In the given figure, ABC is a right angled triangle with ∠BAC = 90°. (i) Prove that : △ADB ~ △CDA. (ii) If BD = 18 cm and CD = 8 cm, find AD. (iii) Find the ratio of the area of △ADB is to area of △CDA.

(i) Let ∠CAD = x. So, ∠DAB = 90° - x. In △ABD, ⇒ ∠DAB + ∠ADB + ∠ABD = 180° [By angle sum property] ⇒ 90° - x + 90° + ∠ABD = 180° ⇒ 180° - x + ∠ABD = 180° ⇒ ∠ABD = x + 180° - 180° ⇒ ∠ABD = x. In △ADB and △CDA, ⇒ ∠CAD = ∠ABD (Both = x) ⇒ ∠CDA = ∠ADB (Both = 90°) ∴ △ADB ~ △CDA [By AA] Hence, proved that △ADB ~ △CDA. (ii) Since, △ADB ~ △CDA and corresponding sides of similar triangles are proportional to each other. ⇒ AD 2 = BD × CD ⇒ AD 2 = 18 × 8 ⇒ AD 2 = 144 ⇒ AD = = 12 cm. Hence, AD = 12 cm. (iii) We know that, Ratio of areas of two similar triangles is same as the square of the ratio between their corresponding sides. Hence, ratio of the area of △ADB to area of △CDA = 9 : 4.

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
(i) △ADE ~ △ACB
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of △ADE : area of quadrilateral BCED.

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In △ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find :
(i) area △APO : area △ABC
(ii) area △APO : area △CQO.

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In the given figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that △ABC ~ △EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of △BED = 9 cm². Calculate the :
(i) length of AB
(ii) area of △ABC

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Mathematics

In the given figure, ABC is a right angled triangle with ∠BAC = 90°.
(i) Prove that : △ADB ~ △CDA.
(ii) If BD = 18 cm and CD = 8 cm, find AD.
(iii) Find the ratio of the area of △ADB is to area of △CDA.

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Related Questions

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
(i) △ADE ~ △ACB
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of △ADE : area of quadrilateral BCED.

In △ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find :
(i) area △APO : area △ABC
(ii) area △APO : area △CQO.

In the given figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that △ABC ~ △EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of △BED = 9 cm². Calculate the :
(i) length of AB
(ii) area of △ABC

Mathematics

In the given figure, ABC is a right angled triangle with ∠BAC = 90°.(i) Prove that : △ADB ~ △CDA.(ii) If BD = 18 cm and CD = 8 cm, find AD.(iii) Find the ratio of the area of △ADB is to area of △CDA.

Similarity

Related Questions

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :(i) △ADE ~ △ACB(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.(iii) Find, area of △ADE : area of quadrilateral BCED.

In △ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find :(i) area △APO : area △ABC(ii) area △APO : area △CQO.

In the given figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that △ABC ~ △EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of △BED = 9 cm2. Calculate the :(i) length of AB(ii) area of △ABC

In the given figure, ABC is a right angled triangle with ∠BAC = 90°.
(i) Prove that : △ADB ~ △CDA.
(ii) If BD = 18 cm and CD = 8 cm, find AD.
(iii) Find the ratio of the area of △ADB is to area of △CDA.

ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
(i) △ADE ~ △ACB
(ii) If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
(iii) Find, area of △ADE : area of quadrilateral BCED.

In △ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find :
(i) area △APO : area △ABC
(ii) area △APO : area △CQO.

In the given figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that △ABC ~ △EBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of △BED = 9 cm². Calculate the :
(i) length of AB
(ii) area of △ABC