In the figure given below, ABCD is a square of side 7 cm. If
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.

Question

In the figure given below, ABCD is a square of side 7 cm. If AE = FC = CG = HA = 3 cm, (i) prove that EFGH is a rectangle. (ii) find the area and perimeter of EFGH.

KnowledgeBoat · Accepted Answer

(i) Given, ABCD is a square of side 7 cm and AE = FC = CG = HA = 3 cm. From figure, DH = DA - HA = 7 - 3 = 4 cm, GD = DC - CG = 7 - 3 = 4 cm, EB = AB - AE = 7 - 3 = 4 cm, FB = BC - FC = 7 - 3 = 4 cm. Since in square, sides are perpendicular to each other, By pythagoras theorem, In right angle triangle HAE, ⇒ HE² = HA² + AE² ⇒ HE² = 3² + 3² ⇒ HE² = 9 + 9 ⇒ HE² = 18 ⇒ HE = cm. In right angle triangle FCG, ⇒ GF² = GC² + FC² ⇒ GF² = 3² + 3² ⇒ GF² = 9 + 9 ⇒ GF² = 18 ⇒ GF = cm. In right angle triangle GDH, ⇒ GH² = GD² + DH² ⇒ GH² = 4² + 4² ⇒ GH² = 16 + 16 ⇒ GH² = 32 ⇒ GH = cm. In right angle triangle EBF, ⇒ EF² = EB² + FB² ⇒ EF² = 4² + 4² ⇒ EF² = 16 + 16 ⇒ EF² = 32 ⇒ EF = cm. In isosceles triangle AEH, ∠E = ∠H = x (let) (As angles opposite to equal side in isosceles triangle are equal). So, ∠A + ∠E + ∠H = 180° 90° + x + x = 180° 2x = 90° x = 45° Similar is the case for triangles EBF, GDH, FCG. From figure, ∠AEH + ∠HEF + ∠FEB = 180° (Linear pairs) 45° + ∠HEF + 45° = 180° ∠HEF = 90° Since, angles of EFGH = 90° and EH = GF and HG = EF. Hence, proved that EFGH is a rectangle. (ii) Area of EFGH = EF × GF = = 24 cm². Perimeter of EFGH = 2(EF + GF) = cm. Hence, area = 24 cm² and perimeter = cm.

Mathematics

In the figure given below, ABCD is a square of side 7 cm. If
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.

Pythagoras Theorem

Related Questions

In figure given below, AB || DC, BC = AD = 13 cm, AB = 22 cm and DC = 12 cm. Calculate the height of the trapezium ABCD.

In figure given below, AB || DC, ∠A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²

Mathematics

In the figure given below, ABCD is a square of side 7 cm. IfAE = FC = CG = HA = 3 cm,(i) prove that EFGH is a rectangle.(ii) find the area and perimeter of EFGH.

Pythagoras Theorem

Related Questions

In figure given below, AB || DC, BC = AD = 13 cm, AB = 22 cm and DC = 12 cm. Calculate the height of the trapezium ABCD.

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In the figure given below, ABCD is a square of side 7 cm. If
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²