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

From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then AABCABCDA(ΔABC)A(ΔBCD) = ? - Geometry Mathematics 2

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प्रश्न

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

बेरीज
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उत्तर

ΔABC and ΔBCD have same base BC.

∴ `("A"(Δ"ABC"))/("A"(Δ"BCD")) = "AB"/"DC"`  ...[Triangles having equal base]

∴ `("A"(Δ"ABC"))/("A"(Δ"BCD")) = 6/8`

∴ `("A"(Δ"ABC"))/("A"(Δ"BCD")) =  3/4`

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पाठ 1: Similarity - Q.2 (B)

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))`


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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 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 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 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))`.


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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.

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

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NR = [ ______ ] cm


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(i) `"A(ΔABD)"/"A(ΔADC)"`

(ii) `"A(ΔABD)"/"A(ΔABC)"`

(iii) `"A(ΔADC)"/"A(ΔABC)"`


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.


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Now, A(ΔPQB)  = `1/2 xx square xx square`

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Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`


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