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

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

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प्रश्न

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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उत्तर

ΔPQS and ΔQRS having seg QS common base.

Areas of two triangles whose base is common are in proportion of their corresponding \[\boxed{\text{heights}}\].

`(A(ΔPQS))/(A(ΔQRS))` = \[\frac{{\boxed{PM}}}{{{NR}}}\],

`100/110` = \[\frac{{\boxed{10}}}{{{NR}}}\],

∴ NR = `(110 xx 10)/100`

∴ NR = \[\boxed{11}\] cm

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पाठ 1: Similarity - Exercise

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))`


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 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 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. 


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

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


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)`


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))` = ?


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))`


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`


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