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महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

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

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प्रश्न

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

 

 

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उत्तर

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.

 

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पाठ 1: Similarity - Practice Set 1.3 [पृष्ठ २२]

संबंधित प्रश्‍न

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 the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`


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"))`

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

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


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


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. 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 `square`.

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

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

NR = `square` cm


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