हिंदी

In the Given, Seg Be ⊥ Seg Ab and Seg Ba ⊥ Seg Ad. If Be = 6 and Ad = 9 Find a ( δ a B E ) a ( δ B a D ) . - Geometry Mathematics 2

Advertisements
Advertisements

प्रश्न

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))`.

योग

उत्तर

`(A(Δ ABE))/(A(Δ BAD)) =(BE)/(AD)`   

....[Ratio of areas of two triangles having qual    base is  equal to the ration of their corresponding heights.]

∴ `(A(Δ ABE))/(A(Δ BAD)) =6/9`   

∴ `(A(Δ ABE))/(A(Δ BAD)) =2/3` 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2013-2014 (October)

APPEARS IN

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 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.


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 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)"`


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.


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 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 [______]

`("A"("PQS"))/("A"("QRS")) = (["______"])/"NR"`,

`100/110 = (["______"])/"NR"`,

NR = [ ______ ] cm


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×