In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that
(i) ∆BCF ≅ ∆ABE.
(ii) area of square ABFG = area of rectangle BENM.

Question

In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that
(i) ∆BCF ≅ ∆ABE.
(ii) area of square ABFG = area of rectangle BENM.

KnowledgeBoat · Accepted Answer

(i) From figure, ⇒ ∠FBC = ∠FBA + ∠ABC ∠FBA = 90° (As each angle of a square = 90°) ⇒ ∠FBC = 90° + ∠ABC ....... (1) Also, ⇒ ∠ABE = ∠CBE + ∠ABC ∠CBE = 90° (As each angle of a square = 90°) ⇒ ∠ABE = 90° + ∠ABC .......(2) From (1) and (2), we get ∠FBC = ∠ABE ........ (3) Now, in ∆BCF and ∆ABE BF = AB (As FBAG is a square) ∠FBC = ∠ABE (Proved above) BC = BE (As BCDE is a square) By using the SAS axiom rule of congruency, ∴ ∆BCF ≅ ∆ ABE. Hence, proved that ∆BCF ≅ ∆ ABE. (ii) We know that, ∆BCF ≅ ∆ABE So, area of ∆BCF = area of ∆ABE ...... (4) Since, ∆BCF and square AGFB have same base FB and are between same parallel lines FB and GC. area of ∆BCF = area of square AGFB ........(5) ∆ABE and rectangle BENM are on same base BE and are between same parallel lines BE and AN. area of ∆ABE = area of rectangle BENM ....... (6) From (4), (5) and (6) ⇒ area of square AGFB = area of rectangle BENM ⇒ area of square AGFB = area of rectangle BENM Hence, proved that area of square AGFB = area of rectangle BENM.

Mathematics

In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that
(i) ∆BCF ≅ ∆ABE.
(ii) area of square ABFG = area of rectangle BENM.

Theorems on Area

Related Questions

In the figure (1) given below, BC || AE and CD || BE. Prove that
area of ∆ABC= area of ∆EBD.

In the adjoining figure, E is the midpoint of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of ∆ ABC is 25 sq. units, find the area of || gm EBCF.

In the figure (2) given below, the perimeter of ∆ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, calculate the lengths of the sides of ∆ABC.

In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP : PC.

Mathematics

In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that(i) ∆BCF ≅ ∆ABE.(ii) area of square ABFG = area of rectangle BENM.

Theorems on Area

Related Questions

In the figure (1) given below, BC || AE and CD || BE. Prove thatarea of ∆ABC= area of ∆EBD.

In the adjoining figure, E is the midpoint of the side AB of a triangle ABC and EBCF is a parallelogram. If the area of ∆ ABC is 25 sq. units, find the area of || gm EBCF.

In the figure (2) given below, the perimeter of ∆ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, calculate the lengths of the sides of ∆ABC.

In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm2 and the area of ∆BCP = 15 cm2. Find(i) area of || gm ABCD(ii) DP : PC.

In figure (2) given below, ABC is a right-angled triangle at A. AGFB is a square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED, prove that
(i) ∆BCF ≅ ∆ABE.
(ii) area of square ABFG = area of rectangle BENM.

In the figure (1) given below, BC || AE and CD || BE. Prove that
area of ∆ABC= area of ∆EBD.

In the figure(3) given below, ABCD is a parallelogram. P is a point on DC such that area of ∆DAP = 25 cm² and the area of ∆BCP = 15 cm². Find
(i) area of || gm ABCD
(ii) DP : PC.