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

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.

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

Given: ΔABC ∼ ΔDEF

Also, AB = 9 cm, DE = 12 cm.

We now know the area theorem of comparable triangles.

`(ar(ΔABC))/(ar(ΔDEF)) = ((AB)/(DE))^2`  ...(i)

Substituting the values in equation (i),

`(ar(ΔABC))/(ar(ΔDEF)) = (9/12)^2`

= `(3/4)^2`

= `9/16`

Hence, the ratio of areas of ΔABC and ΔDEF is 9:16.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2025-2026 (March) Model set 2 by shaalaa.com

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 

 

 


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.


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. 

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


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 `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.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×