English
Maharashtra State BoardSSC (English Medium) 10th Standard

In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios. (i) APQBAPBCA(∆PQB)A(∆PBC) (ii) APBCAABCA(∆PBC)A(∆ABC) (iii) AABCAADCA(∆ABC)A(∆ADC) (iv) AADCAPQCA(∆ADC)A(∆PQC) - Geometry Mathematics 2

Advertisements
Advertisements

Question

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

  1. `("A"(∆"PQB"))/("A"(∆"PBC"))`
  2. `("A"(∆"PBC"))/("A"(∆"ABC"))`
  3. `("A"(∆"ABC"))/("A"(∆"ADC"))`
  4. `("A"(∆"ADC"))/("A"(∆"PQC"))`
Sum
Advertisements

Solution

(i) In ∆PQB and ∆PBC,

ΔPQB and ΔPBC have same height PQ.   ...(Given)

Areas of triangles with equal heights are proportional to their corresponding bases.

∴ `("A"(∆"PQB"))/("A"(∆"PBC")) = "BQ"/"BC"`

(ii) In ∆PBC and ∆ABC,

∆PBC and ∆ABC have same base BC.   ...(Given)

Areas of triangles equal bases are proportional to their corresponding heights.

∴ `("A"(∆"PBC"))/("A"(∆"ABC")) = "PQ"/"AD"`

(iii) In ∆ABC and ∆ADC,

∆ABC and ∆ADC have same height AD.   ...(Given)

Areas of triangles with equal heights are proportional to their corresponding bases.

∴ `("A"(∆"ABC"))/("A"(∆"ADC")) = "BC"/"DC"`

(iv) In ∆ADC and ∆PQC,

Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

∴ `("A"(∆"ADC"))/("A"(∆"PQC")) = "DC × AD"/"QC × PQ"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 1: Similarity - Practice Set 1.1 [Page 6]

RELATED QUESTIONS

In the following figure seg AB ⊥ seg BC, seg DC ⊥ seg BC. If AB = 2 and DC = 3, find `(A(triangleABC))/(A(triangleDCB))`


The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.


In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that:

\[\frac{BD}{CD} = \frac{AB}{AC}\]

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

 

 


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

`"A(∆ ABD)"/"A(∆ ADC)"`


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 `(A(Δ ABC))/(A(Δ DCB))` = ?


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

`(A(triangleABD))/(A(triangleABC))`


Areas of two similar triangles are in the ratio 144: 49. Find the ratio 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 = ?


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


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


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


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)"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×