English
Maharashtra State BoardSSC (English Medium) 10th Standard

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

Question

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))`

Sum
Advertisements

Solution


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
  Is there an error in this question or solution?
Chapter 1: Similarity - Exercise

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 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 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. 


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 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)`


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., 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


In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×