Balbharati solutions for Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board chapter 1 - Similarity [Latest edition]

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles. 

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find \[\frac{A\left( ∆ ABC \right)}{A\left( ∆ ADB \right)} .\] 

In adjoining figure, seg PS ⊥ seg RQ seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then Find QT. 

In the following figure, AP ⊥ BC, AD || BC, then Find A(∆ABC): A(∆BCD). 

In adjoining figure PQ ⊥ BC, AD⊥ BC then find following ratios. 

(i) \[\frac{A\left( ∆ PQB \right)}{A\left( ∆ PBC \right)}\] 

(ii) \[\frac{A\left( ∆ PBC \right)}{A\left( ∆ ABC \right)}\] 

(iii) \[\frac{A\left( ∆ ABC \right)}{A\left( ∆ ADC \right)}\] 

(iv) \[\frac{A\left( ∆ ADC \right)}{A\left( ∆ PQC \right)}\] 

Identify ray PM is the bisector of ∠QPR.

Identify ray PM is the bisector of ∠QPR. 

Identify ray PM is the bisector of ∠QPR.

In ∆PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason. 

In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then Find QP.  

Measures of some angles in the figure are given. Prove that \[\frac{AP}{PB} = \frac{AQ}{QC}\] 

In trapezium ABCD, side AB || side PQ || side ∆C, AP = 15, PD = 12, QC = 14, Find BQ. 

Find QP using given information in the figure.

In the given figure, if AB || CD || FE then Find x and AE.   

In ∆LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT. 

In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x. 

In the given figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF. 

Proof :  In Δ XDE, PQ || DE         ........ ___________

`therefore "XP"/([    ]) = ([    ])/"QE"`  ..... (I) (Basic proportionality theorem)

In Δ XEE, QR || EF                    ........  _________

`therefore ([     ])/([     ]) = ([      ])/([      ])`    .......(II) _________________

`therefore ([     ])/([     ]) = ([      ])/([      ])`   ....... from (I) and  (II)

∴ seg PR || seg DE           ........... (converse of basic proportionality theorem)

In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC. 

In the given figure, ∠ABC = 75°, ∠EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.

Are the triangles in the given figure similar? If yes, by which test ? 

As shown in the given figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time ? 

In ∆ABC, AP ⊥ BC, BQ ⊥ AC B– P–C, A–Q – C then prove that, ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12 then Find AC. 

 In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ 

In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD. 

◻ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE. 

In the given figure, seg AC and seg BD intersect each other in point P and \[\frac{AP}{CP} = \frac{BP}{DP}\] Prove that, ∆ABP ~ ∆CDP

In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA² = CB × CD 

The ratio of corresponding sides of similar triangles is 3 : 5; then Find the ratio of their areas.

If ∆ABC ~ ∆PQR and AB : PQ = 2 : 3, then fill in the blanks. 

\[\frac{A\left( ∆ ABC \right)}{A\left( ∆ PQR \right)} = \frac{{AB}^2}{......} = \frac{2^2}{3^2} = \frac{......}{.......}\]

If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks. \[\frac{A\left( ∆ ABC \right)}{A\left( ∆ . . . . \right)} = \frac{80}{125} \therefore \frac{AB}{PQ} = \frac{......}{......}\] 

∆LMN ~ ∆PQR, 9 × A (∆PQR ) = 16 × A (∆LMN). If QR = 20 then Find MN. 

Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm, then Find corresponding side of the bigger triangle. 

∆ABC and ∆DEF are equilateral triangles. If A(∆ABC) : A (∆DEF) = 1 : 2 and AB = 4, find DE.  

In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then Find A(◻DPQE) by completing the following activity. 

Select the appropriate alternative.
In ∆ABC and ∆PQR, in a one to one correspondence \[\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}\] 

 If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false? 

 In ∆ABC and ∆DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true ?

∆ABC and ∆DEF are equilateral triangles, A (∆ABC) : A (∆DEF) = 1 : 2
If AB = 4 then what is length of DE? 

In the given figure, seg XY || seg BC, then which of the following statements is true?

In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio. 

 \[\frac{A\left( ∆ ABD \right)}{A\left( ∆ ADC \right)}\]  

In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio.

\[\frac{A\left( ∆ ABD \right)}{A\left( ∆ ABC \right)}\]  

In ∆ABC, B – D – C and BD = 7, BC = 20 then Find following ratio. 

\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\] 

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then \[\frac{A \left( ∆ ABC \right)}{A \left( ∆ DCB \right)} = ?\] 

In the given figure, PM = 10 cm A(∆PQS) = 100 sq.cm A(∆QRS) = 110 sq.cm then Find NR. 

∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio \[\frac{A\left( ∆ MNT \right)}{A\left( ∆ QRS \right)}\] 

In the given figure, A – D– C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then Find BE.

In the given figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280 then Find PQ, QR and RS. 

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 

In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find \[\frac{AX}{XY}\] 

In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that \[\frac{AP}{PD} = \frac{PC}{BP}\] 

In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.  

In the given figure, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE² = BD × EC (Hint: Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.)