\] So, \[AB\left( 2.5...

In the given triangle PQR, LM is parallel to QR and PM: MR = 3: 4. Calculate the value of ratio: Area of Δ LQM/ Area of Δ LQN

Class 10, Exercise 15C, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: Because, \[\Delta \text{ }LQM\text{ }and\text{ }\Delta \text{ }LQN\] have common vertex at \[L\]and their bases \[QM\text{ }and\text{ }QN\] are along the same straight line. \[Area\text{...

In the given triangle PQR, LM is parallel to QR and PM: MR = 3: 4. Calculate the value of ratio: (i) PL/PQ and then LM/QR (ii) Area of Δ LMN/ Area of Δ MNR

Class 10, Exercise 15C, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) In \[\Delta \text{ }PLM\text{ }and\text{ }\Delta \text{ }PQR\] As LM || QR, corresponding angles are equal. \[\angle PLM\text{ }=\angle PQR\] \[\angle PML\text{ }=\angle PRQ\] So,...

ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP: AB.

Class 10, Exercise 15C, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

It’s given that, \[Ar\left( \Delta \text{ }APQ \right)\]\[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }Ar\left( \Delta \text{ }ABC \right)\] \[Ar\left( \Delta \text{ }APQ \right)/\text{...

In the given figure, AX: XB = 3: 5. Find: (i) the length of BC, if the length of XY is 18 cm. (ii) the ratio between the areas of trapezium XBCY and triangle ABC.

Class 10, Exercise 15C, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[AX/XB\text{ }=\text{ }3/5\] \[\Rightarrow AX/AB\text{ }=\text{ }3/8\text{ }\ldots .\text{ }\left( 1 \right)\] (i) In \[\Delta \text{ }AXY\text{...

The perimeters of two similar triangles are 30 cm and 24 cm. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.

Class 10, Exercise 15C, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Suppose, \[\vartriangle ABC\text{ }\sim\text{ }\vartriangle DEF\] So, \[AB/DE\text{ }=\text{ }BC/EF\] \[=\text{ }AC/DF\] Or, \[=\text{ }\left( AB+BC+AC \right)/\left( DE+EF+DF \right)\] \[=\text{...

A line PQ is drawn parallel to the base BC of Δ ABC which meets sides AB and AC at points P and Q respectively. If AP = 1/3 PB; find the value of: (i) Area of Δ ABC/ Area of Δ APQ (ii) Area of Δ APQ/ Area of Trapezium PBCQ

Class 10, Exercise 15C, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

According to the given question, \[AP\text{ }=\text{ }\left( 1/3 \right)\text{ }PB\] So, \[AP/PB\text{ }=\text{ }1/3\] In \[\vartriangle \text{ }APQ\text{ }and\text{ }\vartriangle ABC\] As\[PQ\text{...

1. (i) The ratio between the corresponding sides of two similar triangles is 2: 5. Find the ratio between the areas of these triangles. (ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.

Class 10, Exercise 15C, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

As per the given question, The ratio of the areas of two similar triangle are equal to the ratio of squares of their corresponding sides. Thus, (i) The ration is, (ii) The ratio is,

In the given figure, Δ ABC ~ Δ ADE. If AE: EC = 4: 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of ‘x’.

Class 10, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: According to the given question, \[\Delta \text{ }ABC\text{ }\sim\text{ }\Delta \text{ }ADE\] So, we have \[AE/AC\text{ }=\text{ }DE/BC\] \[4/11\text{ }=\text{ }6.6/BC\] Or, \[BC=\left(...

In Δ ABC, D and E are the points on sides AB and AC respectively. Find whether DE ‖ BC, if (i) AB = 9cm, AD = 4cm, AE = 6cm and EC = 7.5cm. (ii) AB = 6.3 cm, EC = 11.0 cm, AD =0.8 cm and EA = 1.6 cm.

Class 10, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(i) In \[\vartriangle \text{ }ADE\text{ }and\text{ }\vartriangle \text{ }ABC\] \[AE/EC\text{ }=\text{ }6/7.5\text{ }=\text{ }4/5\] \[AD/BD\text{ }=\text{ }4/5\] \[\left[ BD\text{ }=\text{ }AB\text{...

A line PQ is drawn parallel to the side BC of Δ ABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.

Class 10, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

In \[\vartriangle \text{ }APQ\text{ }and\vartriangle \text{ }ABC\] \[\angle APQ\text{ }=\angle ABC\] [As PQ || BC, corresponding angles are equal.] \[\angle PAQ\text{ }=\angle BAC\] [Common angle]...

In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. If AP = x, then the value of AC in terms of x.

Class 10, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: As, \[\vartriangle CPQ\text{ }\sim\text{ }\vartriangle CAB\text{ }by\text{ }AA\] criterion for similarity We have, \[CP/AC\text{ }=\text{ }CQ/CB\] \[CP/AC\text{ }=\text{ }4.8/8.4\text{...

In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find: (i) CP/PA (ii) PQ

Class 10, Exercise 1, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) In \[\vartriangle CPQ\text{ }and\text{ }\vartriangle CAB\] \[\angle PCQ\text{ }=\angle APQ\] [As PQ || AB, corresponding angles are equal.] \[\angle C\text{ }=\angle C\] [Common angle]...

In the following figure, point D divides AB in the ratio 3: 5. If BC = 4.8 cm, find the length of DE.

Class 10, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: Because, \[\vartriangle ADE\text{ }\sim\text{ }\vartriangle ABC\text{ }by\text{ }AA\] criterion for similarity So, we have \[AD/AB\text{ }=\text{ }DE/BC\] \[3/8\text{ }=\text{ }DE/4.8\]...

In the following figure, point D divides AB in the ratio 3: 5. Find: (i) AE/AC Also if, (ii) DE = 2.4 cm, find the length of BC.

Class 10, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) In \[\vartriangle ABC,\text{ }as\text{ }DE\text{ }||\text{ }BC\] Using BPT, \[AD/DB\text{ }=\text{ }AE/\text{ }EC\] So, \[AD/AB\text{ }=\text{ }AE/AC\] Now, \[AD/AB\text{ }=\text{...

In the following figure, point D divides AB in the ratio 3: 5. Find: (i) AE/EC (ii) AD/AB

Class 10, Exercise 15B, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Solution: (i) According to the given question, \[AD/DB\text{ }=\text{ }3/5\] And \[DE\text{ }||\text{ }BC\] Using Basic Proportionality theorem, \[AD/DB\text{ }=\text{ }AE/EC\] \[AE/EC\text{...

In the given figure, ∆ ABC and ∆ AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm. (i) ∆ ABC ~ ∆ AMP. (ii) Find AB and BC.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(I) In\[\vartriangle \text{ }ABC\text{ }and\text{ }\vartriangle \text{ }AMP\], we have \[\angle BAC\text{ }=\angle PAM\text{ }\left[ Common \right]\] \[\angle ABC\text{ }=\text{ }\angle PMA\text{...

D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA^2 = CB x CD.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Answer: In \[\Delta \text{ }ADC\text{ }and\text{ }\Delta \text{ }BAC,\] \[\angle ADC\text{ }=\angle BAC\text{ }\left[ Given \right]\] \[\angle ACD\text{ }=\angle ACB\text{ }\left[ Common \right]\]...

Given: ∠GHE = ∠DFE = 90o, DH = 8, DF = 12, DG = 3x – 1 and DE = 4x + 2. Find: the lengths of segments DG and DE.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Answer: In\[\Delta \text{ }DHG\text{ }and\text{ }\Delta \text{ }DFE\], \[\angle GHD\text{ }=\text{ }\angle DFE\text{ }=\text{ }{{90}^{o}}\] \[\angle D\text{ }=\angle D\text{ }\left[ Common \right]\]...

State, true or false: The diagonals of a trapezium, divide each other into proportional segments.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

True

State, true or false: (i)All isosceles triangles are similar. (ii) Two isosceles-right triangles are similar.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(i) True (ii) True

State, true or false: (ii) All equiangular triangles are similar. (ii) All isosceles triangles are similar.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(i) True (ii) False

State, true or false: (i) Two similar polygons are necessarily congruent. (ii) Two congruent polygons are necessarily similar.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(i) False (ii) True

Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and PR = 9 cm; find the length of PB.

Class 10, Exercise 15A, ICSE, Selina, Similarity (With Applications to Maps and Models)

In \[\Delta \text{ }ABC,\] \[AC\text{ }=\text{ }AB\text{ }\left[ Given \right]\] In this way, \[\angle ABC\text{ }=\angle ACB\][Angles inverse to rise to sides are equal.] In \[\Delta \text{...

In the given figure, AB ‖ DC, BO = 6 cm and DQ = 8 cm; find: BP x DO.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Answer: In \[\Delta \text{ }DOQ\text{ }and\text{ }\Delta \text{ }BOP,\] \[\angle QDO\text{ }=\angle PBO\] \[\left[ As\text{ }AB\text{ }\left| \left| \text{ }DC\text{ }in\text{ }this\text{...

In the given figure, AD = AE and AD^2 = BD x EC. Prove that: triangles ABD and CAE are similar.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Answer: In \[\Delta \text{ }ABD\text{ }and\text{ }\Delta \text{ }CAE,\] \[\angle ADE\text{ }=\angle AED\text{ }\left[ Angles\text{ }inverse\text{ }to\text{ }rise\text{ }to\text{ }sides\text{...

In the given figure, DE ‖ BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm. (i) Write all possible pairs of similar triangles. (ii) Find the lengths of ME and DM.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Answer: (I) In\[\Delta \text{ }AME\text{ }and\text{ }\Delta \text{ }ANC\], \[\angle AME\text{ }=\angle ANC\] \[\left[ Since\text{ }DE\text{ }\left| \left| \text{ }BC\text{ }in\text{ }this\text{...

In Δ ABC, BM ⊥ AC and CN ⊥ AB; show that:

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Answer: In \[\Delta \text{ }ABM\text{ }and\text{ }\Delta \text{ }ACN,\] \[\angle AMB\text{ }=\text{ }\angle ANC\] [Since, BM ⊥ AC and CN ⊥ AB] \[\angle BAM\text{ }=\angle CAN\] [Common angle] Thus,...

In Δ ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that : (i) CB : BA = CP : PA (ii) AB x BC = BP x CA

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(I) In\[\Delta \text{ }ABC\], we have \[\angle ABC\text{ }=\text{ }2\angle ACB\][Given] Presently, let \[\angle ACB\text{ }=\text{ }x\] In this way, \[\angle ABC\text{ }=\text{ }2x\] Likewise given,...

In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that: (i) Δ AOB is similar to Δ COD. (ii) OA x OD = OB x OC.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(I) Given, \[AO\text{ }=\text{ }2CO\text{ }and\text{ }BO\text{ }=\text{ }2DO,\] \[AO/CO\text{ }=\text{ }2/1\text{ }=\text{ }BO/DO\] What's more, \[\angle AOB\text{ }=\angle DOC\] [Vertically inverse...

P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that: (i) DP : PL = DC : BL. (ii) DL : DP = AL : DC.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(I) As\[AD||BC\], we have \[AD||\text{ }BP\] too. Along these lines, by \[BPT\] \[DP/PL\text{ }=\text{ }AB/BL\] What's more, since \[ABCD\]is a parallelogram, \[AB\text{ }=\text{ }DC\] Consequently,...

In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that: (i) Δ APB is similar to Δ CPD. (ii) PA x PD = PB x PC.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

(I) In\[\vartriangle APB\text{ }and\text{ }\vartriangle CPD\], we have \[\angle APB\text{ }=\angle CPD\] [Vertically inverse angles] \[\angle ABP\text{ }=\angle CDP\] [Alternate points as, AB||DC]...

In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that: (i) ∆APC and ∆BPD are similar. (ii) If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.

Class 10, Exercise 15A, ICSE, Maths, Selina, Similarity (With Applications to Maps and Models)

Answer: (I) In \[\vartriangle \mathbf{APC}\text{ }\mathbf{and}\text{ }\vartriangle \mathbf{BPD}\]we have \[\angle APC\text{ }=\angle BPD\] [Vertically inverse angles] \[\angle ACP\text{ }=\angle...