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

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))

Advertisements
Advertisements

प्रश्न

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))`

बेरीज
Advertisements

उत्तर


Draw AE ⊥ BC, B-E-C.

BC = BD + DC   ...[B-D-C]

∴ 20 = 7 + DC

∴ DC = 20 – 7

∴ DC = 13

(i) ΔABD and ΔADC have same height AE.

`(A(ΔABD))/(A(ΔADC)) = (BD)/(DC)`   ...[Triangles having equal height]

∴ `(A(ΔABD))/(A(ΔADC)) = 7/13`

(ii) ΔABD and ΔABC have same height AE.

`(A(ΔABD))/(A(ΔABC)) = (BD)/(BC)`   ...[Triangles having equal height]

∴ `(A(ΔABD))/(A(ΔABC)) = 7/20`

(iii) ΔADC and ΔABC have same height AE.

`(A(ΔADC))/(A(ΔABC)) = (DC)/(BC)`   ...[Triangles having equal height]

∴ `(A(ΔADC))/(A(ΔABC)) = 13/20`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Similarity - Complete the following activity

संबंधित प्रश्‍न

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:

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

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


 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 ∆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 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))`


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


In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `(A(ΔABC))/(A(ΔBCD))` = ?


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


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.

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×