SCERT Maharashtra solutions for जियॉमेट्री (माठेमटिक्स २) [इंग्रजी] इयत्ता ९ महाराष्ट्र राज्य मंडळ chapter 1 - Similarity [Latest edition]

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

Which of the following is not a test of similarity?

If ∆XYZ ~ ∆PQR and A(∆XYZ) = 25 cm², A(∆PQR) = 4 cm² then XY : PQ = ?

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

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

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

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

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

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

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

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

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

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°

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

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

ΔPQR ~ ΔSUV. Write pairs of congruent angles.

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

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

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

Write the test of similarity for triangles given in figure.

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 = `square`°   ...(i) 

∠AQC = `square`°   ...(ii)

∠APB ≅ ∠AQC   ...[From (i) and (ii)]

∠PAB ≅ ∠QAC   ...`square`

ΔAPB ~ ΔAQC   ...`square`

Observe the figure and complete the following activity.

In fig, ∠B = 75°, ∠D = 75°

∠B ≅ `square`   ...[each of 75°]

∠C ≅ ∠C   ...`square`

ΔABC ~ Δ `square`   ...[`square` similarity test]

ΔABC ~ ΔPQR, A(ΔABC) = 80 sq.cm, A(ΔPQR) = 125 sq.cm, then complete `(A(ΔABC))/(A(ΔPQR)) = 80/125 = (square)/(square)`, hence `(AB)/(PQ) = (square)/(square)`

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 `square`.

`(A(ΔPQS))/(A(ΔQRS)) = (square)/(NR)`,

`100/110 = (square)/(NR)`,

NR = `square` cm

In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `(A(ΔABC))/(A(ΔBCD))` = ?

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

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

From given information, is PQ || BC?

AP = 2, PB = 4, AQ = 3, QC = 6

Areas of two similar triangles are 225 cm² and 81 cm². If side of smaller triangle is 12 cm, find corresponding side of major triangle. 

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

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 ≅ ∠`square`   ...[each 90°]

∠ACP ≅ ∠`square`   ...[common angle]

ΔCPA ~ ΔCQB   ...[`square` similarity test]

`"AP"/"BQ" = (square)/(BC)`   ...[corresponding sides of similar triangles]

`7/8 = (square)/12`

AC × `square` = 7 × 12

AC = 10.5

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(ΔAPQ))/(A(ΔPQB")) = (square)/(PB)`   ...(i)[areas in proportion of base]

`(A(ΔAPQ))/(A(ΔPQC)) = (square)/(QC)`   ...(ii)[areas in proportion of base]

ΔPQC and ΔPQB have `square` is common base.

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

A(ΔPQC) = A(Δ`square`)    ......(iii)

`(A(ΔAPQ))/(A(ΔPQB)) = (A(Δsquare))/(A(Δsquare))`   ...[(i), (ii), and (iii)]

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

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)/(square)`   ...`[square]`

`(x + 3)/(x - 3) = (x + 5)/(square)`

(x + 3) `square` = (x + 5)(x – 3)

x² + x – `square` = x² + 2x – 15

x = `square`

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 6 m long, then find the length of shadow of longer pole.

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))`

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

In triangle ABC point D is on side BC (B-D-C) such that ∠BAC = ∠ADC then prove that CA² = CB × CD

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

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

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

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.

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.

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.