English
Maharashtra State BoardSSC (English Medium) 10th Standard

In Trapezium Pqrs, Side Pq || Side Sr, Ar = 5ap, as = 5aq Then Prove That, Sr = 5pq - Geometry Mathematics 2

Advertisements
Advertisements

Question

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

 

 

Advertisements

Solution

Given: 
side PQ || side SR
AR = 5AP,
AS = 5AQ
To prove: SR = 5PQ
Proof: In ∆APQ and ∆ARS
∠PAQ = ∠RAS          (Vertically Opposite angles)
∠PQA = ∠RSA          (Alternate angles, side PQ || side SR and QS is a transversal line)
By AA test of similarity
∆APQ ~ ∆ARS  

\[\frac{PQ}{SR} = \frac{AP}{AR} \left( \text{ Corresponding sides are proportional } \right)\]
\[ \Rightarrow \frac{PQ}{SR} = \frac{1}{5} \left( AR = 5AP \right)\]
\[ \Rightarrow SR = 5PQ\] 
Hence proved.

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 1: Similarity - Practice Set 1.3 [Page 22]

RELATED QUESTIONS

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 following figure RP : PK= 3 : 2, then find the value of A(ΔTRP) : A(ΔTPK).


The ratio of the areas of two triangles with common base is 6:5. Height of the larger triangle of 9 cm, then find the corresponding height of the smaller triangle.


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 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"))`

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


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

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


In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.

if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.


A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`


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


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


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


If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.


In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.


Given: PQ ⊥ BC, AD ⊥ BC

Now, A(ΔPQB)  = `1/2 xx square xx square`

A(ΔPBC)  = `1/2 xx square xx square`

Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×