Concise Class-9 Area and Perimeter of Plane Figure ICSE Maths Selina Solutions
Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions Chapter-20. We provide step by step Solutions of Exercise / lesson-20 Area and Perimeter of Plane Figure for ICSE Class-9th Concise Selina Mathematics by R K Bansal.
Our Solutions contain all type Questions with Exe-20 A, Exe-20 B , Exe-20 C and Exe-20 D to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9th Mathematics .
Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions Chapter-20
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Exe-20 A, Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions
Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm.
Since the sides of the triangle are 18 cm, 24 cm and 30 cm respectively.
Hence the area of the triangle is
= 216 sq cm
A =1/2 base x height
216 = 1/2 base x height
216 = 1/2 x 30 x h
h = 14.4 cm
The length of the sides of a triangle are in the ratio 3: 4: 5. Find the area of the triangle if its perimeter is 144 cm.
Let the sides of the triangle are
Given that the perimeter is 144 cm.
⇒ 3x + 4x + 5x =144
⇒ x = 144/ 12
s = (a+b+c )/2
⇒ 6x =72
⇒ x = 12
The sides are a=36 cm, b=48 cm and c=60 cm
Area of the triangle is
= 864 sq cm
ABC is a triangle in which AB = AC = 4 cm and A = 90o. Calculate:
(i) The area of ABC,
(ii) The length of perpendicular from A to BC.
Area of the triangle is given by
Again area of the triangle
The area of an equilateral triangle is sq. cm. Find its perimeter.
Find the area of an isosceles triangle with perimeter is 36 cm and base is 16 cm.
Since the perimeter of the isosceles triangle is 36cm and base is 16cm. hence the length of each of equal sides are
It is given that
Let ‘h’ be the altitude of the isosceles triangle.
Since the altitude from the vertex bisects the base perpendicularly, we can apply Pythagoras Theorem.
Thus we have,
The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.
It is given that
Let ‘a’ be the length of an equal side.
The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.
Find the area and the perimeter of quadrilateral ABCD, given below; if AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and DBC = 90o
Given , AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and DBC = 90o
The base of a triangular field is three times its height. If the cost of cultivating the field at 36.72 per 100 m2 is 49,572; find its base and height.
The sides of a triangular field are in the ratio 5 : 3 : 4 and its perimeter is 180 m. Find:
(i) its area.
(ii) altitude of the triangle corresponding to its largest side.
(iii) the cost of leveling the field at the rate of Rs. 10 per square metre.
Consider the following figure.
Each of equal sides of an isosceles triangle is 4 cm greater than its height. IF the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.
Let the height of the triangle be x cm.
Equal sides are (x+4) cm.
According to Pythagoras theorem,
Area of the isosceles triangle is given by
Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.
Each side of the triangle is
Hence the area of the equilateral triangle is given by
In triangle ABC; angle A = 90o, side AB = x cm, AC = (x + 5) cm and area = 150 cm2. Find the sides of the triangle.
The area of the triangle is given as 150sq.cm
Hence AB=15cm, AC=20cm and
If the difference between the sides of a right angled triangle is 3 cm and its area is 54 cm2; find its perimeter.
Let the two sides be x cm and (x-3) cm.
Hence the sides are 12cm, 9cm
The required perimeter is 12+9+15=36cm.
AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that BOC = 90o. Find the area of quadrilateral ABOC.
Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions Ex. 20(B)
Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively.
The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.
Calculate the area of quadrilateral ABCD, in which ABD = 90o, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.
Consider the figure:
From the right triangle ABD we have
The area of right triangle ABD will be:
Again from the equilateral triangle BCD we have
Therefore the area of the triangle BCD will be:
Hence the area of the quadrilateral will be:
Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm A = 90o and BC = CD = 52 cm.
The figure can be drawn as follows:
Here ABD is a right triangle. So the area will be: