हिंदी

Find the Area of the Pentagon Shown in Fig. 20.48, If Ad = 10 Cm, Ag = 8 Cm, Ah = 6 Cm, Af = 5 Cm, Bf = 5 Cm, Cg = 7 Cm and Eh = 3 Cm. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm.

योग

उत्तर

The given figure is:

Given:
AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm
BF = 5 cm, CG = 7 cm, EH = 3 cm
∴ FG = AG - AF = 8 - 5 = 3 cm
And, GD = AD - AG = 10 - 8 = 2 cm
From given figure:
Area of Pantagon = (Area of triangle AFB) + (Area of trapezium FBCG) + (Area of triangle CGD) + (Area of triangle ADE)
\[= (\frac{1}{2}\times AF \times BF) + [\frac{1}{2} \times (BF + CG) \times (FG)] + (\frac{1}{2} \times GD \times CG) + (\frac{1}{2}\times AD \times EH)\]
\[=(\frac{1}{2} \times 5 \times 5) + [\frac{1}{2} \times (5 + 7) \times (3)] + (\frac{1}{2} \times 2 \times 7) + (\frac{1}{2}\times 10 \times 3)\]
\[= (\frac{25}{2}) + [\frac{36}{2}] + (\frac{14}{2}) + (\frac{30}{2})\]
\[ = 12 . 5 + 18 + 7 + 15\]
\[ {=52.5 cm}^2\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Mensuration - I (Area of a Trapezium and a Polygon) - Exercise 20.3 [पृष्ठ २८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 8
अध्याय 20 Mensuration - I (Area of a Trapezium and a Polygon)
Exercise 20.3 | Q 1 | पृष्ठ २८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as
the sum of the areas of two triangles and one rectangle.


Mohan wants to buy a trapezium shaped field. Its side along the river is parallel and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.


The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.


Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium:


There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some another way of finding its areas?


The following figure shows the cross-section ABCD of a swimming pool which is a trapezium in shape. 

If the width DC, of the swimming pool, is 6.4 m, depth (AD) at the shallow end is 80 cm and depth (BC) at the deepest end is 2.4 m, find its area of the cross-section.


The area of a trapezium is 279 sq.cm and the distance between its two parallel sides is 18 cm. If one of its parallel sides is longer than the other side by 5 cm, find the lengths of its parallel sides.


The area of a trapezium is 180 sq.cm and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm. Find the length of the parallel sides.


A window is in the form of trapezium whose parallel sides are 105 cm and 50 cm respectively and the distance between the parallel sides is 60 cm. Find the cost of the glass used to cover the window at the rate of ₹ 15 per 100 sq.cm


Find the area of the following fields. All dimensions are in metres.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×