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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)"`
उत्तर
Draw AE ⊥ BC, B – E – C.
BC = BD + DC ......[B – D – C]
∴ 20 = 7 + DC
∴ DC = 20 − 7 = 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`
संबंधित प्रश्न
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In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.
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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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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:
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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`
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Therefore,
`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`
= `square/square`
