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

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then A(ΔABC)A(ΔDCB) = ?

Advertisements
Advertisements

प्रश्न

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `(A(Δ ABC))/(A(Δ DCB))` = ?

बेरीज
Advertisements

उत्तर

Given:

∠ABC = ∠DCB = 90°

AB = 6

DC = 8

Now,

`(A(Δ ABC))/(A(Δ DCB)) = (1/2 xx AB xx BC)/(1/2 xx DC xx BC)`

= `6/8`

= `3/4`

∴ `(A(Δ ABC))/(A(Δ DCB)) = 3/4`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Similarity - Problem Set 1 [पृष्ठ २७]

APPEARS IN

बालभारती Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
पाठ 1 Similarity
Problem Set 1 | Q 4 | पृष्ठ २७

संबंधित प्रश्‍न

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.


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:

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

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

`"A(∆ ABD)"/"A(∆ ADC)"`


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


The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.


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

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


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)`


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 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))`


Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:

  1. Draw two triangles, give the names of all points, and show heights.
  2. Write 'Given' and 'To prove' from the figure drawn.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×