PROBLEMS ON DIAGONALOF RHOMUS

Problem 1 :

Find the side length of the rhombus ABCD.

Solution :

Let E be the point of intersection of two diagonals. Considering the triangle DEC,

DE = 20 ft, EC = 30 ft

DC² = DE² + EC²

DC² = 20² + 30²

DC² = 400 + 900

DC² = 1300

DC = 10√13

Problem 2 :

The area of a rhombus is 150 cm². The length of one of its diagonal is 10 cm. The length of their other diagonal is .

(a)  25 cm   (b)  30 cm   (c) 35 cm    (d)  40 cm

Solution :

Area of rhombus = (1/2) x d₁ x d₂

(1/2) x 10 x d₂ = 150

d₂ = 150 / 5

d₂ = 30 cm

Problem 3 :

One of the diagonals of a rhombus is double the other diagonal. Its area is 25 sq.cm. The sum of the diagonal is 

(a)  10 cm   (b)  12 cm   (c) 15 cm    (d)  16 cm

Solution :

d₁ = 2d₂

(1/2) x d₁ x d₂ = 25

(1/2) x 2d₂ x d₂ = 25

d₂² = 25

d₂ = 5

then d1 = 2(5) ==> 10 cm

Sum of the diagonals = 10 + 5 ==> 15 cm

Problem 4 :

If the diagonals of a rhombus are 24 cm and 10 cm, find the area and perimeter of rhombus.

Solution :

Area of rhombus = (1/2) x d₁ x d₂

d₁ = 24 cm and d₂ = 10 cm

= (1/2) x 24 x 10

= 120 square cm

To find perimeter of rhombus, we should find the side length of rhombus.

side² = 12² + 5²

side² = 144 + 25

side² = 169

= √169

= 13

Perimeter of rhombus = 4(13)

= 52 cm

Problem 5 :

Each side of a rhombus is 26 cm and one of its diagonal is 48 cm long. The area of the rhombus is.

(a) 240 cm²       (b) 300 cm²  (c) 360 cm²    (d) 480 cm²

Solution :

The diagonals will bisect each other at right angle.

Half length of given diagonal = 24

half length of another diagonal = x

26² = 24² + x²

676 = 576 + x²

x² = 676 - 576

x² = 100

x = 10

Length of another diagonal = 20

Area of rhombus = (1/2) x 48 x 20

= 480 square cm

Problem 6 :

The length one diagonal of a rhombus is 80% of the other diagonal. The area of the rhombus is how many times the square of length of the other diagonal ?

(a) 4/5   (b)  2/5   (c)  3/4     (d)  1/4

Solution :

Let d₁ and d₂ be length of diagonals.

d₁ = 80% of d₂  ==> 0.80d₂

Area of rhombus = (1/2) x 0.80d₂ x d₂

= (1/2) x 0.80(d₂)²

= 0.40(d₂)²

= (40/100)(d₂)²

= 2/5(d₂)²

So, the answer is 2/5.

Problem 7 :

Use square FNRM or rhombus STPK to find each measure

Solution :

In the square FNRM, given that FR = 24 mm

In sqaure, diagonals will bisect each other and they will be equal in length.

a)  

AR = FR/2

= 24/2

AR = 12 mm

b)  ∠FAN

Diagonals will bisect each other and perpendicular. So, ∠FAN = 90 degree.

d)  MA = 12 mm

c)  BS = BP = 6 cm

e)  TP

Since all four sides are equal in rhombus, the measure of TP is 10 cm.

f)  ∠KTP = 37 

Since diagonals KT and SP are angle bisectors. 

Problem 8 :

The diagonals of a square are (x + 8) feet and 3x feet. Find the measure of the digonals.

Solution :

In a square, diagonals will be equal. Then

x + 8 = 3x

8 = 3x - x

2x = 8

x = 8/2

x = 4

3x = 3(4) ==> 12 feet

So, the length of the diagonals is 12 feet.

Problem 9 :

Refer to rhombus PLAN

a) Classify trangle PLA by its sides.

b)  Classify triangle PEN by its angles.

c) Is triangles PEN and triangle AEL congruent. Explain.

Solution :

a)  In general, in rhombus all four sides will be equal. Then, in triangle PLA, PL and LA will be equal.

b)  Diagonals will be equal, they bisect each other and perpendicular.

∠EPN = ∠ENP and ∠PEN = 90 degree.

c)  In triangle PEN and AEL,

• ∠EPN = ∠EAL
• ∠ENP = ∠ELA
• ∠PEN = ∠LEA (90 degree)

Then triangles PEN and AEL are congruent.

Problem 10 :

The area of a kite is 324 square inches. One diagonal is twice as long as the other diagonal. Find the length of each diagonal.

Solution :

Let x be the length of one diagonal. 2x be the length of other diagonal.

Are of kite = 324 square inches

(1/2) · x · 2x = 324

x² = 324

x = √324

x = 18 inch

2x = 2(18) 

= 36 inch

So, the lengths of diagonal are 18 inches and 36 inches.

Problem 11 :

One diagonal of a rhombus is four times the length of the other diagonal. The area of the rhombus is 98 square feet. Find the length of each diagonal.

Solution :

Let x be the length of one diagonal. 4x be the length of other diagonal.

Are of kite = 98 square inches

(1/2) · x · 4x = 98

2x² = 98

x² = 49

x = √49

x = 7 inch

4x = 4(7) 

= 28 inch

So, the lengths of diagonal are 7 inches and 28 inches.