Area Of A Trapezium And A Polygon
Class 8th Mathematics RS Aggarwal Solution
- Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the…
- Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and…
- The shape of the top surface of a table is trapezium. Its parallel sides are 1…
- The area of a trapezium is 1080 cm^2 . If the lengths of its parallel sides be…
- A field is in the form of a trapezium. Its area is 1586 m^2 and the distance…
- The area of a trapezium is 405 cm^2 . Its parallel sides are in the ration 4:5…
- The area of a trapezium is 180 cm^2 and its height is 9 cm. If one of the…
- In a trapezium-shaped field, one of the parallel sides is twice the other. If…
- The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB…
- In the given figure, ABCD is a trapezium in which ad |bc angle abc = 90^circle…
- The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides…
- The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel…
- In the given figure, ABCD is a quadrilateral in which ac = 24 cm, bl…
- In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD…
- Find the area of pentagon ABCDE in which bl perpendicular ac dm perpendicular…
- Find the area of hexagon ABCDEF in which bl perpendicular ad cm perpendicular…
- Find the area of pentagon ABCDE in which bl perpendicular ac cm perpendicular…
- Find the area enclosed by the given figure ABCDEF as per dimensions given…
- Find the area of given figure ABCDEFGH as per dimensions given in it.…
- Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and…
- The parallel sides of a trapezium measure 14 cm and 18 cm and the distance…
- The length of the parallel sides of a trapezium are 19 cm and 13 cm and its…
- The parallel sides of a trapezium are in the ration 3:4 and the perpendicular…
- The area of a trapezium is 180 cm^2 and its height is 9 cm. If one of the…
- In the given figure, AB||DC and DA AB If dc = 7 cm, bc = 10 cm, ab = 13 cm and…
- The base of a triangular field is three times its height and its area is 1350 m^2 .…
- Find the area of an equilateral triangle of side 6 cm.
- The perimeter of a rhombus is 180 cm and one of its diagonals is 72 cm. Find the length…
- The area of a trapezium is 216 m^2 and its height is 12 m. If one of the parallel sides…
- Find the area of a quadrilateral one of whose diagonals is 40 cm and the lengths of the…
- A field is in the form of a right triangle with hypotenuse 50 m and one side 30m. Find…
- The base of a triangle is 14 cm and its height is 8 cm. The area of the triangle isA.…
- The base of a triangle is four times its height and its area is 50 m^2 . The length of…
- The diagonal of a quadrilateral is 20 cm in length and the lengths of perpendiculars on…
- Each side of a rhombus is 15 cm and the length of one of its diagonals is 24 cm. The…
- The area of a rhombus is 120 cm^2 and one of its diagonals is 24 cm. Each side of the…
- The parallel sides of a trapezium are 54 cm and 26 cm and the distance between them is…
- The area of a trapezium is 384 cm^2 . Its parallel sides are in the ratio 5:3 and the…
- Fill in the blanks. (i) Area of triangle (ii) Area of a ||gm = (............)…
Exercise 18a
Question 1.Find the area of a trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.
Answer:
Given:
Length of parallel sides is 24cm and 20 cm
Height (h) = 15 cm
We know that area of trapezium is 
× (sum of parallel sides) × height
Therefore, Area of trapezium = × (24 +20) × 15 =330 cm2.
Question 2.
Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 16 cm.
Answer:
Given
Length of parallel sides is 38.7cm and 22.3 cm
Height (h) = 16 cm
We know that area of trapezium is 
× (sum of parallel sides) × height
Therefore Area of trapezium = × (38.7 +22.3) × 16 = 488 cm2.
Question 3.
The shape of the top surface of a table is trapezium. Its parallel sides are 1 m and 1.4 m and the perpendicular distance between them is 0.9 cm. Find its area.
Answer:
Given
Length of parallel sides is 1m and 1.4m
Height (h) = 0.9m
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (1 +1.4) × 0.9
=1.08 m2.
Question 4.
The area of a trapezium is 1080 cm2. If the lengths of its parallel sides be 55 cm and 35 cm, find the distance between them.
Answer:
Given
Length of parallel sides is 55cm and 35 cm
Area of trapezium= 1080 cm2
Let Height (h) =y cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (55 + 35) × y =1080 cm2.
× (90) × y =1080
⇒ 45 × y =1080
⇒ y =
= 24
... Distance between the parallel lines is 24 cm.
Question 5.
A field is in the form of a trapezium. Its area is 1586 m2 and the distance between its parallel sides is 26 m. If one of the parallel sides is 84 m, find the other.
Answer:
Given
Let length of parallel sides be 84cm and y cm
Area of trapezium= 1586 cm2
Let Height (h) =26 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (84 +y) × 26 =1586 cm2.
× (84 +y) × 26 =1586
⇒(84 + y) × 13 =1586
⇒ 84 + y =
⇒ y =122— 84 = 38
... Length of the other parallel side is 38 cm.
Question 6.
The area of a trapezium is 405 cm2. Its parallel sides are in the ration 4:5 and the distance between them is 18 cm. Find the length of each of the parallel sides.
Answer:
Given
Lengths of the parallel sides are in the ratio 4:5
Therefore let one of the side length be 4X and other side length be 5X
Area of trapezium= 405 cm2
Let Height (h) =18 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is x (4X +5X) × 18 =405 cm2.
× (4X + 5X) × 18 =405
⇒(9X) × 9 =405
⇒ 81X = 405
⇒ X =
= 5
... Length of the parallel sides is 4X=4 × 5 =20 cm and 5X = 5 × 5 = 25 cm.
Therefore lengths of the parallel sides are 20 cm, 25 cm.
Question 7.
The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.
Answer:
Given
Let length of first parallel side X
Length of other parallel side is X + 6
Area of trapezium= 180 cm2
Let Height (h) =9 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (X + 6 +X) × 9 =180 cm2.
× (X + 6 +X) × 9 =180
⇒ × (2X + 6) × 9 =180
⇒ 2X + 6 = 
⇒ 2X + 6 = 40
⇒2X = 40 – 6 =34
⇒X = 17
... Length of the parallel sides is X=17 cm and X + 6 = 17 + 6 = 23 cm.
Therefore lengths of the parallel sides are 17 cm, 23 cm.
Question 8.
In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is 9450 m2 and the perpendicular distance between the two parallel sides is 84 m, find the length of the longer of the parallel sides.
Answer:
Given
Let length of first parallel side X
Length of other parallel side is 2X
Area of trapezium= 9450 m2
Let Height (h) =84 m
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (X + 2X) × 84 =9450 cm2.
× (X + 2X) × 84 =9450
⇒ (3X) × 42 =9450
⇒ 126X = 9450
⇒ 2X + 6 = 
= 75
⇒X = 17
... Length of the parallel sides is X=75 m and 2X = 150 m.
Therefore length of the longest is 150 m.
Question 9.
The length of the fence of a trapezium-shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If 
m, 
m and 
m, find the area of the field.
Answer:
Given
Length of parallel sides
AD = 42 m
BC = 54 m
Given that total length of fence is 130 m
That is AB + BC +CD +DA = 130
AB + 54 + 19 + 42 = 130
Therefore AB = 15
Height (AB) = 15 m
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (42 + 54) × 15 = 720 m2
Question 10.
In the given figure, ABCD is a trapezium in which 
cm,
cm and 
cm. find the area of the trapezium.
Answer:
Given
AD = 16 cm
BC = 40 cm
AC = 41 cm
ABC = 90
Height = AB =?
Here in 
ABC using Pythagoras theorem
AC2 = AB2 + BC2
412 = AB2 + 402
AB2 = 412 – 402
AB2 = 1681 – 1600 = 81
AB = 9
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (16 + 40) × 9 = 252 cm2.
Question 11.
The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.
Answer:
Let ABCD be the given trapezium in which AB|| DC,
AB = 20 cm, DC = 10 cm and AD=BC=13cm
Draw CL 
AB and CM || DA meeting AB at L and M, respectively.
Clearly, AMCD is a parallelogram.
Now,
AM = DC =10cm
MB = (AB-Am)
= (20-10) = 10 cm
Also,
CM = DA = 13cm
Therefore, CMB is an isosceles triangle and CL MB.
And L is midpoint of B.
⇒ML = LB = 
= 
= 5 cm
From right CLM, we have:
CL2 = (CM2 – ML2)
CL2 = (132 – 52)
CL2 = (169 – 25)
CL2 = 144
CL = 12
Therefore length of CL is 12 cm that is height of trapezium is 12 cm
There fore
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (20 + 10) × 12 = 180 cm2.
Question 12.
The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. find the area of the trapezium.
Answer:
Let ABCD be the given trapezium in which AB|| DC,
AB = 25 cm, CD = 11 cm and AD= 13 cm, BC=15cm
Draw CL AB and CM || DA meeting AB at L and M, respectively.
Clearly, AMCD is a parallelogram.
Now,
MC = AD = 13cm
AM = DC =11cm
MB = (AB—Am)
= (25—11) = 14 cm
Thus, in CMB, we have:
CM = 13 cm
MB = 14 cm
BC = 15 cm
Here let ML = X, hence LB = 14 – X and let CL = Y cm
Now in CML, using Pythagoras theorem
CL2 = (CM2 – ML2)
Y2 = (132 – X2) eq – 1
Again in CLB, using Pythagoras theorem
CL2 = (CB2 – LB2)
Y2 = (152 – (14—X) 2) eq – 2
Sub eq 1 in 2, we get
(132 – X2) = (152 – (14—X) 2)
169 – X2 = 225 – (196 + X2 – 28 X)
169 – X2 = 225 – 196 – X2 + 28 X
28X = 169 + 196 – 225 + X2 – X2
28X = 140
X = 5 cm
Now substitute X value in eq –1
That is Y2 = (132 – X2)
Y2 = (132 – 52)
Y2 = (169 – 25)
Y2 = 144
Y = 12 cm
Therefore CL = 12 cm that is height of the trapezium = 12 cm
Therefore
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (25 + 11) × 12 = 216 cm2.
Exercise 18b
Question 1.In the given figure, ABCD is a quadrilateral in which 
cm, 
and 
such that 
cm and 
cm. find the area of quad. ABCD.
Answer:
Given: A quadrilateral ABCD
and 
AC = 24 cm
BL = 8 cm
DM = 7 cm
Here,
Area (quad. ABCD) = area (ABC) + area (ADC)
Area of triangle = × (base) × (height).
Therefore
Area of quad ABCD = × (AC) × (BL) + × (AC) × (DM)
= × (24) × (8) + × (24) × (7) = 96 + 84 = 180 cm2
Therefore area of the quadrilateral ABCD is 180 cm2
Question 2.
In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, 
and 
such that 
m and 
m. Find the area of the field.
Answer:
Given: A quadrilateral ABCD
and 
AL = 19 cm
BD = 36 cm
CM = 11 cm
Here,
Area (quad. ABCD) = area (ABD) + area (ACD)
Area of triangle = × (base) × (height).
Therefore
Area of quad ABCD = × (BD) × (AL) + × (BD) × (CM)
= × (36) × (19) + × (36) × (11) = 342 + 198 = 540 cm2
Therefore area of the quadrilateral ABCD is 540 cm2.
Question 3.
Find the area of pentagon ABCDE in which 
and 
such that 
cm, 
cm, 
cm, 
cm, 
cm and 
cm.
Answer:
Given: A pentagon ABCDE
and 
cm
cm
cm
cm
cm
cm
MC = AC – AM = 18 – 14 = 4 cm
MN = AM – AN = 14 – 6 = 8 cm
Here,
Area (Pent. ABCDE) = area (AEN) + area (DMC) + area (ABC) + area (Trap. DMNE)
Area of triangle = × (base) × (height).
Area of trapezium is × (sum of parallel sides) × height
Here,
Area (AEN) = × (AN) × (EN) = × (6) × (9) = 27 cm2.
Area (DMC) = × (MC) × (DM) = × (4) × (12) = 24 cm2.
Area (ABC) = × (AC) × (BL) = × (18) × (4) = 36 cm2.
Area (Trap. DMNE) = × (DM + EN) × MN = × (12 + 9) × 8 = 84 cm2.
Area (Pent. ABCDE) = area (AEN) + area (DMC) + area (ABC) + area (Trap. DMNE)
= 27 + 24 + 36 + 84 = 171 cm2.
Area (Pent. ABCDE) = 171 cm2.
Question 4.
Find the area of hexagon ABCDEF in which 
and 
such that 
cm, 
cm, 
cm, 
cm, 
cm, 
cm, 
cm, cm and 
cm.
Answer:
Given: A Hexagon ABCDE
and 
cm
cm
cm
cm
cm
cm
cm
cm
cm
AL = AP + PL = 6 + 2 = 8 cm
PN = PL + LN = 2 + 8 = 10 cm
LM = LN + NM = 8 + 2 = 10 cm
ND = NM + MD =2 + 3 = 5 cm
Here,
Area (Hex. ABCDEF) = area (APF) + area (DEN) + area (ABL) + area (CMD)
+ area (Trap. PNEF) + area (Trap. LMCB)
Area of triangle = × (base) × (height).
Area of trapezium is × (sum of parallel sides) × height
Here,
Area (APF) = × (AP) × (FP) = × (6) × (8) = 24 cm2.
Area (DEN) = × (ND) × (EN) = × (5) × (12) = 30 cm2.
Area (ABL) = × (AL) × (BL) = × (8) × (8) = 32 cm2.
Area (CMD) = × (MD) × (CM) = × (3) × (6) = 9 cm2.
Area (Trap. PNEF) = × (FP + EN) × PN = × (8 + 12) × 10 = 100 cm2.
Area (Trap. LMCB) = × (BL + CM) × LM = × (8 + 6) × 10 = 70 cm2.
Area (Hex. ABCDEF) = area (APF) + area (DEN) + area (ABL) + area (CMD)
+ area (Trap. PNEF) + area (Trap. LMCB) = 24 + 30 + 32 + 9 + 100 + 70 = 265 cm2.
Area (Hex. ABCDEF) = 265 cm2
Question 5.
Find the area of pentagon ABCDE in which 
and such that 
cm, 
cm, 
cm, 
cm and 
cm.
Answer:
Given: A pentagon ABCDE
and 
cm
cm
cm
cm
cm
Here,
Area (Pent. ABCDE) = area (ABC) + area (ACD) + area (ADE)
Area of triangle = × (base) × (height).
Here,
Area (ABC) = × (AC) × (BL) = × (10) × (3) = 15 cm2.
Area (ACD) = × (AD) × (CD) = × (12) × (7) = 42 cm2.
Area (ADE) = × (AD) × (EN) = × (12) × (5) = 30 cm2.
Area (Pent. ABCDE) = area (ABC) + area (ACD) + area (ADE) = 15 + 42 + 30 = 87 cm2.
Area (Pent. ABCDE) = 87 cm2.
Question 6.
Find the area enclosed by the given figure ABCDEF as per dimensions given herewith.
Answer:
Given: A figure ABCDEF
AB = 20 cm
BC = 20 cm
ED = 6 cm
AF = 20 cm
AB || FC
FC = 20 cm
Let distance between FC and ED be h = 8 cm
FC || ED
Here,
From the figure we can see that ABCF forms a square and EFCD forms a trapezium.
Area of square = (side length) 2
Area of trapezium = × (sum of parallel sides) × height
Therefore,
Area of the figure ABCDEF = Area of square (ABCF) + Area of trapezium (EFCD)
Here,
Area of square (ABCF) = (AB) 2 = (20)2 = 400 cm2
Area of trapezium (EFCD) = × (FC + ED) × h = × (6 + 20) × 8 = 104 cm 2
Area (ABCDEF) = Area of square (ABCF) + Area of trapezium (EFCD) = 400 + 104 = 504 cm2.
Area (Fig. ABCDEF) = 504 cm2.
Question 7.
Find the area of given figure ABCDEFGH as per dimensions given in it.
Answer:
Given: A figure ABCDEFGH
BC = FG = 4 cm
AB = HG = 5 cm
CD = EF = 4 cm
ED = 8 cm
ED || AH
AH = 8 cm
Here
ABC and GHF are equal and right angled
AC = AH = ?
In ABC using Pythagoras theorem
AB2 = BC2 + AC2
52 = 42 + AC2
25= 16+ AC2
AC2 = 25 –16 = 9
AC = 3
AH = 3
Area(ABCDEFGH) = area(Rect. ADEH) + 2 X area (ABC)
Area of rectangle = (length × breadth)
Area of triangle = × (base) × (height).
Area(Rect. ADEH) = (DE × AD) = (DE × (AC + AD)) = (8 × (3 + 4)) = 56 cm 2
Area(ABC) = × (BC) × (AC) = × (4) × (3) = 6 cm2
Area(ABCDEFGH) = area(Rect. ADEH) + 2 × area (ABC) = 56 + (2 × 6) = 68 cm2
Area(ABCDEFGH) = 68 cm2.
Question 8.
Find the area of a regular hexagon ABCDEF in which each side measures 13 cm and whose height is 23 cm, as shown in the given figure.
Answer:
Given: a regular hexagon ABCDEF
AB = BC = CD = DE = EF = FA = 13 cm
AD = 23 cm
Here AL = MD
Therefore Let AL = MD = x
Here AD = AL + LM + MD
23 = 13 + 2x
2x = 23 – 13 = 10
x = 5
Now,
In ABL using Pythagoras theorem
AB2 = AL2 + LB2
132 = x2 + LB2
132 = 52 + LB2
169 = 25 + LB2
LB2 = 169 – 25 = 144
LB = 12
Here area (Trap. ABCD) = area (Trap. AFED)
Therefore,
Area (Hex. ABCDEF) = 2 × area (Trap. ABCD)
Area of trapezium = × (sum of parallel sides) × height
Area (Trap. ABCD) = × (BC + AD) × LB = × (13 + 23) × 12 = 216 cm2.
Area(ABCDEFGH) = 2 × area (Trap. ABCD) = 2 × 216 = 432 cm2
Area(ABCDEFGH) = 432 cm2.
Exercise 18c
Question 1.The parallel sides of a trapezium measure 14 cm and 18 cm and the distance between them is 9 cm. The area of the trapezium is
A. 96 cm2
B. 144 cm2
C. 189 cm2
D. 207 cm2
Answer:
Given
Length of parallel sides is 14cm and 18 cm
Height (h) = 9 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (14 +18) × 9 = 144 cm2.
Question 2.
The length of the parallel sides of a trapezium are 19 cm and 13 cm and its area is 128 cm2. The distance between the parallel sides is
A. 9 cm
B. 7 cm
C. 8 cm
D. 12.5 cm
Answer:
Given
Length of parallel sides is 19 cm and 13 cm
Area of trapezium= 128 cm2
Let Height (h) =y cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (19 +13) × y =128 cm2.
× (19 +13) × y =128
⇒ × (32) × y =128
⇒ 16 × y =128
⇒ y =
= 8 cm
... Distance between the parallel lines is 8 cm.
Question 3.
The parallel sides of a trapezium are in the ration 3:4 and the perpendicular distance between them is 12 cm. If the area of the trapezium is 630 cm2, then its shorter length of the parallel sides is
A. 45 cm
B. 42 cm
C. 60 cm
D. 36 cm
Answer:
Given
Lengths of the parallel sides are in the ratio 3:4
Therefore let one of the side length be 3X and other side length be 4X
Area of trapezium= 630 cm2
Let Height (h) =12 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (3X +4X) × 12 =630 cm2.
× (3X + 4X) × 12 = 630
⇒(7X) × 6 =630
⇒ 42X = 630
⇒ X =
= 15
... length of the parallel sides is 3X = 3 × 15 =45 cm and 4X = 4 × 15 = 60 cm.
Therefore shortest length of the parallel sides is 45 cm.
Question 4.
The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, the length of the longer parallel sides is
A. 17 cm
B. 23 cm
C. 18 cm
D. 24 cm
Answer:
Given
Let length of first parallel side X
Length of other parallel side is X + 6
Area of trapezium= 180 cm2
Let Height (h) =9 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (X + 6 +X) × 9 =180 cm2.
× (X + 6 +X) × 26 =180
⇒ × (2X + 6) × 9 =180
⇒ 2X + 6 =
⇒ 2X + 6 = 40
⇒2X = 40 – 6 =34
⇒X = 17
... length of the parallel sides is X=17 cm and X + 6 = 17 + 6 = 23 cm.
Therefore length of the longer parallel side is 23 cm.
Question 5.
In the given figure, AB||DC and DA AB If 
cm, 
cm, 
cm and CL AB the area of trap. ABCD is
A. 84 cm2
B. 72 cm2
C. 80 cm2
D. 91 cm2
Answer:
Given:
AB||DC, DA AB and CL AB
cm
cm
cm
Therefore here AL = DC
That is AL = 7 cm
Hence LB = AB – AL = 13 – 7 = 6cm
In LCB using Pythagoras theorem
BC2 = BL2 + CL2
102 = 62 + CL2
100 = 36 + CL2
CL2 = 100 – 36
CL2 = 64
CL = 8
Here CL = AD = height of the trapezium
Therefore height = 8 cm
Now,
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (7 +13) × 8 = 80 cm2.
Cce Test Paper-18
Question 1.The base of a triangular field is three times its height and its area is 1350 m2. Find the base and height of the field.
Answer:
Given
Area of triangle = 1350 m2
Let the length of the height of triangle be Y cm
Therefore its base is 3Y cm
Area of the triangle = × base × height = 1350
× (3Y) × (Y) = 1350
3Y2 = 1350 × 2 = 2700
Y2 = 
= 900
Y = 30 cm
Therefore height of triangle is 30 cm and base is 3×30 = 90cm
That is
.
Question 2.
Find the area of an equilateral triangle of side 6 cm.
Answer:
Given
Side length of equilateral triangle is 6 cm
We know that area of the equilateral triangle is given by 
a2,where a is side length
Therefore area of the triangle is
⇒ × 62 = × 36 =
× 9 = 
.
Question 3.
The perimeter of a rhombus is 180 cm and one of its diagonals is 72 cm. Find the length of the other diagonal and the area of the rhombus.
Answer:
Given: A rhombus
Diagonal AC = 72 cm
Perimeter = 180 cm
Perimeter of the rhombus = 4x
Therefore 4x = 180
x= 45
hence, the side length of the rhombus is 45 cm
We know that diagonals of the rhombus bisect each other right angles.
AO = AC
⇒AO = ( × 72) cm
⇒AO = 36 cm
From right AOB, we have :
BO2 = AB2 – AO2
⇒BO2 = AB2 – AO2
⇒BO2 = 452 – 362
⇒BO2 = 2025 – 1296
⇒BO2 = 729
BO = 27 cm
BD = 2× BO
BD = 2 × 27 = 54 cm
Hence, the length of the other diagonal is 54 cm.
Area of the rhombus = × 72 × 54 = 1944 cm2
Question 4.
The area of a trapezium is 216 m2 and its height is 12 m. If one of the parallel sides is 14 m less than the other, find the length of each of the parallel sides.
Answer:
Given
Let length of first parallel side X
Length of other parallel side is X – 14
Area of trapezium= 216 m2
Let Height (h) =12 m
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (X – 14 +X) × 12 =216 m2.
× (X – 14 +X) × 12 =216
⇒ × (2X – 14) × 12 =216
⇒ 2X – 14 = 
⇒ 2X – 14 = 36
⇒2X = 36 + 14 =50
⇒X = 25
... length of the parallel sides is X=25 cm and X — 14 = 25 – 14 = m.
Therefore lengths of the parallel sides are 25 m, 11 m.
Question 5.
Find the area of a quadrilateral one of whose diagonals is 40 cm and the lengths of the perpendiculars drawn from the opposite vertices on the diagonal are 16 cm and 12 cm.
Answer:
Given : A quadrilateral
Diagonal AC = 40 cm
Perpendiculars to diagonal AC are: BL = 16 cm and DM = 12 cm
Now,
Area (quad. ABCD) = area (ABC) + area (ADC)
Area of triangle = × (base) × (height).
Therefore
Area of quad ABCD = × (AC) × (BL) + × (AC) × (DM)
= × (40) × (16) + × (40) × (12) = 320 + 240 = 560 cm2
Therefore area of the quadrilateral ABCD is 560 cm2.
Question 6.
A field is in the form of a right triangle with hypotenuse 50 m and one side 30m. Find the area of the field.
Answer:
Given
A right angled triangle with hypotenuse = 50 cm and one of the side = 30 cm
Let base = 30 cm
Height = Y cm
Area = ?
By using hypotenuse theorem
Hypotenuse2 = base2 + height2
502 = 302 + Y2
Y2 = 502 – 302 = 2500 – 900 = 1600
Therefore X2 = 1600
Y = 40cm
Area of the triangle = × base × height
Area = × 30 × Y
= × 30 ×40 = 600 m2.
Question 7.
The base of a triangle is 14 cm and its height is 8 cm. The area of the triangle is
A. 112 cm2
B. 56 cm2
C. 122 cm2
D. 66 cm2
Answer:
Given
Length of the base of the triangle = 14 cm
Length of the heigth of the triangle = 8 cm
Area of the triangle = × base × height
Therefore area = × base × height
= × 14 × 8 = 7 × 8 = 56 cm
Question 8.
The base of a triangle is four times its height and its area is 50 m2. The length of its base is
A. 10 m
B. 15 m
C. 20 m
D. 25 m
Answer:
Given
Area of triangle = 50 m2
Let the length of the height of triangle be Y cm
Therefore its base is 4Y cm
Area of the triangle = × base × height = 50
× (4Y) × (Y) = 50
4Y2 = 50 × 2 = 100
Y2 = 
= 25
Y = 5 cm
Therefore length of base is 4 × 5 = 20 cm
Question 9.
The diagonal of a quadrilateral is 20 cm in length and the lengths of perpendiculars on it from the opposite vertices are 8.5 cm and 11.5 cm. The area of the quadrilateral is
A. 400 cm2
B. 200 cm2
C. 300 cm2
D. 240 cm2
Answer:
Given : A quadrilateral
Diagonal AC = 20 cm
Perpendiculars to diagonal AC are: BL = 11.5 cm and DM = 8.5 cm
Now,
Area (quad. ABCD) = area (ABC) + area (ADC)
Area of triangle = × (base) × (height).
Therefore
Area of quad ABCD = × (AC) × (BL) + × (AC) × (DM)
= × (20) × (11.5) + × (20) × (8.5) = 115 + 85 = 200 cm2
Therefore area of the quadrilateral ABCD is 200 cm2.
Question 10.
Each side of a rhombus is 15 cm and the length of one of its diagonals is 24 cm. The area of the rhombus is
A. 432 cm2
B. 216 cm2
C. 180 cm2
D. 144 cm2
Answer:
Given: A rhombus ABCD
Diagonal AC = 24 cm
Side length : AB = BC = CD = DA = 15 cm
We know that diagonals of the rhombus bisect each other right angles.
AO = AC
⇒AO = ( × 24) cm
⇒AO = 12 cm
From right AOB, we have :
BO2 = AB2 – AO2
⇒BO2 = AB2 – AO2
⇒BO2 = 152 – 122
⇒BO2 = 225 – 144
⇒BO2 = 81
⇒BO = 9 cm
BD = 2 × BO
BD = 2 × 9 = 18 cm
Hence, the length of the other diagonal is 18 cm.
Area of the rhombus = × 24 × 18 = 216 cm2
Question 11.
The area of a rhombus is 120 cm2 and one of its diagonals is 24 cm. Each side of the rhombus is
A. 10 cm
B. 13 cm
C. 12 cm
D. 15 cm
Answer:
Given: A rhombus ABCD
Diagonal AC = 24 cm
Area = 120 cm2
Area of the rhombus = × AC × BD
Therefore,
× AC × BD = × 24 × BD = 120
24 × BD = 120 × 2
BD = 
= 10 cm
OB = 
= 
= 5 cm
OA = 
= 
= 12 cm
Now,
In AOB using Pythagoras theorem
AB2 = OA2 + OB2
AB2 = 122 + 52
AB2 = 144 + 25
AB2 = 169
AB = 13
Therefore length of each side of the rhombus = 13 cm
Question 12.
The parallel sides of a trapezium are 54 cm and 26 cm and the distance between them is 15 cm. The area of the trapezium is
A. 702 cm2
B. 810 cm2
C. 405 cm2
D. 600 cm2
Answer:
Given
Length of parallel sides is 54cm and 26 cm
Height (h) = 15 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (54 +26) × 15 = 600 cm2.
Question 13.
The area of a trapezium is 384 cm2. Its parallel sides are in the ratio 5:3 and the distance between them is 12 cm. the longer of the parallel sides is
A. 24 cm
B. 40 cm
C. 32 cm
D. 36 cm
Answer:
Given
Lengths of the parallel sides are in the ratio 5:3
Therefore let one of the side length be 5X and other side length be 3X
Area of trapezium= 384 cm2
Let Height (h) =12 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium is × (5X +3X) × 12 =384 cm2.
× (5X + 3X) × 12 =384
⇒ (8X) × 6 =384
⇒ 48X = 384
⇒ X =
= 8
... length of the parallel sides is 5X=5 × 8 =40 cm and 3X = 3 × 8 = 24 cm.
Therefore length of the longest side is 40 cm.
Question 14.
Fill in the blanks.
(i) Area of triangle
(ii) Area of a ||gm = (............)× (.............)
(iii) Area of a trapezium
(iv) The parallel sides of a trapezium are 14 cm and 18 cm and the distance between them is 8 cm. The area of the trapezium is ...... cm2.
Answer:
(i) Area of triangle = × (base) × (height).
(ii) Area of || gm = (base) × (height).
(iii) Area of trapezium is × (sum of parallel sides) × (height)
(iv) Given
Length of parallel sides is 14cm and 18 cm
Height (h) = 8 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (14 +18) × 8 = 128 cm2.