Q: Areas of two similar triangles are in the ratio 144: 49. Find the ratio of their corresponding sides.

Let the areas of two similar triangles be A1, A2 and their corresponding sides be S1, S2 respectively. ∴ `"A"_1/"A"_2 = 144/49` ...(i)[Given] ∴ by the theorem of areas of similar triangles, ∴ `"A"_1/"A"_2 = "S"_1^2/"S"_2^2` ∴ `144/49 = "S"_1^2/"S"_2^2` ...[From (i)] ∴ by taking square root of both sides, ∴ `"S"_1/"S"_2 = 12/7` ∴ The ratio of the corresponding sides of the given triangles is 12: 7.

Question 1

A line is parallel to one side of triangle which intersects remaining two sides in two distinct points then that line divides sides in same proportion. Given: In ΔABC line l || side BC and line l intersect side AB in P and side AC in Q. To prove: `"AP"/"PB" = "AQ"/"QC"` Construction: Draw CP and BQ Proof: ΔAPQ and ΔPQB have equal height. `("A"(Δ"APQ"))/("A"(Δ"PQB")) = (["______"])/"PB"` .....(i)[areas in proportion of base] `("A"(Δ"APQ"))/("A"(Δ"PQC")) = (["______"])/"QC"` .......(ii)[areas in proportion of base] ΔPQC and ΔPQB have [______] is common base. Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB. A(ΔPQC) = A(Δ______) ......(iii) `("A"(Δ"APQ"))/("A"(Δ"PQB")) = ("A"(Δ "______"))/("A"(Δ "______"))` ......[(i), (ii), and (iii)] `"AP"/"PB" = "AQ"/"QC"` .......[(i) and (ii)]

Accepted Answer

Proof: ΔAPQ and ΔPQB have equal height.

`("A"(Δ"APQ"))/("A"(Δ"PQB"))` = `"AP"/"PB"` .....(i)[areas in proportion of base]

`("A"(Δ"APQ"))/("A"(Δ"PQC"))` = `"AQ"/"QC"` .......(ii)[areas in proportion of base]

ΔPQC and ΔPQB have PQ is common base.

Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB.

A(ΔPQC) = A(ΔPQB) ......(iii)[Areas of two triangles having equal base and height are equal]

`("A"(Δ"APQ"))/("A"(Δ"PQB"))` = `("A"(Δ"APQ"))/("A"(Δ "PQC"))` ......[(i), (ii), and (iii)]

`"AP"/"PB" = "AQ"/"QC"` .......[(i) and (ii)]

Question 2

ΔABC ~ ΔDEF. Write the ratios of their corresponding sides

Accepted Answer

Answer ΔABC ~ ΔDEF ...[Given] ∴ The ratios of corresponding sides of the given triangles are `"AB"/"DE", "BC"/"EF"` and `"AC"/"DF"`

Question 3

ΔABC ~ ΔPQR, A(ΔABC) = 80 sq.cm, A(ΔPQR) = 125 sq.cm, then complete `("A"(Δ"ABC"))/("A"(Δ"PQR")) = 80/125 = (["______"])/(["______"])`, hence `"AB"/"PQ" = (["______"])/(["______"])`

Accepted Answer

`("A"(Δ"ABC"))/("A"(Δ"PQR")) = 80/125` = `16/25` ......(i)[Given] `("A"(Δ"ABC"))/("A"(Δ"PQR")) = "AB"^2/"PQ"^2` .....(ii)[Theorem of areas of similar triangles] ∴ `"AB"^2/"PQ"^2 = 16/25` .....[From (i) and (ii)] Hence `"AB"/"PQ"` = `4/5` ......[[Taking square root of both sides]

Question 4

ΔABP ~ ΔDEF and A(ΔABP) : A(ΔDEF) = 144:81, then AB : DE = ?

Accepted Answer

`("A"(Δ"ABP"))/("A"(Δ"DEF")) = 144/81` ......(i)[Given] `("A"(Δ"ABP"))/("A"(Δ"DEF")) = "AB"^2/"DE"^2` .......(ii)[Theorem of areas of similar triangles] ∴ `"AB"^2/"DE"^2 = 144/81` .......[From (i) and (ii)] ∴ `"AB"/"DE" = 12/9` or `4/3` .......[Taking square root of both sides]

Question 5

An architecture have model of building. Length of building is 1 m then length of model is 0.75 cm. Then find length and height of model building whose actual length is 22.5 m and height is 10 m

Accepted Answer

Rough Figure -

Actual length/height	Model length/height
1 m	0.75 cm
22.5 m	? (x cm)
10 m	? (y cm)

The actual length of 1 m is shown as 0.75 cm in the model then let the actual length of 22.5 m is shown in the model by 'x' cm.

`therefore 1/22.5 = 0.75/x`

∴ x = `0.75 xx 22.5`

x = 16.875 cm

Now, The actual length of 1 m is shown as 0.75 cm in the model then let the actual height of 10 m is shown in the model by 'y' cm.

`therefore 1/10 = 0.75/"y"`

∴ y = 0.75 × 10

y = 7.5 cm

Question 6

Are triangles in figure similar? If yes, then write the test of similarity.

Accepted Answer

Answer In ΔABC and ΔPQR, `"AB"/"PQ" = 6/3 = 2/1` ......(i) `"BC"/"QR" = 8/4 = 2/1` ......(ii) `"AC"/"PR" = 10/5 = 2/1` ......(iii) ∴ `"AB"/"PQ" = "BC"/"QR" = "AC"/"PR"` ......[From (i), (ii) and (iii)] ∴ ΔABC ∼ ΔPQR ......[SSS test of similarity] ∴ The triangles in the figure are similar by SSS test of similarity.

Question 7

Areas of two similar triangles are 225 cm 2 and 81 cm 2 . If side of smaller triangle is 12 cm, find corresponding side of major triangle.

Accepted Answer

Let the areas of two similar triangles be A1 and A2.

A1 = 225 cm2 and A2 = 81 cm2

Let the corresponding sides of triangles be s1 and s2 respectively.

s1 = 12 cm

`"A"_1/"A"_2 = "s"_1^2/"s"_2^2` ...[Theorem of areas of similar triangles]

∴ `225/81 = "s"_1^2/12^2`

∴ s12 = `(225 xx 12^2)/81`

∴ s1 = `(15 xx 12)/9` ...[Taking square root of both sides]

∴ s1 = 20 cm

∴ The length of the corresponding side of the bigger triangle is 20 cm.

Question 8

Areas of two similar triangles are equal then prove that triangles are congruent

Accepted Answer

Given: ΔABC ~ ΔPQR and A(ΔABC) = A(ΔPQR)

To prove: ΔABC ≅ ΔPQR

Proof:

`("A"(Δ"ABC"))/("A"(Δ"PQR"))` = 1 ......(i) [Given]

Also, `("A"(Δ"ABC"))/("A"(Δ"PQR")) = "AB"^2/"PQ"^2 = "BC"^2/"QR"^2 = "AC"^2/"PR"^2` ......[Theorem of areas of similar triangles]

∴ 1 = `"AB"^2/"PQ"^2 = "BC"^2/"QR"^2 = "AC"^2/"PR"^2` .....[From (i)]

∴ 1 = `"AB"^2/"PQ"^2`

∴ AB2 = PQ2

∴ AB = PQ ......[Taking square root of both sides]

i.e., seg AB ≅ seg PQ

Similarly, seg BC ≅ seg QR and seg AC ≅ seg PR

∴ ΔABC ≅ ΔPQR ......[SSS test of congruency]

Question 9

Areas of two similar triangles are in the ratio 144: 49. Find the ratio of their corresponding sides.

Accepted Answer

Let the areas of two similar triangles be A1 and A2.

A1 = 225 cm2 and A2 = 81 cm2

Let the corresponding sides of triangles be s1 and s2 respectively.

s1 = 12 cm

`"A"_1/"A"_2 = "s"_1^2/"s"_2^2` ...[Theorem of areas of similar triangles]

∴ `225/81 = "s"_1^2/12^2`

∴ s12 = `(225 xx 12^2)/81`

∴ s1 = `(15 xx 12)/9` ...[Taking square root of both sides]

∴ s1 = 20 cm

∴ The length of the corresponding side of the bigger triangle is 20 cm.

Question 10

From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?

Accepted Answer

ΔABC and ΔBCD have same base BC. ∴ `("A"(Δ"ABC"))/("A"(Δ"BCD")) = "AB"/"DC"` ...[Triangles having equal base] ∴ `("A"(Δ"ABC"))/("A"(Δ"BCD")) = 6/8` ∴ `("A"(Δ"ABC"))/("A"(Δ"BCD")) = 3/4`

Question 11

From fig., seg PQ || side BC, AP = x + 3, PB = x – 3, AQ = x + 5, QC = x – 2, then complete the activity to find the value of x. In ΔPQB, PQ || side BC `"AP"/"PB" = "AQ"/(["______"])` ...[______] `(x + 3)/(x - 3) = (x + 5)/(["______"])` (x + 3) [______] = (x + 5)(x – 3) x 2 + x – [______] = x 2 + 2x – 15 x = [______]

Accepted Answer

In ΔPQB,

PQ || side BC

`"AP"/"PB" = "AQ"/bb"QC"` ...[Base proportionality theorem]

`(x + 3)/(x - 3)` = `(x + 5)/bb(x - 2)`

(x + 3) (x – 2) = (x + 5)(x – 3)

x2 + x – 6 = x2 + 2x – 15

∴ x – 6 = 2x – 15

∴ 2x – x = 15 – 6

∴ x = 9

Question 12

From given information, is PQ || BC? AP = 2, PB = 4, AQ = 3, QC = 6

Accepted Answer

`"AP"/"PB" = 2/4 = 1/2` .....(i)[Given]

`"AQ"/"QC" = 3/6 = 1/2` .....(ii)

In ΔABC,

`"AP"/"PB" = "AQ"/"QC" = 1/2` ......[From (i) and (ii)]

∴ Line PQ || side BC .......[Converse of basic proportionality theorem]

Question 13

Given ΔABC ~ ΔDEF, if ∠A = 45° and ∠E = 35° then ∠B = ?

Accepted Answer

35° In ΔABC and ΔDEF, ΔABC ~ ΔDEF ...(Given) ∴ ∠B ≅ ∠E ...(Corresponding angles of similar triangles) But ∠E = 35° ...(Given) ∴ ∠B = 35°.

Question 14

If ΔABC ~ ΔLMN and ∠A = 60° then ∠L = ?

Accepted Answer

60° In ΔABC and ΔLMN, ΔABC ~ ΔLMN ∴ ∠A ≅ ∠L ...(Corresponding angles of similar triangles) But ∠A = 60° ...(Given) ∴ ∠L = 60°.

Question 15

If ΔABC ~ ΔLMN and ∠B = 40°, then ∠M = ? Give reason.

Accepted Answer

Answer ΔABC ~ ΔLMN .......[Given] ∴ ∠B ≅ ∠M ......(i) [Corresponding angles of similar triangles] But ∠B = 40° ......[Given] ∴ ∠M = 40° ......[From (i)]

Question 16

If ∆ABC ~ ∆PQR and AB : PQ = 3 : 4 then A(∆ABC) : A(∆PQR) = ?

Accepted Answer

9: 16

In ∆ABC and ∆PQR,

∆ABC ~ ∆PQR

AB : PQ = 3 : 4 ...(Given)

by theorem of areas of similar triangles,

`"A(∆ABC)"/"A(∆PQR)" = ("AB"^2)/("PQ"^2)`

`"A(∆ABC)"/"A(∆PQR)" = 3^2/4^2`

`"A(∆ABC)"/"A(∆PQR)" = 9/16`.

∴ A(∆ABC) : A(∆PQR) = 9 : 16.

Question 17

If ∆XYZ ~ ∆PQR and A(∆XYZ) = 25 cm 2 , A(∆PQR) = 4 cm 2 then XY : PQ = ?

Accepted Answer

5: 2

In ∆XYZ and ∆PQR,

∆XYZ ~ ∆PQR

A(∆XYZ) = 25 cm2, A(∆PQR) = 4 cm2 ...(Given)

by the theorem of areas of similar triangles,

`"A(∆XYZ)"/"A(∆PQR)" = "XY"^2/"PQ"^2`

∴ `25/4 = "XY"^2/"PQ"^2`

∴ `"XY"/"PQ" = 5/2`

∴ XY : PQ = 5 : 2.

Question 18

If ΔXYZ ~ ΔPQR then `"XY"/"PQ" = "YZ"/"QR"` = ?

Accepted Answer

`"XZ"/"PR"` In ΔXYZ and ΔPQR, ΔXYZ ~ ΔPQR ...(Given) `"XY"/"PQ" = "YZ"/"QR" = "XZ"/"PR"` ...[Corresponding sides of similar triangles.]

Question 19

In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio. (i) `"A(ΔABD)"/"A(ΔADC)"` (ii) `"A(ΔABD)"/"A(ΔABC)"` (iii) `"A(ΔADC)"/"A(ΔABC)"`

Accepted Answer

Diagram: Refer textbook

Question 20

In ΔABC, AP ⊥ BC and BQ ⊥ AC, B−P−C, A−Q−C, then show that ΔCPA ~ ΔCQB. If AP = 7, BQ = 8, BC = 12, then AC = ? In ΔCPA and ΔCQB ∠CPA ≅ [∠ ______] ...[each 90°] ∠ACP ≅ [∠ ______] ...[common angle] ΔCPA ~ ΔCQB ......[______ similarity test] `"AP"/"BQ" = (["______"])/"BC"` .......[corresponding sides of similar triangles] `7/8 = (["______"])/12` AC × [______] = 7 × 12 AC = 10.5

Accepted Answer

In ΔCPA and ΔCQB

∠CPA ≅ ∠CQB ...[each 90°]

∠ACP ≅ ∠BCQ ...[common angle]

ΔCPA ~ ΔCQB ...[AA similarity test]

`"AP"/"BQ"` = `"AC"/"BC"` ...[corresponding sides of similar triangles]

`7/8` = `"AC"/12`

AC × 8 = 7 × 12

AC = `(7 xx 12)/8`

AC = `(7 xx 3)/2`

AC = `21/2`

AC = 10.5.

Question 21

In ΔDEF and ΔXYZ, `"DE"/"XY" = "FE"/"YZ"` and ∠E ≅ ∠Y. _______ test gives similarity between ΔDEF and ΔXYZ.

Accepted Answer

SAS In ΔDEF and ΔXYZ, `"DE"/"XY" = "FE"/"YZ"` ...(Given) ∠E ≅ ∠Y ...(Given) ∴ ΔDEF ≅ ΔXYZ ...(SAS Similarity triangle test)

Question 22

In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?

Accepted Answer

ΔABC and ΔBCD have same base BC. ∴ `"A(ΔABC)"/"A(ΔBCD)" = "AB"/"DC"` .....[Triangles having equal base] ∴ `"A(ΔABC)"/"A(ΔBCD)" = 6/4` .....[Given] ∴ `("A"(Δ"ABC"))/("A"(Δ"BCD")) = 3/2`

Question 23

In fig. BD = 8, BC = 12, B-D-C, then `"A(ΔABC)"/"A(ΔABD)"` = ?

Accepted Answer

3: 2

In ΔABC and ΔABD,

ΔABC and ΔABD have the same height. ...(Given)

The ratio of the areas of two triangles with equal heights is equal to the ratio of their corresponding bases.

∴ `"A(ΔABC)"/"A(ΔABD)" = "BC"/"BD"`

∴ `"A(ΔABC)"/"A(ΔABD)" = 12/8`

∴ `"A(ΔABC)"/"A(ΔABD)" = 3/2`.

Question 24

In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar. In ΔAPB and ΔAQC ∠APB = [ ]° ......(i) ∠AQC = [ ]° ......(ii) ∠APB ≅ ∠AQC .....[From (i) and (ii)] ∠PAB ≅ ∠QAC .....[______] ΔAPB ~ ΔAQC .....[______]

Accepted Answer

In ΔAPB and ΔAQC ∠APB = 90° ......(i) ∠AQC = 90° ......(ii) ∠APB ≅ ∠AQC .....[From (i) and (ii)] ∠PAB ≅ ∠QAC .....[Common angle] ΔAPB ~ ΔAQC .....[AA test of similarity]

Question 25

In fig., line BC || line DE, AB = 2, BD = 3, AC = 4 and CE = x, then find the value of x

Accepted Answer

In ΔADE, line BC || seg DE ......[Given] ∴ by Basic proportionality theorem ∴ `"AB"/"BD" = "AC"/"CE"` ∴ `2/3 = 4/x` ......[Given] ∴ x = `4 xx 3/2` ∴ x = 2 × 3 ∴ x = 6

Question 26

In fig., PM = 10 cm, A(ΔPQS) = 100 sq.cm, A(ΔQRS) = 110 sq.cm, then NR? ΔPQS and ΔQRS having seg QS common base. Areas of two triangles whose base is common are in proportion of their corresponding [______] `("A"("PQS"))/("A"("QRS")) = (["______"])/"NR"`, `100/110 = (["______"])/"NR"`, NR = [ ______ ] cm

Accepted Answer

ΔPQS and ΔQRS having seg QS common base.

Areas of two triangles whose base is common are in proportion of their corresponding heights.

`("A"("ΔPQS"))/("A"("ΔQRS"))` = `"PM"/"NR"`,

`100/110` = `10/"NR"`,

∴ NR = `(110 xx 10)/100`

∴ NR = 11 cm.

Question 27

In fig., PS = 2, SQ = 6, QR = 5, PT = x and TR = y. Then find the pair of value of x and y such that ST || side QR.

Accepted Answer

In ΔPQR,

Line ST || side QR ...[Given]

∴ `"PS"/"SQ" = "PT"/"TR"` ...[Basic proportionality theorem]

∴ `2/6 = "PT"/"TR"` ...[Given]

∴ `1/3 = x/y`

∴ y = 3x

When x = 1, y = 3(1) = 3

∴ x = 1, y = 3

When x = 2, y = 3(2) = 6

∴ x = 2, y = 6

When x = 3, y = 3(3) = 9

∴ x = 3, y = 9

∴ Some of the pairs of values of x and y are (1, 3), (2, 6), (3, 9).

Question 28

In fig., seg AC and seg BD intersect each other at point P. `"AP"/"PC" = "BP"/"PD"` then prove that ΔABP ~ ΔCDP

Accepted Answer

In ΔABP and ΔCDP, `"AP"/"PC" = "BP"/"PD"` ......[Given] ∠APB ≅ ∠CPD .......[Vertically opposite angles] ∴ ΔABP ~ ΔCDP .......[SAS test of similarity]

Question 29

In fig, seg DE || sec BC, identify the correct statement.

Accepted Answer

`"AD"/"DB" = "AE"/"EC"` Explanation: In ΔABC, seg DE || sec BC ...(Given) ∴ By Basic proportionality theorem, `"AD"/"DB" = "AE"/"EC"`.

Question 30

In fig., TP = 10 cm, PS = 6 cm. `"A(ΔRTP)"/"A(ΔRPS)"` = ?

Accepted Answer

Diagram: Refer textbook

Question 31

In given fig., quadrilateral PQRS, side PQ || side SR, AR = 5 AP, then prove that, SR = 5PQ

Accepted Answer

side PQ || side SR and seg SQ is their transversal. ...[Given]

∴ ∠QSR ≅ ∠SQP ...[Alternative angles]

∴ ∠ASR ≅ ∠AQP ......(i) [Q−A−S]

In ΔASR and ΔAQP,

∠ASR ≅ ∠AQP ......[From (i)]

∠SAR ≅ ∠QAP ......[Vertically opposite angles]

∴ ΔASR ∼ ΔAQP ......[AA test of similarity]

∴ `"AR"/"AP" = "SR"/"PQ"` ......(ii)[Corresponding sides of similar triangles]

But, AR = 5 AP ......[Given]

∴ `"AR"/"AP" = 5/1` ......(iii)

∴ `"SR"/"PQ" = 5/1` ......[From (ii) and (iii)]

∴ SR = 5PQ

Question 32

In Quadrilateral ABCD, side AD || BC, diagonal AC and BD intersect in point P, then prove that `"AP"/"PD" = "PC"/"BP"`

Accepted Answer

Proof: seg AD || seg BC and BD is their transversal. ...[Given]

∴ ∠DBC ≅ ∠BDA ...[Alternate angles]

∴ ∠PBC ≅ ∠PDA ...(i)[D−P−B]

In ΔPBC and ΔPDA,

∠PBC ≅ ∠PDA ...[From (i)]

∠BPC ≅ ∠DPA ...[Vertically opposite angles]

∴ ΔPBC ∼ ΔPDA ...[AA test of similarity]

∴ `"BP"/"PD" = "PC"/"AP"` ...[Corresponding sides of similar triangles]

∴ `"AP"/"PD" = "PC"/"BP"` ...[By alternendo]

Question 33

In triangle ABC point D is on side BC (B−D−C) such that ∠BAC = ∠ADC then prove that CA 2 = CB × CD

Accepted Answer

In ΔBAC and ΔADC,

∠BAC ≅ ∠ADC ......[Given]

∠BCA ≅ ∠ACD ......[Common angle]

∴ ΔBAC ∼ ΔADC .....[AA test of similarity]

∴ `"CA"/"CD" = "CB"/"CA"` ......[Corresponding sides of similar triangles]

∴ CA × CA = CB × CD

∴ CA2 = CB × CD

Question 34

Observe the figure and complete the following activity In fig, ∠B = 75°, ∠D = 75° ∠B ≅ [ ______ ] ...[each of 75°] ∠C ≅ ∠C ...[ ______ ] ΔABC ~ Δ [ ______ ] ...[ ______ similarity test]

Accepted Answer

In fig, ∠B = 75°, ∠D = 75° ∠B ≅ ∠D ...[each of 75°] ∠C ≅ ∠C ...[Common angle] ΔABC ~ ΔEDC ...[AA similarity test]]

Question 35

ΔPQR ~ ΔSUV. Write pairs of congruent angles

Accepted Answer

Answer ΔPQR ~ ΔSUV ......[Given] ∴ ∠P ≅ ∠S, ∠Q ≅ ∠U, ∠R ≅ ∠V ......[Corresponding angles of similar triangles]

Question 36

Ratio of areas of two similar triangles is 9 : 25. _______ is the ratio of their corresponding sides.

Accepted Answer

3: 5

Let ΔABC and ΔPQR be two similar triangles.

According to the given condition,

`"A(ΔABC)"/"A(ΔPQR)" = 9/25` ...(Given)

But `"A(ΔABC)"/"A(ΔPQR)" = "AB"^2/"PQ"^2` ...(By the theorem of areas of similar triangles)

∴ `"AB"^2/"PQ"^2 = 9/25`

∴ `"AB"/"PQ" = 3/5`

∴ 3: 5 is the ratio of their corresponding sides.

Question 37

Ratio of corresponding sides of two similar triangles is 4:7, then find the ratio of their areas = ?

Accepted Answer

Let the corresponding sides of similar triangles be s1 and s2.

Let A1 and A2 be their corresponding areas.

s1 : s2 = 4 : 7 ......[Given]

∴ `"s"_1/"s"_2= 4/7` ......(i)

by theorem of areas of similar triangles,

`"A"_1/"A"_2 = "s"_1^2/"s"_2^2`

`"A"_1/"A"_2 = ("s"_1/"s"_2)^2`

`"A"_1/"A"_2 = (4/7)^2` ......[From (i)]

`"A"_1/"A"_2 = 16/49`

∴ Ratio of areas of similar triangles = 16 : 49

Question 38

Side of equilateral triangle PQR is 8 cm then find the area of triangle whose side is half of the side of triangle PQR.

Accepted Answer

Diagram: Refer textbook

Question 39

State whether the following triangles are similar or not: If yes, then write the test of similarity. ∠P = 35°, ∠X = 35° and ∠Q = 60°, ∠Y = 60°

Accepted Answer

Answer In ΔPQR and ΔXYZ, ∠P = 35°, ∠X = 35°, ∠Q = 60° and ∠Y = 60° ......[Given] ∴ ∠P ≅ ∠X and ∠Q ≅ ∠Y ∴ ΔPQR ~ ΔXYZ ...[AA test of similarity] ∴ The triangles in the figure are similar by AA test of similarity.

Question 40

There are two poles having heights 8 m and 4 m on plane ground as shown in fig. Because of sunlight shadows of smaller pole is 6m long, then find the length of shadow of longer pole.

Accepted Answer

Here, AC and PR represent the bigger and smaller poles, and BC and QR represent their shadows respectively.

Now,

ΔACB ∼ ΔPRQ .....[Vertical poles and their shadows form similar figures]

∴ `"CB"/"RQ" = "AC"/"PR"` ......[Corresponding sides of similar triangles]

∴ `x/6 = 8/4`

∴ x = `(8 xx 6)/4`

∴ x = 12 m

∴ The shadow of the bigger pole will be 12 metres long at that time.

Geometry (Math 2) Chapter 1 Similarity Solutions

Q2. ΔABC ~ ΔDEF. Write the ratios of their corresponding sides

Answer

Q3. ΔABC ~ ΔPQR, A(ΔABC) = 80 sq.cm, A(ΔPQR) = 125 sq.cm, then complete `("A"(Δ"ABC"))/("A"(Δ"PQR")) = 80/125 = (["______"])/(["______"])`, hence `"AB"/"PQ" = (["______"])/(["______"])`

Q4. ΔABP ~ ΔDEF and A(ΔABP) : A(ΔDEF) = 144:81, then AB : DE = ?

Q5. An architecture have model of building. Length of building is 1 m then length of model is 0.75 cm. Then find length and height of model building whose actual length is 22.5 m and height is 10 m

Q6. Are triangles in figure similar? If yes, then write the test of similarity.

Answer

Q7. Areas of two similar triangles are 225 cm 2 and 81 cm 2 . If side of smaller triangle is 12 cm, find corresponding side of major triangle.

Q8. Areas of two similar triangles are equal then prove that triangles are congruent

Q9. Areas of two similar triangles are in the ratio 144: 49. Find the ratio of their corresponding sides.

Q10. From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?

Q12. From given information, is PQ || BC? AP = 2, PB = 4, AQ = 3, QC = 6

Q13. Given ΔABC ~ ΔDEF, if ∠A = 45° and ∠E = 35° then ∠B = ?

Q14. If ΔABC ~ ΔLMN and ∠A = 60° then ∠L = ?

Q15. If ΔABC ~ ΔLMN and ∠B = 40°, then ∠M = ? Give reason.

Answer

Q16. If ∆ABC ~ ∆PQR and AB : PQ = 3 : 4 then A(∆ABC) : A(∆PQR) = ?

Q17. If ∆XYZ ~ ∆PQR and A(∆XYZ) = 25 cm 2 , A(∆PQR) = 4 cm 2 then XY : PQ = ?

Q18. If ΔXYZ ~ ΔPQR then `"XY"/"PQ" = "YZ"/"QR"` = ?

Q19. In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio. (i) `"A(ΔABD)"/"A(ΔADC)"` (ii) `"A(ΔABD)"/"A(ΔABC)"` (iii) `"A(ΔADC)"/"A(ΔABC)"`

Q21. In ΔDEF and ΔXYZ, `"DE"/"XY" = "FE"/"YZ"` and ∠E ≅ ∠Y. _______ test gives similarity between ΔDEF and ΔXYZ.

Q22. In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?

Q23. In fig. BD = 8, BC = 12, B-D-C, then `"A(ΔABC)"/"A(ΔABD)"` = ?

Q24. In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar. In ΔAPB and ΔAQC ∠APB = [ ]° ......(i) ∠AQC = [ ]° ......(ii) ∠APB ≅ ∠AQC .....[From (i) and (ii)] ∠PAB ≅ ∠QAC .....[______] ΔAPB ~ ΔAQC .....[______]

Q25. In fig., line BC || line DE, AB = 2, BD = 3, AC = 4 and CE = x, then find the value of x

Q27. In fig., PS = 2, SQ = 6, QR = 5, PT = x and TR = y. Then find the pair of value of x and y such that ST || side QR.

Q28. In fig., seg AC and seg BD intersect each other at point P. `"AP"/"PC" = "BP"/"PD"` then prove that ΔABP ~ ΔCDP

Q29. In fig, seg DE || sec BC, identify the correct statement.

Q30. In fig., TP = 10 cm, PS = 6 cm. `"A(ΔRTP)"/"A(ΔRPS)"` = ?

Q31. In given fig., quadrilateral PQRS, side PQ || side SR, AR = 5 AP, then prove that, SR = 5PQ

Q32. In Quadrilateral ABCD, side AD || BC, diagonal AC and BD intersect in point P, then prove that `"AP"/"PD" = "PC"/"BP"`

Q33. In triangle ABC point D is on side BC (B−D−C) such that ∠BAC = ∠ADC then prove that CA 2 = CB × CD

Q34. Observe the figure and complete the following activity In fig, ∠B = 75°, ∠D = 75° ∠B ≅ [ ______ ] ...[each of 75°] ∠C ≅ ∠C ...[ ______ ] ΔABC ~ Δ [ ______ ] ...[ ______ similarity test]

Q35. ΔPQR ~ ΔSUV. Write pairs of congruent angles

Answer

Q36. Ratio of areas of two similar triangles is 9 : 25. _______ is the ratio of their corresponding sides.

Q37. Ratio of corresponding sides of two similar triangles is 4:7, then find the ratio of their areas = ?

Q38. Side of equilateral triangle PQR is 8 cm then find the area of triangle whose side is half of the side of triangle PQR.

Q39. State whether the following triangles are similar or not: If yes, then write the test of similarity. ∠P = 35°, ∠X = 35° and ∠Q = 60°, ∠Y = 60°

Answer

Q40. There are two poles having heights 8 m and 4 m on plane ground as shown in fig. Because of sunlight shadows of smaller pole is 6m long, then find the length of shadow of longer pole.

Q41. Two triangles are similar. Smaller triangle’s sides are 4 cm, 5 cm, 6 cm. Perimeter of larger triangle is 90 cm then find the sides of larger triangle.

Q42. Which of the following is not a test of similarity?

Q43. Write the test of similarity for triangles given in figure.

Answer

Other Subjects

Q3. ΔABC ~ ΔPQR, A(ΔABC) = 80 sq.cm, A(ΔPQR) = 125 sq.cm, then complete `("A"(Δ"ABC"))/("A"(Δ"PQR")) = 80/125 = ([""])/([""])`, hence `"AB"/"PQ" = ([""])/([""])`

Q24. In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar. In ΔAPB and ΔAQC ∠APB = [ ]° ......(i) ∠AQC = [ ]° ......(ii) ∠APB ≅ ∠AQC .....[From (i) and (ii)] ∠PAB ≅ ∠QAC .....[] ΔAPB ~ ΔAQC .....[]

Q34. Observe the figure and complete the following activity In fig, ∠B = 75°, ∠D = 75° ∠B ≅ [ ] ...[each of 75°] ∠C ≅ ∠C ...[ ] ΔABC ~ Δ [ ] ...[ similarity test]